Chemistry 5 Years Class12

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The Solid State

1

Chapter1 The Solid State

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

1.

What type of substances exhibit antiferromagnetism? [Delhi 2008]

2.

What is the coordination number of each type of ions in a rock salt type crystal structure? [Delhi 2008, AI 2008]

3.

What is the total number of atoms per unit cell in a face centred cubic (fcc) structure? [Delhi 2008, AI 2008]

4.

Name an element with which silicon should be doped to give ntype of semiconductor. [Delhi 2008 C]

5.

What are F centres?

6.

Why is glass considered a super cooled liquid? [AI 2008 C]

[AI 2008 C]

19. “Crystalline solids are anisotropic in nature.” What does this statement mean? [Delhi SetI 2011, Foreign SetII, 2011] 20. Which stoichiometric defect in crystals increases the density of a solid? [Delhi SetII 2011] 21. What is meant by an ‘intrinsic semiconductor’? [ Foreign SetI 2011] 22. How many atoms are there in one unit cell of a body centred cubic crystal? [Foreign SetIII 2011] 23. What is meant by ‘doping’ in semiconductor? [Delhi SetI 2012] 24. Write a point of distinction between a metallic solid and an ionic solid other than metallic lustre. [Delhi SetII 2012]

Which point defect in crystals does not alter the density of relevant solid? [Delhi 2009 ] 25. How may the conductivity of an intrinsic semiconductor be increased? [AI SetI 2012] 8. How do metallic and ionic substances differ in conducting electricity? [AI 2009] 26. Which stoichiometric defect increases the density of a solid? [AI SetII 2012] 9. Which point defect of its crystals decreases the density of a [AI SetIII 2012] solid? [AI 2009, Delhi 2010] 27. What are ntype semiconductors? 7.

10. What is the number of atoms in a unit cell of facecentred cubic crystal? [AI 2009]

SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS)

11. What is the percentage efficiency of packing in case of a simple cubic lattice? [AI 2009 C]

28. Explain the following properties giving suitable examples: (i) Ferromagnetism (ii) Paramagnetism [Delhi 2008]

12. Stability of a crystal is related to the magnitude of its melting point. How? [AI2009 C]

29. How would you account for the following? (i) Frenkel defects are not found in alkali metal halides. 13. Which point defect in crystals of a solid does not change the (ii) Schottky defects lower the density of related solids. density of the solid? [Delhi 2010] (iii) Impurity doped silicon is a semiconductor. 14. What type of interactions hold the molecules together in a [Delhi 2008] polar molecular solid? [AI 2010 ] 30. Ag crystallise in a fcc lattice. The edge length of its unit cell 15. What type of semiconductor is obtained when silicon is doped is 4.077 × 10 –8 cm and its density is 10.5 g cm –3 . Calculate on with arsenic? [AI 2010] this basis the atomic mass of Ag. [AI 2008] 16. Write a distinguishing feature of metallic solids. [AI 2010] 31. Explain the following terms with suitable examples : 17. Write a feature which will distinguish a metallic solid from (i) Schottky defect, (ii) Ferromagnetism. [Delhi 2008 C] an ionic solid. [Delhi 2010, Foreign 2010] 32. What is a semiconductor? Describe the two major types of 18. What is the number of atoms in a unit cell of a simple cubic semiconductors and contrast their conduction mechanism. crystal? [ Foreign 2010] [Delhi 2008 C]

2

Chapterwise XIIth CBSE BOARD 5 Yrs. Papers

10.5 g cm –3 , calculate the atomic mass of silver 33. Account for the following; (N A = 6.02 × 10 23 atoms mol –1 ). [AI 2010, Foreign 2010] (i) Fe 3 O 4 is ferrimagnetic at room temperature but becomes paramagnetic at 850 K. 44. Silver crystallizes in facecentered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the (ii) Zinc oxide on heating becomes yellow. radius of the silver atom. (Assume the atom just touch each (iii) Frenkel defect does not change the density of AgCl other on the diagonal across the face of the unit cell, that crystals. [AI 2008 C] is each face atom is touching the four corner atoms). 34. With the help of suitable diagrams, on the basis of band theory. [Delhi SetI 2011] Explain the difference between –3 45. The density of lead is 11.35 g cm and crystallize with fcc (i) a conductor and an insulator unit cell. Estimate the radius of lead atom. (ii) a conductor and a semiconductor. [AI 2008 C] (At. mass of lead = 207 g mol –1 and N A = 6.02 × 10 23 mol –1 ) 35. (a) In reference to crystal structure, explain the meaning of [Delhi SetIII 2011] coordination number. 46. Explain how we can determine the atomic mass of an (b) What is the number of atoms in a unit cell of (i) a face unknown metal if you know its density and the dimension centred cubic structure, (ii) a body centred cubic structure. of unit cell of its crystal? [AI SetI 2011] [Delhi 2009] 36. Iron has a body centred cubic unit cell with a cell edge of 47. Calculate the packing efficiency of a metal for a simple cubic lattice. [AI SetI 2011] 286.65 pm. The density of iron is 7.87 g cm –3 . Use this

37. 38.

39.

40.

information to calculate Avogadro’s number (Atomic mass 48. Define the following terms in relation to crystalline of Fe = 56 g mol –1 ). [AI 2009 & Foreign, Delhi 2009] solids: (i) Unit cell What are Fcentres? Why are solids containing Fcentres (ii) Coordination number paramagnetic? [AI 2009 C] Give one example in each case. [AI SetII 2011] Silver crystallises with facecentred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius 49. Aluminium crystallises in a cubic closepacked structure. Radius of atom in the metal is 125 pm. of an atom of silver? (Assume that each face atom is touching (i) What is the length of the side of the unit cell? the four corner atoms). [AI 2009, AI 2010] (ii) How many unit cells are there in 1 cm 3 of aluminium? How are the following properties of crystals affected by [ Foreign SetI 2011] Schottky and Frenkel defects? (i) Density, (ii) Electrical conductivity. [Delhi 2009 C] 50. Silver crystallise with facecentred cubic unit cell. Each side of this unit cell has a length of 409 pm. What is the Copper crystallizes into an fcc lattice with edge length radius of silver atom? Assume the atoms just touch each 3.61 × 10 –8 cm. Calculate the density of copper. other on the diagonal across the face of the unit cell. [Given : Cu = 63.5 g mol –1 , N A = 6.022 × 10 23 mol –1 ] [ Foreign SetIII 2011] [Delhi 2009 C]

41. The well known mineral fluorite is chemically calcium 51. Tungsten crystallises in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten fluoride. It is known that in one unit cell of this mineral there atom? are 4 Ca 2+ ions and 8F – ions and that Ca 2+ ions are arranged in a fcc lattice. The F – ion fill all the tetrahedral holes in the OR face centred cubic lattice of Ca 2+ ions. The edge of the unit Iron has a body centred cubic unit cell with a cell dimension cell is 5.46 × 10 –8 cm in length. The density of the solid is of 286.65 pm. The density of iron is 7.874 g cm –3 . Use this –3 3.18 g cm . Use this information to calculate Avogadro’s information to calculate Avogadro’s number. (At. mass of Fe number. (Molar mass of CaF 2 = 78.08 g mol –1 ) = 55.845 u) [Delhi SetI 2012] [Delhi 2010, Foreign 2010] 52. Copper crystallises with face centred cubic unit cell. If the 42. The density of copper metal is 8.95 g cm –3 . If the radius of radius of copper atom is 127.8 pm, calculate the density of copper atom is 127.8 pm. What is copper unit cell a simple copper metal. (Atomic mass of Cu = 63.55 u and Avogadro’s cubic, a body centred cubic or a face centred cubic structure? number, N A = 6.02 × 10 23 mol –1 ) OR [Given : Atomic mass of Cu = 63.54 g mol –1 and 23 –1 Iron has a body centred cubic unit cell with the cell dimension N A = 6.02 × 10 mol ] [AI 2010, Delhi 2010] of 286.65 pm. Density of iron is 7.87 g cm –3 . Use this 43. Silver crystallises in fcc lattice. If the edge length of the unit information to calculate Avogadro’s number. (Atomic mass of cell is 4.07 ´ 10 –8 cm and density of the crystal is Fe = 56.0 u) [AI SetI 2012]

The Solid State

3

answers 1. Those substances show antiferromagnetism in which magnetic domains are aligned in opposite direction. 2. In rock salt type structure coordination number is +

6 : 6 Þ Coordination number of M = 6 Coordination number of A – = 6. 3. No. of atoms per unit cell in fcc 1 1 Z = ´ 8 (corners) + ´ 6 (face centres) 8 2 Z = 1 + 3 = 4.

23. Addition of an appropriate amount of suitable impurity in a crystalline solid is called doping. Doping is done to increasing conductivity of intrinsic semiconductors. 24. Ionic crystals are insulators because their ions are not free to move. Whereas metallic crystals conduct electricity due to the presence of sea of mobile electrons. 25. Conductance of a semiconductor increases with rise in temperature.

4. Phosphorus.

26. Interstitial defect.

5. Voids created by anion vacancy and occupied by free electrons are called F centres. F centres cause colour to the crystal

27. A semiconductor in which doped impurity has more valence electrons than the pure element is called ntype semiconductor. e.g. Ge or Si (Group14) doped with P or As(group15).

6. Glass is an amorphous solid. It has tendency to flow very 28. (i) Ferromagnetism : Materials which are strongly attracted slowly like liquids. Hence, glass is considered as super by magnetic field are called ferromagnetic materials cooled liquid. and the property thus exhibited is caused 7. Frenkel defect. ferromagnetism. 8. Conductivity of metals is due to the motion of free electrons e.g., Fe, Co, Ni show ferromagnetism at room temperature. whereas conductivity of ionic substance is due to movement (ii) Paramagnetism : Substances which are weakly attracted of ions in molten or solution state. by a magnetic field are called paramagnetic substances 9. Schottky defect. and the property is called paramagnetism. e.g. O 2 , S 2 , Cu 2+ , Mn 2+ , etc. 1 1 10. Z = 8 ´ + 6 ´ = 4. 29. (i) This is due to the fact that alkali metal ions have large 8 2 11. Packing efficiency of simple cubic lattice = 52.4%. size which can’t fit into the interstitials sites. (ii) Schottky defects occur when cations and anions are 12. Stability of a crystal increases with increasing magnitude missing from their lattice site. Mass of unit cell decreases of melting point. which decreases the density of the solid. 13. Frenkel defect. (iii) The conductivity of intrinsic semiconductor like silicon 14. Dipoledipole interactions. is too low to be of practical use. This conductivity is 15. ntype semiconductor. increased by adding an appropriate amount of suitable impurity like Al or As which is electron deficient or 16. Malleable, ductile, lustrous and conductor of heat and electron rich. So the electrical conductivity of silicon electricity. is increased. 17. (i) Metallic solids are malleable and ductile whereas ionic 30. Given a = 4.077 × 10 –8 cm, d = 10.5 g cm –3 solids are hard and brittle. For fcc lattice, Z = 4 (ii) Metallic solids are conductors but ionic solids are insulators. Z ´ M Using d = 3 1 a ´ N A 18. For a simple cubic crystal Z = 8 ´ = 1. 8 10.5 ´ (4.077 ´ 10 -8 ) 3 ´ 6.022 ´ 10 23 or M = 19. Crystalline solids are anisotropic in nature means some of 4 their physical properties like electrical conductivity, = 107.09 g mol –1 . refractive index, etc., are different in different directions. 31. (i) Schottky defect : The defect in which cations and anions are missing in the stoichiometric ratio of 20. Interstitial defect increases density of solid. compound is called Schottky defect. Schottky defect is 21. Pure substance which acts as semiconductor is called actually vacancy defect in ionic solids. In this defect intrinsic semiconductor. e.g. Si and Ge at high temperature. electrical neutrality is maintained. In this defect density 22. 2 of solid decreases.


4

Schottky defect is shown by ionic solids in which the cation and anion are of almost similar sizes. Examples : NaCl, KCl, CsCl, AgBr, etc.

Conduction band

NaCl has one Schottky defect per 10 16 lattice points or 10 6 Schottky pairs per cm 3 of solid at room temperature.

Valence band

– + – + – + – +

+ – + – + – + –

– + – + – +

– + – + – + + –

– + – + – + + – + – + – + – + – + – + – + –

Conductor

+ – + – + – + –

Conduction band Small forbidden zone Valence band Semiconductor

35. (a) The number of lattice points touching one lattice point in a crystal lattice is called coordination number. (b) Number of atoms in a unit cell of

1 1 + 6 ´ = 4. 8 2 (ii) Refer Ans. 28 (i). 1 (ii) bcc structure, Z = 8 ´ + 1 ´ 1 = 2. 32. Semiconductors : In semiconductors the energy gap 8 between valence band and conduction band is small. 36. Given : For bcc, Z = 2 Therefore, some electrons may jump from valence band to a = 286.65 pm = 2.87 × 10 –8 cm conduction band and show some conductivity. d = 7.87 g cm –3 , N A = ? Semiconductors are of two types : Z ´ M Using formula d = (i) Intrinsic semiconductors N A ´ a 3 (ii) Extrinsic semiconductors Z ´ M 2 ´ 56 g mol -1 (i) Intrinsic semiconductors : A pure element which = or, N A = d ´ a 3 7.87 ´ (2.87 ´ 10 -8 cm) 3 behaves as semiconductor is called intrinsic 23 semiconductor. or,N A = 6.022 × 10 . (ii) Extrinsic semiconductor : A substance which behaves 37. Electrons trapped in anion vacancy of a crystal are called as semiconductor by adding some impurities is called Fcentres. Solids containing Fcentres are paramagnetic extrinsic semiconductor. Addition of appropriate amount due to the presence of unpaired electron. of suitable impurities increases conductivity of 38. Given : Structure fcc, a = 409 pm, r = ? semiconductors. This process of addition of impurities a 409 pm is called doping. Extrinsic semiconductors are formed Using formula r = = = 144.6 pm. by doping impurity of lower or higher groups. 2 2 2 ´ 1.414 (a) ntype extrinsic semiconductor 39. (i) Density : It decreases due to Schottky defect but remains (b) ptype extrinsic semiconductor. same in Frenkel defect. (ii) Electrical conductivity : It increases in both Schottky 33. (i) On heating at 850 K the magnetic domains undergo and Frenkel defects. realignment and become paramagnetic. (ii) Zinc oxide when heated loses some oxide ion creating 40. Given : for fcc Z = 4, a = 3.61 × 10 –8 cm, d = ? anion vacancy. This defect arises yellow colour. Z ´ M Using formula d = (iii) In Frenkel defect, no ions are missing from the crystal N A ´ a 3 as a whole. Hence, density does not change. 4 ´ 63.5 g mol -1 34. (i) In conductor valence band and conduction band overlap or, d = 6.022 ´ 1023 mol- 1 ´ (3.61 ´ 10 -8 cm)3 with each other whereas in insulator there is a large or,d = 8.965 g cm –3 . forbidden zone between the two. Conduction band Valence band Conductor

(i) fcc structure, Z = 8 ´

Conduction band Large forbidden zone Valence band Insulator

(ii) In conductor, valence band and conduction band overlap together whereas in semiconductor there is small forbidden band between the two.

41. Given for fcc lattice Z = 4, d = 3.18 g cm –3 a = 5.46 × 10 –8 cm, N A = ? Using formula d =

Z ´ M N A ´ a 3

4 ´ 78.08 g mol -1 3.18 g cm -3 ´ (5.46 ´ 10 -8 cm) 3 or,N A = 6.033 × 10 23 mol –1 .

or, N A =

The Solid State

5

42. Let us assume Z = 4 (copper has fcc) Z ´ M d = 3 a ´ N A 4 ´ 63.54 8.95 = 3 a ´ 6.02 ´ 10 23 a 3 = 47.145 × 10 –24 cm 3

494.9 494.9 2 494.9 ´ 1.414 pm = pm = pm 4 4 2 2 \ r = 174.95 pm

\ r =

46. We can determine the atomic mass of an unknown metal by using the formula of density of its unit cell. d (density) =

a = 3 47.145 ´ 10 -24 a = 3.612 × 10 –8 cm = 361.2 pm

By knowing density d, dimension of unit cell we can calculate M, the atomic mass of metal as for a given unit cell, Z is fixed and N A is universal constant.

for fcc, 4r = 2 a It shows Cu has fcc structure, a 361.2 r= = =127.7 pm. 2 2 2 ´ 1.141 43. Given : a = 4.07 × 10 –8 cm, d = 10.5 g cm –3 M = ?, N A = 6.02 × 10 23 mol –1 , Z = 4 (for fcc) Z ´ M N A ´ a 3 d ´ N A ´ a 3 or, M = Z 10.5 g cm - 3 ´ 6.022 ´ 10 23 ´ (4.07 ´ 10 -8 cm) 3 or, M = 4 or,M = 106.57 g mol –1 .

44. a = 400 pm

Z ´M Z ´ M Þ a 3 = a3 ´ N A d ´ N A

For fcc unit cell, Z = 4 Given, M = 207 g mol –1 , N A = 6.02 × 10 23 mol –1 d = 11.35 g cm –3 Substituting these value in equation (i), we get a 3 =

4 ´ 207 g mol -1 11.35 g cm -3 ´ 6.0 ´ 1023 mol -1

a3 =

4 ´ 207 ´ 10 cm 3 11.35 ´ 6.02 ´ 10 24

æ 8280 ö Þ a = ç è 11.35 ´ 6.02 ÷ø

1/3

´ 10 -8 cm

\ a = 4.949 × 10 –8 cm Þ a = 494.9 pm For fcc, r =

For example, unit cell of a cubic crystal is a = b = c, a = b = g = 90°.

a

2 2 400 400 2 400 2 = ´ = = 100 2 \ r = 4 2 2 2 2 2 Þ r = 100 × 1.414 = 141.4 pm

45. d =

Z ´ volume of one atom ´ 100 Volume of cubic unit cell 4 Z ´ pr 3 3 = ´ 100 a 3 For a simple cubic lattice, a = 2r and Z = 1 4 1 ´ pr 3 p \ Packing efficiency = 3 3 ´ 100 = ´ 100 = 52.4% (2r ) 6 48. (i) Unit cell : A unit cell is smallest portion of a crystal lattice which, when repeated in different direction, generates the entire lattice.

47. Packing efficiency =

Using formula d =

For fcc, r =

Z (No. of atoms per unit cell) ´ M (atomic mass) a 3 (cell edge) ´ N A (Avogadro number)

a 2 2

c

a b

...(i)

(ii) Coordination number : The number of nearest neighbours of any constituent particle in a packing is called its coordination number. The coordination number of an atom in the bcc structure is 8. 49. (i) For fcc (or ccp), a = 2 2r = 2 ´ 1.414 ´ 125 pm = 354 pm (ii) a = 354 pm = 3.54 × 10 –8 cm Volume of one unit cell = a 3 = (3.54 × 10 –8 cm) 3 = 4.436 × 10 –23 cm 3 Total volume Number of unit cells = Volume of one unit cell 1 cm 3 = = 2.25 ´ 10 22 4.44 ´ 10-23 cm 3 50. As the atoms just touch each other on the diagonal across the face of unit cell, therefore b 2 = a 2 + a 2 = 2a 2 , b = 2 a ...(i) Also, b = r + 2r + r = 4r ...(iii)


6

From (i) and (ii), we get 4r = 2 a Þ r =

2 1.414 ´ 409 pm a ; r = = 144.58 pm 4 4

52. Given : Structure fcc, hence Z = 4, r = 127.8 pm, M = 63.55 u, N A = 6.02 × 10 23 mol –1 , d = ?

r a

Z ´ M 2 ´ 56 g mol -1 = d ´ a 3 7.87 ´ (2.87 ´ 10 -8 cm) 3 or, N A = 6.022 × 10 23 .

or, N A =

b

Using formula, d =

4 ´ 63.55 g mol -1 6.022 ´ 10 mol -1 ´ (2 2 ´ 127.8 ´ 10 -10 ) 3 = 8.965 g/cm 3 For fcc structure, a = 2 2 r . OR

we get d =

a

51. Given : Structure = bcc, a = 316.5 pm, r = ? Using formula, r =

3a 1.732 ´ 316.5 = = 137.04 pm 4 4 OR

Given : For bcc, Z = 2 a = 286.65 pm = 2.87 × 10 –8 cm d = 7.87 g cm –3 , N A = ? Using formula d =

Z ´ M N A ´ a 3

Z ´ M N A ´ a 3 23

Given : For bcc, Z = 2 a = 286.65 pm = 2.87 × 10 –8 cm, d = 7.87 g cm –3 , N A = ? Using formula d =

Z ´ M N A ´ a 3

Z ´ M 2 ´ 56 g mol -1 = d ´ a 3 7.87 ´ (2.87 ´ 10 -8 cm) 3 23 or, N A = 6.022 × 10 .

or, N A =

JJJ

Solutions

7

Chapter2 Solutions

at 0.320°C, what would be the value of van’t Hoff factor? [K f for water is 1.86°C mol –1 ] [Delhi 2009]


State the main advantage of molality over molarity as the 13. 100 mg of a protein is dissolved in just enough water to make unit of concentration. [Delhi 2009 C] 10.0 mL of solution. If this solution has an osmotic pressure 2. What is meant by ‘reverse osmosis’? [AI SetIII 2011] of 13.3 mm Hg at 25°C, what is the molar mass of the protein? [R = 0.0821 L atm mol –1 K –1 and 760 mm Hg = 1 atm] 3. State Henry’s law about partial pressure of gas in a mixture. [AI 2009, Delhi 2009] [Foreign SetI 2011] 1.

4.

State Raoult’s law in its general form with respect to solution [Foreign SetIII 2011]


14. Calculate the amount of sodium chloride which must be added to one kilogram of water so that the freezing point of water is depressed by 3 K. (Given : K f = 1.86 K kg mol –1 , Atomic mass : Na = 23.0, Cl = 35.5) [Delhi 2009 C]

State Henry’s law correlating the pressure of a gas and its solubility in a solvent and mention two applications for the 15. A solution of urea in water has a boiling point of law. [AI 2008, Delhi 2008] 373.128 K. Calculate the freezing point of the same solution. [Given : For water K f = 1.86 K m –1 , K b = 0.52 K m –1 ] 6. Calculate the temperature at which a solutions containing [Delhi 2009 C] 54 g of glucose. (C H O ) in 250 g of water will freeze. (K 5.

6 12

6

of water = 1.86 K mol –1 kg). 7.

8.

f

[AI 2008]

State Raoult’s law for solutions of volatile liquids. Taking suitable examples explain the meaning of positive and negative deviations from Raoult’s law. [AI 2008, Delhi 2008] Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure. [AI 2008, Delhi 2008]

16. What is meant by positive and negative deviations from Raoult’s law and how is the sign of D mix H related to positive and negative deviation from Raoult’s case. [AI 2009 C] 17. Calculate the freezing point of a solution containing 18 g glucose, C 6 H 12 O 6 and 69.6 g sucrose, C 12 H 22 O 11 in 200 g of water. The freezing point of pure water is 273 K and K f for water is 1.86 K m –1 . [AI 2009 C]

18. A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 of benzene has a boiling 9. The boiling point elevation of 0.30 g acetic acid in 100 g point of 80.31°C. Determine the molar mass of this benzene is 0.0633 K. Calculate the molar mass of acetic acid compound. (Boiling point of pure benzene = 80.10°C and K b from this data. What conclusion can you draw about the for benzene = 2.53°C Kg mol –1 ) [Delhi 2010] molecular state of the solute in the solution? [Given : K b for benzene = 2.53 K kg mol –1 ]

[AI 2008 C]

10. Distinguish between the terms molality and molarity. Under what conditions are the molarity and molality of a solution nearly the same? [AI 2008 C, Delhi 2009, AI 2010]

19. A solution of glycerol (C 3 H 8 O 3 : molar mass = 92 g mol –1 ) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution? K b for water = 0.512 K kg mol –1 . [Delhi 2010, SetII 2012]

11. The freezing point of a solution containing 0.2 g of acetic 20. Nonideal solution exhibit either positive or negative from acid in 20.0 g of benzene is lowered by 0.45°C. Calculate Raoult’s law. What are these deviation and why are they (i) the molar mass of acetic acid from this data. caused? Explain with one example for each type. (ii) van’t Hoff factor. –1 [Delhi 2010 C] [For benzene, K f = 5.12 K kg mol ] What conclusion can you draw from the value of van’t Hoff 21. Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is factor obtained ? [AI 2008 C] the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses 12. Calculate the freezing point depression expected for 0.0711 of solutes in solutions? [AI 2010] m aqueous solution of Na 2 SO 4 . If this solution actually freezes

8


LONG ANSWER TYPE QUESTIONS (5 MARKS) 22. What mass of NaCl (molar mass = 58.5 g mol –1 ) be dissolved in 65 g water of lower the freezing point by 7.5°C ? The freezing 33. (a) The depression in freezing point of water observed for point depression constant K f for water is 1.86 K kg mol –1 . the same molar concentrations of acetic acid, Assume van’t Hoff factor for NaCl is 1.87. [AI 2010] trichloroacetic acid and trifluoroacetic acid increases

23. What mass of ethylene glycol (molar mass = 62.0 g mol –1 ) must be added to 5.50 kg of water to lower the freezing point of water from 0°C to – 10°C? [K f for water = 1.86 K kg mol –1 ] [AI 2010]

in the order as stated above. Explain. (b) Calculate the depression in freezing point of water when 20.0 g of CH 3 CH 2 CHClCOOH is added to 500 g of water. [Given K a = 1.4 × 10 –3 , K f = 1.86 K kg mol –1 ] [Delhi 2008 C]

24. 15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at – 0.34°C. 34. (a) What is meant by : What is the molar mass of the substance. –1 (i) Colligative properties (K f for water = 1.86 K kg mol ) (ii) Molality of a solution. [AI 2010, Delhi SetIII 2012] (b) What concentration of nitrogen should be present in a 25. State the following : glass of water at room temperature. Assume a (i) Raoult’s law in its general form in reference to solutions. temperature of 25°C, a total pressure of 1 atmosphere (ii) Henry’s law about partial pressure of a gas in a mixture. and mole fraction of nitrogen in air of 0.78. [AI SetI 2011] [K H for N 2 = 8.42 × 10 –7 M/mm Hg] [AI 2009] 26. A solution prepared by dissolving 8.95 mg of a gene fragment 35. (a) Differentiate between molarity and molality for a in 35.0 mL of water has an osmotic pressure of 0.335 torr at solution. How does a change in temperature influence 25°C. Assuming that the gene fragment is a nonelectrolyte, their values? calculate its molar mass. [AI SetI 2011] (b) Calculate the freezing point of an aqueous solution 27. What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50°C? The freezing point depression constant (K f ) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g). [AI SetII 2011] 28. Differentiate between molality and molarity values for a solution. What is the effect of change in temperature on molarity and molality values? [AI SetIII 2011] 29. Differentiate between molarity and molality of a solution. Explain how molarity value of a solution can be converted into its molality? [Foreign SetI 2011]

containing 10.50 g of MgBr 2 in 200 g of water. (Molar mass of MgBr 2 = 184 g) (K f for water = 1.86 K kg mol –1 ) OR (a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain. (b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.00 g of water. (K b for water = 0.512 K kg mol –1 ), (Molar mass of NaCl = 58.44 g) [Delhi SetI 2011]

36. (a) State the following : (i) Henry’s law about partial pressure of a gas in a mixture (ii) Raoult’s law in its general form in reference to solutions. (b) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is 31. A 1.00 molal aqueous solution of trichloroacetic acid a nonelectrolyte, determine its molar mass. (CCl 3 COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor OR –1 for trichloroacetic acid. (K b for water = 0.512 K kg mol ). (a) Differentiate between molarity and molality in a solution. What is the effect of temperature change on OR molarity and molality in a solution? Define the following terms: (b) What would be the molar mass of a compound if (i) Mole fraction (ii) Isotonic solutions 6.21 g of it dissolved in 24.0 g of chloroform to form (iii) van’t Hoff factor (iv) Ideal solution a solution that has a boiling point of 68.04°C. The [Delhi SetI 2012] boiling point of pure chloroform is 61.7°C and the 32. Calculate the amount of KCl which must be added to 1 kg of boiling point elevation constant, K b for chloroform is water so that the freezing point is depressed by 2 K (the K f 3.63°C/m. [Delhi SetII 2011] for water = 1.86 K kg mol –1 ). [Delhi SetI 2012] 30. A 0.561 m solution of an unknown electrolyte depresses the freezing point of water by 2.93°C. What is van’t Hoff factor for this electrolyte? The freezing point depression constant (K f ) for water is 1.86°C kg mol –1 . [Foreign SetI 2011]

Solutions

9

37. (a) Define the following terms: (i) Mole fraction (ii) Ideal solution (b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at –0.34°C. What is the molar mass of the material? (K f for water = 1.86 K kg mol –1 ) OR

(a) Explain the following: (i) Henry’s law about dissolution of a gas in a liquid. (ii) Boiling point elevation constant for a solvent. (b) A solution of glycerol (C 3 H 8 O 3 ) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution? (K b for water = 0.512 K kg mol –1 ). [AI SetI 2012]

a n s w e r s type of behaviour of solution is called positive deviation from Raoult’s law. e.g. Ethanol water. (ii) Negative deviation from Raoult’s law : When the If a pressure larger than the osmotic pressure is applied total vapour pressure is less than corresponding vapour on the solution side, the solvent starts to flow from the pressure, expected on the basis of Raoult’s law. This solution into the pure solvent through the semipermeable type of behaviour of solution is called negative deviation membrane. This phenomenon is called reverse osmosis. from Raoult’s law. e.g. Acetone chloroform. Henry’s law : It states that the partial pressure of the gas 8. p = CRT in vapour phase (p) is proportional to the mole fraction of æ W ö n W ´ R ´ T the gas (x) in the solution. n = B ÷ or, p = RT Þ p = B ç M B ø V M B ´ V è p = K H x W B = mass of the solute. Where K H is Henry’s law constant, p = Partial pressure of M B = molecular mass of the solute. gas and x = mole fraction. A known mass of a substance is added to known volume Raoult’s law : It states that for any solution the pressure of a solution and osmotic pressure of solution is determined. of each volatile component of the solution is directly Using formula, molecular mass can be calculated. proportional to its mole fraction. P A = X A .P° A and P B = X B P° B and P T = P A + P B 9. Given, w 1 = 100 g, w 2 = 0.30 g It states that at constant temperature the solubility of a gas DT b = 0.0633 K and K b = 2.53 K kg mol –1 in a liquid is directly proportional to the pressure of the gas. K ´ w 2 ´ 1000 Molar mass of acetic acid = ( M 2 ) = b P = K H .c DTb ´ w1 Applications : 2.53 ´ 0.30 ´ 1000 7.59 = = = 119.90 (i) In the production of carbonated beverages. 0.0633 ´ 100 0.0633 (ii) In the deep sea diving. Actual molar mass of acetic acid = CH 3 COOH (iii) In the functioning of lungs. = 12 + 3 + 12 + 16 + 16 + 1= 60 g mol –1 Molecular mass of glucose, M 2 = 72 + 12 + 96 = 180 Here, there is difference between calculated molar mass

1. Molality is independent of temperature, whereas molarity is a function of temperature. 2.

3.

4.

5.

6.

K f ´ W 2 ´ 1000 1.86 ´ 54 ´ 1000 = M 2 ´ W1 180 ´ 250 100440 = = 2.23 45000 Freezing point of solution = (0 – 2.23)°C = – 2.23°C. DT f =

7. It states that for a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. Nonideal solutions show positive and negative deviation from Raoult’s law. (i) Positive deviation from Raoult’s law : When the total vapour pressure is greater than corresponding vapour pressure expected on the basic of Raoult’s law. This

and actual molar mass 60 Normal molar mass = = 0.5 Abnormal molar mass 119.90 Here, i < 1, so the acetic acid is associated. 10. Molality : It is the number of moles of solute dissolve per kilogram (kg) of the solvent. It is denoted by m. Q i =

Molality ( m ) =

Moles of solute Mass of solvent in kg

w 2 1000 ´ M2 w1 where, w 1 = mass of solvent w 2 = mass of solute m =


10

M 2 = Molar mass of solute. K f = 1.86 K kg mol –1 , w 1 = 1000 g Molarity : It is the number of moles of solute dissolved Q DT f = K f m per liter of the solution. It is denoted by M K f ´ w 2 ´1000 DT f = Moles of solute w1 ´ M 2 M = Volume of solution in liter DT f ´ w1 ´ M 2 3 ´ 1000 ´ 58.5 w 2 1000 = Q w 2 = M = ´ 1.86 ´ 1000 K f ´ 1000 M 2 V 175.5 where, w 2 = mass of solute = = 94.3 g . 1.86 M 2 = molar mass of solute V = volume of solution. 15. DT b = 373.128 – 373 = 0.128 The molarity and molality of a solution will be nearly same, DT b = K b × m if the mass of solvent is nearly equal to the volume of DT b 0.128 solution. m = = K b 0.52 11. Given, w 1 = 20.0 g, w 2 = 0.2 g DT f = 0.45 °C, K f = 5.12 K kg mol –1 Now, DT b = K f × m 0.128 0.23808 K f ´ w 2 ´ 1000 = 1.86 ´ = = 0.458 K (i) Molar mass of acetic acid ( M 2 ) = 0.52 0.52 DT f ´ w1 Freezing point of solution = 273 – 0.458 5.12 ´ 0.2 ´ 1000 1024 = = 272.542 K. = = 113.77 0.45 ´ 20 9 16. (a) Positive deviations from Raoult’s law: When observed (ii) Actual molar mass of solute (acetic acid) vapour pressure of a liquid mixture is higher than the CH 3 COOH = 12 + 3 + 12 + 16 + 16 + 1 = 60 value expected from Raoult’s law it is called positive Normal molar mass deviation from Roult’s law. i = Van’t Hoff factor = Q Abnormal molar mass P T > c A.P ° + c B.P ° A A 60 For positive deviations from Raoult’s law = = 0.52 113.77 D mix H = +ve, DV mix = +ve, DP mix = +ve Here, i < 1, therefore the nonvolatile solute (acetic (b) Negative deviations from Raoult’s law : When observed acid) is associated. vapour pressure of a liquid mixture is lower than the 12. Given, m = 0.0711 molal value expected from Raoult’s law it is called negative DT f = 0.320°C, K f = 1.86°C mol –1 deviations from Raoult’s law. i = ? P T < c A P A° + c B P B° Q DT f = iK f m For negative deviations from Raoult’s law DT f D mix H = –ve, DV mix = –ve, DP = –ve 0.320 0.320 i = = = = 2.4. K f m 1.86 ´ 0.0711 17. Molar mass of C 6 H 12 O 6 = 12 × 6 + 1 × 12 + 16 × 6 0.132 = 72 + 12 + 96 = 180 g mol –1 13. V = 10.0 mL = 0.01 L Molar mass of C 12 H 22 O 11 = 12 × 12 + 1 × 22 + 16 × 11 13.3 p = 13.3 mm Hg = atm = 0.0175 atm = 144 + 22 + 176 = 342 g mol –1 760 18 68.4 T = 25°C = 273 + 25 = 298 K + Total number of moles = 180 342 –1 –1 R = 0.0821 L atm mol K = 0.1 + 0.2 = 0.3 (n B )

M = ? WRT 0.1 ´ 0.0821 ´ 298 = p V 0.0175 ´ 0.01 2.446 2446 1000000 = = ´ 0.000175 175 1000 = 13.977142 × 1000 = 13977.1 g mol –1 14. Given, Q M 2 of NaCl = 23 + 35.5 M 2 = 58.5, DT f = 3 K

Molar mass = M =

Q DT f = K f × m n ´ 1000 DT f = K f ´ B W A 1.86 ´ 0.3 ´ 1000 200 = 1.86 × 0.3 × 5 = 2.79 K DT f =

Solutions

18. Given, w 2 = 1.25 g, M 2 = 78.08 g mol –1 w 1 = 99.0 g Q DT b = (80.31 – 80.10)°C = 0.21°C = 0.21 K Q DT b = K b m K ´ w 2 ´ 1000 w ´ 1000 DTb = K b ´ 2 M 2 = b , M 2 ´ w1 DTb ´ w1 2.53 ´ 1.25 ´ 1000 3162.5 = = = 152.11 1 0.21 ´ 99 20.79 = 152 g mol –1 . Hence freezing point of aqueous solution = 273 – 2.79 = 270.21 K. 19. Given, w 1 = 500 g, M 2 = 92 g mol –1 w 2 = ? DT = (100.42 – 100)°C = 0.42°C = 0.42 K Q b Q DT b = K b.m w ´ 1000 DTb = K b ´ 2 M 2 ´ w1 0.512 ´ w2 ´ 1000 92 ´ 500 0.42 ´ 92 ´ 500 0.42 ´ 46 = Þ w2 = 0.512 ´ 1000 0.512 Q w 2 = 37.7 g So, mass of glycerol = 37.7 g.

Þ 0.42 =

20. When the vapour pressure of a nonideal solution is either higher or lower than that predicted by Raoult’s law, the solution exhibits deviation. These deviation are caused because of unequal intermolecular attractive force between solutesolvent molecules and solutesolute or solventsolvent molecules. Example of positive deviation : Mixture of ethanol and acetone, carbon disulphide and acetone. Example of negative deviation : Chloroform and acetone, nitric acid and water.

11

K f = 1.86 K kg mol –1 Using formula, DT f = i K f m DT f = i K f =

DT f ´ M 2 ´ w 1 w 2 ´ 1000 w 2 = , iK f ´ 1000 M 2 ´ w1

7.5 ´ 58.5 ´ 65 28518.75 = = 8.199 g 1.87 ´ 1.86 ´ 1000 3478.2

Hence mass of NaCl = 8.199 g. 23. Given, M 2 = 62.0 g mol –1 w 1 = 5.50 kg = 5500 g, w 2 = ? Q DT f = 0 – (–10.0) = 10°C or 10 K Now, DT f = K f m Þ DT f = K f ´

w 2 1000 ´ M2 w1

DT f ´ M 2 ´ w 1 K f ´ 1000 10 ´ 62.0 ´ 5500 62 ´ 55 = = 1.86 ´ 1000 1.86 = 1833.33 g = 1.833 kg. 24. Given, w 2 = 15 g, w 1 = 450 g K f = 1.86 K kg mol –1 , M 2 = ? Q DT f = 0 – (– 0.34) = 0.34°C or 0.34 K Now, DT f = K f .m

Þ w 2 =

Þ DT f = K f ´ Þ M 2 =

w 2 1000 ´ M2 w1

K f ´ w 2 ´ 1000 DT f ´ w1

1.86 ´ 15 ´ 1000 Þ M 2 = 0.34 ´ 450 Q M 2 = 182.35 g mol –1 Hence, molar mass of the substance = 182.35 g mol –1 .

25. (i) Refer answer 4 21. Osmosis : The net spontaneous flow of the solvent (ii) Refer answer 3. molecular from the solvent to the solution or from a less 26. Given, W B = 8.95 mg = 8.95 × 10 –3 g, concentrated solution through a semipermeable membrane R = 0.0821 L atm K –1 mol –1 , V = 35 × 10 –3 L is called osmosis. 0.335 Osmotic pressure : The external pressure that must be atm T = (25 + 273) K = 298 K, p = 760 applied to the solution side to stop the passage of solvent W ´ R ´ T into the solution through a semipermeable membrane is Substituting these value in the equation, M B = B known as osmotic pressure. p ´ V The osmotic pressure method has the advantage over other We get, method as pressure measurement is around the room 8.95 ´ 10-3 g ´ 0.0821 L atm K -1 mol-1 ´ 298 K M = B temperature and molarity of the solution is used instead of 0.335 atm ´ 35 ´ 10-3 L molality. As compared to other colligative properties, its 760 magnitude is large even for very dilute solution. M B = 14193.29 g mol –1 –1 22. Given, M 2 = 58.5 g mol , w 1 = 65 g, w 2 = ? DT f = 7.5°C, i = 1.87


12

27. DT f =

i ´ K f ´ W B ´ 1000 M B ´ WA

Given, i = 1.87, W A = 65.0 g, DT f = 7.50 K K f = 1.86 K kg mol –1 , M B = 58.5 g mol –1 Substituting these values in the above equation, we get 7.5 K =

WB =

1.87 ´ 1.86 K kg mol -1 ´ WB ´ 1000 g kg -1 58.5 g mol -1 ´ 65.0 g

7.5 ´ 58.5 ´ 65.0 g = 8.199 g 1.87 ´ 1.86 ´ 1000

(ii) Solutions having same osmotic pressure are called isotonic solution. (iii) (a) Van’t Hoff factor is the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to normal colligative property. (b) p = iCRT or,0.70 = i × 0.0103 × 0.082 × (27 + 273) 0.70 = 2.76 0.0103 ´ 0.082 ´ 300 (iv) Ideal solutions : A solution which obeys Raoult’s law at all concentrations and temperature is called an ideal solution.

or, i =

32. DT f = 2 K K f = 1.86 K kg mol –1 w solvent = 1 kg, DT f = i K b m moles of KCl i= ´ 1000 weight of solvent (g) moles of KCl DT f = ´ 1000 ´ i ´ Kb weight of solvent i = 2 for KCl 29. Molarity (M) is the number of moles of solute dissolved moles of KCl in one litre of solution whereas molality (m) is the number 2= ´ 1000 ´ 2 ´ 1.86 1000 of moles of the solute dissolved per kilogram of solvent. 1 If M B is the molar mass of solute, d is the density of solution Moles of KCl = 1.86 then molarity (M) value of a solution can be converted into g g 1 = = its molarity (m) by using the following formula. M.W. 74.5 1.86 74.5 1000 ´ M Weight = = 40 g m = 1.86 (1000 ´ d ) - ( M ´ M B ) 33. (a) The depression in freezing points are in the order: 30. Given: K f = 1.86°C kg mol –1 , m = 0.561 mol kg –1 acetic acid < trichloroacetic acid < trifluoroacetic acid DT f = 1.86°C kg mol –1 × 0.561 mol kg –1 = 1.04°C Cl F (DT f ) calculated = 1.04°C, (DT f) observed = 2.93°C C COOH < F C COOH CH3 COOH < Cl (DT f ) Observed Cl F Observed colligative property = \ i = Fluorine, being most electronegative, has the highest Calculated colligative property (D T f ) Calculated electron withdrawing inductive effect. Consequently, trifluoroacetic acid is the strongest acid while acetic Substituting these values in the equation, acid is the weakest acid. Hence, trifluoroacetic acid We get, ionises to the largest extent while acetic acid ionizes 2.93° C i= = 2.82 to the minimum extent to give ions in their solution in 1.04° C water. Greater the ions produced, greater is the 31. Given : Molality m = 1.0, T b = 100.18°C, i = ? depression in freezing point. K b = 0.512 K kg mol –1 , T b° = 100°C (known) Hence the depression in freezing point is maximum for Hence, DT b = T b – T b° = 100.18° – 100°C = 0.18°C the fluoroacetic acid and minimum for acetic acid. (b) Mass of CH 3 CH 2 CHClCOOH = 20 g DT b = i K b ∙ m Molar mass of CH 3 CH 2 CHClCOOH DT b 0.18 or i = = = 0.352 = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol –1 Kb × m 0.512 ´ 1.0

28. Molarity is the number of moles of solute dissolved in one litre of solution whereas molality is the number of moles of solute dissolved in one kilogram of the solvent. Molarity changes with increases in temperature as volume increases with increase in temperature. Molality is independent of temperature because mass does not depend on temperature.

OR (i) Molefraction of a component is the number of moles of that component divided by the sum of moles of all the components present in the solution.

= Number of moles of CH 3CH 2CHClCOOH

=

20 g 122.5 g mol-1

8 mol = 0.1632 mol 49

Solutions

13

0.1632 mol × 1000 500 = 0.3264 m If a be the degree of dissociation of CH 3CH 2CHCHClCOOH then

Molality of the solution (m) =

– + H + CH 3CH ƒ CH 3CH 2CHClCOOH 2CHClCOO

Initial conc. C mol L –1 At equilibrium C(1 – a)

Ca Ca

Ca Ca

Ca ´ C a C 2 a 2 = » C a 2 C (1 - a) C (1 - a) Q K a = Ca 2 K a =

K a 1.4 ´ 10 -3 = = 0.004289 = 0.065 C 0.3264 \ a = 0.065 To calculate van’t Hoff factor : a=

– + H + CH 3CH ƒ CH 3CH 2CHClCOOH 2CHClOO

Initial mol. 1 0 0 Moles at eq. 1 – a a a Total number of moles = 1 – a + a + a = 1 + a = 1 + 0.065 = 1.065 Q DT f = 2K f m = (1.065) (1.86)(0.3264) = 0.65° Hence, depression in freezing point of water = 0.65°.

(b) DT f =

i ´ K f ´ W B ´ 1000 M B ´ WA

i = 3, K f = 1.86 K kg mol –1 , W B = 10.5 g M B = 184 g mol –1 , W A = 200 g Substituting these values in above equation, we get 3 ´ 1.86 K kg mol-1 ´ 10.5 g ´ 1000 g kg -1 DT f = 184 g mol -1 ´ 200 g DT f = 1.59 K Therefore, freezing point of aqueous solution = 273.15 K – 1.59 K = 271.56 K Note : This problem can be solved without using van’t Hoff factor because nothing is mentioned in the problem regarding dissociation. OR (a) Osmosis : The spontaneous movement of the solvent molecules from the pure solvent or from a dilute solution to a concentrated solution through a semipermeable membrane is called osmosis. Osmotic Pressure : The minimum excess pressure that has to be applied on the solution to prevent the passage of solvent molecules into it through semipermeable membrane is called osmotic pressure. Osmotic pressure is a colligative property because it depends on the number of solute particles and not on their nature.

34. (a) (i) Colligative properties : Those properties which i ´ K b ´ W B ´ 1000 depend on number of solute particles present in (b) DT b = the solution, not on its nature are called colligative M B ´ WA properties. Given, i = 2, K b = 0.512 K kg mol –1 , W B = 15 g (ii) Molality : The number of moles of solute dissolved M B = 58.44 g mol –1 , W A = 250 g per kilogram (kg) of the solvent is called molality Substituting these values in the above equation, we get of the solution. 2 ´ 0.512 K kg mol -1 ´ 15g ´ 1000 g Kg -1 (b) T = 25°C + 273 = 298 K DTb = Total pressure = 1 atm 58.44 g mol -1 ´ 250 g c N 2 = k H × p N 2 = 1.05 K. Therefore, boiling point of aqueous solution p N 2 = partial pressure of nitrogen = 373.15 K + 1.05 K = 374.20 K k H = Henry’s constant c N 2 = mole fraction of N 2 Note : This problem can be solved without using van’t Hoff factor because nothing is mentioned in the problem c N 2 = 8.42 × 10 –7 × 0.78 × 760 regarding dissociation. p N 2 = 0.78 × 760 mm Hg c N 2 =4991.376 × 10–7 = 4.99 × 10 –4 36. (a) (i) Refer Ans. 3. nN 2 n H 2 (ii) Refer Ans. 4. c N 2 = = (b) Refer Ans. 26. nH2 O + nN2 n H 2 O OR n N or c N 2 = 2 (a) Refer Ans. 28. 1000 (b) K b = 3.63 K kg mol –1 , W B = 6.21 g 18 DT b = 68.04° C – 61.7° C = 6.34° C, W A = 24.0 g 1000 or n N 2 = ´ 4.99 ´ 10-4 = 2.77 ´ 10 -2 M Substituting these values in the equation 18 35. (a) Refer Ans. 28.


14

K b ´ W B ´ 1000 DTb ´ WA we get, M B =

n 2 n1 + n2 (ii) Ideal solutions : A solution which obeys Raoult’s law at all concentrations and temperature is called an ideal solution. (b) Refer Ans. 24. OR (a) (i) Henry’s law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas. (ii) Ebullioscopic constant is the boiling point elevation when one mole of solute is dissolved in 1000 g of solvent. It is denoted by K b . (b) Refer Ans. 19.

Mole fraction of solute c 2 =

3.63 K kg mol-1 ´ 6.21 g ´ 1000 g kg -1 6.34 K ´ 24.0 g M B = 148.15 g mol –1 . M B =

37. (a) (i) Mole fraction : It is the ratio of number of moles of one component to the total number of moles (solvent and solute) present in the solution. It is denoted by c. Let n 1 and n 2 be the number of moles of solute and solvent. Then,

JJJ

Electrochemistry

15

Chapter3 Electrochemistry


1.

What is primary cell? Give an example.

[Delhi 2008]

2.

Express the relation between the conductivity and the molar conductivity (L m ) of a solution. [Delhi 2008, AI 2008]

3.

What is limiting molar conductivity?

4.

Express the relation between conductivity and molar conductivity of a solution held in a cell? [Delhi SetI 2011]

[AI 2010]

10. Given that the standard electrode potential (E°) of metals are K + /K = – 2.93 V, Ag + /Ag = 0.80 V, Cu 2+ /Cu = 0.34 V, Mg 2+ /Mg = – 2.37 V, Cr 3+ /Cr = – 0.74 V, Fe 2+ /Fe = – 0.44 V Arrange these metals in an increasing order of their reducing power. [AI 2010]

11. Determine the value of equilibrium constant (K c ) and DG° for the following reactions: Ni (s) + 2Ag + (aq) ® Ni 2+ (aq) + 2Ag (s) , E° = 1.05 V (1 F = 96500 C mol –1 ) 5. Express the relation among the conductivity of solution in [Delhi SetI, 2011, Foreign SetI 2011] the cell, the cell constant and the resistance of solution in the cell. [Delhi SetIII 2011] 12. The chemistry of corrosion of iron is essentially an SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS)

6.

7.

Formulate the galvanic cell in which the following reaction takes place. Zn (s) + 2Ag + (aq) ® Zn 2+ (aq) + 2Ag (s) State : (i) Which one of the electrodes is negatively charged. (ii) The reaction taking place at each of its electrode. (iii) The carriers of current with in this cell. [Delhi 2008, AI 2008] Two half cell reactions of an electrochemical cell are given below: + – 2+ MnO 4– (aq) + 8H (aq) + 5e ® Mn + 4H 2O (l) E° = + 1.51 V Sn 2+ (aq) ® Sn 4+ (aq) + 2e – , E° + 0.51 V Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or products shown in the equation. [AI 2009]

electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere. [Delhi SetI 2011] 13. What type of a battery is lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery. [Delhi SetIII 2011] 14. Two halfreactions of an electrochemical cell are given below: MnO – + 8H + (aq) + 5e – ® Mn 2+ (aq) + 4H 2 O (l) , 4 (aq) E° = + 1.51V Sn 2+ (aq) ® Sn 4+ (aq) + 2e – , E° = + 0.15 V Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured. [AI 2010, Delhi SetIII 2011]

15. Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity? OR The molar conductivity of a 1.5 M solution of an electrolyte 8. A coppersilver cell is set up. The copper ion concentration 2 is found to be 138.9 S cm mol –1 . Calculate the conductivity in it is 0.10 M. The concentration of silver ion is not known. of this solution. [AI SetI 2012] The cell potential measured is 0.422 V. Determine the concentration of silver ion in the cell. 16. The electrical resistance of a column of 0.05 M NaOH ° ° solution of diameter 1 cm and length 50 cm is 5.55 × 10 3 Given : E + = 0.80 V, E 2+ = + 0.34 V Ag /Ag Cu /Cu ohm. Calculate its resistivity, conductivity and molar [AI 2009] conductivity. [AI SetI 2012] 9. A voltaic cell is set up at 25°C with the following half cells. Al|Al 3+ (0.001 M) and Ni|Ni 2+ (0.50 M) LONG ANSWER TYPE QUESTIONS (5 MARKS) Write the equation for the cell reaction that occurs when the cell generates an electric current and determine the cell 17. (a) Explain why electrolysis of an aqueous solution of NaCl gives H 2 at cathode and Cl 2 at anode. potential. E ° = – 2.71V, E °H 2O/H 2 = – 0.83 V (Given : E ° 2+ = – 0.25 V, E ° 3+ = –1.66 V ). Na + / Na Ni / Ni Al /Al E° = +1.36 V, E ° + = +1.23 V [AI 2008 C] Cl2 /Cl- 2H /H 2 O [Delhi 2009, Foreign SetI 2011, AI SetII 2012]


16

° (b) The resistance of a conductivity cell when filled with Calculate the cell voltage E Ni 2+ / Ni = – 0.25 V, 0.05 M solution of an electrolyte X is 100 ohms at ° E Al = –1.66 V. 3+ [Delhi 2009, AI 2009] /Al 40°C. The same conductivity cell filled with 0.01 M solution of an electrolyte Y has a resistance of 50 ohms. 23. (a) Express the relationship amongst cell constant, resistance of the solution in the cell and conductivity The conductivity of 0.05 M solution of electrolyte X –4 –1 of the solution. How is molar conductivity of a solute is 1.0 × 10 S cm . related to conductivity of its solution? Calculate : (i) Cell constant (b) Calculate the equilibrium constant for the reaction (ii) Conductivity of 0.01 (M) Y solution 2+ 2+ Fe (iii) Molar conductivity of 0.01 M Y [AI 2008] ( s ) + Cd ( aq ) ƒ Fe ( aq ) + Cd ( s ) [Given : E ° 2+ = – 0.40 V, E ° 2+ = – 0.44 V ] 18. (a) State Kohlrausch’s law of independent migration of ions. Cd

/Cd

Fe

/Fe

How can the degree of dissociation of acetic acid in a [Delhi 2009 optional] solution be calculated from its molar conductivity data? 24. (a) Define the term molar conducitivity. How is it related (b) (i) Formulate the electrochemical cell representing the to conductivity of a solution? reaction (b) One half cell in a voltaic cell is constructed from a 2Cr (s) + 3Fe 2+ (aq) ® 2Cr 3+ (aq) + 3Fe (s) silver wire dipped in silver nitrate solution of unknown (ii) Calculate E° cell concentration. Its other half cell consists of a zinc (iii) Calculate E cell at 25°C if electrode dipping in 1.0 M solution of Zn(NO 3 ) 2 . A [Cr 3+ ] = 0.1 M and Fe 2+ = 0.01 M voltage of 1.48 V is measured for this cell. Use this [Given : E° 3+ = – 0.74 V; E ° 2+ = – 0.44 V] information to calculate the concentration of silver Cr /Cr Fe /Fe nitrate solution. [AI 2008 C] é E° = – 0.76 V, E ° + = + 0.80 V ù 2+ 19. Three electrolytic cells A, B and C containing solutions of Ag /Ag ú ëê Zn .Zn û [Delhi 2009] Zinc sulphate, silver nitrate and copper sulphate respectively are connected in series. A steady current of 1.5 ampere is 25. (a) Explain why electrolysis of aqueous solution of NaCl passed through them until 1.45 g of silver is deposited at gives H 2 at cathode and Cl 2 at anode. Write overall reaction. the cathode of cell B. How long did the current flow? What ° [Given E ° + = – 2.71 V, E H = – 0.83 V , mass of copper and what mass of zinc were deposited on 2 O/H 2 Na / Na ° the concerned electodes? E° = +1.23 V ] – = + 1.36 V, E + Cl /Cl H /H 2 /H 2 O 2 (Atomic mass : Ag = 108, Zn = 63.5, Zn = 65.4 all in amu) (b) The resistance of a conductivity cell containing [AI 2008] 0.001 M KCl solution at 298 K is 1500 W. Calculate 20. (a) What type of a cell is the lead storage battery? Write the cell constant if conductivity of 0.001 M. KCl solution the anode and the cathode reactions and the overall at 298 K is 0.146 × 10 –3 S cm –1 . [Delhi 2009 C] reaction occurring in a lead storage battery while 26. (a) Calculate the emf for the given cell at 25°C. operating. [AI 2009, Delhi 2009] Cr|Cr 3+ (0.1 M) | |Fe 2+ (0.01 M)| Fe (s) 3+ (b) A voltaic cell is setup at 25°C with the half cells, Al/Al ° = – 0.74 V, E ° 2+ = – 0.44 V] [Given : E 3+ (0.001 M) and Ni/Ni 2+ (0.50 M). Write the equation for Cr

the reaction that occurs when the cell generates an electric current and determine the cell potential. [Delhi 2009]

/Cr

Fe

/Fe

(b) Calculate the strength of the current required to deposit 1.2 g of magnesium from molten MgCl 2 in 1 hour. [1 F = 96, 500 C mol –1 , Atomic mass Mg = 240] [Delhi 2009 C] optional

21. (a) Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such 27. (a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic change explained? acid at infinite dilution according to Kohlrausch law. (b) A voltaic cell is setup at 25°C with the following half cells. + 2+ Ag (0.001 M) |Ag and Cu (0.1 M)|Cu (b) Calculate : L °m for acetic acid What would be the voltage of this cell (E° cell = 0.46 V). Given that : L °m (HCl) = 426 S cm 2 mol -1 [Delhi 2009] L °m (NaCl) = 126 S cm 2 mol -1 22. (a) State the relationship amongst cell constant of a cell, [Delhi 2010] L °m (CH 3 COONa) = 91 S cm 2 mol -1 resistance of the solution in the cell and conductivity 28. (a) Write the anode and cathode reactions and the overall of the solution. How is molar conductivity of a solute reaction occurring in a lead storage battery. related to conductivity of its solution? (b) A coppersilver cell is set up. The copper ion (b) A voltaic cell is setup at 25°C with the following half cells: concentration is 0.10 M. The concentration of silver Al/Al 3+ (0.001 M) and Ni/Ni 2+ (0.50 M) ion is not known. The cell potential was 0.422 V.

Electrochemistry

17

Determine the concentration of silver ions in the cell. ° ° [Given : EAg + /Ag = + 0.80 V, E Cu 2+ /Cu = + 0.34 V ] [Delhi 2010 Optional] 29. (a) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery. (b) Calculate the potential for halfcell containing 0.10 M K 2 Cr 2 O 7(aq) , 0.20 M Cr 3+ (aq) and 1.0 × 10 –4 M H + (aq) The half cell reaction is 2– + – 3+ Cr 2O 7 (aq) + 14H (aq) + 6e ® 2Cr (aq) + 7H 2O (l) and the standared electrode potential is given as E° = 1.33 V. OR (a) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO 3 ) 2 solution with a current of 2.00 A for 3 hours? (b) A voltaic cell is set up at 25° C with the following half cells Al 3+ (0.001 M) and Ni 2+ (0.50 M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.

(Given : E°Ni2+ / Ni = - 0.25V, E°Al3 + /Al = - 1.66V ) [AI SetI 2011] 30. (a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it. (b) In the button cell, widely used in watches, the following reaction takes place. Zn 2+ (aq) + 2Ag (s) + 2OH – Zn (s) + Ag 2 O (s) + H 2 O (l) (aq)

Determine E° and DG° for the reaction. (Given : E° Ag + /Ag = + 0.80 V, E° Zn 2+ /Zn = – 0.76 V) OR

(a) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte. (b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 W. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10 –3 S cm –1 ? [Delhi SetI 2012]

a n s w e r s 1. Primary cells are those in which the redox reaction occurs only once and the cell becomes dead after sometime, and can not be recharged again. e.g. Dry cell, Mercury cell.

At cathode : Ag + (aq) + e – ® Ag (s) (Reduction) (iii) The current will flow from silver to zinc in the external circuit.

2. The relation between conductivity (k) and molar conductivity (L m ) of a solution is given by 7. (a) Half cell reactions 3 (i) At anode (oxidation) k ´ 10 L m = 5 × [Sn 2+ (aq) ® Sn 4+ (aq) + 2e – , E° = + 0.51 V] M (ii) At cathode (reduction) Where M is concentration of solution in molarity. 2 × [MnO 4 – (aq) + 8H + + 5e – 3. Molar conductance of a solution at infinite dilution is ® Mn 2+ (aq) + 4H 2 O, E° = + 1.51 V] called limiting molar conductivity. It is denoted by L °m . 5Sn 2+ (aq) + 2MnO 4 – (aq) + 16H + k ´ 1000 in CGS units = 5Sn 4+ (aq) + 2Mn 2+ + 8H 2 O 4. L m = M ° (b) Ecell = Ec° - Ea ° = 1.51 V – 0.51 V = 1.0 V k ´ 10 -3 L m = in SI units ° Ecell M is positive, hence the reaction favours formation where k is the conductivity, M is the molar concentration of product. and L m is molar conductivity. 8. The given cell may be represented as 1 ´ G (Cell constant) R (Resistance) 6. (i) The cell reaction is Zn (s) + 2Ag + (aq) ® Zn 2+ (aq) + 2Ag (s) The cell is represented as Zn (s) | Zn 2+ (aq) || Ag + (aq) | Ag (s) (ii) Anode i.e. zinc electrode will be negatively charged. At anode : Zn (s) ® Zn 2+ (aq) + 2e – (oxidation)

5. k (conductivity) =

Cu (s) |Cu 2+ (0.10 M)| | Ag + (C)| Ag (s) ° = E ° - E ° = 0.80 V – 0.34 V = 0.46 V Ecell c a ° Ecell = Ecell -

0.0591 [Cu 2+ ] log 2 [Ag + ] 2

0.0591 0.1 log 2 [Ag + ]2 0.1 - 0.038 V = - 0.0295 log [Ag + ]2

or, 0.422 V = 0.46 V -

18


9.

1 0.1 - 0.038 2 Fe 2 + + O 2 + 2 H 2O ® Fe 2O 3 + 4 H + = = 1.288 2 [Ag + ]2 - 0.0295 Fe 2O 3 + xH 2 O ® Fe 2O 3∙xH 2O (Rust) 0.1 = antilog 1.288 = 19.41 or, + 2 13. Lead storage battery is a secondary cell. [Ag ] Cell reactions 0.1 + 2 = 5.1519 ´ 10 -3 \ [Ag ] = At anode: Pb (s) + SO 4 2– (aq) ® PbSO 4(s) + 2e – 19.41 At cathode: PbO 2(s) + SO 2– + 4H + + 2e – ® [Ag + ] = 7.1 × 10 –2 M. 4 (aq) PbSO 4 + 2H 2 O From electrochemical series we know that Al will act as Overall cell reaction : anode and Ni as cathode. Pb (s) + PbO 2(s) + 2H 2 SO 4(aq) ® 2PbSO 4(s) + 2H 2 O (l) The cell reactions are At anode : 2Al (s) ® 2Al 3+ (aq) + 6e – 14. At anode: Sn2+ (aq) Sn4+(aq ) + 2e – ] × 5 2+ – At cathode : 3Ni (aq) + 6e ® 3Ni (s) At cathode: 2+ 3+ Over all reaction : 2Al (s) + 3Ni (aq) ® 2Al (aq) + 3Ni (s) MnO 4 2– (aq) + 8H + (aq) + 5e – ® Mn 2+ (aq) + 4H 2 O (l) ] × 2 The cell may be represented as Net cell reaction : Al (s) | Al 3+ (aq) | | Ni 2+ (aq) | Ni (s) 2MnO + 5Sn 2+ + 16H + ®

or, log

(aq)

(aq)

° Ecell = Ec° - Ea ° = - 0.25 V - ( - 1.66 V) = 1.41 V

0.0591 [Al3+ ] 2 log 2 + 3 2 ´ 3 [Ni ] -3 2 0.0591 (10 ) or, Ecell = 1.41 V log 6 (5 ´ 10 -1 ) 3 or, E cell = 1.41 + 0.0502 = 1.46 V. ° Using formula Ecell = Ecell -

(aq)

2Mn 2+ (aq) + 5Sn 2+ (aq) + 8H 2 O (l) E° cell = E° cathode – E° anode = 1.51 V – 0.15 V = 1.36 V. As cell potential is positive therefore the reaction is product favoured.

15. Relation between cell constant (G°) = k×R Relation between molar conductivity and conductivity 10. The reducing power increases with decreasing value of k ´ 10 3 L m = electrode potential. Hence the order is M Ag < Cu < Fe < Cr < Mg < K. OR 2 mol –1 , k = ?, M = 1.5 + 2+ Given : L = 138.9 S cm 11. Ni (s) + 2Ag (aq) ® Ni (aq) + 2Ag (s) , E° = 1.05 V m Here, n = 2 Using formula, log K c =

k ´ 10 3 M k ´ 103 cm3 L -1 138.9 S cm 2 mol -1 = 1.5 mol L-1 2 138.9 S cm mol -1 ´ 1.5 mol L -1 or k = 10 3 cm3 L-1 or k = 208.35 × 10 –3 S cm –1 .

Using formula L m =

nE °cell 0.059

2 ´ 1.05 = 35.5932 0.059 K c = antilog 35.5932 or K c = 3.92 × 10 35 Again, DG° = –nFE° cell DG° = –2 × 96500 × 1.05 = – 202650 J DG° = –202.65 kJ

or log K c =

16. Given : Diameter = 1 cm, length = 50 cm R = 5.5 × 10 3 ohm, M = 0.05 r = ? k = ? L m = ? 12. According to electrochemical theory of rusting the impure 2 3.14 2 æ1 ö iron surface behaves like small electrochemical cell. In cm Area of the column, a = pr 2 = 3.14 ´ è cm ø = 2 4 this any point of iron acts as anode and other iron surface Using formula : acts as cathode. Moisture having dissolved CO 2 or O 2 acts a 3.14 cm 2 3 as an electrolyte. The reactions are given below. Resistivity, r = R × = 5.5 ´ 10 ohm × At anode : Fe ® Fe 2 + + 2e - ; E °

= - 0.44 V .

At cathode : 1 2 H + + O 2 + 2e - ® H 2 O; E ° + = 1.23 V H /O 2 /H 2 O 2 Over all reaction: 1 O 2 ® Fe 2 + + H 2 O; E° cell = 1.67 V 2 The Fe 2+ ions are further oxidised by atmospheric oxygen to Fe 3+ ions, which forms hydrated ferric oxide (rust). Fe + 2 H + +

4 ´ 50 cm = 86.35 ohm cm 1 Again, conductivity, k = r 1 = = 1.158 ´ 10-2 ohm -1 cm -1 86.35 10 3 and molar conductivity, L m = k × M 10 3 -2 -1 = 1.158 ´ 10 ohm cm -1 × 5 ´ 10-2 –1 2 –1 = 231.6 ohm cm mol l

Fe 2 + /Fe

Electrochemistry

19

17. (a) Aqueous NaCl ionises as 0.0591 (0.1) 2 NaCl ® Na + + Cl – or Ecell = 0.30 V - log 6 (0.01) 3 (i) Following reactions are possible at cathode 2Na + (aq) + 2e – ® 2Na (s) ; E° = – 2.71 V = 0.30 V – 0.009 log 10 4 – – 2H 2 O (l) + 2e ® H 2(g) + 2OH (aq) ; E° = – 0.83 V = 0.30 V – 4 × 0.009 = 0.264 V. The reaction with higher E° value will take place, hence 19. Reaction for deposition of silver. H 2 is produced at cathode. Ag + + e – ® Ag (ii) Following reactions may take place at anode 1 mol 1 mol 2Cl – (aq) ® Cl 2(g) + 2e – ; E° = 1.36 V 1 F = 96500 C 108 g 1 108 g of silver is deposited by 96500 C of electricity Q + H 2 O (l) ® O 2( g ) + 2 H + 2e , E ° = 1.23 V \ 1.45 g of silver would be deposited by 2 The reaction with higher E° value will take place, hence 96500 ´ 1.45 = 1295.6 C Cl 2 is liberated at anode. 108 (b) (i) For electrolyte X Q = 1295.6 C Molarity M = 0.05 Given, current strength I = 1.5 A R = 100 ohm Q 1295.6 = = 863.7 s = 14.4 min \ t = k = 1.0 × 10 –4 S cm –1 I 1.5 Cell constant G° = k × R Reaction for deposition of zinc : or G° = 1.0 × 10 –4 S cm –1 × 100 ohm Zn 2+ + 2e – ® Zn = 1.0 × 10 –2 cm –1 . 2 mol 1 mol (ii) For electrolyte Y 2 × 96500 C 65.3 g Molarity M = 0.01 Thus 2 × 96500 C electricity deposited 65.3 g Zn Resistance R = 50 ohm \ 1295.6 C electricity would deposit Conductivity k = ? 65.3 ´ 1295.6 = 0.438 g Zn 1 2 ´ 96500 Using formula k = G ° × R Reaction for deposition of copper : 1 -2 - 1 Cu 2+ + 2e – ® Cu or, k = 1.0 ´ 10 cm ´ 50 ohm 2 mol 1 mol = 2 × 10 –4 ohm –1 cm –1 2 × 96500 C 63.5 g Thus 2 × 96500 C electricity deposited 63.5 g Cu k ´ 1000 Molar conductance L m = \ 1295.6 C electricity would deposit M -4 -1 -1 3 3 -1 63.5 ´ 1295.6 2 ´ 10 ohm cm ´ 10 cm L = 0.426 g Cu or, L m = - 1 2 ´ 96500 0.01 mol L or, L m = 20 ohm –1 cm 2 mol –1 18. (a) Kohlrausch law of independent migration of ions: The limiting molar conductance of an electrolyte is equal to the sum of the individual ionic conductances of the ions present in the electrolyte L °Ax B y = x ´ l°A + y ´ l °B Calculation of degree of dissociation of acetic acid. Degree of dissociation of a weak electrolyte (acetic acid) can be calculated using formula a=

L c m L o m

(b) (i) The given cell may be represented as Cr (s) |Cr 3+ (aq) | |Fe 2+ (aq) |Fe (s) (ii) E° cell = E° cathode – E° anode = – 0.44 V – (– 0.74 V) = 0.30 V (iii) E cell = E ° cell -

[Cr 3 + aq ] 2 0.0591 log 2 ´ 3 [Fe 2 + aq ] 3

20. (a) Refer Ans. 13. (b) Refer Ans. 9. 21. (a) (i) Weak electrolyte : When the concentration of weak electrolyte becomes very low, its degree of ionisation rises sharply. There is sharp increase in the number of ions in the solution. Hence the molar conductivity of a weak electrolyte rises steeply at low concentration. (ii) Strong electrolyte : The molar conductivity of strong electrolyte decreases slightly with the increase in concentration. This increase is due to increase in attraction as a result of greater number of ions per unit volume. With dilution the ions are far apart, interionic attractions become weaker and conductance increases. (b) The cell may be represented as Cu (s) |Cu 2+ (0.01 M)| |Ag + (0.001 M)| Ag (s) Using formula E = E °cell -

0.0591 [Cu 2+ ] log 2 [Ag + ] 2


20

E = 0.46 V - 0.0295 log

10 -1 (10 -3 ) 2

= 0.46 – 0.0295 × 10 5 = 0.46 – 0.0295 × 5 = 0.46 – 0.1475 = 0.3125 V. 22. (a) Refer Ans. 15. (b) Refer Ans. 9. 23. (a) Refer Ans. 15.

1 O 2( g ) + 2 H + + 2e - , E ° = 1.23 V 2 The reaction with higher E° value will take place, hence Cl 2 is liberated at anode.. (b) Given, C m = 0.001 M, R = 1500 W, k = 0.146 × 10 –3 S cm –1 , G* = ?

H 2 O (l) ®

Conductivity = Conductivity×Resistance Conductance = (0.146 × 10 –3 S cm –1 ) × (1500 W) = 0.219 cm –1 . 26. (a) (i) The given cell may be represented as Cr (s) |Cr 3+ (aq) | |Fe 2+ (aq) |Fe (s) (ii) E° cell = E° cathode – E° anode = – 0.44 V – (– 0.74 V) = 0.30 V

Cell constant =

° (b) Ecell = Ec° - Ea ° = - 0.40 V - ( - 0.44 V) = 0.04 V

Using formula log K c =

n E ° at 298 K 0.0591

2 ´ 0.04 V 0.0591 V or, K c = antilog 1.356 = 22.70.

or, K c = antilog

0.0591

[Cr 3 + ] 2

aq (iii) E cell = E ° cell 24. (a) Conductance of a solution containing 1 mole of log 2 ´ 3 [Fe 2 + aq ] 3 electrolyte is called molar conductance. It is denoted 0.0591 (0.1) 2 by L m . or Ecell = 0.30 V - log 6 (0.01) 3 Its unit is S cm 2 mol –1 or S m 2 mol –1 . 4 = 0.30 V – 0.009 log 10 1000 In C.G.S. unit L m = k × = 0.30 V – 4 × 0.009 = 0.264 V. M (b) Given, W = 1.2 g, t = 1 hr = 60 × 60 s, C = ? where L m = molar conductance, (24 g) 1 mole Mg is deposited by 2 × 96500 C M = molarity of solution, k = conductivity. Its unit is 2 ´ 96500 \ 1.2 g is deposited by ohm –1 cm 2 mol –1 . ´ 1.2 = 9650 C 24 -3 10 Again Q = I ∙ t In SI unit, L m = k × (where M = mol L –1 ) M Q 9650 C or, L m = k × V , where k = conductivity, V is volume of or, I = = = 2.68 A. t 60 ´ 60 s solution containing one mole of electrolyte. The unit 27. (a) Refer Ans. 18(a). is S m 2 mol –1 . (b) According to Kohlrausch law (b) The cell may be represented as Zn (s) |Zn 2+ (aq) | |Ag + (aq) |Ag (s) L °m (CH 3 COOH) = L °m (CH 3 COONa) + L °m (HCl) - L °m (NaCl) E ° = E - E ° = 0.80 V - ( - 0.76 V)

cell

c

a

or, L °m (CH 3 COOH) = 426 + 91 - 126 Scm 2 mol -1

= 1.56 V ° - Using formula Ecell = Ecell

or, 1.48 = 1.56 -

0.0591 [Zn 2+ ] log 2 [Ag + ] 2

0.0591 1 log 2 [Ag + ]2

or, log [Ag + ] = – 1.356 or, [Ag + ] = antilog (– 1.356) or, [Ag + ] = 4.406 × 10 –2 M 25. (a) Aqueous NaCl ionises as NaCl ® Na + + Cl – (i) Following reactions are possible at cathode 2Na + (aq) + 2e – ® 2Na (s) ; E° = – 2.71 V 2H 2 O (l) + 2e – ® H 2(g) + 2OH – (aq) ; E° = – 0.83 V The reaction with higher E° value will take place, hence H 2 is produced at cathode. (ii) Following reactions may take place at anode 2Cl – (aq) ® Cl 2(g) + 2e – ; E° = 1.36 V

= 391 S cm 2 mol –1 . 28. (a) Refer Ans. 13. (b) The given cell may be represented as Cu (s) |Cu 2+ (0.10 M)| | Ag + (C)| Ag (s) ° = E ° - E ° = 0.80 V – 0.34 V = 0.46 V Ecell c a ° Ecell = Ecell -

0.0591 [Cu 2+ ] log 2 [Ag + ] 2

0.0591 0.1 log 2 [Ag + ]2 0.1 - 0.038 V = - 0.0295 log [Ag + ]2

or, 0.422 V = 0.46 V -

or, log

0.1 - 0.038 = = 1.288 + 2 [Ag ] - 0.0295

Electrochemistry

21

(ii) Calculation of D r G o

0.1 = antilog 1.288 = 19.41 or, [Ag + ]2

o D r G o = – nFE cell

0.0591 [Cr 3 + ] 2 log n [Cr2 O 7 2 - ][H + ] 14

Given, E° cell = 1.33 V, n = 6, [Cr 3+ ] = 0.2 M [Cr 2 O 2– 7 ] = 0.1 M, [H + ] = 1 × 10 –4 M Substituting these values in the above given expression, we get 0.0591 (0.20) 2 Ecell = 1.33 V - log 6 (0.1) (10 -4 ) 14 0.0591 = 1.33 V log (4 ´ 10 55 ) 6 0.0591 = 1.33 V [log 4 + log 1055 ] 6 0.0591 = 1.33 V [log 4 + 55 log 10] 6 0.0591 = 1.33 V [0.620 + 55] 6 = 1.33 V – 0.548 V E cell = 0.782 V OR (a) Quantity of electricity passed Q = I × t = 2.0 A × 3 × 60 × 60 s = 21600 C 2 × 96500 C electricity produces 1 mole Hg 21600 = 0.112 mole of Hg \ 21600 C will produce = 2 ´ 96500 (b) Refer Ans. 9. 30. (a) Refer Ans. 13. (b) The cell reaction in button cell : Zn (s) + Ag 2 O (s) + H 2 O (l) ® Zn 2+ (aq) + 2Ag (s) + 2OH – (aq) o (i) Calculation of E cell

Reactions: Anode : Zn (s) ® Zn 2+ (aq) + 2e – Cathode : Ag 2 O (s) + H 2 O (l) + 2e – ® 2Ag (s) + 2OH – (aq) + 2e – n = 2 o o o Ecell = Ecathode - E anode o o = EAg2 O/Ag - EZn2+ / Zn = + 0.345 – (– 0.76) V = 1.105 V.

JJJ

–1

Ecell = E °cell -

2

29. (a) Refer Ans. 13. (b) For half cell reaction Cr 2 O 2– 7 (aq) + 14H + (aq) + 6e – ® 2Cr 3+ (aq) + 7H 2 O (l)

= – 2 × 96500 C mol –1 × 1.105 V = – 213265 CV mol –1 = – 213265 J mol –1 = – 213.265 kJ mol –1 . OR (a) The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by k (kappa). 1 l r= or k = G ´ k a Hence, conductivity of a solution is defined as the conductance of a conductor of 1 cm length and having 1 sq. cm as the area of cross section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte. Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm 3 of the solution. It is represented by L m . L m = kV Variation of conductivity and molar conductivity with concentration : Conductivity always decreases with decrease in concentration, for both weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution. L m /(S cm mol )

0.1 = 5.1519 ´ 10 -3 \ [Ag ] = 19.41 [Ag + ] = 7.1 × 10 –2 M. + 2

400

CH3 COOH

200 KCl 0

0.2 1/2

0.4

C (mol/L)

1/2

1/2

Molar conductivity versus C for acetic acid (weak electrolyte) and potassium chloride (strong electrolyte) in aqueous solutions. Molar conductivity increases with decrease in concentration. This is because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume. (b) Given, C m = 0.001 M, R = 1500 W, k = 0.146 × 10 –3 S cm –1 , G* = ? Conductivity = Conductivity× Resistance Conductance –3 = (0.146 × 10 S cm –1 ) × (1500 W) = 0.219 cm –1 .

Cell constant =


22

Chapter4 Chemical Kinetics

VERY SHORT ANSWER TYPE QUESTIONS (1 MARKS)

(i) Activation energy of the reaction (ii) Rate constant of the reaction.

[AI 2008 C] Define “order of the reaction”. 13. The data given below is for the reaction [AI, 2008, 2009, AI SetI 2011; Delhi 2009, 2010] 2N 2O 5(g) ƒ 4NO 2(g) + O 2(g) at 298 K 2. Why does the rate of a reaction not remain constant S. No. [N 2 O 5]/ Rate of disappearance throughout the reaction process? [Delhi 2009 C] (mol L –1 ) of N 2O L –1 min –1 ) 5/(mol 3. Define rate of reaction. [Delhi 2010 C] 1. 1.13 × 10 –2 34 × 10 –5 4. Define ‘activation energy’ of a reaction. [AI SetIII 2011] 2. 0.84 × 10 –2 25 × 10 –5 –2 3. 0.62 × 10 18 × 10 –5 SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS) 1.

For this reaction determine : What is meant by the rate constant ‘k’ of a reaction? If the (i) order of the reaction (ii) rate constant concentration be expressed in mol L –1 units and time in (iii) rate law [AI 2008 C] seconds. What would be the unit of k? 14. A reaction is second order with respect to a reactant. How (i) For a zero order reaction, will the rate of reaction be affected if the concentration of (ii) For a first order reaction. [AI 2008] this reactant is 6. Distinguish between order and molecularity of a reaction. (i) doubled (ii) reduced to half. [Delhi 2009] Could order and molecularity of a reaction be the same? 5.

[Delhi 2008 C] 15. A first order reaction has a rate constant of 0.0051 min –1 . If we begin with 0.10 M concentration of the reactant, what 7. A first order decomposition reaction takes 40 minutes for concentration of reactant will remain in solution after 30% decomposition. Calculate t 1/2 value for it. 3 hours? [Delhi 2009, AI 2009] [Delhi 2008, AI 2008] 8. (a) Derive the general form of the expression for the half 16. Define the following : (i) Elementary step in a reaction life of first order reaction. (ii) Rate of a reaction [AI 2009] (b) The decomposition of NH 3 on platinum surface is a zero order reaction. What are the rates of production 17. For a decomposition reaction the values of rate constant k at of N 2 and H 2 if k = 2.5 ´ 10 –4 mol L –1 s –1 ? two different temperatures are given below [Delhi 2008, AI 2008] k 1 = 2.15 × 10 –8 L mol –1 s –1 , T 1 = 650 K k 2 = 2.39 × 10 –7 L mol –1 s –1 , T 2 = 700 K 9. List the factors which influence the rate of a chemical reaction. Calculate the value of activation energy for this reaction [Delhi 2008] Optional [R = 8.314 J K –1 mol –1 ] [AI 2009] 10. The half life for decay of radioactive C14 is 5730 years. An 18. Explain the term “order of a reaction”. Derive the unit for archeological artifact containing wood has only 80% of the first order rate constant. [Delhi 2009 C] C14 activity as found in living trees. Calculate the age of the artifact. [AI 2008] 19. Show that for a first order reaction the time required for half the change (half life period) is independent of initial 11. The decomposition of phosphine concentration. [Delhi 2009] 4PH ® P + 6H has the rate law, 3(g)

4(g)

2(g)

20. What is the effect of temperature on the rate constant of a Rate = k[PH 3 ] –4 –1 reaction? How can this temperature effect on rate constant The rate constant is 6.0 × 10 s at 300 K and activation 5 –1 be expressed quantitatively? [AI 2009 C] energy is 3.05 × 10 J mol . Calculate the value of rate constant at 310 K. [Given : R = 8.314 J K –1 mol –1 ] 21. The half life period for a first order reaction is 5 × 10 4 s. [Delhi 2008 C] What percentage of the initial reactant will react in 2 hours? [AI 2009 C] 12. For a chemical reaction, what is the effect of a catalyst on the following :

Chemical Kinetics

23

22. A first order reaction is 20% complete in 5 minute. Calculate the time taken for the reaction to be 60% complete. [AI 2009 C]

Exp. Initial No. [NO] (M) 1

23. A reaction is of first order in reactant ‘A’ and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times (ii) the concentration of A as well as B are doubled. [Delhi 2010] 24. The rate constant for a reaction of zero order in A is 0.0030 mol L –1 s –1 . How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M? [Delhi 2010] 25. Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction. [Delhi SetI, 2011]

31. 26. Nitrogen pentaoxide decomposes according to equation: 2N 2O 5(g) ® 4NO2(g) + O2(g) This first order reaction was allowed to proceed at 40° C 32. and the data below were collected: [N 2 O 5 ] (M)

Time (min)

0.400 0.289 0.209 0.151 0.109

0.00 20.0 40.0 60.0 80.0

(a) Calculate the rate constant. Include units with answer.

Initial [Cl 2 ] (M)

Initial rate of disapp. of Cl 2 (M/min)

0.15

0.15

0.60

2

0.15

0.30

1.20

3

0.30

0.15

2.40

4

0.25

0.25

?

(a) Write the expression for rate law. (b) Calculate the value of rate constant and specify its units. (c) What is the initial rate of disappearance of Cl 2 in exp. 4? [Delhi SetI 2012] A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half? The reaction, N 2(g) + O 2(g) 2NO (g) contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 × 10 –5 . Suppose in a case [N 2 ] = 0.80 mol L –1 and [O 2 ] = 0.20 mol L –1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K. [AI SetI 2012] LONG ANSWER TYPE QUESTIONS (5 MARKS)

(b) What will be the concentration of N 2 O 5 after 100 33. (a) Explain the following terms : minutes? (i) Rate of a reaction (c) Calculate the initial rate of reaction. (ii) Activation energy of a reaction [Delhi SetI 2011] (b) The decomposition of phosphine (PH 3 ), proceeds according to the following equation 27. What do you understand by the rate law and rate constant of 4PH 3(g) ® P 4(g) + 6H 2(g) a reaction? Identify the order of a reaction if the units of its It is found that the reaction follows the following rate rate constant are : equation : Rate = k[PH 3 ] (i) L –1 mol s –1 (ii) L mol –1 s –1 The half life period of PH 3 is 37.9 s at 120°C. [AI SetI 2011, Delhi SetI 2012] 3 28. The thermal decomposition of HCO 2 H is a first order reaction (i) How much time is required for th of PH 3 to 4 with a rate constant of 2.4 × 10 –3 s –1 at a certain temperature. decompose? Calculate how long will it take for threefourths of initial (ii) What fraction of the original sample of PH 3 quantity of HCO 2 H to decompose? (log 0.25 = – 0.6021) remains behind after 1 minute? [AI SetI 2011] [AI 2010, Delhi 2010] 29. A reaction is of second order with respect to a reactant. How 34. (a) Explain the following terms : is the rate of reaction affected if the concentration of the (i) Order of a reaction reactant is reduced to half? What is the unit of rate constant (ii) Molecularity of a reaction (b) The rate of a reaction increases four times when the for such a reaction? [AI SetII 2011] temperature changes from 300 K to 320 K. Calculate 30. For the reaction : the energy of activation of the reaction assuming that 2NOCl (g) 2NO (g) + Cl 2(g) it does not change with temperature. the following data were collected. All the measurements were [R = 8.314 J K –1 mol –1 ] [AI 2010 Optional] taken at 263 K. 35. (a) Express clearly what you understand by ‘rate expression’


24

and ‘rate constant’ of a reaction. (b) Nitrogen pentoxide decomposes according to the equation 2N 2O 5(g) ® 4NO 2(g) + O 2(g) This first order reaction was allowed to proceed at 40°C and the data given below were collected: [N 2 O 5 ] (M) Time (min) 0.400 0.00 0.289 20.00 0.209 40.00 0.151 60.00 0.109 80.00

(i) Calculate the rate constant for the reaction. Include units with your answer. (ii) Calculate the initial rate of reaction. (iii) After how many minutes will [N 2 O 5 ] be equal to 0.350 M? OR (a) Define (i) Order of a reaction (ii) Elementary step in a reaction. (b) A first order reactions has a rate constant value of 0.00510 min –1 . If we begin with 0.10 M concentration of the reactant, how much of the reactant will remain after 3.0 hours? [Foreign SetI 2011]

a n s w e r s 1. The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.

=

2.303 10 2.303 log = (log 10 - log 7) 40 7 40

2.303 2.303 (0.155) (1 - 0.845) = 40 40 0.693 0.693 ´ 40 27.72 t 1/ 2 = = = k 2.303 ´ 0.155 0.357 3. Change in concentration of reactant or product per unit = 77.7 minutes. time is known as rate of reaction. 8. (a) At t 1/2 4. The extra energy required by the reactant to form the [ R ] activated complex is known as activation energy. Activation [ R] = 0 2 energy = Threshold energy – Average energy of the reactants Applying rate equation of first order reaction 5. Rate constant ‘k’ : The rate constant k is equal to the rate [ R 0 ] 2.303 of reaction when the concentration of the reactant is unity. k = log t1/ 2 [ R0 ] / 2 Unit of k :

2. With the progress of reaction, concentration of reactant decreases, hence rate of reaction also decreases. It can not remain constant.

(i) For a zero order reaction : mol L –1 s –1 (ii) For a 1st order reaction : s –1 . 6.

Order of reaction

Molecularity of reaction

1. Sum of powers of concentration of the reactants present in the rate law expression is called order. 2. It is experimental property. 3. Its value may be fractional or negative.

Number of reacting species which collide simul taneously to give a chemical reaction is called molecularity. It is a theoretical property. It is always a whole number.

In elementary reaction (single step process) order and molecularity are same. 7.

k =

[ R ] 2.303 2.303 100 log log 0 = 40 70 t [ R ]

=

2.303 log 2 0.693 Þ t 1/2 = k k (b) The reaction is 2 NH3( g ) ƒ N 2( g ) + 3H 2( g ) . Here : k = 2.5 × 10 –4 mol L –1 s –1 The order of reaction is 0 i.e., Rate = k [Reactant] 0 Rate = 2.5 × 10 –4 × 1 = 2.5 × 10 –4 mol L –1 s –1 . d [N 2 ] 1 d [H 2 ] = . Q Rate of reaction = dt 3 dt The rate of formation of N 2 = 2.5 × 10 –4 mol L –1 s –1 . 1 d [H 2 ] Again, 2.5 ´ 10 -4 = 3 dt d [H 2 ] -4 \ = 7.5 ´ 10 mol L-1 s -1 dt Therefore, rate of formation of H 2 = 7.5 × 10 –4 mol L –1 s –1 t 1/ 2 =

9. The rate of reaction depends upon the following factors:

Chemical Kinetics

25

(i) Nature of the reactants. (ii) Concentration of the reactants. (iii) Temperature. (iv) Presence of catalyst. (v) Surface area of reactant. (vi) Presence of light. 10. Radioactive processes follow first order kinetics. Let [R] 0 = 100 \ [R] = 100 × 80% = 80 Decay constant k =

0.693 0.693 = years –1 t1/2 5730

[ R ] 2.303 2.303 100 t = log 0 = log k [ R] 0.693 / 5730 80 2.303 ´ 5730 log 1.25 0.693 2.303 ´ 5730 ´ 0.0969 = = 1845.18 years. 0.693 11. k 1 = 6.0 × 10 –4 s –1 , T 1 = 300 K, E a = 3.05 × 10 5 J mol –1 =

k 2 = ?, T 2 = 310 K k2 E a é T2 - T 1 ù = ê ú k1 2.303 R ë T1 T2 û é 310 - 300 ù k 3.05 ´ 105 J mol -1 ´ê or, log 2 = ú -1 -1 k1 2.303 ´ 8.314 J mol K ë 300 ´ 310 û

Using Arrhenius equation; log

or, log

k 2 3.05 ´ 10 3 = = 1.7128 k1 19.147 ´ 93

k 2 = antilog (1.7128) = 51.62 k1 k2 or, 6.0 ´ 10 -4 = 51.62

or,

or k 2 = 3.1 × 10 –2 s –1 . 12. (i) A catalyst decreases the activation energy. (ii) A catalyst does not change the rate constant of a reaction. 13. (i) Let rate law for the disappearance of N 2 O 5 is Rate = k[N 2 O 5 ] a Then

r 1 34 ´ 10-5 k (1.13 ´ 10-2 ) a = = r 2 25 ´ 10 -5 k (0.84 ´ 10 -2 ) a a

34 æ 113 ö =ç ÷ or a = 1 25 è 84 ø Thus order of reaction = 1. (ii) For trial (1) r = k[N 2 O 5 ] or, 34 × 10 –5 = k × 1.13 × 10 –2

or,

34 ´ 10 -5 = 3.0 ´ 10 -2 min -1 1.13 ´ 10 -2 (iii) Rate law = k[N 2 O 5 ].

or, k =

14. Rate = k[R] 2

(i) If [A] = 2R, rate = k[2R] 2 = 4k[R] 2 i.e. rate of reaction increases 4 times when concentration is doubled. 2

1 1 é R ù 2 (ii) If [ A] = R, rate = k ê ú = k [ R ] i.e., 2 4 ë 2 û 1 rate becomes th when concentration is reduced to 4 1 . 2 15. Given : k = 0.0051 min –1 , [R] 0 = 0.10 M, [R] = ? t = 3 hours = 3 × 60 = 180 min

[ R ] 0 2.303 log 10 t [ R ] 2.303 0.10 or, 0.0051 min -1 = log 10 180 min [ R]

Using formula, k =

0.10 0.0051 min -1 ´ 180 min = = 0.3986 [R] 2.303 0.010 = antilog [0.3986] or, [ R]

or, log10

or,

0.010 0.10 , [ R] = = 2.503 or, = 0.039 M . [ R] 2.503

16. (i) Elementary step in a reaction : Each step of a complex reaction (multi step process) is called elementary step in a reaction. (ii) Rate of a reaction : The change in concentration of any one of the reactant or product per unit time is called rate of reaction. 17. Given : k 1 = 2.15 × 10 –8 L mol –1 s –1 , T 1 = 650 K k 2 = 2.39 × 10 –7 L mol –1 s –1 , T 2 = 700 K E a = ?, R = 8.314 J K –1 mol –1 é T2 - T 1 ù ê ú ë T2 ´ T1 û Ea é 700 - 650 ù 2.39 ´ 10 -7 L mol -1 s -1 = or, log ê ú -8 -1 - 1 2.15 ´ 10 L mol s 19.147 ë 700 ´ 650 û k E a Using Arrhenius equation, log 2 = k1 2.303 R

Ea 50 ´ 19.147 455000 or, E a = 1.0457 × 19.147 × 9100 = 182.2 kJ.

or, log11.11 =

18. Order of reaction : Sum of powers of concentration terms present in the rate equation is called order of reaction. Rate = k×C n where n = order of reaction Units for first order rate constant : For first order reaction, Rate = k×C Rate mol L- 1 s - 1 or, k = = = s - 1 . C mol L-1 [ R ] 0 2.303 19. For first order reaction k = log 10 t [ R ] [ R ] 0 when t = t1/ 2 , [ R ] = 2


26

or, k =

[ R ] 0 2.303 2.303 or, k = log 10 log10 2 t1/ 2 [ R0 ] / 2 t1/ 2

[R] 0 = 0.10 M, [R] = 0.075 M Using formula t =

[ R ]0 - [ R ] 0.10 M - 0.075 M = k 0.0030 mol L-1 s -1

2.303 log10 2 2.303 ´ 0.3010 0.693 0.025 = = t = = 8.3 s. k k k 0.0030 Thus half life period of first order reaction is independent 25. Rate expression is an experimentally determined relation of initial concentration. between the rate of reaction and the concentration of the 20. Rate constant of a reaction increases with increasing reactants whereas rate constant is the rate of reaction when temperature. concentration of each reactant in the reaction is unity. Temperature effect on rate constant is expressed The rate expression for the reaction quantitatively by Arrhenius equation e.g. aA + bB ® cC + dD E Rate = k[A] m [B] n k = Ae RT . Where the proportionality constant k is called rate constant. 21. Given : t 1/2 = 5 × 10 4 s, t = 2 hrs = 2 × 60 × 60 s 26. (a) When t = 20 min, [R] = 0.289 mol L –1 [ R ] [R] 0 = 0.400 mol L –1 = ? [ R0 ] For a first order reaction

or, t 1/ 2 =

a

[ R ] 0 2.303 log 10 t [ R ] [ R ] 0 0.693 2.303 = log 10 or, 5 ´ 10 4 s 2 ´ 60 ´ 60 s [ R ] [ R ] 0 0.693 ´ 2 ´ 3600 or, log = = 0.0433 [ R ] 5 ´ 10 4 ´ 2.303 [ R ] 0 = antilog = 0.0433 = 1.105 or, [ R] 1 ´ 100 = 90.49 %. % of reactant reacted = 1.105 22. Given : [R] 0 = 100, [R] 1 = 80 when t 1 = 5 min [R] 2 = 40 then t 2 = ? For a first order reaction k =

[ R ] 2.303 log 0 t [ R ] 2.303 100 2.303 100 \ log10 = log 10 5 80 t 2 40 log 2.5 1 1 or, log1.25 = log 2.5 or, t 2 = 5 ´ log 1.25 5 t2 0.3979 or, t 2 = 5 ´ = 20.53 min . 0.0969 23. Given : Rate, r 1 = k[A] [B] 2 (i) When B increases three times Rate, r 2 = k[A] [3B] 2 = k[A] 9[B] 2 = k[A] [B] 2 × 9 or, r 2 = 9r 1 i.e., rate of reaction increases 9 times (ii) when A and B both are doubled then Rate r 3 = k[2A] [2B] 2 or, r 3 = k 2[A] × 4[B] 2 or, r 3 = k A[B] 2 × 8 or, r 3 = 8 r 1 i.e., rate of reaction increases 8 times. For a first order reaction k =

24. Given : k = 0.0030 mol L –1 s –1 , n = 0

k =

[ R ] 2.3030 log 0 t [R]

\ k =

2.303 0.400 2.303 4.00 log log Þ k = 20 0.289 20 2.89

Þ k =

2.303 [log4.00 - log 2.89] 20

Þ k =

2.303 [0.6021 - 0.4609] 20

2.303 ´ 0.1412 20 Þ k = 2.303 × 0.00706 Þ k = 1.626 × 10 –2 min –1

Þ k =

(b) t =

[ R ] 2.303 log 0 k [ R]

Here, [R] 0 = 0.400 mol L –1 , t = 100 min, k = 1.626 × 10 –2 min –1 , [R] = ? 2.303 0.400 log \ 100 = -2 1.626 ´ 10 [ R] 100 ´ 1.626 ´ 10 -2 0.4 = log 2.303 [ R ] 0.4 0.7060 = log [ R]

or Antilog (0.7060) =

0.4 [ R]

0.4 0.4 = 0.079 M Þ [ R] = [ R] 5.082 (c) Initial rate, i.e., rate of reaction when t = 0 When t = 0.00 min, [R] = 0.400 mol L –1 Also, k = 1.626 × 10 –2 min –1 5.082 =

Chemical Kinetics

27

(a) Let rate of this reaction r = k[NO] m [Cl 2 ] n

\ Initial rate = k[R] = 1.626 × 10 –2 min –1 × 0.400 mol L –1 = 6.504 × 10 –3 mol L –1 min –1 27. An experimentally determined expression which relates the rate of reaction with the concentration of reactants is called rate law. The rate of reaction when concentration of each reactant is unity in a rate law expression is called rate constant. (i) For n th order reaction unit of rate constant is (mol L –1 ) 1–n s –1 ,= mol 1–n L n–1 s –1 Given unit of k = mol L –1 s –1 Thus, 1 = 1 – n Þ n = 0 i.e.,zero order reaction (ii) Again given unit is L mol –1 s –1 28. For a first order reaction, [ R ] 2.303 t = log 0 k [ R] Given k = 2.4 × 10 –3 s –1 [ R ] 0 [ R] = , t = ? 4 Substituting these values in the equation, we get [ R ] 2.303 t = log 0 -3 -1 2.4 ´ 10 s [ R ] 2.303 2.303 t = log 4 = ´ 0.6021 s -3 - 1 2.4 ´ 10 s 2.4 ´ 10 -3 t = 577.7 s = 578 s.

n

or,

1 æ 1 ö = 2 è 2 ø

Þ n = 1 r 2

m 1 1.20 k (0.15) (0.30)

Again from r = 2.40 = k (0.30)m (0.15) 1 3 m

or

1 æ1ö 2 = × 2 è2ø 1

m

or

1 æ 1 ö = 4 è 2 ø

Þ m = 2

Hence expression for rate law is r = k[NO] 2 [Cl 2 ] 1 (b) Substituting the values in experiment (i), 0.60 M min –1 = k(0.15 M) 2 (0.15 M) 1 or k =

0.60 M min -1 0.0225 ´ 0.15 M 3

= 177.77 M -2 min -1

(c) r = 177.7 M –2 min –1 × (0.25 M) 2 (0.25 M) = 2.77 M min –1 . 31. Since the reaction is second order with respect to a reactant say A. The rate law is : Rate = k[A] 2 Let the concentration of A = R mol L –1 \ Rate = r 1 = kR 2 (i) When the concentration of A is doubled, [ A] = 2R

29. Consider the reaction R ® Products As the reaction is of second order with respect to the reactant \ Rate, r = k [R] 2 ...(i) If the concentration of the reactant reduced to half, then 2

é R ù Rate, r ¢ = k ê ú ë 2 û 2 [ R ] or r ¢ = k . 4 Dividing equation (ii) by (i), we get

m n 0.60 k (0.15) (0.15)

r 1

then r = 1.20 = k (0.15)m (0.30) n 2

...(ii)

Rate = r 2 = k(2R) 2 = 4kR 2 = 4r 1 Thus, rate is increased to 4 times to the original rate. (ii) When the concentration of A is reduced to half, 1 [ A] = R 2 2 r 1 æ1 ö Rate = r3 = k ç R ÷ = kR 2 = 1 è 2 ø 4 4 Thus, the rate is decreased to one fourth of the original rate. 32. Given reaction is N 2(g) + O 2(g)

r¢ [ R ] 2 1 = = r 4[ R ] 2 4

1 of the initial rate. 4 th The unit of rate constant is mol –1 L s –1 . 30. Given reaction : 2NO (g) + Cl 2(g) ® 2NOCl (g)

2NO (g) ; K = 1.0 × 10 –5

Initially : 0.80 mol L –1 0.20 mol L –1 At equil.: 0.80 – x 0.20 – x

i.e., rate of reaction becomes

Exp. Initial No. [NO] (M)

Initial [Cl 2 ] (M)

Initial rate of disapp. of Cl 2 (M/min)

1

0.15

0.15

0.60

2

0.15

0.30

1.20

3

0.30

0.15

2.40

4

0.25

0.25

?

According to law of mass action, K =

0 2x

[NO] 2 [N 2 ][O 2 ]

(2 x ) 2 (0.80 - x)(0.20 - x ) x is very small hence can be neglected in the denominator.

or 1.0 ´ 10 -5 =

4 x2 4 x 2 or 1.0 ´ 10 -5 = 0.8 ´ 0.2 0.16 or, 4x 2 = 1.0 × 10 –5 × 0.16 or 4x 2 = 16 × 10 –7 or, x 2 = 4 × 10 –7 or x = 6.32 × 10 –4 –1 \ [N 2] = 0.80 – 0.000632 = 0.7994 mol L –1 [O 2] = 0.20 – 0.000632 = 0.199 mol L [NO] = 2x = 2 × 0.000632 = 0.001264 mol L –1 .

Thus, 1.0 ´ 10

-5

=


28

33. (a) (i) Refer Ans. 3. (ii) Refer Ans. 4. (b) Given : Rate = k[PH 3 ], t 1/2 = 37.9 s

or, 0.6020 =

or, E a = 0.6020 × 19.147 × 4800 E a = 55.327 kJ mol –1 . 35. (a) Refer Ans. 25. (b) (i) Refer Ans. 26 (a). (ii) Refer Ans. 26 (c).

1 (i) t = ? when [ R ] = [ R0 ] 4 n [ R ] æ 1 ö Using formula =ç ÷ [ R ]0 è 2 ø 1 n n [ R] 0 1 æ 1 ö æ 1 ö 4 = ç ÷ or, = çè ÷ø or, n = 2 4 2 è 2 ø [R]

(iii) t =

0

Again t = n × t 1/2 = 2 × 37.9 s = 75.8 s. (ii) t = 60 s (1min .) =

[ R ] = ? [ R] 0

[ R ] 0 0.693 2.303 = log 10 37.9 s 60 s [ R ]

=

[ R ] [ R ] = antilog ( - 0.476) = - 0.476 or, [ R ] 0 [ R] 0

[ R ] = 0.3342. [ R ] 0

34. (a) (i) Refer Ans. 1. (ii) Molecularity of reaction : The number of reacting species which must collide simultaneously to give a chemical reaction is called molecularity of a reaction. (b) Given : r 2 = 4r 1 , T 1 = 300 K, T 2 = 320 K E a = ?

or, log

2.303 [log 40log 35] 1.6259 ´ 10 -2

= 141.583 [1.6021 – 1.5441] = 141.583 × 0.0580 = 141.583 × 0.0580 = 8.21 min.

[ R ] 0 0.693 60 s or, log = ´ = 0.476 [R] 37.9 s 2.303

Using Arrhenius equation log 10

[ R ] 2.303 log o k [R ]

Here, k = 1.6259 × 10 –2 min –1 , [R] ° = 0.400 M, [R] = 0.350 M Substituting these values in the equation, we get 2.303 0.400 t = log - 2 1.6259 ´ 10 0.350

[ R ] 0 2.303 Using formula k = log 10 t [ R]

or, log

Ea 1 ´ -1 -1 19.147 J K mol 4800

é T2 - T 1 ù k2 E a = ê ú k1 2.303 R ë T2 ´ T1 û

OR (a) (i) Refer Ans. 1. (ii) Elementary step: Each step of a complex reaction is called an elementary step. (b) Given : k = 5.10 × 10 –3 min –1 , t = 3 × 60 min = 180 min [R 0] = 0.1 M Substituting these values in the equation [ R ] 2.303 t = log ° k [ R] We get 2.303 0.1 180 min = log 5.1 ´ 10 -3 min -1 [ R ] log

0.1 180 min ´ 5.1 ´ 10 -3 min -1 = = 0.398 [ R] 2.303

0.1 = Antilog 0.398 = 2.50 [ R] [R] = 0.1/2.5 = 0.04 M.

4 k 1 E a é 20 ù = ú -1 -1 ê k1 2.303 ´ 8.314 J K mol ë 320 ´ 300 û

JJJ

Surface Chemistry

29

Chapter5 Surface Chemistry


1.

Which has a higher enthalpy of adsorption, physisorption or chemisorption? [AI 2008]

2.

Describe ‘electrophoresis briefly. [AI 2008, Delhi SetI 2011]

3.

What causes Brownian movement in a colloidal solution? [AI 2008]

4.

Why does physisorption decrease with the increase of temperature? [Delhi 2008 C]

5.

x why initially increases and then decreases m with rise in temperature? [Delhi 2008]

In chemisorption

6.

Define the term “Tyndall effect”.

[Delhi 2009]

7.

What is “coagulation” process?

8.

Name the two types of adsorption phenomenon. [Delhi 2009 C]

9.

What is meant by ‘shapeselective catalysis’ of reactions? [AI 2009 C, Delhi SetII 2011, AI SetI 2011, AI SetII 2012]

[AI 2009]

10. Why is adsorption always exothermic?

[AI 2009 C]

11. What is an emulsion?

[Delhi 2010]

12. Give an example of ‘shapeselective catalyst’. [Delhi 2010] 13. What are lyophobic colloids? Give an example for them. [AI SetII 2011] 14. Define peptization.

[AI SetI 2012]


18. Distinguish between micelles and colloidal particles. Give one example of each. [AI 2008 C] 19. How are the following colloids different from each other in respect to dispersion medium and dispersed phase? Give one example of each type (i) An aerosol (ii) A hydrosol (iii) An emulsion [Delhi 2009, 2010] 20. What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids? [Delhi 2009, 2010] 21. What are emulsions? State one application of emulsification. [Delhi 2009 C] 22. Explain the following terms : (i) Tyndall effect (ii) Coagulation 23. Describe the following : (i) Tyndall effect (ii) Shape selective catalysis.

[AI 2009 C]

[AI 2010]

24. What is meant by coagulation of a colloidal solution? Name any method by which coagulation of lyophobic solution can be carried out. [AI 2010] 25. Explain how the phenomenon of adsorption finds application in each of the following processes : (i) Production of vacuum (ii) Heterogeneous catalysis (iii) Froth Floatation process OR Define each of the following terms : (i) Micelles (ii) Peptization (iii) Desorption [Delhi SetI 2011]

15. What are lyophilic and lyophobic sols ? Giving one example of each type. Which one of these two types of sols is easily coagulated and why? [AI 2008] 26. Classify colloids where the dispersion medium is water. State their characteristics and write an example of each of these 16. Explain what is observed when classes. (i) an electrolyte, KCl, is added to hydrated ferric oxide solution. OR (ii) an electric current is passed through a colloidal solution. Explain what is observed when (iii) a beam of strong light is passed through a colloidal (a) an electric current is passed through a sol. solution. [AI 2008, Delhi 2008 C] (b) a beam of light is passed through a sol 17. (a) What is the difference between a colloidal solution and (c) an electrolyte (say NaCl) is added to ferric hydroxide an emulsion? Give one example of each. sol. [AI SetI 2011] (b) What are emulsifiers? [Delhi 2008 C]


30

27. Define the following terms giving an example of each: 32. Describe a conspicuous change observed when (i) Emulsion (ii) Hydrosol [Foreign SetI 2011] (i) a solution of NaCl is added to a sol of hydrated ferric oxide. 28. Explain how the phenomenon of adsorption finds application (ii) a beam of light is passed through a solution of NaCl in the following processes: and then through a sol. [Delhi SetII 2012] (i) Production of vacuum [Foreign SetI 2011] 33. Explain the following terms giving one example for each. (i) Micelles (ii) Aerosol [Delhi SetIII 2012] 29. Write four distinguishing features operating between chemisorption and physisorption. [Foreign SetII 2011] 34. Explain the following terms giving a suitable example for each: 30. Define the following terms : (i) Aerosol (ii) Emulsion (iii) Micelle (i) Aerosol [AI SetI 2012] (ii) Coagulation of colloids. [Foreign SetIII 2011] (ii) Heterogeneous catalysis

31. Name the two groups into which phenomenon of catalysis 35. Write three distinct features of chemisorptions which are not found in physisorptions. [AI SetII 2012] can be divided. Give an example of each group with the chemical equation involved. [Delhi SetI 2012]

answers 1. Enthalpy of chemisorption is high (80240 kJ mol –1 ) as it involves chemical bond formation.

product molecules is called shape selective catalysis. e.g., zeolites are good shape selective catalyst.

2. Electrophoresis : When electric potential is applied across two platinum electrodes dipped in a colloidal solution, the colloidal particles move towards one of the electrodes. The movement of colloidal particles under an applied electric potential is called electrophoresis. 3. The molecules of dispersion medium due to their kinetic motion strike against the colloidal particles (dispersed phase) from all sides with different forces causing them to move. However, colloidal particles being comparatively heavier, move with a slower speed.

10. Adsorption occurs because of attraction between adsorbate and adsorbent molecules and the energy is always released during adsorption. Hence adsorption is an exothermic process.

4.

5.

6.

7. 8. 9.

11. Emulsions are liquidliquid colloidal systems. 12. Zeolite (ZSM5). 13. Refer Ans. 9

14. Lyophobic sols are those sols in which the particles of the dispersed phase have little affinity for the particles of the dispersion medium, e.g., sols of metal and their sulphides and hydroxides. Since adsorption process is exothermic hence according to 14. The process of conversion of freshly prepared precipitate Le Chatelier principle the physical adsorption occurs readily into colloidal solution by adding suitable electrolyte is at low temperature and decreases with increase in called peptisation. temperature. 15. (i) Solution in which dispersed phase and dispersion x In chemisorption, adsorption isobar æç ö÷ shows an initial medium attract each other are known as lyophilic sol è m ø e.g., proteins, starch, rubber etc. They directly pass increase with temperature and then the expected decrease. into the colloidal state when brought in contact with This initial increase is because the heat supplied acts as the solvent. activation energy required in chemisorption. Solution in which dispersed phase and dispersion Tyndall effect : When a beam of light is passed through a medium repel each other are called lyophobic sols e.g., colloidal solution and viewed perpendicular to the path of the metals, their sulphides, hydroxides, etc. They do not incident light, the path of the light becomes visible due to form colloidal solution readily which mixed with the scattering of light. This phenomenon is called Tyndall effect. dispersion medium. Their colloidal solution can be Coagulation : The process of setting of colloidal particles prepared only by special methods. is called coagulation or precipitation of the solution. (ii) Hydrophobic solutions get easily coagulated on the (i) Physical adsorption or physisorption. addition of small amount of electrolyte or by heating (ii) Chemical adsorption or chemisorption. or even shaking as they are not stable. The catalytic reaction which depends upon the pore 16. (i) When an electrolyte like KCl is added to Fe(OH) 3 structure of the catalyst and the size of the reactant and solution, the positively charged colloidal particles of

Surface Chemistry

31

Fe(OH) 3 get coagulated by the oppositely charged Cl – path of the incident light, the path of the light becomes ions provided by KCl. visible due to scattering of light. This phenomenon is (ii) On passing the electric current, colloidal particles move called Tyndall effect.. towards the oppositely charged electrode where they (ii) Coagulation : The process of setting of colloidal particles lose their charge and get coagulated. is called coagulation or precipitation of the solution. (iii) When a beam of strong light is passed through a colloidal 23. (i) Refer Ans. 22(i). solution scattering of light by colloidal particles takes (ii) Refer Ans. 9. place and the path of light becomes visible. This 24. (i) The process of setting of colloidal particles into phenomenon is called Tyndall effect. precipitate is called coagulation of colloidal solution. 17. (a) In a colloidal solution, the dispersal phase is a solid and (ii) Coagulation of lyophobic sol can be carried out by the dispersion medium is a liquid. In an emulsion, both adding electrolyte. the dispersal phase and dispersion medium are liquid. 25. Application of adsorption : e.g., colloidal solution cell fluids, muddy water, (i) Production of Vacuum : Adsorption can be applied emulsionmilk, hair cream. to create condition of high vacuum. Vessel which has (b) Emulsifiers : The substances which are added to stabilise already been exhausted by vacuum pump is connected the emulsion are called emulsifiers. e.g., various kinds to a bulb containing charcoal. The remaining traces of of soaps, lyophilic colloids (proteins, gum etc.). air inspite of low pressure are adsorbed by the charcoal 18. Micelles : When small particles (ions) of an electrolyte almost completely. molecules form the aggregate particles which behave like (ii) Heterogeneous Catalysis : The gaseous reactants are colloidal particles, these aggregated particles are known as adsorbed on the surface of the solid catalyst. As a result, micelles. the concentration of the reactants increases on the surface of the catalyst and hence the rate of reaction increases. Examples : Soap and detergents. Manufacture of ammonia using iron as a catalyst, Colloidal particles Colloidal particles have an enormous manufacture of H 2 SO 4 by process using V 2 O 5 catalyst surface area per unit mass as a result of their small size. and use of finely divided nickel in the hydrogenation of Its size ranges between 1 nm to 100 nm. e.g., sulphur sol. vegetable oils are the excellent examples. 19. (i) An aerosol : It is a colloidal dispersion of a liquid in (iii) Froth Floatation Process : In froth floatation process, a gas. e.g., fog. the powdered ore is mixed with water. It is then mixed (ii) A hydrosol : It is a colloidal solution of a solid in water with pine oil (a frother). . The oil particles are adsorbed as the dispersion medium. e.g., starch solution. on the surface of ore particles which stick to the bubbles (iii) An emulsion : It is a colloidal system when both the of the air and rises to surface along with the foam while dispersed phase and the dispersion medium are in the the gangue particles which are wetted by water settle liquid state. e.g. milk. at the bottom. The foam is separated out and is collected 20. Multimolecular colloids Macromolecular colloids from which the ore particles are separated out. OR The particles of this type of The particles of this type of (i) Micelles : There are some substances which at low colloids are aggregates of colloids are themselves concentration behave as normal strong electrolytes but atoms or molecules with large molecules of colloidal at higher concentration exhibits colloidal behaviour diameter less than 1 nm. dimensions. e.g., starch, due to formation of aggregated particles. The aggregated e.g., sulphur sol consists of protein etc. particles thus formed are called micelles. The formation colloidal particles which are of micelles take place only above a particular aggregate of S 8 molecules. temperature called Kraft temperature and above a Associated colloids are different from these two. Those particular concentration called critical micelle electrolytes which dissociate into ions and these ions concentration (CMC). Surface active agents such as associate together to form ionic micelles whose size lies soap and synthetic detergents belong to this class. in the colloidal range. e.g., soap. (ii) Peptization : The process of converting a fresh precipitate into colloidal sol by shaking it with 21. Emulsions are the colloidal solutions in which both the dispersion medium in the presence of a small amount dispersed phase and the dispersion medium are liquids. of suitable electrolyte is called peptization. During Application : The various pharmaceuticals and cosmetics peptization, the precipitate adsorbs one of the ions of available in liquid form such as codliver oil, ointments the electrolyte on its surface. This causes development etc. are emulsions of water in oil type. of positive or negative charge on precipitates, which 22. (i) Tyndall effect : When a beam of light is passed through ultimately break up into particles of colloidal dimension. a colloidal solution and viewed perpendicular to the


32

(iii) Desorption : The process of removing an adsorbed substance from a surface of adsorbent is called desorption. 26. These are of two types: (i) Hydrophilic Stability : More stable as the stability is due to charge on the particle and water envelop surrounding the sol particle. Nature : Reversible Examples : Starch, gum, etc. (ii) Hydrophobic Stability : Less stable as the stability is due to charge only. Nature : Irreversible Examples : Metal hydroxide like Fe(OH) 3 and metal sulphides like As 2 S 3 .

Esterification : CH 3 COOH (l) + C 2 H 5 OH (l) H SO

4 ¾ ¾2 ¾¾ ® C H 3 C OO C 2 H 5 ( l ) + H 2 O ( l )

Hydrolysis of ester : NaOH

( l ) CH 3 COOC2 H 5(l ) + H 2 O ( l ) ¾¾¾¾ ®

CH3 COONa ( l ) + C2 H 5 OH ( l ) The process in which reactants and catalysts are in different physical state. e.g. Manufacture of ammonia by Haber’s process Fe ( s ) N 2( g ) + 3H 2( g ) ¾¾¾ ® 2 NH 3( g )

Hydrogenation of oils : OR Ni( s ) Vegetable oil (l) + H 2(g) ¾¾¾ ® vegetable ghee (s) . (a) On passing electric current through a sol, colloidal Theory of homogeneous catalysis : Theory of particles start moving towards oppositely charged homogeneous catalysis is also known as intermediate electrode where they lose their charge and get coagulated compound formation theory. (electrophoresis). According to this theory, catalyst combines with one of the (b) Scattering of light by the colloidal particles takes place reactants and forms an unstable intermediate. This and the path of light becomes visible (Tyndall effect). intermediate reacts with another reagent or decomposes (c) The positively charged colloidal particles of ferric itself to give product. e.g. oxidation of SO 2 to SO 3 . hydroxide sol get coagulated by the oppositely charged 2NO + O 2 ® 2NO 2 Cl – ions provided by NaCl. Catalyst 27. (i) Emulsion : An emulsion is a colloidal system in which SO 2 + NO 2 ® SO 3 + NO both the dispersed phase and the dispersion medium are – 32. (i) Cl ions are provided by NaCl for coagulation of liquids (e.g., milk, Cod liver oil, etc.). positive ions of Fe(OH) 3 . (ii) Hydrosol : A sol in which water is dispersion medium (ii) Beam of light passes straight when passed through is called hydrosol. NaCl solution but scattering of light is observed when 28. (i) Refer answer 25 (i). it is passed through the colloidal solution. The path of (ii) Refer answer 25 (ii). light is illuminated. This is called Tyndall effect. 29. Distinction between chemisorption and physisorption 33. (i) Aggregated particles of associated colloids at high Physisorption Chemisorption concentration are called micelles. e.g. soaps. It arises due to chemical bond It arises because of van der (ii) Colloid of a liquid in a gas is called aerosol e.g. fog, formation between adsorbent Waals’ forces between sprays etc. and adsorbate. adsorbent and adsorbate. 34. (i) Refer answer 33 (ii) It is irreversible in nature. It is reversible in nature. (ii) Emulsions are colloidal systems in which both It is highly specific in nature. It is not specific in nature. dispersed phase and dispersion medium are liquids. It result into unimolecular It result into multimolecular Types of emulsions : layers on adsorbent surface. layers on adsorbent surface – oil dispersed in water e.g. milk. under high pressure. – water dispersed in oil e.g. butter. 30. (i) Aerosols: These are the colloidal system in which (iii) Refer answer 33 (i) dispersion medium is gas and dispersed phase is either 35. solid or liquid e.g., smoke, insecticide sprays, etc. Chemical adsorption Physical adsorption (ii) Coagulation : The process of aggregating together the 1. Forces of attraction Forces between adsorbent and colloidal particles into large sized particle which between adsorbent and adsorbate are strong chemical ultimately settle down under the force of gravity as a adsorbate are weak van bonds. precipitate is called coagulation. der Waal’s forces. 31. The process in which reactants and catalysts are in same 2. Heat of adsorption is low Heat of adsorption is high physical state. e.g. oxidation of SO 2 to SO 3 in presence of NO. (20100 kcal mol –1 ). (510 kcal mol –1 ). NO( g ) 3. It is temporary and It is permanent and irreversible. 2SO 2( g ) + O 2( g ) ¾¾¾® 2SO 3( g ) reversible.

JJJ

General Principles and Processes of Isolation of Elements

33

Chapter6 General Principles and Processes of Isolation of Elements


(iii) Cryolite in the metallurgy of aluminium. [Delhi 2008 C]

Why is the froth floatation method selected for the concentration of sulphide ores? [Delhi 2009] 15. Explain the role of (i) cryolite in the electrolytic reduction of alumina 2. What is meant by term pyrometallurgy? [AI 2009] (ii) carbon monoxide in the purification of nickel. 3. Which of the two scraps, zinc or iron would be preferred for [Delhi 2009, Delhi Set II 2012] the recovery of copper from the leached copper ore and why? 16. Describe the role of the following : [AI 2009 C] (i) NaCN in the extraction of silver from a silver ore. 1.

4.

Why is electrolytic reduction preferred over chemical reduction for the isolation of certain metals? [AI 2009 C]

(ii) Cryolite in the extraction of aluminium from alumina. [AI 2009]

5.

Differentiate between a mineral and an ore. [AI SetI 2011]

6.

Why is it that only sulphide ores are concentrated by froth floatation process? [AI SetII 2011]

17. Describe the underlying principles of each of the following metal refining methods : (i) Electrolytic refining of metals (ii) Vapour phase refining of metals. [AI 2009

7.

What type of ores can be concentrated by magnetic separation method? [AI SetIII 2011]

18. Give reasons for the following : (i) Alumina is dissolved in cryolite for electrolysis instead 8. What is the role of graphite in the electrometallurgy of of being electrolysed directly. aluminium? [Delhi SetI 2012] (ii) Zinc oxide can be reduced in the metal by heating with carbon not Cr 2 O 3 . 9. How is copper extracted from a low grade ore of it? (iii) Extraction of copper directly from sulphide ores is less [AI SetI 2012] favourable than that from its oxide ore through 10. What is the role of collectors in Froth Floatation process? reduction. [Delhi 2009 C] [AI SetII 2012] 19. Describe how the following changes are brought about (i) Pig iron into steel SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARK) (ii) Zinc oxide into metallic zinc 11. What are the chief ores of zinc? Write chemical reactions (iii) Impure titanium into pure titanium. [Delhi 2010] taking place in the extraction of zinc from zinc blende. 20. Describe the role of [AI 2008] (i) NaCN in the extraction of gold from gold ore. 12. State briefly the principles which serve as basics for the (ii) SiO 2 in the extraction of copper from copper matte. following operations in metallurgy : (iii) Iodine in the refining of titanium. Write chemical (i) Froth floatation process equations for the involved reactions. [Delhi 2010] (ii) Zone refining 21. Describe the role of the following : (iii) Refining by liquation. [Delhi 2008, AI 2010] (i) NaCN in the extraction of silver from a silver ore 13. What chemical principle is involved in choosing a reducing (ii) Iodine in the refining of titanium agent or getting the metal from its oxide. Consider the metal (iii) Cryolite in the metallurgy of aluminium. [AI 2010] oxides; Al 2 O 3 and Fe 2 O 3 and justify the choice of reducing 22. Describe the principle behind each of the following processes: agent in each case. [AI 2008] (i) Vapour phase refining of a metal. 14. Describe the role of the following : (ii) Electrolytic refining of a metal. (i) Depressant in froth floatation process (iii) Recovery of silver after silver ore was leached with (ii) Silica in the extraction of copper from copper pyrites ore NaCN. [Delhi SetI 2011, Foreign SetI 2011]


34

23. Explain the role of each of the following in the extraction of 27. Write the reactions involved in the following processes: (i) Leaching of bauxite ore to prepare pure alumina. metal from their ores : (ii) Refining of zirconium by van Arkel method. (i) CO in the extraction of nickel (iii) Recovery of gold after gold ore has been leached with (ii) Zinc in the extraction of silver NaCN solution. [Foreign SetII 2011] (iii) Silica in the extraction of copper [Delhi SetII 2011] 24. Describe the principle controlling each of the following 28. Describe the principle involved in each of the following processes. processes: (i) Mond process for refining of nickel. (i) Vapour phase refining of titanium metal. (ii) Column chromatography for purification of rare (ii) Froth floatation method of concentration of a sulphide elements. [Delhi SetI 2012] ore. [AI SetI 2011] 25. Describe the principle controlling each of the following 29. Which methods are usually employed for purifying the following metals? processes: (i) Nickel (ii) Germanium (i) Zone refining of metals Mention the principle behind each one of them. (ii) Electrolytic refining of metals [AI SetII 2011] [AI SetI 2012] 26. Describe the principle controlling each of the following 30. Explain the role of each of the following. process : (i) NaCN in the extraction of silver. (i) Preparation of cast iron from pig iron. (ii) SiO 2 in the extraction of copper. [AI SetIII 2012] (ii) Preparation of pure alumina (Al 2 O 3 ) from bauxite ore. [AI SetIII 2011]

answers 1. Sulphide ore is lighter, it is not wetted by water and floats with the froth. 2. The process of extraction of metal from its ore by heating the ore with a suitable reducing agent is known as pyrometallurgy. 3. Zinc scrap would be preferred because zinc is more electropositive than iron.

9. Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing Cu 2+ is treated with scrap iron or H 2 . Cu 2+ (aq) + H 2(g) ® Cu (s) + 2H + (aq) . 10. Collectors increase the natural hydrophobicity of the surface, increasing the separability of the hydrophobic and hydrophilic particles.

11. The chief ores of zinc are (i) Zinc blende, ZnS (ii) Calamine, ZnCO 3 5. The naturally occurring chemical substance present in the (iii) Zincite, ZnO. earth’s crust which can be obtained by mining are called Extraction of zinc : From zinc blende zinc is extracted by minerals. Those minerals from which metals can be roasting followed by reduction with coke. extracted economically are called ores. (a) Roasting : The concentrated ore is heated with oxygen 6. This is because the sulphide ore particles are lighter and at 900°C in reverberatory furnace to convert zinc preferentially wetted by oil whereas gangue particles are sulphide to zinc oxide. preferentially wetted by water. 2 ZnS + 3O 2 ® 2 ZnO + 2SO 2 7. If either the ore or the gangue particles are capable of (Zinc blende) (Zinc oxide) being attracted by magnetic field, then such ore is (b) Reduction : The reduction of zinc oxide is done using concentrated by magnetic separation method. coke. 4. Electropositive metals which cannot be reduced by chemical reduction are easily reduced electrolytically.

673 K 8. Graphite anode is used in the electrometallurgy of ZnO + C ¾¾¾ ® Zn + CO aluminium from alumina, Al 2 O 3 , Oxygen liberated at high The metal is distilled off and collected by rapid chilling. temperature reacts with graphite to form both CO 2 and CO gases and prevent the liberation of O 2 gas at the anode 12. (i) Froth floatation process : This method is based on the preferential wetting properties with the frothing agent which may react with Al metal to give Al 2 O 3 again. and water. At anode : C (s) + O 2– (melt) ® CO (g) + 2e – The powdered ore is added to water containing pine C (s) + 2O 2– (melt) ® CO 2(g) + 4e – oil or turpentine oil (Frothing agent).


A stream of air is passed which agitates the mixture and produces froth. The sulphide particles of the ore stick to the oil droplets and rise to the surface and floats with air bubbles. Water wets the gangue particles which sink to bottom. Foam carrying ore particles

Air

Water

35

e.g., NaCN is used as depressant in the purification of PbS in presence of ZnS. (ii) During the metallurgy of copper from copper pyrites, CuFeS 2 , its roasting gives FeO besides Cu 2 O and SO 2 , respectively. 4 CuFeS2( s ) + 11O 2( g ) ® 4 FeO ( s ) + 2 Cu 2 O ( s ) + 8SO 2( g ) (Copper pyrites)

To remove FeO, SiO 2 is added to form slag. FeO ( s ) + SiO 2( s ) ® FeSiO 3(l ) (Basic oxide)

Ore + Water + Pine oil + Sodium ethyl xanthate

Water

(Acidic oxide)

Slag

(iii) Cryolite, Na 3 AlF 6 , dissolves alumina, Al 2 O 3 , lowers the melting temperature of electrolyte and also increase its conductivity.

15. (i) Cryolite (Na 3 AlF 6 ) is added as flux in the electrolytic reduction of alumina. It decreases the melting point of Fig. 6.3 the alumina and increases its conductivity. This method is used for concentration of sulphide ores. (ii) Carbon monoxide combines with nickel forming a e.g. Copper pyrite (ore of copper) argentite (ore of volatile complex nickel tetracarbonyl which silver) etc. decomposes on further heating to give pure nickel. (ii) This method is based on the principle that the impurities are more soluble in the melt than in the solid state of 16. (i) NaCN combines with argentite ore of silver forming the metal. A circular mobile heater is fixed at one end a soluble complex. of a rod of the impure metal. The molten zone moves Ag 2S + 4 NaCN ® 2 Na[Ag(CN) 2 ]+ Na 2 S Sodium dicyano argentate along with the heater which is moved forward. As the This complex displaced Ag when more reactive metal heater moved forward, the pure metal crystallises out is added to it. of the melt and the impurities pass on into the adjacent (ii) Refer Ans. 15 (i). molten zone. The process is repeated several times and the heater is moved in the same direction. At one end, 17. (i) Electrolytic refining : In this method, the impure metal impurities get concentrated. This end is cut off. This is made to act as anode. A strip of the same metal in method is very useful for producing semiconductor and pure form is used as cathode. They are put in a suitable other metals of very high purity, e.g., germanium, silicon, electrolytic bath containing soluble salt of the same boron, gallium and indium. metal. The more basic metal remains in the solution and the less basic ones go to the anode mud. Anode : M ® M n+ + ne – Cathode : M n+ + ne – ® M Noblegas atmosphere Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken Metal rod as cathode. The electrolyte is acidified solution of Inductioncoil Molten zone copper sulphate and the net result of electrolysis is the heaters moving as shown transfer of copper in pure form from the anode to the Zone refining process cathode: (iii) Refining by liquation : In this method a low melting Anode : Cu ® Cu 2+ + 2e – metal like tin is heated on a sloping surface of a Cathode : Cu 2+ + 2e – ® Cu reverberatory furnace. Metal melts and flows down, Impurities from the blister copper deposit as anode impurities leave behind. mud which contains antimony, selenium, tellurium, 13. (i) Thermodynamic factor helps us in choosing a suitable silver, gold and platinum; recovery of these elements reducing agent for the reduction of a particular metal may meet the cost of refining. oxide to the metallic state. Zinc may also be refined this way. (ii) Any metal will reduce the oxides of other metals if D f G° (ii) Vapour phase refining : In this method, the metal is of the oxide of the element is less than D f G° of the metal converted into its volatile compound and collected oxide to be reduced. Thus both Al and Zn can reduce elsewhere. It is then decomposed to give pure metal. Fe 2 O 3 and Mg can reduce Al 2 O 3 to Al. So, the two requirements are : (a) The metal should form a volatile compound with an 14. (i) The role of depressant in froth floatation process is to available reagent. prevent some of the sulphide from forming froth. Gangue

Concentrated ore


36

(b) The volatile compound should be easily decomposable, so that the recovery is easy. Following example will illustrate this technique. Mond process for Refining Nickel : In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl : 330 - 350 K

Ni + 4 CO ¾¾¾¾¾® Ni(CO) 4

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal : 450 - 470 K

Ni(CO) 4 ¾¾¾¾¾® Ni + 4 CO

18. (i) Alumina is dissolved in cryolite to lower the melting point and improve its electrical conductivity. (ii) Carbon is suitable reducing agent for reduction of zinc oxide. Reduction of Cr 2 O 3 by carbon is not thermodynamically favourable. (iii) Free energy change for the reduction of copper sulphide to copper by carbon is positive. Where as D r G° for the reduction of copper oxide to copper by carbon is negative and hence feasible. 19. (i) Pig iron is mixed with scarp iron and heated in a furnace. After removal of impurities required quantity of spiegel is added to make steel. (ii) Zinc oxide is mixed with coke and clay and brickettes are made. The brickettes are heated at 673 K to give metallic zinc 673 K

ZnO ( s) + C ( s) ¾¾¾® Zn ( s ) + CO ( g ) (iii) Van Arkel de Boer process : Impure titatinum is heated with iodine to form volatile complex TiI 4 which on further heating at higher temperature decomposes to give pure titanium. D

Ti ( s ) + 2 I 2( s) ® TiI 4 ¾¾ ® Ti ( s ) + 2 I 2 Impure

Pure

20. (i) Refer Ans. 16(i). (ii) SiO 2 is added in the extraction of copper from copper matte to remove remaining FeS and FeO as silicate (slag) 2FeS + 3O 2 ® 2FeO + 2SO 2(g) 2FeO + SiO 2 ® 2FeSiO 3 (slag) (iii) Iodine forms a volatile compound with titanium which on further heating decomposes to give pure titanium. D Ti + 2I 2(s) ® TiI 4 ¾¾ ® Ti + 2I 2 . (Impure)

(Pure)

21. (i) Refer Ans. 16 (i). (ii) Refer Ans. 20 (iii). (iii) Reger Ans. 14 (iii). 22. (i) Refer Ans. 17(ii). (ii) Refer Ans. 17(i). (iii) Recovery of silver after silver ore was leached with NaCN : During leaching Ag is oxidised to Ag + which

then combines with CN – ions to form soluble complex. [Ag(CN) 2 ] – . Silver is then recovered from this complex by displacement method using more electropositive zinc metal. 2[Ag(CN) 2 ] – (aq) + Zn (s) ® 2Ag (s) + [Zn(CN) 4 ] 2– (aq) 23. (i) CO in the extraction of nickel : The role of CO is to convert impure nickel to volatile nickel tetracarbonyl whereas the impurities are left behind in the solid state. The vapour of nickel tetracarbonyl on heating at 450 470 K decomposes to give pure nickel metal and carbon monoxide. Ni + 4CO

330 350 K

Impure Nickel

Ni(CO) 4 Nickel tetracarbonyl

Ni + 4CO ( ) g

Pure nickel

(ii) Zinc in the extraction of silver : Role of zinc is to recover silver from the complex by displacement reaction. Silver ore is leached with dilute solution of NaCN in the presence of air or oxygen to form a soluble complex. – 4Ag(s) + 8CN (aq) + 2H2 O(l) + O 2 (g ) –

–

4[Ag(CN)2 ] + 4OH (aq ) Soluble Complex – 2– + 2Ag 2[Ag(CN) 2] [Zn(CN) 4] + Zn (iii) Silica in the extraction of copper : Role of silica in the extraction of Cu is to remove ferrous impurities as slag during smelting and bessemerisation. The impurity ferrous sulphide is oxidised to ferrous oxide which is then reacted with silica (flux) to form ferrous silicate (slag). FeO + SiO 2 FeSiO 3

Ferrous Silicate (Slag)

Flux

24. (i) In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal. 500 K TiI 4 Ti + 2I 2 Impure metal

TiI 4 Ti

700 K

Ti + 2I 2

Pure metal

(ii) This method of concentration of ore is based upon the principle that sulphide ore is lighter and its surface is preferentially wetted by oils while that of a gangue is preferentially wetted by water. 25. (i) Refer Ans. 12(ii). (ii) Refer Ans. 17(ii). 26. (i) Pig iron is melted with scrap iron and coke using hot air blast. Due to this, impurities such as C, S and P present in the pig iron are removed as CO 2 , SO 2 and P 2 O 5 and carbon reduced to about 3%. (ii) Al 2 O 3 present in bauxite is soluble in concentrated


NaOH solution whereas impurities are not, Al 2 O 3 is reprecipitated from the solution. 27. (i) Leaching of bauxite ore to prepare pure alumina: Al 2 O 3 + 2NaOH (aq) + 3H 2 O (l) ® 2Na[Al(OH) 4 ] (aq)

37

Ni(CO) 4

450 - 470 K ¾¾¾¾ ® Ni + 4 CO

Nickel tetracarbonyl

Pure metal

(ii) Chromatographic method : This method is based on the extent of adsorption of different components on an adsorbent.

2Na[Al(OH) 4 ] (aq) + CO 2(g) ® Al 2 O 3 ∙xH 2 O (s) + 2NaHCO 3(aq) 1470 K Al2 O 3 ∙x H2 O (s) Al 2 O 3 (s ) + x H 2 O (g )

(ii) Refining of zirconium by van Arkel de Boer method. Zr (s) + 2I 2(s) ZrI 4(g)

D

ZnI 4(g)

2075 K Tungsten filament

Mixture of compounds

Coloured bands

Stationary phase

Zr (s) + 2I 2(s) Pure

(iii) Extraction of gold: 4Au (s) + 8CN – (aq) + 2H 2 O (l) + O 2(g) ® 4[AuCN) 2 ] – (aq) + 4OH – (aq) Gold is recovered from [Au(CN) 2 ] – complex by displacement method using a more electropositive metal zinc. 2[Au(CN) 2 ] – (aq) + Zn (s) ® 2Au (s) + [Zn(CN) 4 ] 2– (aq)

Column chromatography

For example, the solid of Al 2 O 3 is packed in a glass tube and moving medium containing a solution of the components in liquid form is run down. This method is very useful for purification of the elements which are available in small quantities and the impurities are not very different in chemical properties from the element to be purified.

28. (i) Mond’s process for refining of nickel : Nickel is refined by Mond’s process. In this process, Nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl. 29. (i) Refer Ans. 28(i). 330 - 350 K (ii) Refer Ans. 12(ii). Ni + 4CO ¾¾¾¾ ® Ni(CO)4 The carbonyl is subjected to higher temperature so that 30. (i) Refer Ans. 16(i). it is decomposed giving the pure nickel metal. (ii) Refer Ans. 23(iii).

JJJ


38

Chapter7 The pBlock Elements


1.

Which is a stronger acid in aqueous solutions HF or HCl? [2008]

(i) Chlorine reacts with a hot concentrated solution of sodium hydroxide. (ii) Orthophosphorus acid is heated. (iii) PtF 6 and xenon are mixed together. [Delhi 2008]

In which one of the two structure NO 2 + and NO 2 – the bond 20. Complete the following chemical equations : angle has a higher value? [Delhi 2008] (i) Ca 3P 2(s) + H 2O (l) ® –––––– 2+ (ii) Cu + NH (aq) 3(aq) ® – – – – – – 3. Why is the bond angle in PH 3 molecule lesser than that in (excess) NH 3 Molecule? [AI 2008] (iii) F 2(g) + H 2O [Delhi 2008] (l) ® –––––– 4. Why is dinitrogen very unreactive as compared to phosphorus? [AI 2008] 21. Give reason for the following : (a) Oxygen molecule has the formula O 2 while sulphur is S 8 . 5. When HCl reacts with finely powdered iron, it forms ferrous (b) H 2 S is less acidic than H 2 Te. [Delhi 2008] chloride and not ferric chloride. Why? [AI 2008] 22. Answer the following : 6. Fluorine exhibits only – 1 oxidation state in its compounds (i) Which neutral molecule would be isoelectronic with whereas other halogens exhibit many other oxidation states. ClO – ? Why? [AI 2008 C] (ii) Of Bi (V) and Sb (V) which may be a stronger oxidising agent and why? [AI 2008] 7. Why is Bi (V) a stronger oxidant than Sb(V). ? [Delhi 2009] 23. Assign a reason for each of the following statements : (i) Phosphorus shows marked tendency for catenation but 8. Why does NO 2 dimerise? [AI 2010] nitrogen shows little tendency for catenation. 9. Draw the structure of O 3 molecule. [Delhi 2010] (ii) The electron gain enthalpy with negative sign for oxygen [– 141 kJ mol –1 ] is less than that for sulphur 10. Nitrogen is relatively inert as compared to phosphorus. Why? (– 200 kJ mol –1 ). [AI 2008] [AI 2010] 24. (a) How are interhalogen compounds formulated and how 11. Draw the structure of XeF 2 molecule. [Delhi SetI 2011] are they prepared? 12. Draw the structure of XeF 4 molecule. [Delhi SetII 2011] (b) With which neutral molecule of this group is ClO – isoelectronic? [AI 2008 C] 13. Draw the structure of BrF 3 molecule. [Delhi SetIII 2011] 25. (i) Draw the structure of SO 2 molecule. Comment on the + 14. Which one of PCl 4 and PCl 4 – is not likely to exist and why? nature of two S – O bonds formed in it. Are the two [Delhi SetI 2012] S – O bonds in this molecule equal? 15. Of PH 3 and H 2 S which is more acidic and why? (ii) What happens when white phosphorus is heated with [Delhi SetIII 2012] conc. NaOH solution in an inert gas atmosphere? [AI 2008 C] 16. Which is a stronger reducing agent, SbH 3 or BiH 3 and why? [AI SetI 2012] 26. Draw the structures of the following molecules : (i) XeF 4 (ii) BrF 3 [Delhi 2009] 17. What is the basicity of H 3 PO 2 acid and why? [AI SetIII 2012] 27. Complete the following chemical reaction equations : SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS) (i) P 4(s) + NaOH (aq) + H 2O (l) ® – (ii) I + H O + O [Delhi 2009] (aq) 2 (l) 3(g) ® 18. Draw the structure of the following molecules : (i) SF 4 (ii) XeF 4 [Delhi 2008] 28. Why is red phosphorus less reactive than white phosphorus? 2.

19. Write chemical equations for the following processes:

[Delhi 2009]

The pBlock Elements

39

29. Complete the following chemical equations : 41. Complete the following chemical reaction equations : (i) XeF 2 + H 2O ® (i) I 2 + HNO 3 ® (ii) PH 3 + HgCl 2 ® [Delhi 2009] (ii) HgCl 2 + PH 3 ® [AI 2010] 30. (a) Draw the structure of the following : (i) H 2 S 2 O 8 (ii) HClO 4

[AI 2009]

42. Draw the structural formula of the following compounds: (i) H 4P (ii) XeF 4 2O 7 [AI 2010]

31. Write the formula and the structures of noble gas species (one each) which are isostructural with 43. How would you account for the following? (i) ICl 4– (i) NCl 3 is an endothermic compound while NF 3 is an (ii) BrO 3 – [Delhi 2009 C] exothermic one. (ii) XeF 2 is linear molecules without a bend? 32. Give reason for following : (iii) The electron gain enthalpy with negative sign for (i) PH 3 is a weaker base than NH 3 . fluorine is less than that for chlorine, still fluorine is (ii) OF 6 compound is not known. a stronger oxidising agent than chlorine. [AI 2010] (iii) Fluorine provides the largest variety of interhalogen compounds amongst the halogens.

[Delhi 2009 C] 44. How would you account for the following? (i) The electron gain enthalpy with negative sign is less 33. Arrange the following in order of the property mentioned: for oxygen than that for sulphur. (i) PH 3 , NH 3 , SbH 3 , AsH 3 (Increasing basic strength). (ii) Phosphorus shows greater tendency for catenation than (ii) HCl, HBr, HI, HF (Increasing acid strength). nitrogen. (iii) HClO 4 , HClO, HClO 2 (Increasing oxidation power). (iii) Fluorine never act as the central atom in polyatomic [Delhi 2009 C] interhalogen compounds. [AI 2010] 34. Suggest possible reason for the following observations. 45. State reasons for each of the following: (i) In the solid state, PCl 5 behaves as an ionic species. (i) The N – O bond in NO – 2 is shorter than the N – O bond (ii) H 2 S is more acidic than water. [Delhi 2009 C] in NO – . 3 (iii) Fluorine forms the largest number of interhalogen (ii) SF is kinetically an inert substance. 6 compounds amongst the halogens. [Delhi 2010] OR 35. Draw the structure of (i) XeF 2 (ii) (HPO 3 ) 3 and predict their State reasons for each of the following : shapes. [AI 2010] (i) All the P–Cl bonds in PCl 5 molecule are not equivalent. 36. Give reasons for the following observations : (ii) Sulphur has greater tendency for catenation than oxygen. (i) Hydrogen iodide in a stronger acid than hydrogen [Delhi SetI 2011] fluoride in aqueous solution. (ii) The basic character of the hydrides of group 15 elements 46. How would you account for the following? decreases with increasing atomic number. (i) H 2 S is more acidic than H 2 O. (iii) Sulphur exhibits tendency for catenation but oxygen (ii) The N O bond is NO – 2 is shorter than the N O does not do. How. [AI 2010] bond in NO – 3 . 37. Complete the following reactions : (iii) Both O 2 and F 2 stabilise high oxidation states but the (i) I 2 + NaClO 3 ® ability of oxygen to stabilise the higher oxidation state (ii) NH 3 + NaOCl ® exceeds that of fluorine. [AI SetI 2011] (iii) XeF 6 + KF ® [AI 2010] 47. How would you account for the following? 38. Draw the structure of (i) NF 3 is an exothermic compounds but NCl 3 is not. (ii) The acidic strength of compounds increases in the (i) XeF 4 order: PH 3 < H 2 S < HCl (ii) H 2S [AI 2010] 2O 8 and predict their shapes. (iii) SF 6 is Kinetically inert. [AI SetIII 2011] 39. Complete the following reactions : 48. How would you account for the following? (i) XeF 4 + O 2F 2 ® (i) The following order of increase in strength of acids: (ii) XeF 4 + SbF 5 ® PH 3 < H 2 S < HCl (iii) 6NaOH (hot conc.) + Cl 2 ® [AI 2010] (ii) The oxidising power of oxoacids of chlorine follows 40. Draw the structure of white phosphorus and red phosphorus. the order: Which one of these two types of phosphorus is more reactive HClO 4 < HClO 3 < HClO 2 < HClO and why? [AI 2010] [Foreign SetI 2011]


40

57. (a) Complete the following chemical reaction : (i) F 2 + H 2 O ® (i) NaOH + Cl 2 ® (ii) XeF4 + O 2 F2 ¾¾¾ ® (hot and conc.) (ii) Ca 3 P 2 + H 2 O ® (iii) XeF 4 + H 2 O ® (iii) Br2 + F2 ® [Foreign SetI 2011] (excess) (b) Draw structure of the following species : 50. (a) Mention the optimum conditions for the industrial (i) H 2 S 2 O 7 (ii) NO 3 – [AI 2008] manufacture of ammonia by Haber’s process. 58. Assign reason for each of the following : (b) Explain the following giving appropriate reasons: (i) Sulphur vapour is paramagnetic. (i) Sulphur vapour exhibits paramagnetic behaviour. (ii) Ammonia (NH 3 ) has greater affinity for protons than (ii) Red phosphorus is less reactive than white phosphorus. phosphine (PH 3 ). OR (iii) The negative value of electron gain enthalpy of fluorine Draw the structure of the following molecules: is less than of chlorine. (i) NF 3 (ii) H 2S (iii) H 3PO 2O 8 3 (iv) SF 6 is much less reactive than SF 4 . [Foreign SetI 2011] (v) Of the noble gases only xenon is known to form well established chemical compounds. [AI 2008] 51. Complete the following chemical equations: 49. Complete the following chemical equations:

143 K

59. (a) Describe the favourable condition for the manufacture of (ii) Fe 3+ + SO 2 + H 2 O ® (i) Ammonia by Haber’s process (ii) Sulphuric acid by contact process. [Foreign SetII 2011] (iii) XeF6 + H2 O (excess) (b) Draw the structure of the following : 52. Explain giving reason for each of the following situation. (i) PCl 5(g) (ii) S 8(g) (iii) ClF 3(g) [AI 2008] (i) In aqueous medium HCl is stronger acid than HF. (ii) White phosphorus is more reactive than red phosphorus. 60. (a) Arrange HClO 3 , HClO 2 , HOCl and HClO 4 in order of increasing acid strength. Give reason for your answer. [Foreign SetIII 2011] (b) Write the balanced chemical equation for the reaction 53. Complete the following chemical equations: of Cl 2 with hot and conc. NaOH solution. Justify that (i) C + H 2SO this reaction is a disproportionation reaction. 4 (conc.) ® (ii) P 4 + NaOH + H 2O (c) Give use of ClF 3 . [Delhi 2008 C] ® Cl + F [Foreign SetIII 2011] 61. (a) Why do some noble gases form compounds with fluorine (iii) 2 2 (i) P 4 + SO 2 Cl 2 ®

(excess)

and oxygen only? 54. Explain the following giving an appropriate reason in each case. (b) How are the following compounds prepared from XeF 6 ? (i) O 2 and F 2 both stabilize higher oxidation states of (i) XeOF 4 and (ii) XeO 3 metals but O 2 exceeds F 2 in doing so. (c) Draw the structure of BrF 3 . [Delhi 2008 C] (ii) Structures of xenon fluorides cannot be explained by valence bond approach. [Delhi SetI 2012] 62. (a) Draw the structure of the following : (i) H 2 S 2 O 7 (ii) HClO 3 55. Explain the following facts giving appropriate reason in each (b) Explain the following observations. case: (i) In the structure of HNO 3 ; the N – O bond (121 (i) NF 3 is an exothermic compound whereas NCl 3 is not. pm) is shorter than the N – OH bond (140 pm). (ii) All the bonds in SF 4 are not equivalent. (ii) All the P – Cl bonds in PCl 5 are not equivalent. [AI SetI 2012] (iii) ICl is more reactive than I 2 . 63. (a) Draw the structures of the following : (i) H 3 PO 2 (ii) BrF 3 56. Assign a reason for each of the following statements : (b) How would you account for the following observations: (i) Phosphorus (P 4 ) is more reactive than nitrogen (N 2 ). (i) Phosphorus has a greater tendency for catenation (ii) All the bonds in PCl 5 are not equal in length. than nitrogen. (iii) Sulphur in vapour state exhibits paramagnetism. (ii) Bond dissociation energy of fluorine is less than (b) Write the structural formulae of the following that of chlorine. compounds : (iii) No chemical compound of helium is known. (i) BrF 3 (ii) XeF 2 [AI 2008] [AI 2009] LONG ANSWER TYPE QUESTIONS (5 MARKS)

The pBlock Elements

64. (a) Draw the structures of the following : (i) N 2 O 5 (ii) XeOF 4 (b) Explain the following observations : (i) The electron gain enthalpy of sulphur atom has a greater negative value than that of oxygen atom. (ii) Nitrogen does not form pentahalides. (iii) In aqueous solutions HI is a stronger acid than HCl. [AI 2009]

41

(i) SO 32– and the angle O – S – O (ii) ClF 3 and the angle F – Cl – F (iii) XeF 2 and the angle F – Xe – F OR (a) Complete the following chemical equations: (i) NaOH + Cl 2 ® (hot and conc.) (ii) XeF 4 + O 2F 2 ® (b) Draw the structures of the following molecules: (i) H 3PO 2 (ii) H 2S 2 O 7 (iii) XeOF 4 [Delhi SetI 2012]

65. (a) Draw the structures of the following : (i) XeF 4 (ii) H 2 S 2 O 7 [AI 2009] (b) Explain the following observations : (i) Phosphorus has a greater tendency for catenation than nitrogen. (ii) The negative value of electron gain enthalpy is 70. (a) Draw the molecular structures of following compounds. (i) XeF 6 less for fluorine than that for chlorine. (ii) H 2S 2O 8 (iii) Hydrogen fluoride has much higher boiling point (b) Explain the following observations: than hydrogen chloride. (i) The molecules NH 3 and NF 3 have dipole moments 66. (a) Draw the structure of following : which are of opposite direction. (i) PCl 5 (ii) SO 3 2– (ii) All the bonds in PCl 5 molecule are not equivalent. (b) Explain the following observations : (iii) Sulphur in vapour state exhibits paramagnetism. (i) Ammonia has higher boiling point than phosphine. OR (ii) Helium does not form any chemical compound. (a) Complete the following chemical equations: (iii) Bi(V) is a stronger oxidising agent than Sb(V). (i) XeF 4 + SbF 5 [AI 2009] (ii) Cl 2 + F 2 (excess) (b) Explain each of the following: 67. (a) Complete the following chemical equations. (i) Nitrogen is much less reactive than phosphorus. (i) NaOH (aq) + Cl 2(g) (Hot and conc.) ® (ii) The stability of +5 oxidation state decreases down (ii) XeF 6(s) + H 2 O (l) ® group 15. (b) How would you account for the following? (iii) The bond angles (O – N – O) are not of the same (i) The value of electron gain enthalpy with negative + value in NO 2– [Delhi SetI 2012] and NO 2 sign for sulphur is higher than that for oxygen. (ii) NF 3 is an exothermic compound but NCl 3 is 71. (a) Draw the molecular structures of the following compounds: endothermic compound. (i) N 2O (ii) XeOF 4 (iii) ClF 3 molecule has a Tshaped structure and not a 5 (b) Explain the following observations. trigonal planar one. [Delhi2010, AI 2010] (i) Sulphur has a greater tendency for catenation than 68. (a) Complete the following chemical equation. oxygen. (i) P 4 + SO 2 Cl 2 ® (ii) ICl is more reactive than I 2 . (ii) XeF 4 + H 2 O ® (iii) Despite lower value of its electron gain enthalpy (b) Explain the following observations giving appropriate with negative sign, fluorine (F 2 ) is a stronger reasons : oxidising agent than Cl 2 . (i) The stability of + 5 oxidation state decreases down OR the group in group 15 of the periodic table. (a) Complete the following chemical equations: (ii) Solid phosphorus pentachloride behaves as an ionic (i) Cu + HNO 3 (Dilute) ® compound. (ii) XeF 4 + O 2 F 2 ® (iii) Halogens are strong oxidising agents. [Delhi 2010] (b) Explain the following observations: (i) Phosphorus has greater tendency for catenation than 69. (a) Complete the following chemical reactions equations: nitrogen. (i) P 4 + SO 2 Cl 2 ® (ii) Oxygen is a gas but sulphur is a solid. (ii) XeF 6 + H 2 O ® (iii) The halogens are coloured. Why? [AI SetI 2012] (b) Predict the shape and the asked angle (90° or more or less) in each of the following cases.


42

72. (a) Draw the structures of the following molecules: (i) H 3 PO 2 (ii) ClF 3 (b) Explain the following observations: (i) Nitrogen is much less reactive than phosphorus. (ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water. (iii) Sulphur has greater tendency for catenation than oxygen in the same group.

OR (a) Draw the structures of the following molecules: (i) N 2O 5 (ii) HClO 4 (b) Explain the following observations: (i) H 2 S is more acidic than H 2 O. (ii) Fuorine does not exhibit any positive oxidation state. (iii) Helium form no real Chemical Compounds. [AI SetII 2012]

a n s w e r s 1. HCl is stronger acid than HF. The acid strength of hydrogen halides is HF < HCl < HBr < HI. 2. The bond angle of NO 2 + is higher than NO 2 – . 3. The bond angle in PH 3 is much lower [93.6°] than that in NH 3 [107.8°] due to less repulsion between bond pairs. 4. Due to the presence of a triple bond between the two nitrogen atoms, the bond dissociation energy of N 2 is very high as compared to phosphorus. Therefore, N 2 is less reactive than phosphorus.

8 - 2 = 3 2 No. of orbitals = 2 + 3 = 5 Hybridisation – sp 3 d Therefore, the shape of molecule would be linear. F

No. of lone pairs =

Xe

5. The reaction of HCl with iron produces H 2 F Fe + 2HCl ® FeCl 2 + H 2 12. Total number of electrons in the valence shell of central Liberation of hydrogen prevents the formation of ferric Xe atom = 8 chloride. No. of bond pairs = 4 6. This is due to nonavailability of dorbitals in valence shell No. of lone pairs = 2. of fluorine. Hybridisation – sp 3 d 2 Therefore, the shape of molecule would be square planar. 7. Bismuth (Bi) in the pentavalent state easily accepts two F F electrons and gets reduced to trivalent bismuth due to inert pair effect Bi 5+ + 2e – ® Bi 3+ Xe Therefore, it acts as a strong oxidising agent. 8. Because NO 2 contains odd number of valence electrons F F and on dimerisation, it is converted to stable N 2 O 4 molecule 13. Total number of electrons in the valence shell of central with even number of electrons. Br atom = 7 9. Structure of ozone. Number of bond pairs = 3 Number of lone pairs = 2 O Hybridisation = sp 3 d O O Therefore, the shape of molecule would be see saw shape. 10. Due to the presence of a triple bond between the two nitrogen atoms, the bond dissociation energy of N 2 is very F high as compared to phosphorus. Therefore, N 2 is less reactive than phosphorus. Br

11. Total no. of electrons in the valence shell of central Xe atom = 8 No. of bond pairs = 2

F

F

See saw shape

The pBlock Elements

43

14. PCl 4 – because PCl 3 cannot form bond with Cl – ions. 15. H 2 S is more acidic because H – S bond is more polar. 16. Among hydrides of group15 elements the bond length increases from N – H to Bi – H with increasing size of element. Bi – H bond is largest and weakest, it can break more easily and evolve H 2 gas which acts as the reducing agent. 17. Phosphinic acid, H 3 PO 2 has one P – OH group and two hydrogen directly attached to phosphorous. It has only one ionisable hydrogen and hence behaves as monoprotic acid. O H

P

OH

H

18. (i)

F F

less stable than + 5 oxidation state of Sb. Therefore, Bi(V) is stronger oxidising agent than Sb (V). 23. (i) The property of catenation depends upon the strength of the element element bond. Since, P—P (213 kJ mol –1 ) bond strength is much more than N—N (159 kJ mol –1 ) bond strength. So, phosphorus shows marked catenation properties than nitrogen. (ii) The electron gain enthalpy of oxygen is less than that of sulphur due to its small size and high interelectronic repulsion. 24. (a) Preparation of interhalogen compounds :The interhalogen compounds can be prepared by the direct combination of the halogens. The product formed depends upon some specific conditions. For example Cl2 + F2

S

437 K

¾¾¾® 2ClF

(Equal volume)

F

F

SF 4

(b) Refer Ans. 22(i). 25. (i) Structure of SO 2 molecule :

Shape : Seasaw 3 Hybridisation : sp d

S

(ii) Refer Ans. 12. 19. (i) 3Cl 2 + 6NaOH ® 5NaCl + NaClO 3 + 3H 2 O Heat (ii) 4H3 PO3 ¾¾¾ ® 3H3 PO 4 + PH3

(iii) PtF 6 + Xe ® Xe + PtF 6 – Red colour compound.

20. (i) Ca 3 P 2(s) + H 2 O (l) ® 3Ca(OH) 2(aq) + 2PH 3(g) 2+ 2 + (ii) Cu ( aq ) + 4 NH 3( aq ) ƒ [Cu(NH 3 ) 4 ] ( aq )

(Blue)

(iii) 2F 2(g) + 2H 2 O (l) ® 4H + (aq)

deep blue

+ 4F – (aq) + O 2(g)

21. (a) Due to small size oxygen forms pp pp multiple bonds. As a result oxygen exist as diatomic (O 2 ) molecule. Due to its bigger size sulphur does not forms pp pp multiple bonds. Consequently, sulphur because of its higher tendency of catenation and lower tendency for pp pp multiple bonds forms octaatomic (S 8 ) molecules having eight membered puckered ring structure. S S S S S S O O S S (b) Due to decrease in bond (E — H) dissociation enthalpy down the group, acidic character increases. Hence H 2 S is less acidic than H 2 Te.

O

S O

O

O

In SO 2 , S is sp 2 hybridised. Two of the three sp 2 orbitals form two s bonds while the third contains the lone pair of electrons. Now S has one half filled porbital and one half filled dorbital. These form one pp–pp and one pp–dp double bond with oxygen atom. Due to resonance, the two pbonds are equal in length. (ii) When white phosphorus is heated with conc. NaOH solution in inert gas atmosphere, phosphine gas is produced. P4 + 3 NaOH + 3 H 2 O ® PH 3 + 3 NaH 2 PO 2 Phosphine

26. (i) Refer Ans. 12. (ii) Refer Ans. 13. 27. (i) P4( s ) + NaOH (aq ) + H 2 O (l ) ® PH3 ( g ) + 3 NaH 2 PO 2 Phosphine

(ii) 4I – (aq)

+ 4H + (aq)

+ O 2(g) ® 2I 2(s) + 2H 2 O (l) .

28. White phosphorus is more reactive than red phosphorus under normal conditions because of angular strain in the P 4 molecule where the angles are only 60°. 29. (i) 2XeF 2 + 2H 2 O ® 2Xe + 4HF + O 2 (ii) 2PH 3 + 3HgCl 2 ® Hg 3 P 2 + 6HCl.

30. (a) (i) H 2 S 2 O 8 (Peroxodisulphuric acid) 22. (i) ClO – has 17 + 8 + 1 = 26 electrons O O A neutral molecule with 26 electrons is OF 2 (8 + 2 × 9) = 18 + 8 = 26 electrons. S S (ii) On moving down the group, the stability of + 5 oxidation O O O O state decreases while + 3 oxidation state increases due OH OH to inert pair effect. Thus + 5 oxidation state of Bi is (ii) HClO 4 (Perchloric acid)


44

H

36. (i) Because F atom is very small as compared to I atom. H – F bond is smaller and stronger than H – I bond. (ii) All hydrides of group 15 (NH 3 , PH 3 , AsH 3 , SbH 3 and BiH 3 ) have one lone pair of electrons on their central atom. Therefore, they act as lewis bases. They can donate an electron pair to electron deficient species (Lewis acid). As we go down the group size of central atom increases, electron density on the atom decreases, the basic character of these hydride decrease. (iii) Sulphur has a stronger tendency for catenation than oxygen because stronger S – S bonds as compared to O – O bond due to interelectronic repulsions of smaller O atoms.

O Cl O

O O

31. (i) Refer Ans. 12. (ii) XeO 3 is pyramidal.

: Xe 3

sp hybridisation

O

O

O 32. (i) Due to the presence of lone pair of electrons on the central atom both NH 3 and PH 3 are lewis bases. Due to smaller size of N, than P, electron density on N of NH 3 is greater than P in PH 3 . As a result NH 3 has more tendency than PH 3 to accept a proton. Therefore NH 3 is stronger base than PH 3 . (ii) Due to small size of O, six F atoms can not be accommodated around oxygen atoms. Hence OF 6 is not known or oxygen can not expand its valency beyond 4 due to absence of dorbital. (iii) Since fluorine is the most electronegative element among halogens.

38. (i) Refer Ans. 12. (ii) Refer Ans. 32(i). 143 K 39. (i) XeF4 + O 2 F2 ¾¾¾® XeF6 + O 2

(ii) XeF 4 + SbF 5 ® [XeF 3 ] + [SbF 6 ] – (iii) 6NaOH (hot, conc.) + 3Cl 2 ® 5NaCl + NaClO 3 + 3H 2 O 40. Structure of white and red phosphorus. P

(a)

33. (i) The order of increasing basic strength. SbH 3 < AsH 3 < PH 3 < Least basic

37. (i) I 2 + 2NaClO 3 ® 2NaI + 2ClO 2 + O 2 (ii) NH 3 + NaOCl ® NaNH 2 + HClO (iii) XeF 6 + KF ® K + [XeF 7 ] – .

NH 3

Weakest acid

HI

Strongest acid

(iii) The order of increasing acid strength. HClO < HClO 2 < HClO 4 Weakest acid

Strongest acid

P P

White phosphorus

Most basic

(ii) The order of increasing acid strength. HF < HCl < HBr <

P

P

(b)

P

P P

P

P

P P

P

P

P P

Red phosphorus

White phosphorus is more reactive due to its discrete tetrahedral structure and angular strain.

34. (i) PCl 5 is ionic in the solid state because it exists as [PCl 4 ] + [PCl 6 ] – in which the cation is tetrahedral and 41. (i) I 2 + 10HNO 3 ® 2HIO 3 + 10NO 2 + 4H 2 O anion is octahedral. (ii) 3HgCl 2 + 2PH 3 ® Hg 3 P 2 + 6HCl. (ii) The size of central atom is H 2 S is larger than that in 42. (i) Structure of H 4 P 2 O 7 H 2 O. S – H bond is larger and weaker than O – H bond H 2 O, hence H 2 S is more acidic than H 2 O. O O (iii) Because fluorine is most electronegative element and can form compounds of higher valency also. P P O OH OH 35. (i) Refer Ans. 11. OH OH (ii) Structure of (HPO 3 ) 3 (ii) Refer Ans. 12. O O O 43. (i) Because bond energy of F 2 is lower than Cl 2 and P P N – F bond is smaller and stronger than N – Cl bond. HO OH (ii) Since there are two Xe – F covalent bonds and three O O lone pairs in XeF 2 . According to VSEPR theory, the P shape of XeF 2 is linear. O

OH

Shape Cyclic trimer with a hexagonal ring

The pBlock Elements

(iii) Standard electrode potential value of F 2 is more than chlorine. 44. (i) Because of the compact nature and large interelectronic repulsion of oxygen atom, oxygen has less negative electron gain enthalpy than sulphur. (ii) Phosphorus cannot form pp pp multiple bonds. (iii) Fluorine being smaller in size and highly electronegativity can not show higher valence state, hence fluorine never acts as the central atom in polyatomic interhalogen compounds. 45. (i) This is because the N – O bond in NO 2 – is an average of a single bond and a double bond whereas the N – O bond in NO – 3 is an average of two single bonds and a double bond. (ii) In SF 6 , S atom is sterically protected by six F atoms and does not allow any reagent to attack on the S atom. Due to these reasons, SF 6 is kinetically an inert substance. OR (i) In PCl 5 the two axial bonds are longer than the three equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs. (ii) The property of catenation depends upon E – E bond strength of the element. As S – S bond is much stronger (213 kJ mol –1 ) than O – O bond (138 kJ mol –1 ), thus sulphur has greater tendency for catenation than oxygen. 46. (i) Refer Ans. 34(ii). (ii) Refer Ans. 45(i). (iii) This is due to ability of oxygen to form multiple bonds with metal atom. 47. (i) N – F bond enthalpy is higher than N – Cl bond enthalpy so, NF 3 is exothermic compound but NCl 3 is endothermic compound. (ii) Larger the difference in electronegativity greater will be polarity and hence greater will be acidic character. Electronegativity of Cl > S > P, hence acidity order is HCl > H 2 S > PH 3 . (iii) Refer Ans. 45(ii).

45

Pressure : 200 bar Temperature : 723 – 773 K Catalyst : Finely divided iron oxide, K 2O as promoter. (b) (i) In vapour state sulphur partly exists as S 2 molecule which has two unpaired electrons in the antibonding p* molecular orbitals. (ii) Refer Ans. 28. OR (i) Total no. of electrons around the central N atom = 5 No. of bond pairs = 3 No. of lone pairs = 1 Hybridisation = sp 3 Therefore, according to VSPER theory; NF 3 should be pyramidal.

N

F

F

F

(ii)

O

O

S

S

O

O

O

OH

O OH

Peroxodisulphuric acid (H2S 2O8 )

O P OH

(iii) H OH

Orthophosphorus acid (H3 PO3 )

51. (i) P 4 + 10SO 2 Cl 2 ® 4PCl 5 + 10SO 2 48. (i) Refer Ans. 47(ii). – (ii) 2Fe 3+ + SO 2 + 2H 2 O ® 2Fe 2+ + 4H + + SO 2 4 (ii) Oxidising power increases with increasing oxidation state of element. Oxidation state of Cl decreases from (iii) XeF 6 + 3H 2O XeO 3 + 6HF (excess) HClO 4 to HClO. Hence oxidising power order is HClO 4 > HClO 3 HClO 2 > HClO. 52. (i) Because bond dissociation enthalpy of H Cl is lower than that of H F . 49. (i) 6NaOH + 3Cl2 5NaCl + NaClO3+ 3H2 O (ii) Refer Ans. 28. Hot and conc. 143 K (ii) XeF4 + O2F2 XeF6 + O 2 (iii) Br2 + 5F 2 2 BrF 5 (excess)

50. (a) Optimum conditions for the industrial manufacture of ammonia by Haber’s process.

53. (i) C + 2H 2 SO 4 (conc.) ® CO 2(g) + 2H 2 O (l) + 2SO 2(g) (ii) P 4 + 3NaOH + 3H 2 O ® PH 3 + 3NaH 2 PO 2 573 K

(iii) Cl2 + 3F2 2ClF 3 (excess)

F S F

F F

Seesaw shape

:

54. (i) The highest oxidation number coincides with the group number in the periodic table. e.g. Sc 2 O 3 , TiO 2 , V 2 O 5 ,. CrO 3 , Mn 2 O 7 . Beyond group7 no higher oxides are known except Fe 2 O 3 and Co 3 O 4 . Some oxocations such as TiO 2+ and VO 2 + are also known. (ii) According to the valence bond approach orbitals containing unpaired electrons only take part in covalent bond formation. Xenon has no unpaired electrons. Hence structure of its fluorides cannot be explained by valence bond approach. 55. (i) Refer Ans. 47(i). (ii) In SF 4 two axial bonds experience greater repulsion from the lone pair of electrons, hence bonds are slightly elongated.

:


46

P

N

H H H H H H Because size of P is larger than N atom. Therefore N atom carries more negative charge density than carried by P. Hence NH 3 has more proton affinity than PH 3 . (iii) The electron gain enthalpy of fluorine is less negative than that of chlorine due to the small size of the fluorine atom and high interelectronic repulsion. (iv) In SF 6 six F atoms protect the sulphur atom from attack by the reagent to such an extent that even thermodynamically most favourable reactions like hydrolysis do not occur. But in SF 4 , S is not stearically protected since it is surrounded by only four F atoms hence SF 4 is reactive. (v) Except radon which is radioactive, Xe has least ionisation energy among noble gases and hence it readily forms chemical compounds particularly with oxygen and fluorine.

56. (a) (i) Phosphorus (P 4 ) is a tetratomic molecule. Since the P–P single bond is much weaker (213 kJ mol –1 ) 59. (a) (i) On a large scale ammonia is manufactured by Haber’s process. than N N triple bond [941.4 kJ mol –1 ]. Therefore, N 2(g) + 3H 2(g) ® 2NH 3(g) ; D rH = – 46.1 kJ phosphorus is more reactive than nitrogen (N 2 ). 3 The optimum condition for the production of ammonia (ii) As a result of sp d hybridisation there are two are a pressure about 200 atm, a temperature of about axial bonds and three equatorial bonds in PCl 5 700 K and the use of catalyst such as iron oxide with molecule. small amount of K 2 O and Al 2 O 3 as promoter. Axial bond pairs suffer more repulsive interactions (ii) Low temperature about 720 K and high pressure about from the equatorial bond pairs. Therefore the axial 2 bar and a catalyst V 2 O 5 are favourable condition for bonds are slightly elongated and hence slightly the manufacture of H 2 SO 4 by contact process. weaker than equatorial bonds. (b) (i) Structure of PCl 5(g) (iii) Due to presence of unpaired electron in antibonding molecular orbitals. Cl Cl (b) (i) Refer Ans. 13. 240 pm (ii) Refer Ans. 11 Trigonal bipyramidal P 202 p 3 + – m 57. (a) (i) 2F 2 + H 2 O ® 4H (aq) + 4F (aq) + O 2(g) sp d hybridisation Cl Cl (ii) Ca 3 P 2 + 6H 2 O ® 3Ca(OH) 2 + PH 3 Cl (iii) 6XeF + 12H O ® 4Xe + 2XeO + 24HF + 3O 4

2

O

S

S

2

(ii) Structure of S 8(g) S

pm

O OH

:

O O OH

4

(b) (i)

S

O :

O

N

N

: :

218 pm

(ii) – : O

:

: O : O 120° O 2 sp hybridisation Planar structure 58. (i) Refer Ans. 50(b) (i). (ii) PH 3 and NH 3 both are Lewis base, since they have a lone pair of electrons on ‘N’ and ‘P’ atom respectively.

S

S

20

O

3

107°

S (Crown shape)

S S

S

(iii) Structure of ClF 3(g) F F Cl F

(Bent T shaped)

The pBlock Elements

47

:

60. (a) Acid strength of oxoacids of the same halogen increases 64. (a) (i) Structure of N 2 O 5 with increase in oxidation number of the halogen. O O O N N 134° Thus the increasing order of acid strength is 120° O O HOCl < HClO 2 < HClO 3 < HClO 4 Planar +1 +3 +5 +7 (ii) Structure of XeOF 4 (b) 3Cl 2 + 6NaOH ® 5NaCl + NaClO 3 + 3H 2 O O This reaction is a disproportionation reaction as chlorine F F from zero oxidation state is changed to – 1 and + 5 Xe oxidation states. (c) ClF 3 is used for the production of UF 6 in the enrichment F F of 235 U. (b) (i) The electron gain enthalpy of oxygen is least U (s) + 3ClF 3(l) ® UF 6(g) + 3ClF (g) . negative in this group, due to its small size. 61. (a) Fluorine and oxygen are the most electronegative elements and hence are very reactive. Therefore, they form compounds with noble gases particularly xenon. (b) Hydrolysis of XeF 6 with water gives XeOF 4 and XeO 3 . (i) XeF 6 + H 2 O ® XeOF 4 + 2HF (ii) XeF 6 + 3H 2 O ® XeO 3 + 6HF. (c) Refer Ans. 13. 62. (a) (i) H 2 S 2 O 7 (Oleum) O

O

S

S O

O

OH (ii) HClO 3 (Chloric acid)

O

(b) (i)

(ii) (iii)

63. (a) (i)

O HO

H

Cl O O N – O has some double bond character. Therefore, the N – O bond length in HNO 3 is smaller than N – OH bond length. Refer Ans. 45 or (1). ICl is more reactive than I 2 (except F 2 ). Because X – X¢ bond in interhalogens is weaker than X – X bond in halogen except F – F bond. Thus I – Cl bond is weaker than I – I bond. So ICl is more reactive than I 2 H 3 PO 2 O P OH

H H

(ii) Nitrogen does not form pentahalide due to the absence of dorbitals in its valence shell. (iii) The strength of an acid depends upon its bond strength. Since bond dissociation energy of HCl is greater than HI. Therefore, HI is stronger acid than HCl. 65. (i) Refer Ans.12. (ii) Refer Ans. 57 b(i). (b) (i) Phosphorus cannot form pp pp multiple bonds. (ii) The electron gain enthalpy of fluorine is less negative than that of chlorine due to the small size of fluorine atom. (iii) Hydrogen fluoride has a much higher boiling point (b.p. 293 K) due to strong hydrogen bonding. 69. (a) (i) Structure of PCl 5(s) + Cl

Cl

Cl P Cl

Cl

–

Cl P

Cl

Cl

Cl Cl

(ii) Structure of SO 3 2– 2–

O S O

O

(b) (i) NH 3 molecules are held together by strong inter molecular hydrogen bonds whereas PH 3 is held together by weak van der Waals bond. Thus, NH 3 has a higher boiling point than PH 3 . (ii) Since helium has completely filled ns 2 electronic configurations in its valence shell. (iii) Refer Ans. 7.

(ii) Refer Ans. 13. (b) (i) Phosphorus cannot form pp pp multiple bonds. (ii) Because F atom is very small as compared to Cl atom. Interelectronic repulsion in F 2 is larger than 67. (a) (i) 6NaOH + 3Cl 2 ® 5NaCl + NaClO 3 + 3H 2 O (ii) XeF 6 + 2H 2 O ® XeOF 4 + 2HF that in chlorine. or (iii) Since helium has completely filled ns 2 electronic XeF + 2H O ® XeO 2 F 2 + 4HF 6 2 configurations in its valence shell.


48

(b) (i) Because of large size of sulphur atom than oxygen atom inter electronic repulsion on S is less than that on oxygen. (ii) Because bond energy of F 2 is lower than Cl 2 and N – F bond is smaller and stronger than N – Cl bond. (iii) Because it has sp 3 d hybridization.

(ii) Refer Ans. 57 b(ii). O F Xe

(iii) XeOF 4

68. (a) (i) P 4 + 10SO 2 Cl 2 ® 4PCl 5 + 10SO 2

F

F

F

Hybridisationsp3d 2 StructureOctahedral ShapeSquare pyramidal

(ii) 6XeF 4 + 12H 2 O ® 2XeO 3 + 4Xe + 24HF + 3O 2 (b) (i) Due to inert pair effect. (ii) Due to the formation of [PCl 4 ] + [PCl 6 ] – . F (iii) Because halogens readily accept an electron. F F 3 2 (b) (i) NH 3 molecules are held together by strong inter Hybridisation sp d Xe molecular hydrogen bonds whereas PH 3 is held Structure Pentagonal bipyramidal together by weak van der Waals bond. Thus, NH 3 70. (a) (i) F F Shape Octahedral has a higher boiling point than PH 3 . F (ii) Since helium has completely filled ns 2 electronic configurations in its valence shell. (ii) H 2 S 2 O 8 (Peroxodisulphuric acid) (iii) Bismuth (Bi) in the pentavalent state easily accepts O O two electrons and gets reduced to trivalent bismuth due to inert pair effect S S O Bi 5+ + 2e – ® Bi 3+ O O O OH OH Therefore, it acts as a strong oxidising agent. 69. (a) (i) P 4 + 10SO 2 Cl 2 ® 4PCl 5 + 10SO 2

(b) (i)

(ii) XeF 6 + 2H 2 O ® XeOF 4 + 2HF or XeF 6 + 2H 2 O ® XeO 2 F 2 + 4HF

N H

O

(b) (i)

Cl

O –

O –

Hybridisation = sp 3 Shape : pyramidal O – S – O > 90°

(ii)

F

F

F

In NH 3 the bond moment and lone pair moment work in same direction. Hence net dipole moment is higher.

In NF3 the bond moment and lone pair moment work in opposite direction. Hence net dipole moment is lower.

F Hybridisation = sp 3 d Shape : Tshaped F – Cl – F < 90°

F Xe

(iii)

F

H

F S

N H

F Hybridisation = sp 3 d Shape : Linear F – Xe – F > 90°

OR (a) (i) 6NaOH + 3Cl 2 ® 5NaCl + NaClO 3 + 3H 2 O 143 K (ii) R XeF4 + O 2 F2 ¾¾¾® XeF6 + O 2

(ii) Refer Ans. 45 or (i). (iii) Refer Ans. 50 (b) (i). OR (a) (i) XeF 4 + SbF 5 ® [XeF 3 ] + [SbF 6 ] –1 . (ii) Cl 2 + 3F 2 (excess) ® 2ClF 3 (b) (i) Due to the presence of a triple bond between the two nitrogen atoms, the bond dissociation energy of N 2 is very high as compared to phosphorus. Therefore, N 2 is less reactive than phosphorus. (ii) The stability of +5 oxidation state decreases down the group of 15 due to inert pair effect. The inertness of a pair of selectron decreases down the group with increasing atomic size. (iii) In NO 2 – ‘N’ atom has sp 2 hybridisation whereas in NO 2 + , ‘N’ atom has sphybridisation.

O N

P

(b) (i) H

OH H

O

Å

O –

O N O

Here bond angles are not of the same value

The pBlock Elements

49

71. (a) (i) The structure of N 2 O 5 is O O N O N O O

O

72. (a) (i) H

Covalence of nitrogen in N2O5 = 5

(ii) XeOF 4 is square pyramidal. O

F 3

2

Hybridisation sp d

P OH H

(ii) Structure of ClF 3(g)

F Xe

F

:

F F (b) (i) O – O bond is weaker than the single S – S bond due to high interelectronic repulsion of non bonding electrons of oxygen. (ii) ICl is more reactive than I 2 because I – Cl bond is weaker than I – I bond. Consequently, ICl breaks easily to form halogen atoms which readily bring about the reactions. (iii) Fluorine has higher standard reduction potential than chlorine, so it is more easily reduced and hence it is stronger oxidising agent than chlorine. OR (a) (i) Copper reacts with dil. HNO 3 to give NO and with concentrate HNO 3 to give NO 2 . 3Cu + 8HNO 3(dil.) ® 3Cu(NO 3) 2 + 2NO + 4H 2 O Cu + 4HNO 3(conc.) ® Cu(NO 3) 2 + 2NO 2 + 2H 2 O 143 K (ii) XeF4 + O 2 F2 ¾¾¾® XeF6 + O 2

F Cl

(Bent T shaped)

F

(b) (i) Refer Ans. 70 or (b) (i). (ii) Hydrogen fluoride is more polar but boils at lower temperature than water because H 2 O molecules are highly associated with intermolecular Hbond. H –– O H –– O H –– O H

H

H

(iii) Refer Ans. 71 (b) (i). OR (a) (i) The structure of N 2 O 5 is O O N O N O O Covalence of nitrogen in N2O5 = 5

(b) (i) The single N–N bond is weaker than the single P– P bond because of high interelectronic repulsion of the nonbonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen. (ii) O 2 molecules are held together by weak van der Waal’s forces because of the small size and high electronegativity of oxygen. Sulphur shows catenation and the molecule is made up of eight atoms, (S 8 ) with strong intermolecular attractive forces. Hence sulphur exists as solid at room temperature. (iii) Halogens absorb radiations in visible region which results in excitation of outer electrons to higher level resulting in different colours.

JJJ

(ii) HClO 4 (Perchloric acid) H O Cl O O (b) (i) Refer Ans. 34(ii). (ii) Since fluorine is the most electronegative element, it shows only a negative oxidation state of –1, and does not show any positive oxidation state. (iii) Since helium has completely filled ns 2 electronic configurations in its valence shell. O


50

Chapter8 The d and f Block Elements


found to be good catalysts in chemical reactions. [AI 2010]

What is meant by ‘lanthanoid contraction’? 9. Describe how the following changes are brought about: [AI SetI 2011] (i) Pig iron into steel. (ii) Zinc oxide into metallic zinc. SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS) (iii) Impure titanium into pure titanium. [AI 2010] 2. Write complete chemical equations for : 10. Describe the role of (i) Oxidation of Fe 2+ by Cr 2 O 7 2– in acid medium. (i) NaCN in the extraction of gold from gold ore. (ii) Oxidation of S 2 O 3 2– by MnO 4 – in neutral aqueous (ii) SiO 2 in the extraction of copper from copper matte. medium. [AI 2008] (iii) Iodine in the refining of zirconium. 3. How would you account for the following situations? Write chemical equations for the involved reactions. (i) The transition metals generally form coloured compounds. [AI 2010] (ii) With 3d 4 configuration Cr 2+ acts as a reducing agent 11. How would you account for the following? but Mn 3+ acts as an oxidising agent. (i) The atomic radii of the metals of the third (5d) series [Atomic no. Cr = 24, Mn = 25]. of transition elements are virtually the same as those (iii) The actinoids exhibit a large number of oxidation states of the corresponding members of the second (4d) series. than the corresponding lanthanoids. [AI 2008] (ii) The E° value for the Mn 3+ /Mn 2+ couple is much more 4. (a) What may be the possible oxidation states of the transition positive than that for Cr 3+ /Cr 2+ couple or metals with the following d electronic configuration in Fe 3+ /Fe 2+ couple. the ground state of their atoms : (iii) The highest oxidation state of a metal is exhibited in 3d 3 4s 2 3d 5 4s 2 and 3d 6 4s 2 . Indicate relative stability of its oxide or fluoride. [AI 2010] oxidation states in each case. 12. Explain the following observations : (b) Write steps involved in the preparation of (i) Na 2 CrO 4 (i) Transition elements generally form coloured from chromite ore and (ii) K 2 MnO 4 from pyrolusite ore. compounds. [Delhi 2008] (ii) Zinc is not regarded as a transition element. 5. Account for the following : [Delhi 2010] (i) There are irregularities in the electronic configuration 13. Assign reasons for the following: of actinoids. (i) Copper (I) ion is not known in aqueous solution. (ii) Compounds of transition elements are often coloured. (ii) Actinoids exhibit greater range of oxidation states than [Delhi 2008 C] lanthanoids. [Delhi SetI 2011] 6. (a) Describe the commercial preparation of potassium 14. Complete the following chemical equations: permanganate from pyrolusite ore. (i) MnO – + C 2 O 2– + H + ® 4 (b) Write ionic equation to represent the reaction of acidified heated 4 (ii) KMnO 4 KMnO 4 solution with oxalic acid. [Delhi 2008] (iii) Cr 2 O 2– + H 2 S + H + ® [Delhi SetI 2011] 7 7. Account for the following : 15. Assign reasons for each of the following: (i) Transition metals are well known to form complex (i) Transition metals generally form coloured compounds. compounds. (ii) Manganese exhibits the highest oxidation state of +7 (ii) The second and third members in each group of the among the 3rd series of transition elements. transition elements have very similar atomic radii. [Delhi SetIII 2011] [Delhi 2008 C] 16. How would you account for the following: 8. Explain the following observations : (i) Cr 2+ is reducing in nature while with the same (i) Generally there is an increase in density of elements dorbital configuration (d 4 ) Mn 3+ is an oxidising agent. from titanium (Z = 22) to copper (Z = 29) in the first (ii) In a transition series of metals, the metal which exhibits series of transition elements. (ii) Transition elements and their compounds are generally 1.

The dand fBlock Elements

51

the greatest number of oxidation states occurs in the 24. How would you account for the following: middle of the series. [AI SetI 2011] (i) Among lanthanoids, Ln(III) compounds are predominant. However, occasionally in solutions or in 17. Complete the following chemical equations : – 2– solid compounds, +2 and +4 ions are also obtained. (i) MnO 4 (aq) + S 2 O 3 (aq) + H 2 O (l) ® (ii) The E° M 2+ /M for copper is positive (+0.34 V). Copper (ii) Cr 2 O 7 2– (aq) + Fe 2+ (aq) + H + (aq) ® is the only metal in the first series of transition elements OR showing this behaviour. State reasons for the following: (iii) The metallic radii of the third (5d) series of transition (i) Cu (I) ion is not stable in an aqueous solution. metals are nearly the same as those of the corresponding (ii) Unlike Cr 3+ , Mn 2+ , Fe 3+ and the subsequent other M 2+ members of the second series. [AI SetI 2012] ions of the 3d series of elements, the 4d and the 5d 25. Explain the following observations: series metals generally do not form stable cationic (i) Many of the transition elements are known to form species. [AI SetI 2011] interstitial compounds. 18. Explain giving a suitable reason for each of the following: (ii) There is a general increase in density from titanium (i) Transition metals and their compounds are generally (Z = 22) to copper (Z = 29). found to be good catalysts. (iii) The members of the actinoid series exhibit a larger (ii) Metalmetal bonding is more frequent for the 4d and number of oxidation states than the corresponding the 5d series of transition metals than that for the 3d members of the lanthanoid series. [AI SetII 2012] series. [AI SetII 2011] 26. Explain each of the following observations. 19. Explain giving reasons; (i) With the same dorbital configuration (d 4 ), Cr 2+ is a (i) Transition metals and their compounds generally exhibit reducing agent while Mn 3+ is an oxidising agent. a paramagnetic behaviour. (ii) Actinoids exhibit a much larger number of oxidation (ii) The chemistry of actinoids is not as smooth as that of states than the lanthanoids. lanthanoids. [AI SetIII 2011] (iii) There is hardly any increase in atomic size with increasing atomic numbers in a series of transition 20. Complete the following chemical equations: 2– + – – – + metals. [AI SetIII 2012] (i) Cr 2 O 7 + H + I (ii) MnO 4 + NO 2 + H [Delhi SetI 2012, AI SetI 2012] LONG ANSWER TYPE QUESTIONS (5 MARKS)

21. How would you account for the following? (i) Many of the transition elements are known to form 27. Assign reason for the following : (i) The enthalpies of atomisation of transition elements interstitial compounds. are high. (ii) The metallic radii of the third (5d) series of transition (ii) The transition metals and many of their compounds act metals are virtually the same as those of the as good catalyst. corresponding group member of the second (4d) series. (iii) From element to element the actinoids contraction is (iii) Lanthanoids from primarily +3 ions, while the actinoids greater than the lanthanoid contraction. usually have higher oxidation states in their compounds, (iv) The E° value for the Mn 3+ /Mn 2+ couple is much more +4 or even +6 being typical. [Delhi SetI 2012] positive than that for Cr 3+ /Cr 2+ . 22. How would you account for the following? (v) Scandium (Z = 21) does not exhibit variable oxidation (i) With the same dorbital configuration (d 4 ) Cr 2+ is a states and yet it is regarded as transition element. reducing agent while Mn 3+ is an oxidising agent. [Delhi 2008] (ii) The actinoids exhibit a larger number of oxidation states 28. (a) Give reason for the following : than the corresponding members in the lanthanoid series. (i) Actinoids exhibit greater number of oxidation (iii) Most of the transition metal ions exhibit characteristic states as compared to lanthanoids in general. colours in aqueous solutions. [Delhi SetII 2012] (ii) Though a transition element, scandium (Z = 21) 23. Explain the following observations giving an appropriate does not exhibit variable oxidation states. reason for each. (b) (i) Describe how potassium dichromate is prepared (i) The enthalpies of atomization of transition elements from sodium chromate. are quite high. (ii) The colour of potassium dichromate solution (ii) There occurs much more frequent metalmetal bonding changes with the changes of pH of the solution. in compounds of heavy transition metals (i.e., 3 rd series). Explain how. [Delhi 2008, AI 2009] (iii) Mn 2+ is much more resistant than Fe 2+ towards oxidation. 29. (a) On what ground can you say that scandium [Delhi SetIII 2012]


52

(Z = 21) is a transition element but zinc (Z = 30) is not. (b) Use Hund’s rule to derive the electronic configuration of Ce 3+ ion and calculate its magnetic moment by using ‘spin only’ formula : [Z for Ce = 58]. (c) What is lanthanoid contraction and what is its consequence? [AI 2008 C] 30. (a) Complete the following chemical equations : (i) MnO 4 – (aq) + C 2 O 4 2– (aq) + H + (aq) ® (ii) Cr 2 O 7 2– (aq) + Fe 2+ (aq) + H + (aq) ®

(b) Explain the following observations : (i) In general the atomic radii of transition elements decrease with atomic number in the given series. (ii) The E° M 2+ /M for copper is positive (+0.34 V). It is only metal in the first series of transition elements showing this type of behaviour. (iii) The E° value for Mn 3+ /Mn 2+ couple is much more positive than for Cr 3+ /Cr 2+ or Fe 3+ /Fe 2+ couple. [Delhi 2009]

34. (a) Give reason for the following observations : (b) Explain the following observations about the transition/ (i) Of the d 4 species, Cr 2+ is strongly reducing whereas inner transition element: manganese (III) is strongly oxidising. (i) There is in general an increase in density of element (ii) The enthalpies of atomisation of the transition from titanium (Z = 22) to copper (Z = 29). metals are quite high. (ii) There occurs much more frequent metalmetal (iii) Interstitial compounds are well known for bonding in compounds of heavy transition transition metals. elements. (3 rd series). (b) What is lanthanoid contraction? List any two (iii) The members in the actinoid series exhibit a larger consequence of lanthanoid contraction. [AI 2009] number of oxidation states than the corresponding 35. (a) Complete the following chemical equations : members in the lanthanoid series. [Delhi 2009] (i) Cr O 2– + H S + H + ® 31. (a) Complete the following chemical equations for reactions. (i) MnO 4 – (aq) + S 2 O 3 2– (aq) + H 2 O (l) ®

2 7

(aq)

2 (g)

(aq)

(ii) Cu 2+ (aq) + I – (aq) ® (b) How would you account for the following : (i) The oxidising power of oxoanoins are in the order VO 2 + < Cr 2 O 7 2– < MnO 4 – (ii) The third ionization enthalpy of manganese (Z = 25) is exceptionally high. (iii) Cr 2+ is a stronger reducing agent than Fe 2+ . [Delhi 2010]

(ii) Cr 2 O 7 – (aq) + H 2 S (g) + H + (aq) ® (b) Give an explanation for each of the following observations. (i) The gradual decrease in size (actinoid contraction) from element to element is greater among the actinoids than among the lanthanoids (lanthanoid 36. (a) Complete the following chemical equations : contraction). (i) MnO 4 – (aq) + S 2 O 3 2– (aq) + H 2 O (l) ® (ii) The greater number of oxidation states are exhibited by the members in the middle of a (ii) Cr 2 O 7 2– (aq) + Fe 2+ (aq) + H + (aq) ® transition series. (b) Explain the following observations : (iii) When the same dorbitals configuration (d 4 ) Cr 2+ (i) La 3+ (Z = 57) and Lu 3+ (Z = 71) do not show any ion is a reducing agent but Mn 3+ ion is an oxidising colour in solutions. agent. [Delhi (ii) Among the divalent cations in the first series of 2009] transition elements, manganese exhibits the 32. (a) Complete the following chemical reaction equation: maximum paramagnetism. (i) Fe 2+ (aq) + MnO 4 – (aq) + H + (aq) ® (iii) Cu + ion is not known in aqueous solutions. [Delhi 2010] (ii) Cr O 2– + I – + H + ® 2 7

(aq)

(aq)

(aq)

(b) Explain the following observation: (i) Transition elements are known to form many interstitial compounds. (ii) With the same dorbital configuration Cr 2+ ion is reducing while, Mn 3+ ion is oxidising. (iii) The enthalpies of atomisation of the transition element are quite high. [Delhi 2009] 33. (a) Complete the following chemical equation : (i) Cr 2 O 7 2– (aq) + I – (aq) + H + (aq) ® (ii) MnO 4 – (aq) + Fe 2+ (aq) + H + (aq) ®

37. (a) Complete the following reactions in an aqueous medium: (i) MnO – 4 + C 2 O 2– + H + ® 4 (ii) Cr 2 O 2– + H 2 S + H + ® 7 (b) How would you account for the following : (i) Metalmetal bonding is more extensive in the 4d and 5d series of transition elements than the 3d series. (ii) Mn (III) undergoes disproportionation reaction easily. (iii) Co (II) is easily oxidised in the presence of strong ligands. OR


53

answers 1. The steady decrease in the atomic and ionic radii (having the same charge) with increase in atomic number across the series from lanthanum to lutetium is known as lanthanoid contraction. 2. (i) Acidified potassium dichromate oxidises iron (II) salt to iron (III). Cr 2 O 7 2– + 14H + + 6Fe 2+ ® 2Cr 3+ + 6Fe 3+ + 7H 2 O (ii) In neutral aqueous medium thiosulphate is oxidised almost quantitatively to sulphate. 8MnO 4 – + 3S 2 O 3 2– + H 2 O ® 8MnO 2 + 6SO 4 2– + 2OH –

acidified with sulphuric acid to give a orange solution of sodium dichromate Na 2 Cr 2 O 7 which is crystallised. 2 Na 2 CrO4 + H 2SO 4 ® Na 2 Cr2 O 7 + Na 2SO4 + H 2 O Sod.Chromate

Sodium dichromate

The solution of sodium dichromate is treated with potassium chloride to obtain potassium dichromate. Na 2 Cr2 O7 + 2 KCl ® K 2 Cr2 O7 + 2 NaCl Potassium dichromate .

(ii) Preparation of potassium permanganate : Potassium permanganate is prepared by the fusion of MnO 2 (pyrolusite) with potassium hydroxide and an oxidising agent like KNO 3 to form potassium manganate which disproportionates in a neutral or acidic solution to form permanganate. 2MnO 2 + 4KOH + O 2 ® 2K 2 MnO 4 + 2H 2 O + – 3MnO 42– + 4H ® 2MnO 4 + MnO 2 + 2H 2 O or, 3K 2MnO 4 + 4HCl ® 2KMnO 4 + MnO 2 + 2H 2O + 4KCl.

3. (i) Since transition elements contain partially filled d subshells. Therefore, electrons in these subshells go from lower dsubshells to higher dsubshells. This is called dd transition. This transition takes place by absorbing energy from the visible light. The mixture of the wavelength which is not absorbed is transmitted out. This accounts for the colour of a transition elements. (ii) It can be explained on the basis of E° values of Cr 3+ /Cr 2+ (–0.4 V) and Mn 3+ /Mn 2+ (+1.5 V). On the 5. (i) The irregularities in the electronic configuration of the actinoids are due to the extra stabilities of the f 0 , f 7 basis of these values it is clear that Cr 2+ has a greater and f 14 occupancies of the 5forbitals. tendency to oxidise into Cr 3+ , thus it acts as reducing (ii) Refer Ans. 3 (i). agent 2+ 3+ – Cr (aq) ® Cr (aq) + e 6. (a) Refer Ans. 4 (b) (ii). While Mn 3+ has a greater tendency to reduce into Mn 2+ (b) 5C 2 O 4 2– + 2MnO 4 – + 16H + ® 2Mn 2+ + 8H 2 O + 10CO 2 . thus it acts an oxidising agent. 7. (i) Transition metals form complex compounds because Mn 3+ (aq) + e – ® Mn 2+ (aq) of comparatively smaller size of their metal ions, high (iii) Actinoids show a large number of oxidation states ionic charge and availability of vacant dorbitals. because of small energy gap between 5f, 6d and 7s (ii) Due to lanthanoid contraction. subshells. 8. (i) As we move along transition metal series from left to 4. (a) The most stable oxidation state for 3d 3 4s 2 right (i.e. Ti to Cu), the atomic radii decrease due to = +5, +4, +3, +2. increase in nuclear charge. Hence the atomic volume The most stable oxidation state for 3d 5 4s 2 = +7, +6, +5, decreases. At the same time, atomic mass increases. +4, +3, +2 Hence the density from titanium (Ti) to copper (Cu) The most stable oxidation state for 3d 6 4s 2 = +2, +3. increases. In a transition series the oxidation states which lead to (ii) Many transition metals and their compounds act as exactly half filled or completely filled dorbitals are catalysts. The catalytic activity is due to their ability more stable. to exhibit multiple oxidation states. For example, V 2 O 5 (b) (i) Preparation of potassium dichromate K 2 Cr 2 O 7 : in contact process and finely divided iron in Haber Potassium dichromate is prepared from chromite ore process. FeCr 2 O 4 in the following steps. The powdered chromite ore is heated with sodium 9. (i) Pig iron is heated with iron in a suitable converter to carbonate and quick lime in presence of air to get remove impurities by oxidation and slag formation. yellow coloured sodium chromate. Then steel of required quality is made by adding C and spiegel in suitable quantity. 7 2 FeCr2 O 4 + 4 Na 2 CO3 + O 2 (ii) Metallic zinc is obtained from zinc oxide by reduction 2 with coke. ® Fe 2 O3 + 4 Na 2 CrO 4 + 4 CO2 Heat The yellow solution of sodium chromate is filtered and ZnO + C ¾¾¾ ® Zn + CO


54

(iii) Impure titanium is heated with iodine to form volatile complex TiI 4 which on further heating to higher temperatures decomposes to give pure titanium. D Ti + 2 I 2 ® TiI 4 ¾¾ ® Ti + 2 I 2

Impure

(ii) As manganese has maximum number of unpaired electrons (5) in 3d subshell in addition to 2 electrons in the 4s subshell. It can use the 7 electrons for bonding purpose.

Pure

16. (i) Cr 2+ is reducing as its configuration changes from 10. (i) Role of NaCN in the extraction of gold is the leaching d 4 to d 3 , a more stable half filled t 2g configuration while of gold ore in the presence of air from which the gold Mn 3+ is oxidising as Mn 2+ result a more stable half is obtained later by replacement. filled d 5 conguration. – 4Au (s) + 8CN (aq) + 2H 2 O + O 2(g) (ii) Middle of the transition series contains greater number ® 4[Au(CN 2 )] + + 4OH – of unpaired electrons in (n –1)d and ns orbitals. (ii) SiO 2 is added in copper matte to convert the remaining 17. (i) 8MnO – 4(aq) + 3S 2 O 2– 3 (aq) + H 2 O (l) FeS, FeO to slag FeO + SiO 2 ® FeSiO 3 (slag) – ® 8MnO 2(s) + 6SO 42– (iii) Iodine is heated with Zirconium to form a volatile (aq) + 2OH (aq) (ii) Cr 2 O 2– + 14H + (aq) + 6Fe 2+ (aq) compound which on further heating decomposes to 7 (aq) give pure zirconium ® 2Cr 3+ (aq) + 6Fe 3+ (aq) + 7H 2O (l) Zr (impure) + 2I 2 ® ZrI 4 D OR ZrI4 ¾¾ ® Zr (pure) + 2 I2 (i) Refer Ans. 13 (i). (ii) This is because due to lanthanoid contraction the 11. (i) Due to Lanthanoid contraction. expected increase in size does not occur hence they (ii) Mn 3+ (3d 4 ) is much less stable than Mn 2+ (3d 5 ) which have very high value of ionisation enthalpy. has stable half filled electronic configuration. 3+ 5 2+ 6 Fe (3d ) is more stable than Fe (3d ) due to same 18. (i) Refer Ans. 8 (ii). reason. Cr 3+ (3d 3 ) has stable t 2g 3 configuration than (ii) The 4d and 5d orbitals of transition elements has larger Cr 2+ (3d 4 ). This is why E° Mn 3+ /Mn is much more positive than E° Fe 3+ /Fe 2+ or E° Cr 3+ /Cr 2+ .

size than that of 3d orbitals. Thus, these orbitals can form metalmetal bond more frequently by dd overlap.

(iii) Because oxygen and Fluorine have high 19. (i) Refer Ans. 3 (i). electronegativity and small size. (ii) The chemistry of actinoids is not as smooth as lanthanoid because they show greater number of oxidation states 12. (i) Refer Ans. 3(i). due to comparable energies of 5f, 6d and 7s orbitals. 10 (ii) Because Zn has completely filled dorbitals (3d ). 2– – + 13. (i) In aqueous solution Cu + undergoes disproportionation 20. (i) Cr 2 O 7 (aq) + 6I (aq) + 14H (aq) ® 2Cr 3+ (aq) + 7H 2O to form a more stable Cu 2+ ion. (l) + 3I 2(g) – + – (ii) MnO 4 + 8H + 5e ® Mn 2+ + 4H 2 O 2Cu + (aq) ® Cu 2+ (aq) + Cu (s) Cu 2+ in aqueous solution is more stable than Cu + ion 21. (i) Transition metals form a large number of interstitial because hydration enthalpy of Cu 2+ is higher than that compounds because small atoms of certain non metallic of Cu + . It compensates the second ionisation enthalpy elements (H, B, C, N, etc.) get trapped in voids or of Cu involved in the formation of Cu 2+ ions. vacant spaces of lattice of the transition metals. (ii) Actinoid exhibits greater range of oxidation states than As a result of filling up of the interstitial spaces such lanthanoids. This is because there is less energy interstitial compounds are hard and rigid difference between 5f and 6d orbitals in actinoids than (ii) Due to Lanthanoid contraction. the energy difference between 4f and 5d orbitals in case (iii) Actinoids show a greater range of oxidation states, of lanthanoids. which is in part attributed to the fact that the 5f, 6d and – + – 2+ 7s levels are of comparable energies. 14. (i) MnO 4 + 8H + 5e ® Mn + 4H 2O] × 2 2– – C 2O 22. (i) E° values for the Cr 3+ /Cr 2+ and Mn 3+ /Mn 2+ couples are 4 ® 2CO 2 + 2e ] × 5 – 2– + 2+ 2MnO 4 + 5C 2O 4 + 16H ® 2Mn + 10CO 2 + 8H 2 O Cr 3+ ( aq ) + e - ® Cr 2 + ( aq ) ; E °= - 0.41 V

heat

(ii) 2KMnO4 K2MnO4 + MnO2 + O 2 + – 3+ (iii) CrO 2– 7 + 14H + 6e ® 2Cr + 7H 2 O + – H 2S ® 2H + S + 2e ] × 3 2– + 3+ Cr 2O 7 + 3H 2S + 8H ® 2Cr + 3S + 7H 2 O

15. (i) Refer Ans. 3(i).

Mn 3+ ( aq ) + e - ® Mn 2 + ( aq ) ; E ° = + 1.551 V These E° values indicate that Cr 2+ is strongly reducing while Mn 3+ is strongly oxidising. (ii) Refer Ans. 3 (iii). (iii) Refer Ans. 3 (i). 23. (i) Because of large number of unpaired electrons in their


55

atoms they have stronger interatomic attraction and (ii) Removal of lonely electron from 3dorbital of Sc hence stronger metallic bonding between atoms resulting requires less energy. Hence it show valency 3 only. in higher enthalpies of atomisation. (b) (i) Refer Ans. 4 (b) (i). (ii) The metals of 4d and 5dseries have more frequent (ii) The following equilibrium exists in aq. K 2 Cr 2 O 7 metal bonding in their compounds than the 3dmetals solution. because 4d and 5dorbitals are more exposed in space Cr2 O 27 - + H 2 O Û 2 CrO 2 4 - + 2 H + than the 3d orbitals. When pH is less than 7 (acidic medium), it exists as (iii) Electronic configuration of Mn 2+ is 3d 5 which is half Cr 2 O 7 2– and has orange colour. filled and hence stable. Therefore, third ionization When pH is greater than 7 (basic medium), it exists as enthalpy is very high, i.e., 3 rd electron cannot be lost CrO 4 2– and has yellow colour. easily. In case of Fe 2+ , electronic configuration is 3d 6 . Hence it can lose one electron easily to give the stable 29. (a) On the basis of incompletely filled dorbitals : Scandium (Z = 21), atom has incompletely filled d configuration 3d 5 . orbitals (3d 1 ) in its ground states, so it is regarded as 24. (i) Lanthanum and all the lanthanoids predominantly show transition element. +3 oxidation state. However, some of the lanthanoids On the other hand zinc (Z = 30) atom has completely also show +2 and + 4 oxidation states in solution or filled dorbitals (3d 10 ) in its ground states as well as in solid compounds. This irregularity arises mainly due most common oxidation state of +2. to attainment of stable empty (f o ), halffilled (f 7 ) and (b) Ce(Z = 58) = [Xe] 4f 1 5d 1 s 2 14 full filled (4f ) sub shell. \ Ce 3+ = [Xe] 4f 1 5d 0 6s 0 e.g. Ce 4+ 4f 0 Eu 2+ 4f 7 Therefore it has only one unpaired electron, i.e., n = 1 Tb 4+ 4f 7 Yb 2+ 4f 14 m s = n ( n + 2) = 1(1 + 2) = 3 = 1.73 B.M . (ii) Due to high enthalpy of atomisation D a H° and low 2+ hydration enthalpy (D hyd H°) of Cu ion. (c) Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with (iii) Due to Lanthanoid contraction. increase in atomic number is called lanthanoid 25. (i) Refer Ans. 21(i). contraction. (ii) As we move along transition metal series from left to It is caused due to imperfect shielding of nuclear charge right (i.e. Ti to Cu), the atomic radii decrease due to by 4felectrons. increase in nuclear charge. Hence the atomic volume Lanthanum (La) has the largest ionic radius while decreases. At the same time, atomic mass increases. lutetium (Lu) has the smallest among the 4fseries Hence the density from titanium (Ti) to copper (Cu) elements. increases. 30. (a) (i) 5C 2 O 4 2– + 2MnO 4 – + 16H + (iii) Refer Ans. 3 (iii). ® 2Mn 2+ + 8H 2O + 10 CO 2 26. (i) Refer Ans. 22 (i). 2– 2+ + (ii) Cr 2 O 7 + 6Fe + 14H ® 2Cr 3+ + 6Fe 3+ + 7H 2 O. (ii) Refer Ans. 3 (iii). (b) (i) As we move along transition metal series from left (iii) As one proceeds along a transition series, the nuclear to right (i.e. Ti to Cu), the atomic radii decrease due charge increases which tends to decrease the size but to increase in nuclear charge. Hence the atomic volume the addition of electrons in the dsubshell increases the decreases. At the same time, atomic mass increases. screening effect which counter balances the effect of Hence the density from titanium (Ti) to copper (Cu) increased nuclear charge. increases. 27. (i) Refer Ans. 23(i). (ii) The metals of 4d and 5dseries have more frequent (ii) Refer Ans. 8(ii). metal bonding in their compounds than the 3dmetals (iii) The actinoid contraction is more than lanthanoid because 4d and 5dorbitals are more exposed in space contraction because shielding by 5felectrons in the than the 3d orbitals. actinoids is more than 4felectrons in the lanthanoids. (iii) Refer Ans. 3 (iii). (iv) Much larger third ionisation energy of Mn(where change 31. (a) (i) 8MnO 4 2– (aq) + 3S 2 O 3 2– (aq) + 8H 2 O (l) ® is d 5 to d 4 ) is mainly responsible for this. This also – 8MnO 2(aq) + 6 SO 42– (aq) + 2OH (aq) explains that +3 state of Mn is of little importance. 2– + (v) Scandium (Z = 21) has incompletely filled 3dorbitals (ii) Cr 2 O 7 (aq) + 3H 2 S (g) + 8H (aq) ® in the ground state (3d 1 ). Hence it is considered as a 2Cr 3+ (aq) + 7H 2O (l) + 3S (s) transition element. (b) (i) The actinoid contraction is similar to lanthanoid 28. (a) (i) Refer Ans. 3 (iii). contraction, this contraction is due to poor shielding by 5felectrons in the actinoids than that


56

by 4felectrons in the lanthanoids. Shielding power lutetium hydroxide Lu(OH) 3 is least basic. of 5forbitals is less than that of 4f orbitals. Hence (iii) Atomic and ionic sizes of 4d transition series elements effective nuclear charge of lanthanoids is greater and 5d series elements are similar. e.g. Atomic radii of than that of lanthanoids. Zirconium(Zr) is same as that of Hafnium Hf. (ii) The highest oxidation in the middle of the transition 35. (a) (i) Cr O 2– + 3H S + 8H + 2 7 (aq) 2 (g) (aq) series is due to the presence of more unpaired 3+ ® 2Cr (aq) + 7H 2 O (l) + 3S (s) electrons in the middle of series. 2+ – (ii) 2Cu + 4I ® Cu 2 I 2(s) + I 2(g) . (iii) Refer Ans. 22 (i). (aq) (aq) 32. (a) (i) 5Fe 2+ (aq) + MnO 4 – (aq) + 8H + (aq) ® 3+ Mn 2+ (aq) + 4H 2O (l) + 5Fe (aq) (ii) Cr 2 O 7 2– (aq) + 6I – (aq) + 14H + (aq)

® 2Cr 3+ (aq) + 7H 2O (l) + 3I 2(g) . (b) (i) Refer Ans. 21 (i). (ii) Refer Ans. 21 (i). (iii) Refer Ans. 23 (i).

(b) (i) Change in Cr 2 O 7 2– to Cr(III) is 3 and in MnO 4 to Mn (II) is 5. Change in oxidation state is large and the stability of reduced product V(III) < Cr(III) < Mn(II). This is why oxidising power of VO 2 + < Cr 2 O 7 2– < MnO 4 – . (ii) Third ionization enthalpy of Mn is very high because the third electron has to be removed from the stable halffilled 3dorbi tals [ Q Mn 2+ (Z = 25) = 3d 5 ]. (iii) Since E° (Cr 3+ /Cr 2+ ) is negative (– 0.41 V) whereas E° (Fe 3+ /Fe 2+ ) is positive (+ 0.77 V). Thus Cr 2+ is easily oxidized to Cr 3+ but Fe 2+ cannot be easily oxidized to Fe 3+ . Hence Cr 2+ is stronger reducing agent than Fe 2+ .

33. (a) Refer Ans. 32 (a). (b) (i) The atomic radii of transition elements decrease with the increase in atomic number as the effective nuclear charge increases because shielding effect of delectron is small. (ii) Due to high enthalpy of atomisation D a H° and low hydration enthalpy (D hyd H°) of Cu 2+ ion. (iii) Much larger ionisation energy of Mn (where the 36. (a) (i) 8MnO – + 3S O 2– + H O ® 4 (aq) 2 3 (aq) 2 (l) required change is d 5 to d 4 ) is mainly responsible – 8MnO 2(s) + 6SO 42– (aq) + 2OH (aq) for this. (ii) Cr 2 O 7 2– (aq) + 6Fe 2+ (aq) + 14H + (aq) ® 2Cr 3+ (aq) + 34. (a) (i) Cr 2+ has the configuration d 4 and easily changes 6Fe 3+ (aq) + 7H 2O (l) to d 3 which has t 23 g configuration and hence stable. (b) (i) Because they have empty 4f subshell. Therefore, Cr 2+ is reducing. On the other hand (ii) Mn 2+ has maximum number of unpaired electrons. Mn 2+ is more stable due to half filled d 5 (iii) Refer Ans. 13 (i). configuration and Mn 3+ easily changes to Mn 2+

hence it is oxidising. 37. (a) (i) Refer Ans. 30 (a) (i). (ii) Refer Ans. 23 (i). (ii) Refer Ans. 33 (a) (ii). (iii) In a crystall lattice transition elements having (b) (i) Refer Ans. 18 (ii). interstitial vacant spaces into which small sized (ii) Mn 3+ is less stable and changes to Mn 2+ which is nonmetal atom (H, C, N, B) can fit resulting in more stable due to halffilled dorbital the formation of interstitial compound. configuration. That is why, Mn 3+ undergoes (b) Lanthanoid contraction : The steady decrease in the disproportionation reaction. atomic and ionic radii of lanthanoid elements with (iii) Co (II) has electronic configuration 3d 7 , 4s 0 , i.e., increase in atomic number is called lanthanoid it has three unpaired electrons. In the presence of contraction. strong ligands, two unpaired electrons in 3d It is caused due to imperfect shielding of nuclear charge subshell pair up and third unpaired electron shift by 4felectrons. to higher energy subshell from where it can be Lanthanum (La) has the largest ionic radius while easily lost and hence oxidised to Co (III). lutetium (Lu) has the smallest among the 4fseries OR elements. (a) (i) Refer Ans. 13 (i). Consequences of lanthanoid contraction : (ii) 2CrO 4 2– + 2H + ® Cr 2 O 7 2– + H 2 O (i) The basic strength of oxides and hydroxides of (b) (i) Refer Ans. 13 (i). lanthanoids decrease with increasing atomic number. (ii) Refer Ans. 22 (i). (ii) Lanthanum hydroxide La(OH) 3 is most basic and (iii) Refer Ans. 8 (ii).

JJJ

Coordination Compounds

57

Chapter9 Coordination Compounds

VERY SHORT ANSW ER TYPE QUESTIONS (1 MARK)

1.

Give an example of linkage isomerism.

[Delhi 2010]

2.

Give an example of coordination isomerism. [Delhi 2010]

3.

Give an example of ionization isomerism.

4.

What do you understand by ‘denticity of a ligand’? [Foreign SetI 2011]

5.

Why is CO a stronger ligand than Cl – ? [Foreign SetII 2011]

6.

Name the following coordination compound : K 3 [CrF 6 ] [Foreign SetIII 2011]

[Delhi 2010]


7. (a) What is a ligand ? Give an example of bidentate ligand. (b) Explain as to how the two complexes of nickel [Ni(CN) 4 ] 2– and [Ni(CO) 4 ] have different structures but don’t differ in their magnetic behaviour (Ni = 28). [Delhi 2008]

(b) What is meant by chelate effect?

[AI 2009]

13. Giving a suitable example for each, explain the following: (i) Crystal field splitting (ii) Linkage isomerism (iii) Ambidentate ligand. [AI 2009] 14. Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units : [Co(NH 3 ) 6 ] 3+ , [Cr(NH 3 ) 6 ] 3+ , Ni(CO) 4 [At. no. Co = 27, Cr = 24, Ni = 28) [AI 2009] 15. Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units : (i) [Ni(CN) 4 ] 2– (ii) [NiCl 4 ] 2– 3– (iii) [CoF 6 ] [At. no. Ni = 28, Co = 27] [AI 2009]

16. For the complex [Fe(en) 2 Cl 2 ]Cl, (en = ethylene diamine), identify : 8. (a) What is the basis of formation of the spectrochemical (i) The oxidation number of iron series? (ii) The hybrid orbitals and the shape of the complex (b) Draw the structures of geometrical isomers of the (iii) The magnetic behaviour of the complex following coordination complexes : + (iv) The number of geometrical isomers [Co(NH 3 ) 3 Cl 3 ] and [CoCl 2 (en) 2 ] (v) Whether there is an optical isomer also, and (en = ethylenediamine and atomic number of Co is 27) (vi) Name of the complex (Atomic number of Fe = 26). [AI 2008] [Delhi 2009, Delhi SetII 2011] 9. Using the IUPAC norms, write the systematic name of 17. Compare the following complexes with respect to their shape, following : magnetic behaviour and the hybrid orbitals involved: (i) [CoCl(NO 2 )(NH 3 ) 4 ]Cl (i) [CoF 4 ] 2– (ii) [Cr(H 2 O) 2 (C 2 O 4 ) 2 ] – (ii) [Co(NH 3 ) 6 ][Cr(CN) 6 ]. [Delhi 2008 C] (iii) [Ni(CO) 4 ] [At. no. : CO = 27, Cr = 24, Ni = 28] 10. (a) Draw the structure and write the hybridisation state of [Delhi 2009] Co in cis[Co(NH 3 ) 4 Cl 2 ] + . (b) Using IUPAC nomenclature name the following 18. Name the following coordination compounds according to IUPAC system of nomenclature: complex. [AI 2008] (i) [Co(NH 3 ) 4 (H 2 O)Cl]Cl 2 11. (a) Square planar complexes (of MX 2 L 2 type) with co (ii) [CrCl 2 (en) 2 ]Cl, (en = ethane1,2diamine)] [AI 2010] ordination number of 4 exhibit geometrical isomerism whereas tetrahedral complexes with similar composition 19. Describe the shape and magnetic behaviour of following complexes : do not. Why? (i) [Co(NH 3 ) 6 ] 3+ (ii) [Ni(CN) 4 ] 2– (b) Describe the type of hybridisation, shape and magnetic [Atomic number Co = 27, Ni = 28] [Delhi 2010] properties of [Co(NH 3 ) 4 Cl 2 ]Cl. [Given : At. no. of Co = 27] [AI 2009] 20. Explain the following cases giving appropriate reasons: 12. (a) Write the IUPAC name and indicate the shape of complex ion [Co(en) 2 Cl(ONO)] + . [At. no. of Co = 27]

(i) Nickel does not form low spin octahedral complexes. (ii) The pcomplexes are known for the transition metals only.


58

(iii) Co 2+ is easily oxidised to Co 3+ in the presence of a 28. Give the name, the stereochemistry and the magnetic strong ligand. [AI 2010] behaviour of the following complexes : (i) [Co(NH 3 ) 5 Cl]Cl 2 21. Write the name, the state of hybridisation, the shape and the (ii) K 2 [Ni(CN) 4 ] [Foreign SetIII 2011] magnetic behaviour of the following complexes: [CoCl 4 ] 2– , [Ni(CN 4 )] 2– , [Cr(H 2 O) 2 (C 2 O 4 ) 2 ] – (At. no. Co = 27, Ni = 28, Cr = 24)

29. Give the formula of each of the following coordination entities: (i) Co 3+ ion is bound to one Cl – , one NH 3 molecule and 22. Write the name, stereochemistry and magnetic behaviour of two bidentate ethylene diamine (en) molecules. the following : (At. nos. Mn = 25, Co = 27, Ni = 28) (ii) Ni 2+ ion is bound to two water molecules and two (i) K 4 [Mn(CN) 6 ] oxalate ions. (ii) [Co(NH 3 ) 5 Cl]Cl 2 Write the name and magnetic behaviour of each of the (iii) K 2 [Ni(CN) 4 ] [Delhi SetI 2011] above coordination entities. (At. nos. Co = 27, Ni = 28) [Delhi SetI 2011] 23. Explain the following terms giving a suitable example in each case: 30. State a reason for each of the following situations. (i) Ambident ligand (ii) Denticity of a ligand (i) Co 2+ is easily oxidised to Co 3+ in presence of a strong (iii) Crystal field splitting in an octahedral field. ligand. [AI SetI 2011] (ii) CO is a stronger complexing reagent than NH 3 . (iii) The molecular shape of Ni(CO) 4 is not the same as that 24. Write the structures and names of all the stereoisomers of the of [Ni(CN) 4 ] 2– . [Delhi SetII 2011] following compounds: [AI 2010]

(i) [Co(en) 3 ]Cl 3 (ii) [Pt(NH 3 ) 2 Cl 2 ] 31. Write the name, the structure and the magnetic behaviour of (iii) [Fe(NH 3 ) 4 Cl 2 ]Cl [AI SetII 2011] each one of the following complexes: (i) [Pt(NH 3 )Cl(NO 2 )] 25. Write the state of hybridization, the shape and the magnetic (ii) [Co(NH 3 ) 4 Cl 2 ]Cl behaviour of the following complex entities : (iii) Ni(CO) 4 (i) [Cr(NH 3 ) 4 Cl 2 ]Cl (ii) [Co(en) 3 ]Cl 3 (At. nos. Co = 27, Ni = 28, Pt = 78) [Delhi SetIII 2 011] (iii) K [Ni(CN) ] [AI SetIII 2011] 2

4

26. Name of the following coordination compounds and draw 32. Name the following coordination entities and draw the structures of their stereoisomers: their structures. (i) [Co(en) 2 Cl 2 ] + (en = ethane1,2diamine) (i) [CoCl 2(en) 2 ]Cl (ii) [Cr(C 2 O 4 ) 3 ] 3– (ii) [Pt(NH 3) 2 ]Cl(NO 2)] (iii) [Co(NH 3 ) 3 Cl 3 ] (At. No. Co = 27, Pt = 78) [Foreign SetI 2011] (Atomic numbers Cr = 24, Co = 27) [AI SetI 2012] 27. Draw the structure of isomers, if any and write the name of 33. Name the following coordination entities and describe their the following complexes: structures. (i) [Cr(NH 3 ) 4 Cl 2 ] + (ii) [Co(en) 3 ] 3+ 4– + (i) [Fe(CN) 6] (ii) [Cr(NH 3) 4Cl 2] [Foreign SetII 2011] 2– (iii) [Ni(CN) 4] [AI SetIII 2012]

a n s w e r s 1. [Co(NH 3 ) 5 (NO 2 )]Cl 2 and [Co(NH 3 ) 5 (ONO)]Cl 2 .

6. Potassium hexafluoridochromate (III).

2. [Cr(NH 3 ) 6 ][CoF 6 ] and [Co(NH 3 ) 6 ][CrF 6 ].

7. (a) Ligands : The ions or molecules bound to the central atom or ion in the coordination entity are called ligands. Example of a bidenate ligand : H2 N CH2 CH2 NH 2 4. The number of coordinating groups present in ligand is (ethylenediamine) called the denticity of ligand. For example, bidentate ligand 2– 2+ (b) [Ni(CN) ] contains Ni ion 3d 8 system dsp 2 ethane1, 2diamine has two donor nitrogen atoms which 4 hybridisation and square planar geometry. can link to central metal atom. Ni 2+ [Ar] 3d 8 4s 0 4p 0 H2 N CH2 CH2 N H 2 Ni 2+ in the presence of strong (CN – ) ligand Ethane1, 2diamine 3. [Co(NH 3 ) 5 Cl]SO 4 and [Co(NH 3 ) 5 SO 4 ]Cl.

5. Because CO has vacant molecular orbitals with which it can form pbond with metal through back donation

2+

Ni – 8

3d

2

dsp


59

[Ni(CO) 4 ] contains Ni 0 3d 8 4s 2 system. sp 3 hybridisation and tetrahedral geometry, Ni(0) – 3d 8 4s 2

L

X

L

L

X

8

4s

(b) Hybridisation Shape Magnetic properties

4p

In the presence of strong ligand (CO) two electrons of 4s orbital shift to 3d orbitals. 10

4s 4p 3d It has no unpaired electrons, hence it is diamagnetic. Thus [Ni(CN) 4 ] 2– and [Ni(CO) 4 ] have different structures but same magnetic behaviour. 8. (a) Spectrochemical series : The arrangement of ligands in order of their increasing field strength i.e. increasing crystal field splitting energy (CFSE) value is called spectrochemical series. (b) (i) The complex in facial and meriodional isomers of [Co(NH 3 ) 3 Cl 3 ] may be represented as. Cl

NH 3

NH 3

NH 3 NH 3

Co NH 3

Octahedral field

(ii) Linkage isomerism : Linkage isomerism arises in a coordination compound containing ambidentate ligand. Same ligand is linked with metal through different donor sites. e.g. [Co(NO 2 ) 6 ] 4– , [Co(ONO) 6 ] 4– . (iii) Ambidentate ligand : Unidentate ligands containing more than one coordinating atoms are called ambidentate ligands. e.g., NO 2 – , CN – , SCN – .

Cl

(ii) [CoCl 2 (en) 2 ] + ion exists in two geometrical isomers as shown below.

en

Cl

en en

Co

Cl

Co

en

Cl trans form

cis form

9. (i) Tetramminechloridonitrocobalt (III) chloride. (ii) Hexamminecobalt (III) hexacyanochromate (III). 10. (a) Structure of cis[Co(NH 3 ) 4 Cl 2 ] + Cl + Cl NH 3 Co NH 3

NH 3 NH 3 2

3

Hybridisation (d sp )

d 2 sp 3 Octahedral Diamagnetic

Tetrahedral field

NH 3 meridional

Cl

: : :

Cl

Cl

Cl facial isomer

cis

12. (a) IUPAC name : [Chloridobis(ethane1,2diamine)nitrito Ocobalt(III)ion]. Shape : Octahedral (b) Chelate effect : Chelate ligands form more stable complexes than similar complexes with monodentate ligand. This extrastability is called chelate effect. 13. (i) Crystal field splitting : In a free transition metal ion, the five orbitals are degenerate, but when it is involved in a complex formation, the degeneracy splits. This is called crystal field splitting.

Co

Cl

X

L

trans

3d

X M

M

14.

Complexes

Magnetic Hybridisation behaviour

[Co(NH 3 ) 6 ] 3+ diamagnetic [Cr(NH 3 ) 6 ] 3+ paramagnetic [Ni(CO) 4 ] diamagnetic 15. Complexes

Magnetic Hybridisation behaviour

[Ni(CN) 4 ] 2– Diamagnetic [NiCl 4 ] 2– [CoF 6 ] 3

d 2 sp 3 d 2 sp 3 sp 3

Paramagnetic Paramagnetic

dsp 2 sp 3 sp 3 d 2

Shape octahedral octahedral tetrahedral Shape Square planar Tetrahedral Octahedral

16. (i) Oxidation number : + 3 (ii) d 2 sp 3 (octahedral) (iii) Paramagnetic (iv) Two/cis and trans isomers (v) Yes (vi) Dichloridobis(ethane1,2 diamine) iron(III)chloride. 17. (i) Tetrahedral, paramagnetic, sp 3 (ii) Octahedral, paramagnetic d 2 sp 3 (iii) Tetrahedral, diamagnetic, sp 3 .

(b) TetraamminechloridonitritoOcobalt(III)chloride. 11. (a) Isomerism in complexes with coordination number 4: The tetrahedral complexes do not show geometrical isomerism, because the relative position of the ligand with respect to each other will be same. The square 18. (i) Tetraammineaquachloridocobalt(III) chloride. (ii) Di ch l or i dobi s(et h a n e1, 2 di a m i n e)ch r om i um planar complexes show geometrical isomerism due to (III)chloride different relative arrangement of ligands.


60

19. (i) [Co(NH 3 ) 6 ] 3+ contains Co 3+ – 3d 6

e g

Co(III) 6 4s 4p 3d In presence of NH 3 ligand pairing of electrons in 3d orbital takes place.

Co(III) with 6 NH 3 2

3

d sp 2

3

Hybridisation d sp Structure Octahedral Diamagnetic No unpaired electron (ii) [Ni(CN) 4 ] 2– contains Ni 2+ – 3d 8

t 2g Hence it does not form low spin octahedral complexes. (ii) The transition metals/ions have empty d orbitals into which the electron pair can be donated by ligands containing p electrons. For example : CH 2 = CH 2 and C 6 H 6 , C 5 H 5 – . (iii) In presence of strong ligand Co(II) has electronic configuration, t 2g 6 e g 1. e g 1

Ni(II) 8 4 s 4 p 3 d In presence of strong CN – ligand pairing of electron takes place in 3d orbital. Ni(II) with 4CN – ligand

D o > P t 2g It can easily lose only electron of e g orbital to give stable t 2g 6 configuration. This is why Co 2+ is easily oxidised to Co 3+ in the presence of strong ligand.

21. Names : 2

[CoCl 4 ] 2– Tetrachloridocobaltate(II)ion

dsp

Hybridisation dsp 2 Structure square planar No unpaired electron hence diamagnetic. 20. (i) Nickel forms octahedral complexes mainly in +2 oxidation state which has d 8 configuration. In presence of strong ligand also it has two unpaired electrons in e g orbital

[Ni(CN) 4 ] 2– Tetracyanonickelate(II)ion [Cr(H 2 O) 2 (C 2 O 4 ) 2 ] – Diaquadioxalatochromate(III)ion Complexes

Hybrid isation

Shape

Magnetic behaviour

[CoCl 4 ] 2–

sp 3

[Ni(CN 4 )] 2–

dsp 2 Square planar Diamagnetic

[Cr(H 2 O) 2 (C 2 O 4 ) 2 ] – d 2 sp 3

Tetrahedral Paramagnetic

Octahedral Paramagnetic

22. Formulae K 4 [Mn(CN) 6 ] [Co(NH 3 ) 5 Cl]Cl 2

K 4 [Ni(CN) 4 ]

Name of the Complex Potassium hexacyano manganate (II) Pentaammine chloridocobalt (III) Chloride Potassium tetra cyanonikelate (II)

Hybridisation of metal Geometry of complex Magnetic behaviour ion involved (stereochemistry) Paramagnetic d 2 sp 3 Octahedral d 2 sp 3

Octahedral

Diamagnetic

dsp 2

Square planar

Diamagnetic

23. (i) Refer Ans. 13(iii). 24. (i) [Co(en) 3 ]Cl 3 en (ii) Refer Ans. 4 (iii) The splitting of the degenerate dorbitals into three orbitals of lower energy, t 2g set and two orbitals of en Co higher energy e g , set due to the interaction of ligand in an octahedral crystal field is known as crystal field en dTris(ethane 1,2diamine) splitting in an octahedral field. cobalt (III) e g Energy t 2g Degenerate

en Co

en

en lTris(ethane 1,2diamine) cobalt (III)


NH 3

H3 N

NH 3

Cl

Cl

H3 N

+

H3 N

NH 3 Fe

Fe

(iii)

Transdiamminedichlorido platinum (II)

Cis diamminedichlorido platinum (II)

Cl

+

H3 N

Pt

Pt Cl

NH 3

Cl

NH 3

Cl

(ii)

61

Cl

H3 N

Cl NH 3

NH 3 Cis tetraamminedichlorido iron (III)

Transtetraamminedichlorido iron (III)

25. Central metal ion/atom Hybridisation of metal Geometry of complex Magnetic behaviour ion involved Octahedral 3+ Paramagnetic d 2 sp 3 [Cr(NH 3 ) 4 Cl 2 ]Cl Cr Complex

[Co(en) 3 ]Cl 3

Co 3+

d 2 sp 3

Octahedral

Diamagnetic

K 2 [Ni(CN) 4 ]

Ni 2+

dsp 2

Square planar

Diamagnetic

26. (i) Dichloridobis(ethane1, 2diamine) cobalt (III) chloride. + + Cl Cl Cl en

Co

en

en

Cl

Octahedral + cis[CoCl2(en )2 ]

Octahedral + trans[CoCl2( en )2 ]

(ii) DiamminechloridonitritoN platinum (II) NH 3

en

Co

en

Co

en

(ii) Optical isomers of [Co(en) 3 ] 3+ – Trisethane1,2 diammine cobalt (III) ion. en en

dextro Tris(ethane 1,2diamine cobalt (III) ion

Co

en

en laevoTris(ethane 1,2diamine) cobalt (III) ion

28. (i) [Co(NH 3 ) 5 Cl]Cl 2 : Pentaamminechloridocobalt (III) chloride. Co(III) : [Ar] 3d 6 4s 0 – d 2 sp 3 hybridisation leads to octahedral shape. Magnetic behaviour: Diamagnetic. (ii) K 2 [Ni(CN) 4 ] : Potassium tetracyanonickelate (II) Ni(II): [Ar] 3d 8 4s 0 – dsp 2 hybridisation leads to square planar shape. Magnetic behaviour: Diamagnetic.

29. (i) [Co(en) 2 Cl(NH 3 )] 2+ Amminechloridobis(ethane1,2diamine) cobalt(III) ion NH 3 O2 N In presence of NH 3 and en Co 3+ (3d 6 ) form low spin complex. Hence complex is diamagnetic. Cl (ii) [Ni(ox) 2 (H 2 O) 2 ] 2– : Diaquadioxalato nickel(II) ion Tetrahedral In the presence of weak ox and H 2 O ligand Ni(II) form 27. (i) Geometrical isomers of [Cr(NH 3 ) 4 Cl 2 ] + – Tetra high spin complex. (sp 3 d hybridisation. It is amminedichlorido chromium (III). paramagnetic). + + Cl Cl 30. (i) Refer Ans. 20(iii). H3 N H3 N Cl (ii) In CO both lone pair of electrons and vacant p* orbitals NH 3 are present. Hence it acts as electron pair donor as well Cr Cr as acceptor by back bonding. Hence M – CO bond is H3 N H3 N stronger. NH 3 NH 3 :NH 3 is electron pair donor only. Accumulation of Cl NH 3 negative charge on the metal ion takes place, hence cisTetraamminedichlorido transTetraamminedichlorido chromium (III) ion chromium (III) ion M – NH 3 bond is weaker. Pt


62

(iii) [Ni(CN) 4 ] 2– contains Ni 2+ ion 3d 8 system dsp 2 hybridisation and square planar geometry. Ni 2+ [Ar] 3d 8 4s 0 4p 0 Ni 2+ in the presence of strong (CN – ) ligand

(ii) [Cr(C 2 O 4 ) 3 ] 3– Trioxalatochromium(III) ion It will show optical isomerism. ox

ox ox

Cr

Cr

ox

2+

Ni – 2

dsp

Two enantiomeric forms

(iii) [Co(NH 3 ) 3 Cl 3 ] Triamminetrichloridocobalt(III) It will exist in two geometrical isomeric forms.

[Ni(CO) 4 ] contains Ni 0 3d 8 4s 2 system. sp 3 hybridisation and tetrahedral geometry, Ni(0) – 3d 8 4s 2 8

3d

4s

NH 3 NH 3 Co Cl Cl Cl

H3 N

4p

In the presence of strong ligand (CO) two electrons of 4s orbital shift to 3d orbitals.

10

4s 4p 3d It has no unpaired electrons, hence it is diamagnetic. Thus [Ni(CN) 4 ] 2– and [Ni(CO) 4 ] have different structures but same magnetic behaviour.

en

Co en Cl Trans

en

Cl

Cl

en

Cl Cl

NH 3

Cl

Co Cl

Cl NH 3

Meriodional (mer)

4–

33. (i) [Fe(CN) 6 ] Hexacyanidoferrate(II) ion Hybridisation d 2 sp 3 Structure : Inner orbital octahedral complex

en

4–

CN CN

CN Fe

CN

CN CN

Octahedral

(ii) [Cr(NH 3 ) 4 Cl 2 ] + Tetraamminedichlorido chromium(III) ion Hybridisation sp 3 d 2 Structure : Outer orbital octahedral complex +

H3 N H3 N

NH 3 Cr

NH 3 Cl

Cl

(iii) [Ni(CN) 4 ] 2– Tetracyanido nickellate(II) ion Hybridisation dsp 2 Structure Square planar

Co

Co

H3 N

Facial (fac)

31. (i) [Pt(NH 3 )Cl(NO 2 )] : Ammine chloridonitroplatinum(II) Question is wrong. (ii) [Co(NH 3 ) 4 Cl 2 ]Cl : Tetramminedichloridocobalt(III) chloride Octahedral, Paramagnetic (iii) Ni(CO) 4 : Tetracarbonylnickel(0) Tetrahedral, Diamagnetic 32. (i) [Co(en) 2 Cl 2 ] + Dich loridobis(et hane1,2 diamine)cobalt(III) ion It will exist in two geometrical isomeric forms. Cl

ox

ox

8

3d

en

2–

cisform exists in two enantiomeric forms

CN

CN Ni

CN

JJJ

CN

Haloalkanes and Haloarenes

63

Chapter10 Haloalkanes and Haloarenes


1.

Draw the structure of the compound whose IUPAC name is 4chloropentan2one. [AI 2008]

11. Write the IUPAC name of the following: H H

H

H

[AI SetII 2011, AI SetIII 2012]


Br

(b) Complete the following chemical equation.

CH3CH2 CH CH2 + HBr

H H Br

2. (a) State the IUPAC name of the following compound. CH 3

CH 3

Peroxide

...........

12. Which compound in each of the following pairs will react faster in S N 2 reaction with –OH? Why? (i) CH 3 Br or CH 3 I (ii) (CH 3 ) 3 CCl or CH 3 Cl. [AI 2008]

[Delhi 2008]

13. Explain as to why haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions? 3. A solution of KOH hydrolyses CH 3 CH(Cl)CH 2 CH 3 and [Delhi 2008] CH 3 CH 2 CH 2 CH 2 Cl. Which one of these is more easily hydrolysed? [Delhi 2010] 14. (i) Why is it that haloalkanes are more reactive than haloarenes towards nucleophiles? 4. In the following pairs of organic compounds identify the (ii) Which one of the following reacts faster in S N 1 reaction? compound which will undergo S N 1 reaction faster? Why? Cl Cl CH2 Cl Cl or [Delhi 2008, AI 2008] (a) (b) Cl Cl 15. Account for the following : [AI 2010] Cl Cl reacts faster than (i) 5. Write the IUPAC name of the following compound : (ii) The treatment of an alkyl chloride with aqueous KOH (CH 3 ) 3 CCH 2 Br [Dehi SetI 2011] leads to the formation of an alcohol whereas in the 6. Write the IUPAC name of the following compound : presence of alcoholic KOH, alkene is the major product. [AI SetI 2011] CH2 CHCH2 Br . [AI 2008] 7. Which will react faster in S N 2 displacement, 16. (a) Suggest a mechanism for the reaction 1bromopentane or 2bromopentane and why? C2H5 OH, H2 O nBuBr + KCN nBuCN [Foreign SetI 2011] (b) Describe carbylamine reaction. [AI 2008, 2009] 8. Which will react faster in S N 1 displacement reaction : 17. Which one in the following pairs of substances undergoes 1Bromobutane or 2bromobutane and why? S N 2 substitution reaction faster and why? [Foreign SetIII 2011] 9.

Give the IUPAC name of the following compound.

(i)

CH 2 Cl

(ii)

I

CH2 C – CH 2 Br CH 3

10. What happens when bromine attacks CH 2 CH– CH 2 – C CH?

[Delhi SetI 2012]

[AI SetI 2012]

or

or

Cl

Cl

[AI 2009]

18. Which one of the following pairs undergoes S N 1substitution reaction faster and why?


64

Cl

(i)

C

Cl or

Cl

(ii)

or

Cl

[AI 2009]

19. Complete the following reaction equations:

CH 3 (i)

+ HI

(b) Allyl chloride is more reactive than npropyl chloride towards nucleophilic substitution reaction. Explain why? (c) Haloalkanes react with KCN to give alkyl cyanide as main product while with AgCN they form isocyanide as main product. Give reason. [AI 2010] 27. (a) Which will have higher boiling point? 1chloroethane or 2methyl2chlorobutane. Give reasons. (b) pnitro chlorobenzene undergoes nucleophilic substitution faster than chlorobenene. Explain giving the resonating structure as well. [AI 2010]

[AI 2009] 28. Answer the following: (i) Haloalkanes easily dissolve in organic solvents, why? 20. Suggest a possible reason for the following observation: (ii) What is known as a racemic mixture? Give an example. (i) The order of reactivity of haloalkanes is RI > RBr > RCl. (iii) Of the two bromoderivatives, C 6 H 5 CH(CH 3 )Br and (ii) Neopentyl chloride (CH 3 ) 3 CCH 2 Cl does not follow S N 2 C 6 H 5 CH(C 6 H 5 )Br, which one is more reactive in S N l mechanism. [AI 2009] substitution reaction and why? [Delhi SetI 2011] 21. Give reason for the following observations : 29. Rearrange the compounds of each of the following sets in (i) pdichlorobenzene has higher melting point than those order of reactivity towards S N 2 displacement: of oand misomers. (i) 2Bromo2methyl butane, 1Bromopentane, (ii) Haloarenes are less reactive than haloalkanes towards 2Bromopentane nucleophilic substitution reaction. (ii) 1Bromo3methyl butane, 2Bromo2methyl butane, (iii) The treatment of alkyl chloride with aqueous KOH 3bromo2methyl butane leads to the formation of alcohol but in the presence (iii) 1Bromobutane, 1Bromo2, 2dimethyl propane, of alcoholic KOH, alkenes is the major product. 1Bromo2methyl butane [AI SetI 2011] [AI 2009] 30. Compute the following reaction equations: 22. (i) State one use each of DDT and iodoform. (ii) CH3CH2 CH CH2 + HBr

(ii) Which compound in the following couples will react faster in S N 2 displacement and why? (a) 1Bromopentane or 2bromopentane. (b) 1Bromo2methylbutane or 2bromo2methylbutance. [AI 2009]

(i)

(ii)

CH 3 + HI

H

+ HBr H 23. How are the following conversions carried out? H (i) Benzyl chloride to benzyl alcohol. (iii) CH 3 CH 2 CH CH 2 + HBr ® [Foreign SetI 2011 1] (ii) Methyl magnesium bromide to 2methylpropan 2ol. [AI 2010] 31. Although chlorine is an electron withdrawing group, yet it is orthoparadirecting in electrophilic aromatic substitution 24. How would you differentiate between S N 1 and S N 2 mechanisms reactions. Explain why it is so? [Delhi SetI 2012] of substitution reaction ? Give one example of each. [AI 2010] 32. Answer the following questions: (i) What is meant by chirality of a compound? Give an 25. An optically active compound having molecular formula example. C 7 H 15 Br reacts with aqueous KOH to give a racemic mixture (ii) Which one of the following compounds is more easily of products. Write the mechanism involved for this reaction. hydrolysed by KOH and why? [AI 2010] CH 3 CHClCH 2 CH 3 or CH 3 CH 2 CH 2 Cl (iii) Which one undergoes S 2 substitution reaction faster 26. (a) Which of the following two compounds would react N 2 and why? faster by S pathway : 1bromobutane or N

2bromobutane and why?

Cl

I or

[AI SetI 2012]


65

answers 1. The structure of 4chloropentan2one is Cl O CH3 CH CH2 C CH 3

CH 3

H

11. H

3bromo2methylpropene

H

H

Br

2. (a) 2Bromo2methyl pent3ene. –

(b) CH3 CH2CH CH2 + HBr

Peroxide Anti Markownikoff's addition

CH 3CH 2CH 2 CH 2 Br lBromobutane

3.

CH3CH2 CHCH 3 hydrolyses easily with KOH because it is Cl secondary halide.

4. (a) In case of alkyl halides, 3° alkyl halides undergo S N 1 reaction very fast. Cl

will undergo S N 1 reaction faster. .

12. (i) Since I ion is a better leaving group than Br – ion therefore CH 3 I reacts faster than CH 3 Br in S N 2 reaction with OH – ion. (ii) On steric grounds 1° alkyl halides are more reactive than tertalkyl halides in S N 2 reaction. Therefore CH 3 I will react at a faster rate than (CH 3 ) 3 CCl in a S N 2 reaction with OH – ion. 13. Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions due to the following reasons. (i) Resonance effect : In haloarenes the electron pairs on halogen atom are in conjugation with pelectrons of the ring and the following resonating structures are possible. Cl

+ Cl

+ Cl –

–

(b) Benzylic halide shows high reactivity towards the S N 1 reaction because carbocation is more stable. CH2 Cl

will undergo S N 1 reaction faster. .

CH 3

5. CH3 C CH2 Br CH 3

+ Cl

–

C – Cl bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarene is difficult than haloalkane and therefore, are less reactive towards nucleophilic substitution reaction. (ii) Difference in hybridisation of carbon atom in C – X bond. 14. (i) Refer Ans. 13. (ii) Due to greater stability of 2° carbocation over 1° Cl

carbocation,

will react faster than

1Bromo2,2dimethylpropane

6. CH2 CH CH 2 Br 3Bromoproplene

Cl

in S N 1 reaction.

7. 1–Bromopentane is a primary alkyl halide, hence reacts faster in S N 2 displacement than secondary halide 2bromopentane.

Cl 15. (i) Tertiary halide reacts faster than the secondary

8. 2Bromobutane will react faster in S N l displacement reaction because it will form more stable secondary carbocation intermediate.

halide because of the greater stability of tert carbocation. (ii) In aqueous solution, KOH is almost completely involved to give OH ion which being a better nucleophile gives a substitution reaction on alkyl halides to form alcohol. But an alcoholic solution of KOH containing alkoxide (RO – ) ions which being a much stronger base than OH – ion preferentially snatches a H + ion from an alkyl chloride to form alkenes. 16. (a) Normal butyl bromide will give S N 2 reaction.

9.

CH2 C – CH 2 Br CH 3

3Bromo2methylpropene 10. CH2 – CH – CH2 – C CH Br

Br


66

+

d +

–

CH2 Cl

d –

CH2 OH

K CN + CH3 CH2CH2CH2 Br nbutyl bromide

CH3CH2CH2 CH2 CN + KBr

+ aq KOH

23. (i)

+ KCl

nbytyl cyanide

Benzyl chloride

D

Benzyl alcohol

(b) CH3NH2 + CHCl3 + 3KOH CH3NC + 3 KCl + 3 H 2 O

O CH3Mg Br + CH3 C CH 3

(ii) CH2 Cl is primary halide and therefore

17. (i)

Methyl magnesium bromide

OMgBr

undergoes S N 2 reaction faster than the secondary halide

CH3 C CH 3

Cl .

OH H 2 O

CH3 C CH 3

CH 3 CH 3 2methyl propan2ol (ii) As iodide is a better leaving group because of its large I undergoes S 2 reaction faster 24. In S 1 mechanism of substitution reaction, the rate of size, therefore, N N reaction depends upon the concentration of only reactant. Cl . than It is two steps process in which carbocation is formed.

18. (i)

Cl Cl reacts faster than because of greater stability of secondary carbocation than primary.

CH 3

19. (i)

+ HI

CH 3 I

(CH3 )3 COH + Br

2Bromo2methyl propane

C : Tertiary halide reacts faster than secondary halide because of the greater stability of tert carbocation.

(ii)

–

e.g. (CH3)3 C Br + OH –

Cl

2methyl propan2ol

In S N 2 mechanism of substitution, the rate of reaction depends upon the concentration of both the reactants and reagents. It is a single step process. H H

OH + H H

Cl

H

HO H

+ Cl

H H

H

25. Clearly C 7 H 15 Br is a tertiary halide C2H 5

(ii) CH CH CH CH + HBr 3 2 2

OH

Cl

CH CH CH CH 3 2 3 Br

20. (i) Among the various halides with same alkyl group the order of reactivity is RI > RBr > RCl. Due to increasing bond strength of C – I, C – Br and C – Cl the reactivity decreases. (ii) Neopentyl chloride being a primary halide reacts slowly through S N 1 and the carbon carrying halogen is sterically more hindered. Hence it does not follow S N 2 mechanism. 21. (i) The melting point of pdichlorobenzene is higher than that of oand m isomers. This is because, pisomer has symmetrical structure and therefore, its molecules can easily pack closely in crystal lattice. Hence it has stronger intermolecular force of attraction than oisomers. (ii) Refer Ans. 13. (iii) Refer Ans. 15 (ii). 22. (i) DDT is used as an insecticide and iodoform is used as a mild antiseptic. (ii) (a) 1Bromopentane, as it is a primary alkyl halide. (b) 1Bromomethylbutane, as it is a primary alkyl halide.

C2H 5

C3H 7 C Br + KOH

C3H 7 C OH + KBr

CH 3

CH 3

Mechanism : C2H 5 CH 3

C

C2H 5 –

+

Br

C

C3H 7

+ Br C3H 7

CH 3

Carbocation

C2H 5 HO C C3H 7 One enantiomer

HO Rear attack

CH 3

C2H 5

C2H 5 –

–

+

C

HO Frontal attack

CH 3 C3H 7 Carbocation (planar) Racemic mixture

C OH CH 3 C H 3 7

Other enantiomer


67

than C 6H 26. (a) 1bromobutane is 1° alkyl halide while 2bromobutane 5 (CH 3 )Br for S N 1 reaction because its carbocation is resonance stabilised by two phenyl groups. is 2° alkyl halide. Due to steric hindrance in 2° alkyl halides, 1° alkyl halide will react faster than 2° alkyl 29. (i) 1Bromopentane > 2Bromopentane > 2Bromo2 halide in S N 2 reaction. methyl butane. (b) Allyl chloride is more reactive than npropyl chloride (ii) 1Bromo3methyl butane > 3Bromo2methyl butane towards nucleophilic substitution reaction. Since allyl > 2Bromo2methyl butane carbocation is more stable due to resonance. (iii) 1Bromobutane > 1Bromo2methyl butane > (c) Cyanide ion is a resonance hybrid of the following two 1Bromo2,2dimethyl propane contributing structures. CH 3 Markownikov's CH 3 addition 30. (i) C N C N I + HI Therefore, either the carbon or the nitrogen atom can act as the electron donor to alkyl halide. (ambident Markownikov's H nucleophilies) (ii) + HBr addition In principle, the reaction can occur either through carbon H H or nitrogen. But in practice the reaction occurs through carbon since C – C bonds are stronger than C– N bonds CH CH 3 and hence alkyl cyanide is the chief product –

R X + CN R CN + X On the other hand, silver cyanide is predominantly covalent, therefore only nitrogen electron pair is available for bond formation. Hence, alkyl isocyanides are the chief products. R X + AgNC R NC + Ag X 27. (a) The boiling points of isomeric haloalkanes decrease with increase in branching. Hence boiling point of 1chloroethane is higher than 2methyl2chlorobutane. (b) The presence of NO 2 group at para positions withdraws 31. electrons from the benzene ring and thus facilitates the attack of the nucleophile on haloarenes. The carbocation thus formed is stabilized by resonance. Cl

+

+

Cl

Cl –

Br (iii) CH3 CH2 CH CH2 + HBr Markownikov's addition

CH CH CH 3 2

CH 3

Br +

+

Cl

Cl

+

Cl –

Cl –

–

From above resonating structures we can see that electron density is maximum at o/ppositions. Hence halobenzene is o/p directing.

32. (i) An object which has no plane of symmetry (can’t be divided into two identical halves) is called chiral (Greek; ChiralHand) or dissymmetric or asymmetric. A Chiral NO 2 N N object is not superimposable on its mirror image. e.g. O O O O left and right hand of a person are mirror images of 28. (i) Haloalkanes dissolve in organic solvents because the each other and are not superimposable. intermolecular attractions between haloalkanes and organic solvent molecules have the same strength as in the separate haloalkanes and solvent molecules. (ii) An equimolar mixture of a pair of enantiomers is called (ii) CH3CH2 CHCH 3 hydrolyses easily with KOH because it racemic mixture. A racemic mixture is optically inactive due to external compensation. Cl Example: CH CH CH CH 3 2 3 OH Butan2ol

(iii) Of the two bromo derivatives, C 6 H 5 CH(CH 3 )Br and C 6H the C 6H is more reactive 5CH(C 6H 5)Br, 5 CH(C 6H 5)Br

JJJ

is secondary halide (iii) As iodide is a better leaving group because of its large I undergoes S 2 reaction faster size, therefore, N

than

Cl .


68

Chapter11 Alcohols, Phenols and Ethers

11. How are the following conversions carried out? (i) Benzyl chloride to benzyl alcohol. How would you convert ethanol to ethene? [AI SetI 2011] (ii) Methyl magnesium bromide to 2methyl propan2ol. [AI 2010] Draw the structure of 2,6dimethyl phenol. [AI SetII 2011] 12. How are the following conversion carried out? Give the IUPAC name of the following compound : (i) Propene to propan2ol. CH 3 C C CH2 OH (ii) Ethylmagnesium chloride to propan 1ol. [AI 2010] [Foreign SetI 2011] CH 3 Br 13. How would you convert the following: Draw the structural formula of 2methylpropan2ol molecule. (i) Phenol to benzoquinone. [Delhi SetI 2012] (ii) Propanone to 2methylpropan2ol. (iii) Propene to propan2ol. [AI 2010] Draw the structure of hex1en3ol compound


1. 2. 3.

4. 5.

[Delhi SetIII 2012] 14. Explain the following reactions with an example for each: (i) ReimerTiemnn reaction. SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS) (ii) FriedelCrafts reaction. [Delhi 2010] 6. Describe a chemical test each to distinguish between the 15. Explain the following giving one example for each: following pairs. (i) ReimerTiemann reaction. (i) Ethanol and phenol. (ii) Friedel Craft’s acetylation of anisole. (ii) 1propanol and 2propanol. [AI 2008] [Delhi SetI 2011] 7. Give reason for the following : 16. How would you obtain (i) maminophenol is stronger acid than oaminophenol. (i) Picric acid (2, 4, 6trinitrophenol) from phenol. (ii) Alcohol acts as weak base. [AI 2008] (ii) 2Methyl propene from 2methyl propanol? [Delhi SetI 2011] 8. Give reason for the following : (i) Propanol has higher boiling point than that of the 17. How would you obtain the following: hydrocarbon butane. (i) Benzoquinone from phenol (ii) Preparation of ether by acid dehydration of secondary (ii) 2Methyl propan2ol from methyl magnesium bromide and tertiary alcohols is not a suitable method. (iii) Propan2ol from propene? [Delhi 2008] [AI SetI 2011, Foreign SetI and II 2011] Name the reagents and write the chemical equations for the 18. Draw the structure and name the product formed if the preparation of the following compounds by Williamson’s following alcohols are oxidized. Assume that an excess of synthesis : oxidising agent is used. (i) Ethoxy benzene. (i) CH 3 CH 2 CH 2 CH 2 OH (ii) 2butenol (ii) 2Methyl2methoxy propane. (iii) 2methyl1propanol [Delhi SetI 2012] (iii) Why do phenols not give the protonation reaction 19. Explain the mechanism of acid catalysed hydration of an readily? [AI 2008] alkene to form corresponding alcohol. [AI SetI 2012] 10. Complete the following reaction equations: 20. Explain the following behaviours: (i) Alcohols are more soluble in water than the (i) OH + SOCl 2 hydrocarbons of comparable molecular masses. (ii) Orthonitrophenol is more acidic than ortho CH2 OH methoxyphenol. [AI SetI 2012] + HCl (ii) [AI 2009] OH 9.

Alcohols, Phenols and Ethers

69

answers 1.

CH 3CH 2 OH

Conc.H2 SO 4

443 K

Ethanol

Ethene

OH

2. CH 3

C3H 7 C3H 7

CH 3

(ii) Acid dehydration of 2° and 3° alcohol gives alkenes rather than ethers. Due to steric hindrance the nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not occur. The protonated 2° and 3° alcohols lose a molecules of water to form stable 2° and 3° carbocations.

2, 6Dimethylphenol

3. 2–Bromo3methylbut2en1ol OH

4.

are held together by stronger intermolecular hydrogen bonding. H O H O

CH 2 + H 2 O

CH 2

CH3 – C – CH 3

2Methylpropan2ol

ONa

CH 3

5. Hex1en3ol

OCH2CH 3 + CH3CH2 Br

9. (i)

Sod. phenoxide

OH H H H

heat

Bromo ethane

+ NaBr

Ethoxy benzene

H – C C – C – C – C – C – H H H H

CH 3

H H H

6. (i) Distinction between ethanol and phenol. FeCl 3 test : Phenol gives a violet colouration with FeCl 3 solution while ethanol does not. 3C6H5OH + FeCl 3 (C6H5O)3 Fe + 3HCl Phenol

Violet colouration

C 2H 5OH + FeCl 3

No violet colouration.

(ii) CH3 C ONa + CH3 Br heat

Yellow ppt.

CH3 CH2CH2 OH + NaOH + I 2

CH3 C OCH3 + NaBr

Bromo methane

CH 3 Sodium2methyl 2propoxide

CH 3 2methyl2methoxy propane

(iii) In phenols, the lone pairs of electron on the oxygen atom are delocalised over the benzene ring due to resonance and hence are not easily available for protonation.

(ii) Distinction between 1 propanol and 2 propanol : Iodoform test : When 2propanol is warmed with sodium hypoiodite, it gives yellow precipitate of iodoform. 10. (i) But 1propanol does not respond to iodoform test. CH3CH(OH)CH3 + I2 + NaOH 2propanol CHI3 + CH3COONa + H2 O

CH 3

(ii) OH

OH + SOCl 2

Cl + SO 2 + HCl

CH 2 OH Heat + HCl OH

CH 2 Cl + H 2 O

No yellow ppt.

CH2 Cl CH2 OH 7. (i) In substituted phenols, the presence of electron releasing aq. NaOH group decreases the acidic strength of phenol. This 11. (i) effect is more pronounced when such a group is present Benzyl chloride Benzyl alcohol at ortho and parapositions. It is due to the increased negative charge in phenoxide ion. Thus maminophenol O OMgBr is stronger acid than oaminophenol. (ii) Alcohols acts as weak base because of the presence of (ii) CH3 MgBr + CH3 C CH 3 CH3 C CH 3 lone pairs of electrons on oxygen atom. H + ions from CH 3 an acid adds on to oxygen of OH bond forming an oxonium ion. H OH +

R O H + H

+

R O H

8. (i) The molecules of butane are held together by weak van der Waals forces of attraction while those of propanol

H 2 O Hydrolysis

CH3 C CH 3 CH 3 2methyl propan2ol


70

Cl

OH

Cl

+

12. (i) H3C CH CH 2

H2 O/H

+ CH3 Cl

CH3 CH CH 3

propene

propan2ol

1chloro2 methyl benzene (Minor)

(ii) Ethyl magnesium chloride on addition to formaldehyde followed by hydrolysis gives propan1ol. O CH CH MgCl + H C H 3 2 Ethyl magnesium chloride

Dry ether

CH CH CH OH 3 2 2 Propan1ol

O

CH 3 1chloro4 methylbenzene (Major)

Anhyd.AlCl 3

+ CH C Cl 3

Na2 Cr2 O 7

13. (i)

+

15. (i) Refer Ans. 14 (i). (ii) Friedel Craft’s acetylation of anisole : Anisole on treatment with acylchloride in the presence of anhydrous AlCl 3 undergo electrophilic substitution in the ring at ortho and para positions. OCH 3 O

CH CH CH OMgCl 3 2 2 + H O 3

OH

CH 3

Anhyd. AlCl 3

Chlorobenzene

Cl

Ethanoyl chloride

Anisole

H2 SO 4 Phenol

OCH 3 O

O

C CH 3

Benzoquinone

O

OMgBr

(ii) CH3 MgBr + CH3 C CH 3

OCH 3 +

oMethoxy acetophenone

CH3 C CH 3

C CH 3 O pMethoxy acetophenone (major product)

CH 3

16. (i) Picric acid from phenol.

OH H 2 O

CH3 C CH 3

Hydrolysis

OH

CH 3

OH SO3 H

H2SO4 (conc.)

2methyl propan2ol

(iii) Refer Ans. 12(i). 14. (i) ReimerTiemann Reaction : Treatment of phenol with chloroform in the presence of aqueous alkali at 340 K followed by hydrolysis of resulting product to give salicylaldehyde as a major product is known as Reimer Tiemann reaction. OH

Phenol

SO3 H Phenol2,4 disulphonic acid

OH O N 2

HNO 3

CHCl3 + NaOH 340 K Phenol

NO

NO

ONa

Picric acid (2,4,6trinitrophenol)

OH CHO

H3 O +

CH 3

CHO

(ii) CH3 CH CH2 OH

+ SOCl 2

2Methyl propanol

Salicylaldehyde

(ii) FriedelCrafts reaction : When chlorobenzene reacts with alkyl halides in the presence of anhydrous. AlCl 3 , the alkyl groups are introduced at ortho and para positions.

CH

CH 3 CH CH CH Cl 3 2

alc. KOH

CH C CH 3 2 2Methyl propene

Alcohols, Phenols and Ethers

71

18. (i) CH 3CH 2CH 2CH 2 OH

H OH

H

17. (i) Refer Ans. 13 (i). (ii) Refer Ans. 11 (ii). (iii) Refer Ans. 13 (iii).

C C O H + H2 O +

+

C C + H3 O

Oxidation

CH 3 – CH 2 CH 2 COOH 20. (i) The solubility of alcohols in water is due to their ability CH 3 CH CH – COOH (ii) CH 3 – CH CH – CH 2 OH to form hydrogen bonds with water molecules. 2Butenol Hydrocarbons cannot form such hydrogen bond, hence they are insoluble in water. (iii) CH3 – CH – CH2 OH CH 3 – CH – COOH (ii) As we know that the electron withdrawing groups CH 3 CH 3 enhance the acidic character of phenols because they + + help in the stabilisation of phenoxide ion. Nitro group H3 O 19. H2 O + H is an electron withdrawing group whereas methoxy group is an electron releasing group, so onitrophenol H H + is more acidic than omethoxyphenol. C C + H O H C C + H O 2

+

H

OH H

+

C C + H2 O

OH NO 2

H

OCH 3

C C O H +

oNitrophenol

JJJ

oMetoxyphenol


72


1.

Write Cannizzaro reaction giving example.

[AI 2008]

(i) A primary alcohol to an aldehyde. (ii) Butan2one to butan2ol. (iii) Phenol to 2,4,6tribromophenol.

[Delhi 2008]

Draw the structural formula of 1phenyl propan1one 13. Write the chemical tests to distinguish between the following molecule. [AI 2010] pairs of compounds : (i) Acetophenone and benzophenone. 3. What is Tollen’s reagent? Write one usefulness of this reagent. (ii) Ethanal and propanal. [AI 2008] [AI 2010] 2.

Draw the structure of 3methyl butanal. [Delhi SetI 2011] 14. An organic compound A (C 7 H 6 Cl 2 ) on treatment with NaOH solution gives another compound B (C 7 H 6 O). B on oxidation 5. Draw the structure of 4chloropentan2one. [AI SetI 2011] gives an acid C (C 7 H 6 O 2 ) which on treatment with a mixture 6. Write the structure of the following compound : of conc. HNO 3 and H 2 SO 4 gives a compound D (C 7 H 5 NO 4 ). 3Oxopentanal. [Foreign SetI 2011 B on treatment with conc. NaOH gives a compound E (C 7 H 8 O) and C 6 H 5 COONa. Deduce the structures of A, B, C, D and E. 7. Arrange the following compounds in an increasing order of [AI 2008] their reactivity in nucleophilic addition reactions : ethanal, 4.

propanal, propanone, butanone.

[Delhi SetI 2012]

15. Explain the mechanism of the following reactions : (i) Addition of Grignard reagent to the carbonyl group of 8. Write the IUPAC name of the following: a compound forming an adduct followed by hydrolysis. O (ii) Acid catalysed dehydration of an alcohol forming an CH3 – CH2 – CH CH – C – H alkene. [AI SetI 2012] (iii) Acid catalysed hydration of an alkene forming an 9. Write the IUPAC name of Ph – CH CH – CHO. alcohol. [AI 2009] [AI SetII 2012] 16. Illustrate the following name reactions by giving example: SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS) (i) Cannizzaro reaction. (ii) Clemmensen reduction. [AI 2009] 10. Arrange the following compounds in an increasing order of 17. How will you bring about the following conversions: their acid strength. (i) Ethanol to 3hydroxy butanal. (CH 3 ) 2 CHCOOH, CH 3 CH 2 CH(Br)COOH, (ii) Benzaldehyde to benzophenone. [AI 2009] CH 3 CH(Br)CH 2 COOH. [Delhi 2008] 11. (a) Complete the following reaction statements by giving LONG ANSWER TYPE QUESTIONS (5 MARKS) the missing starting material, reagent or product as 18. (a) How will you bring about the following conversions? required : (i) Ethanol to acetone. O 3 (i) .......... . (ii) Benzene to acetophenone. O Zn H2 O 2 (iii) Benzoic acid to benzaldehyde. .......... (b) Describe the following giving a suitable example in (ii) CH 2 CHO . each case : CH2CH 3 KMnO (i) Decarboxylation. 4 (iii) .......... . KOH, heat (ii) Cannizzaro’s reaction. [Delhi 2008] (b) Describe the following reactions : 19. (a) Giving a chemical equation for each, illustrate the (i) Cannizzaro reaction. following processes : (ii) Cross aldol condensation. [Delhi 2008] (i) Cannizzaro reaction (ii) Acetylation 12. Name the reagents which are used in the following (iii) Decarboxylation conversions :

Aldehydes Ketones and Carboxylic Acids

73

(b) State chemical tests to distinguish between the following 25. (a) Write chemical equations to illustrate the following name bearing reactions. pairs of compounds : (i) Cannizzaro reaction. (i) Propanal and propanone. (ii) HellVolhardZelinsky reaction. (ii) Phenol and benzoic acid. [AI 2008] (b) Give chemical tests to distinguish between the following 20. (a) How would you account for the following : pairs of compound. (i) Aldehydes are more reactive than ketones towards (i) Propanal and propanone. nucleophiles. (ii) Acetophenone and benzophenone. (ii) The boiling points of aldehydes and ketones are (iii) Phenol and benzoic acid. [AI 2009] lower than the corresponding acids. (iii) The aldehydes and ketones undergo a number of 26. (a) Arrange the following compounds in an increasing addition reaction. order of their indicated properties. (b) Give chemical tests to distinguish between : (i) Benzoic acid, 4nitrobenzoic acid, 3,4 (i) Acetaldehyde and benzaldehyde. dinitrobenzoic acid, 4methoxy benzoic acid (acid (ii) Propanone and propanol. [Delhi 2008] strength). (ii) CH 3 CH 2 CH(Br)COOH, CH 3 CH(Br)CH 2 COOH, 21. (a) An organic compound contains 69.77% carbon, 11.63% (CH 3 ) 2 CHCOOH, CH 3 CH 2 CH 2 COOH (acid hydrogen and the rest is oxygen. The molecular mass of strength). the compound is 86. It does not reduce Tollen’s reagent (b) How would you bring about the following conversions. but forms an addition compound with sodium hydrogen (i) Propanone to propene. sulphite and gives a positive iodoform test. On vigorous (ii) Benzoic acid to benzaldehyde. oxidation it gives ethanoic and propanoic acids. Deduce (iii) Bromobenzene to 1phenyl ethanol. [AI 2009] the possible structure of organic compound. (b) State reason for the following : 27. An organic compound (A) on treatment with acetic acid in (i) Monochloroethanoic acid has a higher pK a value the presence of sulphuric acid produces an ester (B). (A) on than dichloroethanoic acid. mild oxidation gives (C). C with 50% KOH followed by (ii) Ethanoic acid is a weaker acid than benzoic acid. acidification with dilute HCl generates (A) and (D). (D) with [Delhi 2008] PCl 5 followed by reaction with ammonia gives (E). (E) on dehydration produced hydrocyanic acid. Identify the 22. An organic compound A (C 3 H 6 O) is resistant to oxidation but compound A, B, C, D and E. forms compound B (C 3 H 8 O) on reduction. B reacts with HBr [AI 2009] to form the compound C. C with Mg forms Grignard reagent D which reacts with A to form a product which on hydrolysis 28. Compound (A) C H O on reduction with LiAlH yield two 6 12 2 4 gives E. Identify A to E. [AI 2008, 2009, 2010] compounds ‘B’ and ‘C’. The compound ‘B’ on oxidation gives ‘D’ which on treatment with aqueous alkali and subsequent 23. Identify A to E in the following series of reactions : heating furnished ‘E’. The latter on catalytic hydrogenation [O] Ca(OH) 2 Distillation LiAlH 4 A B C D gave ‘C’. The compound ‘D’ on further oxidation gave [2H] –CaCO 3 HBr CH 3 COOH. Deduce the structure of A, B, C, D and E. CH3 CH CH 3 E [AI 2008] [AI 2009] Br 29. Identify A to E in the following sequences of operations: 24. (a) How are the following obtained? Br2 /KOH NH 3 (i) Benzoic acid from ethyl benzene. HONO C CH3CH2 COOH D A B (ii) Benzaldehyde from Toluene. (b) Complete each synthesis by giving the missing material Ca(OH) 2 LiAlH 4 (O) CH3 COOH Distill D E reagent or product. H 2 [AI 2009] (i) C6H5 COCl PdBaSO 4 30. (a) Explain the mechanism of a nucleophilic attack on the O carbonyl group of an aldehyde or a ketone. C (b) An organic compound (A) (molecular formula C 8 H 16 O 2 ) Anhyd. AlCl 3 CH 3 (ii) + ........ was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of O (C) with chromic acid also produced (B). On C dehydration (C) gives but 1ene. Write the equations + ........ (iii) for the reactions involved. [AI 2010] [AI 2009]

74

31. (a) Give chemical tests to distinguish between the following pairs of compounds : (i) Ethanal and propanal (ii) Phenol and benzoic acid (b) How will you bring about the following conversions? (i) Benzoic acid to benzldehyde (ii) Ethanal to but2enal (iii) Propanone to propene. Give complete reaction in each case. [Delhi 2010] 32. (a) Illustrate the following name reaction giving chemical equation in each case. (i) Clemmensen reaction (ii) Cannizzaro’s reaction (b) Describe how the following conversions can be brought about : (i) Cyclohexanol to cyclohexanone (ii) Ethylbenzene to benzoic acid (iii) Bromobenzene to benzoic acid [AI 2010]


(i) Cannizzaro’s reaction (ii) Clemmensen reduction (b) How would you obtain the following: (i) But2enal from ethanal (ii) Butanoic acid from butanol (iii) Benzoic acid from ethylbenzene? OR (a) Give chemical tests to distinguish between the following: (i) Benzoic acid and ethyl benzoate (ii) Benzaldehyde and acetophenone. (b) Complete each synthesis by giving missing reagents or products in the following : COOH

(i)

SOCl 2 heat

COOH

(ii) C6H5 CHO H2NCONHNH 2

(iii) 1] CH 2 CHO [AI SetI 2011 33. (a) Illustrate the following name reactions : (i) HellVolhardZelinsky reaction 36. (a) Illustrate the following reactions giving a suitable (ii) WolffKishner reduction reaction example for each. (b) How are the following conversions carried out : (i) Clemmensen reduction (i) Ethylcyanide to ethanoic acid (ii) HellVolhardZelinsky reaction (ii) Butan1ol to butanoic acid (b) How are the following conversions carried out? (iii) Methylbenzene to benzoic acid. (i) Ethyl cyanide to ethanoic acid. Write chemical equations for the involved reactions. (ii) Butan1ol to butanoic acid. [AI 2010] (iii) Benzoic acid to mbromobenzoic acid. 34. (a) Give chemical tests to distinguish between OR (i) Propanal and propanone (a) Illustrate the following reactions giving a suitable (ii) Benzaldehyde and acetophenone. example for each. (b) How would you obtain (i) Cross aldol condensation (i) But2enal from ethanal (ii) Decarboxylation (ii) Butanoic acid from butanol, (b) Give simple tests to distinguish between the following (iii) Benzoic acid from ethyl benzene? pairs of compounds. OR (i) Pentan2one and pentan3one (a) Describe the following giving linked chemical (ii) Benzaldehyde and acetophenone equations: (iii) Phenol and benzoic acid (i) Cannizzaro reaction [Delhi SetI, 2012] (ii) Decarboxylation 37. (a) Give a plausible explanation for each one of the (b) Complete the following chemical equations : following: CH 2CH 3 (i) There are two –NH 2 groups in semicarbazide. KMnO 4 (i) However, only one such group is involved in the KOH, heat formation of semicarbazones. (ii) Cyclohexanone forms cyanohydrins in good yield COOH SOCl 2 but 2,4,6trimethylcyclohexanone does not. heat (ii) (b) An organic compound with molecular formula C H O 9 10 COOH forms 2,4DNP derivative, reduces Tollen’s reagent + and undergoes Cannizzaro’s reaction. On vigorous (iii) C 6H5CONH 2 H3 O heat oxidation it gives 1,2benzenedicarboxylic acid. [Delhi SetI 2011, Foreign SetI 2011] Identify the compound. 35. (a) Illustrate the following name reactions: OR


75

(ii) 4Methylacetophenone to benzene1,4 dicarboxylic acid (b) An organic compound with molecular formula C 9 H 10 O forms 2,4DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2benzenedicarboxylic acid. Identify the compound. OR (a) Give chemical tests to distinguish between (i) Propanol and propanone (ii) Benzaldehyde and acetophenone (b) Arrange the following compounds in an increasing order of their property as indicated: (i) Acetaldehyde, acetone, methyltertbutyl ketone (reactivity towards HCN) (ii) Benzoic acid, 3,4dinitrobenzoic acid, 4 methoxybenzoic acid (acid strength) (iii) CH 3 CH 2 CH(Br)COOH, CH 3 CH(Br)CH 2 COOH, (CH 3 ) 2 CHCOOH (acid strength) [AI SetI 2012]

(a) Give chemical tests to distinguish between (i) Phenol and benzoic acid (ii) Benzophenone and acetophenone (b) Write the structures of the main products of the following reactions: (i)

anhydrous AlCl 3

+ C 6H5 COCl

(ii) H3 C – C C – H

CS 2

Hg2+ , H2SO 4

CH 3 1. CrO2Cl 2

(iii)

[Delhi SetII 2012]

2. H 2 O

NO 2

38. (a) Write a suitable chemical equation to complete each of the following transformations: (i) Butan1ol to butanoic acid

a n s w e r s 1. Cannizzaro’s reaction : Aldehydes which do not contain a H atom undergo disproportionation when heated with 8. concentrated (50 %) NaOH.

HCHO + HCHO

50 % NaOH

Methanal

CH3CH2 – CH CH – C – H 5

4

2

3

1

Pent2enal

HCOONa + CH3 OH Sodium formate Methanol

3

9.

O

2.

O

1

2

Ph – CH CH – CHO 3Phenylprop2enal

C CH2 CH2CH 3

10. We know that + Ieffect decreases while –Ieffect increases the acid strength of carboxylic acids. The overall acid strength increases in the order. 3. Tollen’s reagent is an ammoniacal silver nitrate solution. (CH 3 ) 2 HCOOH < CH 3 CH(Br )CH 2 COOH < Tollen’s reagent is used to test an aldehyde. Both aliphatic CH 3 CH 2 CH(Br)COOH. and aromatic aldehydes reduce Tollen’s reagent and give silver mirror. O 3 2 O 11. (a) (i) Zn H 2 O CH 3 O Cyclohexylidene Cyclohexanone 1phenylpropan1one

4. Structure of 3methyl butanal CH3 CH CH2 C H

CH 2

(ii) Cl

5.

CH 3

CH

O CH 2

C

CH B 3

Methylene cyclohexane

CH 3

H2 O 2 NaOH

CH 2 OH

PCC

CHO

Cyclohexane carbaldehyde

6. 3Oxopentanal

O

B 2H 6 /THF

CH2CH 3

O (iii)

CH3 CH2 C CH2 C H 7. Butanone < Propanone < Propanal < Ethanal

Ethylbenzene

(b) (i) Refer answer 1

KMnO 4 KOH, heat

COOK

Potassium benzoate


76

(ii) CH3CHO + CH3CH2 CHO dil. NaOH

CHO

CH3 CH CH CHO + CH3 CH2 CH CH2 CHO

NaOH solution

2

OH CH 2

OH

1

A (C7 H6 Cl2 ) Benzal chloride

Cross aldol condensation

OH

B (C7 H6 O) Benzaldehyde

COOH

OH

CH3 CH CH2 CHO + CH3CH2 CH CH CHO 3

Oxidation

conc. H2 SO 4 conc. HNO 3

4

CH 3

12. (i) Pyridinium chlorochromate (PCC), a complex of chromium trioxide with pyridine and HCl is used to convert a primary alcohol to an aldehyde. PCC

CH3 CH CH CH2 OH CH3 CH CH CHO (ii) Ni/H 2 or LiAlH 4 or NaBH 4 converts butan2one to butan2ol. (iii) Aqueous bromine or bromine water.

COOH

C (C 7H 6O 2 ) Benzoic acid

NO 2

D (C7 H5 NO4 ) m Nitrobenzoic acid

CHO

CH2 OH +

+ conc. NaOH B (C7 H6 O) Benzaldehyde

COONa

E (C7 H8 O) Benzyl alcohol

sod. benzoate

Therefore, (A) : Benzal chloride OH OH (B) : Benzaldehyde Br Br H 2 O (C) : Benzoic acid + 3 Br 2 (D) : mnitrobenzoic acid Phenol (E) : Benzyl alcohol. Br 2,4,6Tribromophenol 15. (i) The first step of the reaction is the nucleophilic addition 13. (i) Acetophenone and Benzophenone can be distinguished of Grignard reagent to the carbonyl group to form an by the iodoform test. Acetophenone when treated with adduct. Hydrolysis of the adduct yields an alcohol. NaOI (I 2 /NaOH) gives yellow ppt. of iodoform but +d –d – d +d – + benzophenone does not. C O + R Mg X C O Mg X ...(i) C6H 5 COCH3 + 3 NaOI C6H 5 COONa + CHI3 + 2 NaOH R Acetophenone Sod. benzoate Iodoform Adduct

(yellow ppt.)

C6H5COC6H 5

NaOI

No yellow ppt. of CHI3 .

H2 O

Benzophenone

(ii) Ethanal and propanal can be distinguished by the iodoform test. Ethanal when treated with NaOI (I 2 /NaOH) gives yellow ppt. of iodoform but propanal does not give yellow ppt. CH3 CHO + 3 I2 + 4NaOH Ethanal HCOONa + CHI3 + 3NaI + 3H2 O Sod. formate

CH3CH2 CHO Propanal

NaOI

Iodoform (yellow ppt.)

No yellow ppt. of CHI3 .

14. The relevant equations for all the reactions involved may be explained as follows : A NaOH B Oxidation C conc. HNO D 3 C7H6Cl 2 C7H6 O C7H6O 2 conc. H SO C7H5NO 4 2 4 NaOH (an acid) E + C6H5 COONa C7H8 O

C OH + Mg (OH)X

Hydrolysis

R (ii) Mechanism of dehydration of ethanol involves the following steps: Step :1. Formation of protonated alcohol. H H

H H H +

H C C O H + H

Fast

+

H C C O H

H H

H H

Ethanol

Protonated alcohol

Step : 2. Formation of carbocation. H H H H C C O H

H H Slow

+

H C C + H2 O

+

H H

H H Carbocation

Step : 3. Formation of ethene by elimination of a proton.


77

H H H + H C C + H H Ethene H

+

H C C

H H

Ethanol

(iii) Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols. Mechanism : The mechanism of the reaction involves the following three steps. Step : 1. Protonation of alkene to form carbocation by electrophilic attack of H 3 O + . +

+

H + H2 O

H3 O

H

H +

C C + H O H

K 2Cr2O 3/H 2SO 4 Oxidation

CH3CH2 OH

C C + H2 O

Ca(OH) 2

CH3 COO

– H2 O

CH3 COO

Acetic acid Dry distillation

Ca

– CaCO 3

CH3COCH 3 Acetone

Calcium acetate

(ii) Benzene to acetophenone. O + CH3 C Cl Benzene

COCH 3

Anhyd. AlCl 3 FriedelCraft's acylation Acetophenone

(iii) Benzoic acid to benzaldehyde. COOH COCl SOCl 2 – SO2 , – HCl Benzoyl chloride

+

CHO

Step : 2. Nucleophilic attack of water on carbocation.

H2 /Pd + BaSO4 + S Boiling xylene

H

H +

(Rosenmund's reduction) Benzaldehyde

+

C C + H2 O

C C O H

(b) (i) Decarboxylation : The process of removal of a molecule of CO 2 from a carboxylic acid is called decarboxylation. It is carried out by heating a carboxylic acid with soda lime (NaOH + CaO) at 630 K. COOH

H

Step : 3. Deprotonation to form an alcohol. H

CH3CH2 OH

H

H OH +

C C O H + H2 O

C C + H3 O

+

CaO, 630 K

+ 2 NaOH

+ Na2CO3 + H2 O

Benzoic acid

Benzene

(ii) Refer Ans. 1.

16. (i) Refer Ans. 1. (ii) Clemmensen reduction : The carbonyl group of 19. (a) (i) Refer Ans. 1. (ii) Acetylation : The replacement of an active aldehyde and ketones is reduced to CH 2 group on hydrogen of alcohols, phenols or amines with an treatment with zinc amalgam and concentrated acyl group to form the corresponding esters or hydrochloric acid. amides is called acylation. CH 3 Zn Hg CH 3 Pyridine C O HCl CH2 + H2 O CH3COCl + CH3CH2 OH CH3COOC2H5 + HCl CH 3 CH 3 Propanone

17. (i) CH 3CH 2 OH Ethanol

Acetyl chloride Ethyl alcohol

Propane

K 2Cr2O 7/H 2SO 4 oxidation

CH 3 CHO

OH dil. NaOH CH3CHO + CH3 CHO CH3 CH CH2 CHO Aldol 3hydroxy butanal condensation

K2 Cr2 O 7

(ii) C 6H 5 CHO H SO Benzaldehyde 2 4

C6H 5 COOH

(C6H 5COO)2 Ca

18. (a) (i) Ethanol to acetone.

CaCO 3

dry distil. –CaCO 3

C 6H 5 CO C 6H 5 Benzophenone

Ethyl acetate

(iii) Decarboxylation : The process of removal of a molecule of CO 2 from a carboxylic acid is called decarboxylation. It is carried out by heating a carboxylic acid with soda lime (NaOH + CaO) at 630 K. COOH CaO, 630 K

+ 2 NaOH Benzoic acid

+ Na2CO3 + H2 O Benzene

(b) (i)Propanal reduces Tollen’s reagent into silver mirror while propanone does not gives this test. +

–

CH3CH2 CHO + 2 [Ag (NH3)2 ] + 2OH Propanal

Tollen's reagent

CH3CH2 COONH4 + 2Ag + H2O + 2NH 3 Silver mirror


78

(ii) Benzoic acid being a stronger acid than phenol, decomposes NaHCO 3 to evolve CO 2 but phenol does not. C6H5COOH + NaHCO 3 C6H5COONa + H2O + CO 2- C 6H 5 OH

NaHCO 3

No evolution of CO 2

20. (a) (i) Ketones are less reactive than aldehydes towards nucleophilic addition reactions because : – The two electron releasing alkyl groups decrease the magnitude of positive charge on carbonyl carbon and make it less susceptible to nucleophilic attack. R R C O C O R H Ketone

Aldehyde

– The two bulkier alkyl groups hinder the approach of the nucleophile to the carbonyl carbon. This is called steric factor. (ii) The boiling points of aldehydes and ketones are lower than corresponding acids and alcohols due to absence of intermolecular hydrogen bonding. (iii) Aldehydes and ketones undergo a number of addition reactions as both possess the carbonyl functional group which reacts with a number of nucleophiles such as HCN, NaHSO 3 , alcohols, ammonia derivatives and Grignard reagents. (b) (i) Acetaldehyde and benzaldehyde can be distinguished by Fehling solution. Acetaldehyde gives red coloured precipitate with Fehling solution while benzaldehyde does not. 2+

CH3 CHO +

–

2 Cu + 5 OH Fehling solution

Emperical formula mass = 5 × 12 + 10 × 1 + 1 × 16 = 86 86 = 1 \ n = 86 Molecular formula of the given compound = 1 × (C 5 H 10 O) = C 5 H 10 O. Determination of structure : Since the compound does not reduce Tollen’s reagent and gives a positive iodoform test, so the given organic compound is ketone and not aldehyde. Since the given compound gives iodoform test, on vigorous oxidation gives a mixture of ethanoic acid and propanoic acid. Therefore given organic compound is methyl ketone and its structure would be : CH 3 COCH 2 CH 2 CH 3 K 2Cr2O 7

CH3COCH2 CH2CH 3 Pentan2one (Methyl ketone)

H2 SO 4

CH3COOH + CH3CH2 COOH Ethanoic acid

Propanoic acid

(b) (i) The strength of an acid is indicated by pK a value where, pK a = – log k a Since monochloroethanoic acid is weaker than dichloroethanoic acid. So it has lower value of dissociation constant k a . Therefore, it has higher value of pK a . (ii) The – COOH group in benzoic acid is attached to sp 2 carbon of the phenyl ring and is more acidic than acetic acid in which – COOH group is attached to sp 3 – carbon atom of CH 3 group. So benzoic acid is stronger than acetic or acetic acid is weaker acid than benzoic acid.

–

CH3COO + Cu2O + H2 O Red ppt.

(ii) Propanone (CH 3 COCH 3 ) and propanol (CH 3 CH 2 CH 2 OH) can be distinguished by iodoform test. Propanone when warmed with sodium hypoiodite (NaOI) i.e. I 2 in NaOH, it gives yellow precipitate of iodoform while propanol does not respond to iodoform test. CH3COCH3 + 3NaOI

22. CH3CH2 CHO Reduction CH3CH2CH2 OH (A ) (C 3H 6 O) propanal

HBr –H 2 O

(B ) (C 3H 8 O) 1propanol

Mg

CH3CH2CH2 Br (C ) (Propyl bromide)

CH3CH2 CH2 MgBr (D ) (Propyl magnesium bromide)

CH3CH2CHO + CH3CH2CH2 MgBr (A )

(B )

OMgBr

Propanone

CHI3 ¯ + CH3COONa + 2NaOH

CH3 CH2 C C3H 7

Yellow ppt.

21. (a) Calculation of the molecular formula; C = 69.77%, H = 11.63% \ O = 100 – (69.77 + 11.63) = 18.6%. Element Percentage Molar Percentage/ Simplest mass molar mass ratio C 69.77 12 5.88 5 H 11.63 1 11.63 10 O 18.6 16 1.16 1 Empirical formula of given compound = C 5 H 10 O

H +

H2 O, H

Hydrolysis

C2H5 C

OH + Mg(OH)Br C3H 7

(E ) (Hexan3ol)

Therefore, (A) : (B) : (C) : (D) : (E) :

Propanal Propanol Propyl bromide Propyl magnesium bromide Hexan3ol.


23. CH3CH2 OH

[O]

Ca(OH) 2

CH3 COOH

(A )

79

Cl2 CHCOOH + Cl 2

(B )

P

Cl3 CHCOOH + HCl Trichloro acetic acid

O (CH3COO)2 Ca (C )

Distillation –CaCO 3

CH3 C CH 3 (D )

OH

(b) (i) Refer Ans. 19 (b) (i). (ii) Refer Ans. 13 (i). (iii) Refer Ans 19 (b) (ii).

LiAlH 4 [2H]

Br

CH3 CH CH 3

HBr

CH3 CH CH 3

(E )

24. (a) (i) Benzoic acid from ethyl benzene. C2H 5

COOK

KMnO 4 – KOH Heat

Ethyl benzene

Pot. benzoate

COOH

+

H3 O

26. (a) (i) The overall acid strength increases in the order. 4methoxy benzoic acid < benzoic acid < 4nitro benzoic acid < 3,4dinitrobenzoic acid. (ii) The overall acid strength increases in the order. (CH 3 ) 2 CHCOOH < CH 3 CH 2 CH 2 COOH < CH 3 CH(Br) CH 2 COOH < CH 3 CH 2 CH(Br)COOH. (b) (i) Propanone to propene. O OH NaBH 4, CH3 OH Reduction

CH3 C CH 3 Propanone

Benzoic acid

(ii) Benzaldehyde from Toluene. CH 3

CH3 CH CH 2 Propene

(ii) Benzoic acid to benzaldehyde. COOH COCl

(ii) Hydrolysis Benzaldehyde

Toluene

(b) (i)

2propanol

conc. H2 SO4 , 443 K Dehydration

CHO

(i) CrO 2Cl 2, CH 3 COOH

CH3 CH CH 3

C6H5 COCl Benzoyl chloride

SOCl 2

CHO + HCl

H 2 PdBaSO 4

–SO 2 , –HCl Benzoic acid

Benzoyl chloride

CHO

Benzaldehyde H 2/Pd. + BaSO 4 + S

O O (ii)

+ CH3 C Cl

C

Anh. AlCl 3

boiling xylene

CH 3

Benzaldehyde

(iii) Bromobenzene to 1phenyl ethanol. Acetophenone

C6H5 Br

C6H5 MgBr

Bromobenzene

COCl

(iii)

Mg/Dry ether

Phenyl magnesium bromide

+

OH Benzoyl chloride

CH3 CHO

O

+

H 3 O

1phenyl ethanol

C Anhyd. AlCl 3

+ HCl

F.C acylation Benzophenone

25. (a) (i) Refer Ans. 1. (ii) HellVolhardZelinsky reaction : Carboxylic acids react with chlorine or bromine in the presence of phosphorous to give compounds in which ahydrogen atom is replaced by halogen atom. CH 3COOH + Cl 2

P

Cl CH 2 COOH + HCl

Acetic acid

Cl CH 2 COOH + Cl 2

Chloro acetic acid P

Cl2 CHCOOH + HCl Dichloro acetic acid

27. CH3COOH + CH3 OH

conc. H2 SO 4

CH3COOCH 3

Acetic acid Methyl alcohol (A )

CH3 OH

mild [O]

Methyl alcohol (A )

C 6H 5 CH CH 3

Methyl acetate (B ) (i) 50% KOH

HCHO

Formaldehyde (ii) HCl (C )

CH3 OH + HCOOH Methanol (A )

HCOOH Formic acid (D )

PCl 5

HCOCl

Methanoic acid (D )

NH 3

D dehydration HCN Methanamide P2 O 5 Hydrocyanic

HCONH 2 (E)

acid


80

Therefore, (A) : (B) : (C) : (D) : (E) :

A nucleophile attacks the electrophilic carbon atom from a direction perpendicular to the plane of sp 2 hybridised orbital of carbonyl carbon. The hybridisation of carbon changes from sp 2 to sp 3 in this process and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product. The net result is addition of Nu – and H + across the carbon oxygen double bond.

Methyl alcohol Methyl acetate Formaldehyde Formic acid Formamide.

O

28. (i) CH3 C O CH2 CH2 CH2 CH 3

LiAlH 4

Butylethanoate

+

CH3 CH2 OH + CH3CH2 CH2CH2 OH (B ) Ethanol

(ii) CH3CH2 OH

(C ) Butan1ol [O]

[O]

CH 3 CHO

(B )

[O]

Ethanoic acid

CH 3 COOH

(D )

CrO

CH

4 NaOH Methane (E )

O

Hence, (A ) CH3 C O CH2 CH2 CH2 CH 3 (B) CH 3 CH 2 OH (C) CH 3 CH 2 CH 2 OH (D) CH 3 CHO (E) CH 4 . NH

29. CH3CH2 COOH D 3 CH3NH 2

HONO

Methyl amine (B )

CH3CH2 CH2CH2 OH 3 CH3CH2CH2 COOH (C ) (B ) CrO 3 CH3CH2 CH2 CH2 OH CH3CH2CH2 COOH (C ) (B )

NaOH

CH3 COONa Heat

31. (a) (i) Refer answer 13 (ii). (ii) Refer answer 19 (b)(ii) (b) Conversions COOH

Br2 /KOH

Ethanamide (A)

CH3 OH

[O]

CH3 COOH

Methanol (C )

CHO H2 , PdBaSO 4 Rosenmund reduction

Benzoic acid

(ii)

Benzaldehyde

OH

O

Ca(OH) 2 Distill.

COCl

PCl 5

(i) CH3CONH 2

H 2 O/H

CH3CH2 CH2COOH + CH3CH2 CH2CH2 OH (B ) (C )

CH 3 COOH

Ethanal (D )

(iii) CH3 CHO

(b) CH3CH2 CH2COOCH2 CH2CH2 CH 3 (A )

OH – 2CH3 – C – H Aldol Ethanal condensation

CH3 – CH – CH2 – CHO D H +

HCOO Ca

LiAlH 4

HCHO

HCOO

CH3 – CH CH – CHO

(E )

But2enal

(D )

Hence, (A) CH 3 CONH 2 (B) CH 3 NH 2 (C) CH 3 OH (D) (HCOO) 2 Ca (E) HCHO. 30. (a) Mechanism of nucleophilic addition reactions : Nucleophile attack from the top face Nu – +

a

d

C O

b

O

OH

Step 1

(b) (i) Nu –

+

fast,H

CH3 CH CH 3

conc. H2 SO4 , 443 K Dehydration

Planar Show

OH

NaBH 4, CH3 OH CH3 C CH 3 Reduction Propanone

2propanol

CH3 CH CH 2

32. (a) (i) Refer Ans. 32 (a) (i). (ii) Refer Ans. 1.

–

d

(iii) Propanone to propene.

Nu –

a a Step 2 b b O – OH Tetrahedral intermediate Addition product

O CrO 3, H 2SO 4 Oxidation

Cyclohexanol

Cyclohexan1one

Propene


81

CH2CH 3

ketone on treatment with I 2 /NaOH undergoes iodoform reaction to give yellow ppt. of iodoform but benzaldehyde does not give C 6H 5 COCH 3 + 3NaOI

COOH –

KMnO4 /OH

(ii)

Vigorous oxidation Ethyl benzene Benzoic acid

Br

Acetophenone

C6H5 COONa + CHI3 + 2NaOH Iodoform (Yellow ppt.)

Mg Br I + NaOH

2 C 6H 5 CHO No reaction

Mg/dry ether

(iii)

Grignard reaction Bromobenzene Phenylmagnesium bromide

OH dil. NaOH

(b) (i) 2CH CHO CH CH CH CHO 3 2 3 Ethanal Aldol condensation

COOH

D – H 2 O

+ 3

(i) Dry ice, (ii) H O

33. (a) (i) Refer Ans. 25 (a) (ii). (ii) Wolff Kishner reduction reaction : The carbonyl group of aldehydes and ketones is reduced to CH 2 group on treatment with hydrazine followed by heating with potassium hydroxide in high solvent such as ethylene glycol.

(ii) CH3CH2CH2OH CH3CH2CH2 COOH Butan1ol

Butanoic acid

– + COOK

CH CH 2 3 KMnO –KOH 4 D

(iii) Ethyl benzene

COOH

NH2 NH 2 CH 3 CH 3 C O C NNH 2 –H2 O CH 3 CH 3

H3 O +

Acetone

CH 3 CH2 + N 2 CH 3

heat

Benzoic acid

OR

propane

O

(b) (i) CH CH CN 3 2 Ethyl cyanide Br2 – NaOH

Conc. HCl Partial hydrolysis

CH 3CH 2NH 2 Ethylamine

–N2 , –H2 O

Oxidation

Butan1ol

KMnO 4 KOH, heat

(b) (i) CH 3CH 2 OH

Ethyl benzene

Ethanol

COOH

CH3 COOH Ethanoic acid

K 2Cr2O 7/H 2SO 4 Oxidation

CH 3CH 2CH 2 COOH

Butanoic acid

(ii)

(iii)

COOH K 2Cr2O 7 + H 2SO 4 Oxidation

Methylbenzene

Benzoic acid

34. (a) (i) Refer Ans. 19 (b) (i). (ii) Benzaldehyde and acetophenone Iodoform test : Acetophenone being a methyl

Pot. benzoate

COCl

SOCl 2 heat

COCl

COOH

Phthalic acid

Phthaloyl chloride +

H3 O (iii) C6 H 5 CONH2 C 6 H 5 COOH + NH 3 Benzamide

CH 3

– + – + COOK

CH2 CH 3

propionamide

K 2Cr2O 7/H 2SO 4

(ii) CH3CH2 CH2CH2 OH

(a) (i) Refer Ans. 1. (ii) Refer Ans. 19 (a) (iii).

CH CH C NH 3 2 2

HONO

But2enal

CrO3–H2SO 4

Benzoic acid

KOH/ethylene glycol

CH 3 CH CHCHO

heat Benzoic acid

35. (a) (i) Refer Ans. 1. (ii) Refer Ans. 16 (ii). (b) Refer Ans. 34 (b). OR (a) (i) Benzoic acid on warming with sodium hydrogen carbonate gives brisk effervescence of CO 2 gas while ethyl benzoate does not respond to this test.


82

+ C6 H5 COOH + NaHCO3 C6 H– 5 COO Na + CO2 + H2 O Benzoic acid

–

O

O

Brisk effervescence

H 2NCONHNH 2

C 6H 5 CH

CH 2

(iii)

NNHCONH 2 + H 2 O

(i) B2H 6,H 2O 2 /OH –

CHO

(ii) PCC

36. (a) (i) Refer Ans. 16 (ii). (ii) Refer Ans. 25 (a) (ii). (b) (i) Refer Ans. 33 (b) (i). (ii) Refer Ans. 33 (b) (ii). COOH

NH

+

:

(ii) C 6H 5 CHO

C

:

:

H2 N

C 6H 5COOC 2 H 5 + NaHCO 3 ® No reaction (ii) Refer Ans. 34 (a) (ii). (b) (i) Refer Ans. 34 OR (b) (ii).

NH 2

H2N C NHNH 2

So, it is the other —NH 2 group which acts as nucleophile and is involved in condensation with aldehydes and ketones to give semicarbazone derivatives. (ii) Formation of cyanohydrin involves the nucleophilic attack of cyanide ions (CN – ) at the carbonyl carbon. In cyclohexanone, reaction proceeds but in 2,2,6trimethylcyclohexanone, the methyl groups cause steric hindrance and yields are poor.

O

O

COOH

CH 3

H3 C

Br2/FeBr 3

(iii)

CH 3 Br

Benzoic acid

OR (a) (i) Cross aldol condensation. When two different aldehydes or ketones having an a hydrogen atoms are mixed and treated with aqueous sodium hydroxide, they undergo aldol condensation with each other. This reaction gives a mixture of four products and is known as cross aldol condensation. e.g. A mixture of ethanal and propanal yields the following four products. CH3 CHO + CH 3 CH 2 CHO OH O OH O

(b) The compound forms 2,4DNP derivative. It shows that it is a carbonyl compound. Further it reduces Tollens’ reagent which shows that it contains aldehydic group. It undergoes Cannizzaro reaction indicating that aldehyde group is without any a hydrogen. On vigorous oxidation, it gives 1,2benzenedicarboxylic acid which shows that there are two carbon residues on benzene ring. Since the molecular formula is C 9 H 10 O, it fits into the structure, 2ethylbenzaldehyde. CHO COOH Oxidation

CH3 CH CH2 C H + CH 3 CH CH C H

CH 3

COOH

OR (a) (i) Refer Ans. 19 (b) (iii). (ii) Refer Ans. 13 (i).

O

+ CH3 CH2 CH CH2 C H OH

CH 2 2 Ethylbenzaldehyde

CH 3 OH

2,2,6Trimethylcyclohexanone

Cyclohexanone

3bromobenzoic acid

O

+ CH3 CH2 CH CH C H CH 3 (ii) Refer Ans. 19 (a) (iii). (b) (i) Pentan2one and pentan3one can be distinguished by iodoform test. Pentan2one being a methyl ketone, forms yellow precipitate of iodoform, on treatment with iodine and sodium hydroxide. (ii) Refer Ans. 34 (a) (ii). (iii) Refer Ans. 19 (b) (ii). 37. (a) (i) In semicarbazide,two —NH 2 group is attached to carbonyl group which is an electron withdrawing group. So this nitrogen is not able to act as an effective nucleophile.

(b) (i)

1, 2 Benzenedicarboxylic acid

anhy. AlCl 3

+ C 6H5 COCl

Cs 2

O C Benzophenone

O 2+

(ii) CH 3

C

CH

Hg , H 2SO 4

CH 3

C

CH 3

Propanone

(iii) O 2 N

CH 3

1. CrO2 Cl 2 +

2. H 3 O

O2 N

CHO

p Nitrobenzaldehyde


83

38. (a) (i) CH3 CH 2CH2CH2 OH Butan1ol

CrO 3 H 2SO 4 Jones reagent

(b) (i) CH3 – C O > CH3 – C O > (CH3)3 C – C O CH 3CH2CH 2 COOH Butanoic acid

O C – CH 3

(ii)

oxidation KMnO 4

CH 3

H Acetaldehyde

COOH

HOOC

(b) Refer Ans. 34 (a) (ii). OR (a) (i) Refer Ans. 20 (b) (ii). (ii) Refer Ans. 34 (a) (iii).

JJJ

CH 3 Acetone

CH 3 Methyl tertbutyl ketone

(ii) The overall acid strength increases in the order. 4methoxy benzoic acid < benzoic acid < 4nitro benzoic acid < 3,4dinitrobenzoic acid. (iii) We know that + Ieffect decreases while –Ieffect increases the acid strength of carboxylic acids. The overall acid strength increases in the order. ` (CH 3 ) 2 CHCOOH < CH 3 CH(Br)CH 2 COOH < CH 3 CH 2 CH(Br)COOH.


84


1.

Why is an alkylamine more basic than ammonia? [AI 2009]

C 6 H 5 NH 2 , (C 2 H 5 ) 2 NH, (C 2 H 5 ) 3 N and NH 3 . (ii) Arrange the following in increasing order of basic strength in gas phase : C 2 H 5 NH 2 , (C 2 H 5 ) 2 NH, (C 2 H 5 ) 3 N and CH 3 NH 2 . [AI 2008]

Arrange the following compounds in an increasing order of basic strength of their aqueous solutions : NH 3 , CH 3 NH 2 , (CH 3 ) 2 NH, (CH 3 ) 3 N [AI 2009] 13. Write one chemical reaction each to illustrate the following: (i) Hofmann’s bromamide reaction. 3. Complete the following reaction equation : (ii) Gabriel phthalimide synthesis. [AI 2008] C 6 H 5 N 2 Cl + KI ® .......... [Delhi 2008] 14. Give one chemical test each to distinguish between the 4. Write a chemical reaction in which iodide ion displaces following pairs of compounds : diazonium group from a diazonium salt. [Delhi 2008, AI 2008] (i) Ethyl amine and aniline. 5. Arrange the following compounds in an increasing order of (ii) Aniline and Nmethylaniline. [AI 2009, 2010] their solubility in water. 15. Complete the following reaction equation. C 6 H 5 NH 2 , (C 2 H 5 ) 2 NH, C 2 H 5 NH 2 [Delhi SetI 2011] O 6. Rearrange the following in an increasing order of their basic (i) R C NH LiAlH 4 strengths: 2 H O 2 C 6 H 5 NH 2 , C 6 H 5 N(CH 3 ) 2 , (C 6 H 5 ) 2 NH and CH 3 NH 2 (ii) C 6 H 5 N 2 Cl + H 3 PO 2 + H 2 O [AI SetI 2011] (iii) C 6 H 5 NH 2 + Br 2(aq) [AI 2009] 7. How will you differentiate between aniline and ethylamine? [AI SetIII 2011] 16. Complete the following reaction equations : (i) C 6 H 5 NH 2 + CH 3 COCl 8. Why is an alkylamine more basic than ammonia? (ii) C 2 H 5 NH 2 + C 6 H 5 SO 2 Cl [Foreign SetI 2011] (iii) C 2 H 5 NH 2 + HNO 2 [AI 2009] 9. Arrange the following in the decreasing order of their basic 17. Assign reason for the following : strength in aqueous solutions: (i) Ethylamine is soluble in water whereas aniline is CH 3 NH 2 , (CH 3 ) 2 NH, (CH 3 ) 3 N and NH 3 insoluble. [Delhi SetI 2012] (ii) The pK b of aniline is higher than that of methylamine. SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS) [AI 2009] 2.

18. In the following cases rearrange the compounds as directed : 10. Account for the following observations : (i) In an increasing order of basic strength : (i) pK b value for aniline is more than that for methylamine. C 6 H 5 NH 2 , C 6 H 5 N(CH 3 ) 2 , (C 2 H 5 ) 2 NH and CH 3 NH 2 (ii) Methylamine in water reacts with ferric chloride to (ii) In a decreasing order of basic strength : give a precipitate of ferric hydroxide. Aniline, pnitroaniline and ptoluidine (iii) Aniline does not undergo FriedelCrafts reaction. (iii) In an increasing order of basic strength : [Delhi 2008] C 2 H 5 NH 2 < C 6 H 5 NHCH 3 , (C 2 H 5 ) 2 NH and C 6 H 5 NH 2 . 11. Assign reason for the following : [AI 2010] (i) Amines are less acidic than alcohols of comparable 19. Complete the following chemical equations molecular masses. (i) C 6 H 5 N 2 Cl + C 6 H 5 NH 2 ® (ii) Aliphatic amines are stronger bases than aromatic amines. [AI 2009] (ii) C 6 H 5 N 2 Cl + CH 3 CH 2 OH ® 12. (i) Arrange the following in an increasing order of basic (iii) RNH 2 + CHCl 3 + KOH ® [AI 2010] strength in water : 20. Give the chemical test to distinguish the following pairs of

Amines

85

compounds. (i) Ethylamine and aniline (ii) Aniline and benzylamine.

[AI 2010]

21. Identify A and B each of the following processes : (i) CH CH Cl 3 2

NaCN

A

reduction Ni/H 2

(i) pK b value for aniline is more than that for ethylamine. (ii) Ethylamine is soluble in water whereas aniline is not soluble in water. (iii) Primary amines have higher boiling points than tertiary amines. [AI SetI 2011]

B

25. Illustrate the following reactions giving a chemical equation in each case: (ii) [AI 2010] (i) Gabriel phthalimide synthesis (ii) A coupling reaction 22. (a) Explain why an alkylamine is more basic than ammonia. (iii) Hofmann’s bromamide reaction [Foreign SetI 2011] (b) How would you convert (i) Aniline to nitrobenzene 26. Write chemical equations for the following conversions: (ii) Aniline to iodobenzene? [Delhi SetI 2011] (i) Nitrobenzene to benzoic acid. (ii) Benzyl chloride to 2phenylethanamine. 23. Complete the following chemical equations: (iii) Aniline to benzyl alcohol. [Delhi SetI 2012] NaCN reduction (i) CH3CH2 Cl (A ) ( B ) 27. Describe the following giving the relevant chemical equation Ni/H 2 in each case. (ii) C 6 H 5 N 2 Cl + H 3 PO 2 + H 2 O ® (i) Carbylamine reaction O (ii) Hofmann’s bromamide reaction [AI SetI 2012] LiAlH 4 [Delhi SetIII 2011] 28. Complete the following reaction equations: (iii) R C NH 2 C H NH NaNO /HCl 6 5 2 2 C H NH 6 5 2 A OH – B

H O 2

(i) C 6 H 5 N 2 Cl + H 3 PO 2 + H 2 O ® (ii) C 6 H 5 NH 2 + Br 2(aq) ®

24. State reasons for the following:

[AI SetI 2012]

a n s w e r s

3.

C6 H5 N 2 Cl

Benzene diazonium chloride

+ KI ® C6 H 5 I + KCl + N 2 Iodo Benzene

4. Refer Ans. 3. 5.

Increasing order of solubility of amines in water C 6H < C 2 H 5NH 5NH 2 < (C 2H 5) 2NH 2

6.

(C 6 H 5 ) 2 NH < C 6 H 5 NH 2 < C 6 H 5 N(CH 3 ) 2 < CH 3 NH 2

7.

Aniline being an aromatic primary amine on treatment with HNO 2 [NaNO 2 + HCl (dil.)] at 273–278 K followed by treatment with an alkaline solution of bnaphthol gives an orange coloured azo dye. Ethylamine does not give this test.

8. Refer Ans. 1. 9. Refer Ans. 2. 10. (i) In aniline, the lone pair of the electrons on the N – atom is delocalized over the benzene ring.

As a result, electron density on the nitrogen decreases. In contrast in CH 3 NH 2 +I effect of CH 3 increases the electron density on the Natom. Therefore, aniline is a weaker base than methylamine and hence its pK b value is higher than that of methylamine. (ii) Methylamine being more basic than water, accepts a proton from water liberating OH – ions. CH3 NH 2 + H OH

_

+

:

1. Electron density of Natom increases due to the +I effect of the alkyl group. Hence alkylamines are stronger base than ammonia. NH3, R NH 2 2. The order of basic strength of methyl substituted amines in aqueous solution is : (CH 3 ) 2 NH > CH 3 NH 2 > (CH 3 ) 3 N > NH 3 .

CH3 NH 2 + OH

These OH – ions combine with Fe 3+ ions present in H 2 O to form brown ppt. of hydrated ferric oxide. FeCl3 ® Fe3 + + 3Cl2 Fe3 + + 6 OH - ®

2 Fe(OH) 3

.

Hydrated ferric oxide (Brown ppt.)

(iii) Aniline being a Lewis base reacts with Lewis acid AlCl 3 to form a salt. C 6 H 5 NH 2 + AlCl 3 ® C 6 H 5 N + H 2 AlCl 3 – . As a result, N of aniline acquires positive charge and hence it acts as strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel Crafts reaction. 11. (i) Loss of proton from amine gives amide ion whereas


86

NH 2

loss of a proton from alcohol gives an alkoxide ion. – + RNH2 R NH + H Amine

ROH R O– + H + Alcohol

Alkoxide ion

:

+

NH 2

+

NH 2

NH 2

+

NH 2

Br

+ 3Br 2

Amide ion

Since O is more electronegative than N, therefore, RO – can accommodate the –ve charge more easily than RNH – . Hence amines are less acidic than alcohol. (ii) Due to resonance in aniline, the lone pair of electron on nitrogen gets delocalised over the benzene ring and becomes less available for protonation. In alkyl amine, alkyl group releases electrons and increase density of electron of nitrogen, making it stronger base

NH 2 Br

+ 3HBr

Br Ethyl amine does not react with bromine water (ii) Each compound is warmed separately with few drops of chloroform and alcoholic KOH. Aniline (primary amine) gives offensive smell of isocyanide while N methyl aniline (secondary amine) does not give carbylamine test. C 6H 5NH 2 + CHCl 3 + 3KOH (alc.)

Warm

C 6 H 5 NC

C6H 5NHCH 3 CHCl3 + KOH (alc.) No reaction.

O

:

15. (i) R C NH 2

NH 2

R

NH 2 N

12. (i) The increasing order of basic strength in water of the given amines and ammonia follows the following order. C 6 H 5 NH 2 < NH 3 < (C 2 H 5 ) 3 N < (C 2 H 5 ) 2 NH (ii) The increasing order of basic strength in gas phase of the given amines follows the order. CH 3 NH 2 < C 2 H 5 NH 2 < (C 2 H 5 ) 2 NH < (C 2 H 5 ) 3 N.

LiAlH 4 H O 2

R CH NH 2 2

(ii) C 6 H 5 N 2 Cl + H 3 PO 2 + H 2 O C 6 H 6 + N 2 + H 3 PO 3 + HCl NH 2

(iii) C 6 H 5 NH 2 + Br 2(aq)

Br

Br + 3 HBr Br

13. (i) Hofmann bromamide reaction : Primary amines when 16. (i) C 6 H 5 NH 2 + CH 3 COCl C 6H 5NHCOCH3 + HCl heated with Br 2 and (aqueous or ethanoic solution of) Acetanilide NaOH lose a carbon atom and are converted to the (ii) C 2 H 5 NH 2 + C 6 H 5 SO 2 Cl C 6 H 5 SO 2 NHC 2 H 5 corresponding amines. It is an example of stepdown + – [C 2H 5N 2 Cl ] (iii) C 2H 5NH 2 + HNO 2 reaction. H2 O C 2H 5 OH + N 2 + HCl CH3CONH 2 + Br2 + 4NaOH D 17. (i) Solubility of ethylamine in water is attributed to its Acetamide ability to form hydrogen bonds with water molecules. CH3NH2 + 2NaBr + Na2CO3 + 2H2 O In aniline the non polar hydrocarbon part is relatively (ii) Gabriel phthalimide synthesis : In this reaction larger and has no interaction with polar water molecules. phthalimide is converted into its potassium salt by (ii) Refer Ans. 10 (i). treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide 18. (i) Increasing order of basic strength : to yield an Nalkylpthalimide which is hydrolysed to C 6 H 5 NH 2 < C 6 H 5 N(CH 3 ) 2 < CH 3 NH 2 < (C 2 H 5 ) 2 NH phthalic acid and primary amine by alkaline hydrolysis. (ii) Decreasing order of basic strength : ptoluidine > aniline > pnitroaniline CO CO – + C 2H 5 I KOH (alc.) NK –KI NH –H O (iii) Increasing order of pK b values : 2 CO CO (C 2 H 5 ) 2 NH < C 2 H 5 NH 2 < C 6 H 5 NHCH 3 < C 6 H 5 NH 2 . CO NaOH (aq ) – NC2H 5 (Hydrolysis) 19. (i) C 6 H 5 6N 2 Cl + C 6 6H 5 5NH 2 2 OH CO 5 2 COOH + C2H5NH 2 NH2 + Cl– + H2 O N N COOH Ethyl amine Phthalic acid

This synthesis is very useful for the preparation of pure aliphatic primary amines. However, aromatic primary amines cannot be prepared by this method. 14. (i) Aniline gives white or brown precipitate with bromine water.

(ii) C 6 H 5 N 2 Cl + CH 3 CH 2 OH (iii) R NH 2 + CHCl3 + KOH

+ N 2 + HCl + CH 3 CHO R NC + KCl + H 2O

Amines

87

20. (i) Aniline gives white or brown precipitate with bromine water. NH 2

NH 2 Br

NaNO /HCl 2

Br

+ 3Br 2

+ – N Cl 2

NH 2

+ 3HBr

I + KI

273 – 278 K

(ii)


Aniline

Br Ethyl amine does not react with bromine water

Iodobenzene

23. (i) Refer Ans. 21 (i). (ii) Aniline and benzylamine : Benzylamine reacts with (ii) Refer Ans. 15 (ii). nitrous acid to form a diazonium salt which is unstable (iii) Refer Ans. 15 (i). at low temperature, decomposes with evolution of N 2 24. (i) In aniline, the lone pair of electrons on Natom are gas delocalised over benzene ring due to resonance. As a NaNO 2 + – result, electron density on the nitrogen atom decreases. C H CH NH [C H CH N NCl ] 6 5 2 2 HCl 6 5 2 In contrast, in methylamine, +Ieffect of CH 3 group Benzylamine unstable increases electron density on the nitrogen atom. H O 2 C H CH OH + N + HCl 6 5 2 2 - Therefore, aniline is a weaker base than methylamine, Benzene alcohol hence its pK b value is more than that for methylamine. Aniline reacts with HNO 2 to form benzene diazonium (ii) Ethylamine is soluble in water due to formation of chloride which is stable at 273 278 K and hence does intermolecular hydrogen bonds with water molecules. not evolve N 2 gas. It forms orange dye with 2naphthol. However, in aniline due to large hydrophobic aryl group NaNO 2 , HCl NH 2 N+ NCl – the extent of hydrogen bonding decreases considerably 273 278 K and hence aniline is insoluble in water. Aniline Benzenediazonium (iii) Due to the presence of two Hatoms, primary amines chloride undergo extensive intermolecular hydrogen bonding. NaCN Reduction 21. (i) CH CH Cl CH CH C N Due to the absence of N–H bond t–amine do not undergo 3 2 3 2 Chloroethane propanenitrile Hbonding. As a result, primary amines have higher ( ) A boiling points than tertiary amines. CH3CH2CH2NH 2 propan1amine 25. (i) Refer Ans. 13 (ii). (B ) (ii) Coupling reactions : Diazonium salts react with aromatic amines in weakly (ii) Refer Ans. 19 (i). acidic medium and phenols in weakly alkaline medium 22. (a) Due to electron releasing nature, the alkyl group (R) to form coloured compounds called azo dyes by pushes electrons towards nitrogen in alkyl amine and coupling at pposition of amines or phenols. The thus makes the unshared electron pair more available mechanism is basically that of electrophilic aromatic for sharing with the proton of the acid. Therefore alkyl substitution where the diazonium ion is electrophile. – amines are more basic than ammonia. + OH – NCI + H OH N H H (pH 910) R N

Phenol

+

+

R

H

+

N H H

H + – N Cl 2

NH 2

273 – 278 K Aniline

+ – N BF 2 4


NO 2

N

+ N

+

– NCI + H

NH 2

Cu, D Nitrobenzene

H

(pH 45)

Phenol

N

N

p Amino azobenzene (Yellow dye)

NaNO 2

–

OH + Cl + H O 2

pHydroxyazobenzene (orange dye)

HBF 4

NaNO /HCl 2

(b) (i)

N

–

NH + Cl + H O 2 2


88

(iii) Refer Ans. 13(i). NO 2

CH2 NH 2 NH 2

Sn / HCl

26. (i)

CH2 OH NaNO2 /HCl 273 278 K

LiAlH 4 NaNO 2 / HCl 273 278 K

Benzyl alcohol

Nitrobenzene

CN

+ –

N Cl 2

COOH +

CuCN

H2 O / H

Benzoic acid

CH Cl 2

(ii)

KCN

CH CN 2

2phenylethanamine + – N Cl 2

NaNO /HCl 2 273 278 K

(iii)

Heat

R NH2 + CHCl3 + 3KOH

LiAlH 4

Benzyl Chloride

NH 2

CH CH NH 2 2 2

27. (i) Carbylamine reaction : Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is carbylamine reaction or isocyanide test and is used as a test for primary amines.

CN

R NC + 3KCl + 3H2 O Carbylamine (foul smell)

(ii) Refer Ans. 13(i). 28. (i) Refer Ans. 15 (ii). NH 2

CuCN

(ii) C 6 H 5 NH 2 + Br 2(aq)

Br

Br + 3 HBr

Aniline

Br

JJJ

Biomolecules

89


1.

Describe the following giving example for each. (i) Glycosidic linkage. [AI 2008]

2.

Name the bases present in DNA. Which one of these is not present in RNA? [AI 2009]

3.

What are the expected products of hydrolysis of lactose? [AI 2009]

4.

What is the biological effect of denaturation of protein? [AI 2009]

5.

Explain the following terms. (i) Invert sugar (ii) Polypeptide.

[AI 2010]

6.

What is meant by reducing sugars?

[AI 2010]

7.

Amino acids may be acidic, alkaline or neutral. How does this happen. [AI 2010]

8.

What are monosaccharides?

9.

What are the products of hydrolysis of sucrose?

[AI 2010] [AI 2010]

10. Explain what is meant by (i) a peptide linkage, (ii) a glycosidic linkage? [Delhi SetII 2011] 11. Name the bases present in RNA. Which one of these is not present in DNA? [Delhi SetII 2011] 12. Write the structure of the product obtained when glucose is oxidised with nitric acid. [AI SetI, 2012]

while in alkaline solution they migrate towards the anode. Give reason. [AI 2008] 18. Mention the type of linkage responsible for the formation of the following : (i) Primary structure of protein. (ii) Cross linkage of polypeptide chains. (iii) ahelix formation. (iv) bsheet structure. [AI 2008] 19. When RNA is hydrolysed, there is no relationship among the quantities of different bases formed. What does this fact suggest about the structure of RNA? [AI 2008] 20. List the reaction of glucose which cannot be explained by its open chain structure. [Delhi 2008] 21. What happens when Dglucose is treated with the following reagents? (i) HI (ii) Bromine water (iii) HNO 3 [AI 2008] 22. Name two water soluble vitamins, their sources and the diseases caused due to their deficiency in diet. [AI 2009] 23. Name two fat soluble vitamins, their sources and the diseases caused due to their deficiency in diet? [AI 2009] 24. (a) What is the structural difference between a nucleoside and a nucleotide. (b) The two strands in DNA are not identical but are complementary explain. [AI 2009] 25. What are essential and nonessential amino acids in human food? Give one example of each type. [AI 2010]

13. Write a reaction which shows that all the carbon atoms in glucose are linked in a straight chain. [AI SetIII 2012]

26. Differentiate between fibrous protein and globular protein. What is meant by the denaturation of a protein? [AI 2010]


27. What is essentially the difference between aform and bform of glucose? Explain. [Delhi SetI 2011]

14. Describe in terms of reference of protein : (i) Primary structure (ii) Denaturation.

[AI 2008]

15. Define the following terms in relation to proteins: (i) Peptide linkage (ii) Denaturation. [AI 2008, 09, 10]

28. Describe what you understand by primary structure and secondary structure of proteins? [Delhi SetI 2011] 29. Explain what is meant by the following: (i) peptide linkage (ii) pyranose structure of glucose? [AI SetI 2011, Foreign SetI 2011]

16. How are the vitamins classified? Mention the chief source of vitamins A and C. [AI 2008] 30. Write the main structural difference between DNA and RNA. Of the four bases, name those which are common to both 17. What are the amino acids ? On electrolysis in an acidic DNA and RNA. [AI SetI 2011] solution, the relevant amino ions migrate towards the cathode


90

31. Write such reactions and facts about glucose which can not 35. Write any two reactions of glucose which cannot be explained be explained by open chain structure. [AI SetIII 2011] by the open chain structure of glucose molecule. [Delhi SetI 2012] 32. Name the products of hydrolysis of (i) sucrose and (ii) lactose. 36. Write the main structural difference between DNA and RNA. Of the two bases, thymine and uracil, which one is present in Mention three such properties of glucose which cannot be DNA? [Delhi SetI 2012] explained by its open chain structure. [Foreign SetI 2011] 37. What is essentially the difference between aglucose and 33. State what you understand by primary and secondary structure bglucose? What is meant by pyranose structure of glucose? of proteins. [Foreign SetII 2011] [AI SetI 2012] 34. What is meant by (i) peptide linkage, (ii) biocatalysts? [Delhi SetI 2012] OR

a n s w e r s (i) Glycosidic linkage : The two monosaccharide units are O joined together through an oxide linkage formed by loss of a molecule of H 2 O. Such a linkage between two 10. (i) Peptide linkage : The amide linkage ( C NH monosaccharide units through oxygen atom is called formed between two aamino acid molecules with the glycosidic linkage. loss of a water molecule in a polypeptide chain is called a peptide linkage. 2. DNA contains four bases viz adenine (A) guanine (G), (ii) Refer Ans. 1(i). cytosine (C), and thymine (T). RNA also contains four bases but thymine is not present in RNA. 11. The bases present in RNA are adenine (A), guanine (G), 3. Lactose on hydrolysis gives bD galactose and cytosine (C) and uracil (U). Uracil is not present in DNA bD glucose. 12. On oxidation with nitric acid, D glucose yields saccharic C12 H22 O11 + H2 O C6H12O6 + C6H2O 6 acid. 1.

Lactose

4. 5.

6.

7.

8.

9.

Galactose

Glucose

CHO

COOH

Protein loses its biological activity due to denaturation. conc. HNO

3 (i) Hydrolysis of sucrose brings about a change in a sign (CHOH) 4 (CHOH) 4 of rotation from dextro (+) to laevo (–) and the product is known as invert sugar. CH2 OH COOH (ii) Macro molecules formed by combination of ten or D Glucose D Saccharic acid more amino acids is called polypeptide. 13. Glucose when heated with red P and HI gives nhexane. Carbohydrates which reduce Tollen’s reagent are reducing Red P - HI C 6 H12 O 6 ¾¾¾¾® CH 3 - CH 2 - CH 2 - CH 2 - CH 2 - CH 3 D sugars. All monosaccharide, aldoses or ketones are reducing ( n - hexane) sugars. It indicates presence of straight chain of six carbon atoms in Amino acids are classified as acidic, basic or neutral glucose. depending upon the relative number of amino and carboxyl 14. (i) Proteins may contain one or more polypeptide chains. groups in their molecule. Each polypeptide chain has a large no. of aamino acids (a) Equal number of amino and carboxyl group makes it which are linked to one another in a specific manner. neutral (b) more number of amino groups than carboxyl The specific sequences in which the various aamino groups makes it basic and (c) more carboxyl groups as acids are present in a polypeptide chain of a protein compared to amino groups make it acidic. is called its primary structures. These are the simplest carbohydrates which can not be (ii) Denaturation : The loss of biological activity of a hydrolysed to smaller molecules. Their general formula is protein by changing the pH, temperature or by adding (CH 2 O) n . some salt due to disruption of the native structure of protein is called denaturation. Glucose and fructose. + H During denaturation secondary and tertiary structure of C12 H22 O11 + H2 O C6 H12 O6 + C6 H12 O 6 protein is destroyed but primary structure remains intact. Sucrose Glucose Fructose

Biomolecules

91

15. (i) Peptide linkage : Chemically, a peptide bond is an 20. The following reactions of Dglucose cannot be explained amide linkage formed between –COOH group of one on the basis of its open chain structure : (i) Dglucose does not react with sodium bisulphite aamino acid and NH 2 group of the other aamino acid (NaHSO 3 ). by loss of a water molecules. (ii) It does not give 2, 4DNP test and Schiff’s test. (ii) Denaturation : The loss of biological activity of a (iii) The pentaacetate of Dglucose does not react with protein by changing the pH, temperature or by adding hydroxylamine. some salt due to disruption of the native structure of protein is called denaturation. (iv) Dglucose shows the phenomenon of mutarotation, i.e., During denaturation secondary and tertiary structure of when its aqueous solution is kept for sometime its protein is destroyed but primary structure remains intact. optical activity changes. (v) On reaction with 1 mole of methanol, it yield two 16. Vitamins are classified into two groups depending upon monomethyl derivatives which are known as methyl their solubility in water or fat. aDglucoside and methylbDglucoside. (i) Fat soluble vitamins. (ii) Water soluble vitamins.

21. (i) Refer Ans. 13. (ii) D glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with bromine water. COOH CHO Source of vitamin C : Citrus fruits, amla and green leafy vegetables. Br2 , Water (CHOH) 4 (CHOH) 4 17. Amino acids contain amino –(NH 2 ) and carboxyl (–COOH) functional groups. Depending upon the relative position of CH2 OH CH2 OH amino group with respect to carboxyl group, the amino D Glucose Gluconic acid acids can be classified as a, b, g, d, and so on. (iii) Refer Ans. 12. Sources of vitamin A : Fish, liver oil, carrots, butter and milk.

–

H2 N CH COO

R (II)

+

+

H

–

H3 N CH COO

R

22. Example of water soluble vitamins : B group vitamins and vitamin C.

+

H3 N CH COOH

Name of vitamin Vitamin B

R (I)

Source

Deficiency diseases

Yeast, milk, green Beriberi vegetable, cereals etc. Citrus fruit, amla, Scurvy and leafy (bleeding gums) vegetable.

Vitamin C In acidic medium : aamino acids exist as cation (I) and thus migrate towards cathode under the influence of an electric field. On other hand, in alkaline medium, aamino acids exist as 23. Examples of fat soluble vitamins are vitamins A and D. anions (II) and thus migrate towards anode under the Name of vitamin Source Deficiency diseases influence of an electric field. Vitamin A Fish liver oil, Night blindness 18. carrots, butter, Biomolecule Type of linkage milk (i) Primary structure of Peptide bond (linkage). Vitamin D Fish and egg yolk Rickets and protein. osteomalacia (ii) Cross linkage of Hydrogen bonds polypeptide chain. disulphide linkage, 24. Nucleoside is formed by condensation of a purine or pyrimidine base with pentose sugar at position 1. When electrostatic force of nucleoside is linked to phosphoric acid at position 5 of attraction. sugar moiety we get a nucleotide. (iii) ahelix formation. Hydrogen bond. (i) Nucleoside has two units pentose sugar and a base. (iv) bsheet structure. Intermolecular hydrogen (ii) Nucleotide has three units phosphate group, pentose bonds. sugar and a base. 19. When a RNA molecule is hydrolysed then there is no DNA is a double helix in which the two strands of DNA relationship between the quantities of four bases viz. adenine are held by the hydrogen bonds between the bases on the (A), guanine (G), cytosine (C), and uracil (U). This fact two strands. Thymine (T) pairs with adenine through two suggests that RNA has single stranded structure


92

hydrogen bonds and cytosine (C) pairs with guanine (G) 28. Primary Structure : The specific sequence in which the through three H–bonds. Hence the two strands of DNA are various aamino acids present in a protein are linked to complementary to each other. one another is called its primary structure. Any change in the primary structure creates a different protein. 25. Amino acids which cannot be synthesised in the body and must be obtained through diet are known as essential amino Secondary Structure : The conformation of the acids, e.g. valine and leucine. There are ten essential amino polypeptide chain is known as secondary structure. The acids. Amino acids which can be synthesised in the body two types of secondary structure are ahelix and bpleated are known as nonessential amino acids, e.g., alanine and sheet structure. glutamic acids. In ahelix structure, the polypeptide chain forms all the 26. Difference between fibrous and globular protein : Characteristic differences between globular and fibrous proteins can be given as : Globular proteins 1. These are cross linked proteins and are condensation product of acidic and basic amino acids. 2. These are soluble in water, mineral acids and bases. 3. These proteins have three dimensional folded structure. These are stabilised by internal hydrogen bonding. e.g., egg albumin, enzymes.

Fibrous proteins These are linear condensation polymer.

These are insoluble in water but soluble in strong acids and bases. These are linear polymers held together by intermolecular hydrogen bonds. e.g., hair, silk.

Denaturation of protein : The loss of biological activity of a protein by changing the pH, temperature or by adding some salt due to disruption of the native structure of protein is called denaturation. During denaturation secondary and tertiary structure of protein is destroyed but primary structure remains intact.

possible hydrogen bonds by twisting into a right handed screw (helix) with the –NH groups of each amino acid residue hydrogen bonded to the

C O group of an

adjacent turn of the helix. In bpleated sheet structure, all peptide chains are stretched out nearly maximum extension and then laid side by side which are held together by inter molecular hydrogen bonds. 29. (i) Refer Ans. 15(i). (ii) The six membered cyclic structure of glucose is called pyranose structure (a–or b–), in analogy with heterocylic compound pyran. CH2 OH H H H OH OH HO OH H aD(+)Glucopyranose 30. Structural difference between DNA and RNA DNA RNA 1. The sugar present in DNA 1. The sugar present in RNA is D(–) ribose. is 2deoxy D(–) ribose.

2. DNA has double stranded 2. RNA has single stranded ahelix structure. ahelix structure. 3. DNA contain thymine base 3. RNA contain uracil base. Common base in DNA and RNA are adenine, guanine and cytosine.

27. In aD Glucose, the –OH group at C 1 is toward right 31. The following facts and reactions can not be explained whereas in bglucose, the –OH group at C 1 is towards left. by open chain structure of glucose. Such a pair of stereoisomers which differ in the (i) Despite having the aldehyde group, glucose does not configuration only at C 1 are called anomers. give 2, 4DNP test, Schiff’s test and it does not form the hydrogen sulphite addition product with NaHSO 3 . 1 HO C H H 1 C OH (ii) The pentaacetate of glucose does not react with 2 hydroxylamine indicating the absence of free aldehydic H C OH H 2 C OH group. 3 3 HO C H (iii) Glucose shows mutarotation HO C H O O 4 4 32. (i) Refer Ans. 9. H C OH H C OH (ii) Refer Ans. 3. 5 5 H C H C OR 6 CH OH 2 Refer Ans. 31. CH2 OH a – D – (+) – Glucose

bD(+)Glucose

33. Refer Ans. 28.

Biomolecules

93

37. The only difference between a and bglucose is in the 34. (i) Refer Ans. 15 (i). configuration at C 1. In aglucose – OH group at C 1 is on the (ii) Substances which catalyses chemical reactions taking right whereas in bglucose – OH group at C 1 is in the left hand place in living organisms are called biocatalysts. e.g. side. enzymes CH2 OH

35. Refer Ans. 31. 36. Differences between DNA and RNA. DNA 1. It usually occurs, inside nucleus and some cell organelles. 2. DNA is the genetic material.

3. It is double stranded with ahelix structure in which two strands are coiled spirally. 4. It contains deoxyribose sugar. 5. Nitrogen base thymine occurs in DNA along with three other adenine, cytosine and guanine.

HO H

RNA Very little RNA occurs inside nucleus. Most of it is found in the cytoplasm. RNA is not the genetic material except in certain viruses, e.g., TMV, Reovirus. RNA is single stranded with the exception of some viruses (e.g., double stranded in Reovirus). It contains ribose sugar. Thymine is replaced by uracil in RNA. The other three are adenine, cytosine and guanine.

JJJ

H H OH

O

CH2 OH OH

H H OH

bglucose

HO H H H OH

O

H

H OH OH

aglucose

Pyranose structure of glucose means it has a six membered ring including oxygen like pyran.


94

Chapter15 Polymers


1.

Define the term polymerisation.

2.

Name the monomers of nylon2nylon6.

3.

Give an example of elastomer.

4.

What are biodegradable polymers?

5. 6.

16. What is difference between elastomers and fibres. Give one example each. [Delhi 2009]

[Delhi 2008]

17. Differentiate between molecular structure and behaviours of thermoplastic and thermosetting polymers. Give one example of each type. [AI 2009] [Delhi 2009] [AI 2008]

[Delhi SetI 2011] 18. Draw the structures of the monomers of the following polymers. Indicate the type of polymerisation of each polymer. In nylon 6,6 what does the designation ‘6,6’ mean? (i) Teflon, Polythene, Terylene, Bakelite. [Delhi SetIII 2011] [Delhi AI 2009] Define the term, ‘homopolymerisation’ giving an example. (ii) Neoprene, BunaS, PVC, Nylon6. [Delhi SetI 2012] [Delhi AI 2010

19. Give one example each of (i) Addition polymers 7. Distinguish between homopolymers and copolymers and give (ii) Condensation polymers an examples of each class? [Delhi 2008] (iii) Copolymers. [Delhi 2010] 8. Write the names and structures of the monomers of the 20. Mention two important uses of each of the following: following polymers: (i) Bakelite (i) BunaS (ii) Neoprene (ii) Nylon 6 [Delhi SetI 2011] (iii) Nylon6. [Delhi 2008] 21. Name the subgroup into which polymers are classified on 9. What are thermoplastic and thermosetting polymers? Give the basis of magnitude of intermolecular forces. one example of each. [AI 2008, Delhi 2010] [Delhi SetIII 2011] 10. Differentiate the following polymers based on the property 22. Draw the structure of the monomers of the following mentioned against each polymers: (i) Novolac and Bakelite (structure) (i) Polythene (ii) PVC (ii) BunaS and Terylene (intermolecular force) (iii) Teflon [AI SetI 2011] [AI 2008] 23. Write the names and structures of the monomers of the 11. Write names of monomers of the following polymers and following polymers : classify them as addition or condensation polymers. (i) BunaS (ii) Dacron (a) Teflon (b) Bakelite (iii) Neoprene [AI SetIII 2011] (c) Natural rubber. [AI 2008] 24. Mention two important uses for each of the following 12. (a) What is the role of benzoyl peroxide in the polymers: polymerisation of ethene? (i) Bakelite (ii) Nylon 6,6 (b) What are LDPE and HDPE? How are they prepared? (iii) PVC [Foreign SetI 2011] [AI 2008] 25. (a) What does the designation ‘6,6’ in nylon 6, 6 polymer 13. What is step growth polymerisation? Explain the steps mean? involved in this process. [AI 2008 C] (b) Which polymer is obtained when free radical polymerisation of chloroprene occurs? Write the 14. What is the difference between the two notations structure of polymer thus obtained. nylon6 and nylon6,6? [Delhi 2009, AI 2009] [Foreign SetII 2011] 15. What are biodegradable polymers? Give two examples. [Delhi 2009 C] 26. What are addition polymers? How are the two types of SHORT ANSWER TYPE QUESTIONS (2 OR 3 MARKS)

Polymers

95

addition polymers different from each other? Give one example of each type. [Foreign SetIII 2011]

polymers. Give one example of each. [AI SetI 2012]

27. Draw the structure of the monomer for each of the following 29. Explain the following terms giving a suitable example for polymers. each: (i) Nylon6 (ii) Polypropene (i) Elastomers (ii) Condensation polymers (iii) Addition polymers [AI SetII 2012] [Delhi SetI 2012] 28. Differentiate between thermoplastic and thermosetting

a n s w e r s 1. Polymerisation : It is a process of formation of a big molecule from small molecules by linking together a through covalent bonds. e.g. Formation of polythene from ethylene.

(ii) Neoprene : CH C CH CH 2 2

2. Monomer of nylon2nylon6 Glycine and aminocaproic acid. H 2 N CH 2 COOH, H 2 N (CH 2 )5 COOH

(iii) Nylon6 : HOOC (CH ) NH . 2 5 2

Cl

Chloroprene

Amino caproic acid

9. Thermoplastic : Thermoplastics are linear polymers which can be repeatedly softened on heating and hardened on 4. Polymers which are degraded in biological system by cooling and hence can be used again and again without any enzymatic hydrolysis or by oxidation are called change in chemical composition and mechanical strength. biodegradable polymers. An example of biodegradable Example : Polythene and polypropene. polymer is PHBV, i.e., polybhydroxy butyratecob Thermosetting : Thermosetting polymers on heating in a hydroxy valerate. mould get hardened and set and cannot be softned again. This 3. BunaS.

5. In nylon 6, 6, designation ‘6, 6’ mean that both the monomers hexamethylenediamine and adipic acid contain six carbon atoms each.

hardening on heating is due to crosslinking between different polymer chains to give a three dimensional network solid. Example : Bakelite.

6. Polymerisation of only one type of monomer molecule is 10. (i) Novolac is a linear polymer and Bakelite is a cross called homopolymerisation. linked polymer of phenol and formaldehyde. (ii) BunaS is elastomer in which polymer chains are held e.g. nCH 2 CH 2 Polymerisation – CH ( ) n 2 – CH 2 – together by the weakest intermolecular forces. Polythene Ethylene Terylene is a fibre in which polymer chains are held together by strong intermolecular forces like hydrogen 7. Homopolymer : The addition polymers formed by the bonding. polymerisation of a single compound are called homopolymers. e.g. polyethene. ).It 11. (a) Monomers of teflon :Tetrafluroethylene ( CF CF 2 2 nCH 2 CH 2 ( CH CH ) is addition polymer. 2 2 n Polyethene

Ethene

Copolymer : A polymer made by polymerisation of two or more different monomers is called copolymer. When styrene and butadiene are polymerised together, a polymer called styrenebutadiene rubber is formed. CH CH 2

(HCHO).It is condensation polymer. (c) Natural rubber : Monomer isoprene CH C CH CH 2 2 2Methyl buta1,3diene

Styrene

[ CH CH CH CH CH CH ] 2 2 2 n Styrene Butadiene Rubber

8. (i) BunaS : CH CH CH CH and C H CH CH 6 5 2 2 2 Butadiene

OH and formaldehyde

CH 3

nCH CH CH CH + 2 2 2 2 1,3Butadiene

(b) Bakelite : Phenol

Styrene

It is addition polymer. 12. (a) Role of benzoyl peroxide : Benzoyl peroxide produces free radical which acts as chain initiator in the polymerisation of ethene to polythene. (b) LDPE (Low density polythene) : It is a branched chain polymer. Preparation : Polymerisation of ethene at high


96

temperature (350 570 K) and pressure (1000 2000 19. (i) Addition polymers : Polythene, rubber. (ii) Condensation polymers : Terylene, Nylon66. atm) in the presence of a peroxide initiator. (iii) Copolymers : SBR, BunaN. HDPE (High density polythene) : It is a straight chain polymer. 20. (i) Bakelite is used for making combs, electrical switches, Preparation : It is prepared by heating ethylene in a handles of various utensils and phonograph records. hydrocarbon solvent in the presence of ZieglerNatta (ii) Nylon 6 is used for making tyre cords, ropes and fabrics. catalyst at a temperature of 333 to 343 K and under 21. On the basis of magnitude of intermolecular forces, a pressure of 67 atmosphere. polymers are classified into following four subgroups. 13. The process in which each step of polymerisation is (i) Elastomers (ii) Fibres independent of other is called step growth polymerisation. (iii) Thermoplastic polymers and All condensation polymerisation are step growth (iv) Thermosetting polymers polymerisation. 22. Name of Name of Structure Step : It involves condensation reaction of bifunctional polymer monomer molecules with elimination. (i) Polythene Ethene CH 2 CH 2 O O (ii) PVC Vinyl chloride CH 2 CH Cl (iii) Teflon Tetrafluoroethene F 2C CF 2 H N (CH ) NH + HO C (CH ) C OH 2

2 6

2

2 4

O

O

H N (CH ) NH C (CH ) C OH + H O 2 2 6 2 4 2 14. Nylon6 is obtained from the monomer caprolactum which contains 6 carbon atoms. Nylon66 is a condensation polymer of hexamethylene diamine and adipic acid both the monomer have 6 carbon atoms each.

15. Biodegradable polymers : Refer Ans. 4. Examples : (i) PHBV : It is a copolymer of bhydroxy butyric acid and bhydroxy valeric acid. (ii) PGPL : Polyglycolic acid and poly lactic acid. 16. (i) In elastomers polymer chains are held together by weakest intermolecular forces. These have elastic properties. e.g. BunaN, BunaS. (ii) In fibres polymer chains are held together by strong intermolecular forces like hydrogen bonding. These have high tensile strength. e.g. Terylene, Nylon6,6, etc. . 18. (i) Teflon : CF2 CF 2 : Addition polymer. Polythene : CH 2 CH 2 : Addition polymer. . Terylene : HO CH2 CH2 – OH + HO C

O C OH :

Condensation polymer. Bakelite : HCHO +

Ethylene glycol

COOH

COOH Terephthalic acid

(iii) Refer Ans. 8 (ii). 24. (i) It is used for making combs, fountain pen barrels, phonograph records, making electrical goods such as switches, plugs, handles of various utensils. (ii) It is used in making bristles for brushes, ropes, making carpets and fabrics in textile industry. (iii) It is used in the manufacture of rain coats, hand bags, water pipes, vinyl flooring, insulating electric wires. 25. (a) Refer Ans. 5. (b) Neoprene is obtained by free radical polymerisation of chloroprene.

17. Refer Ans. 9.

O

23. (i) Refer Ans. 8 (i). (ii) Dacron : HO CH2 CH2 OH and

CH2 C CH CH 2

Peroxide or O 2

Cl

Chloroprene

[CH 2 C CH CH 2 ] n

. OH : Condensation polymer.

(ii) Neoprene : CH2 C CH CH 2 : Addition polymer. .

Cl Neoprene (Polymer)

26. Polymers which are formed by the repeated addition reaction of unsaturated monomer molecules are called the BunaS : CH2 CH CH CH 2 + CH CH 2 : addition polymers. Addition polymer. The two types of addition polymers are: PVC : CH 2 CHCl : Addition polymer. . (i) Homopolymers : (Refer Ans. 7). Nylon6 : H 2N (CH 2) 5 COOH : Condensation polymer. . Cl

Polymers

97

Example : Terylene (dacron), nylon 6,6, nylon 6, is formed by the condensation of hexamethylene diamine and adipic acid. O O

(ii) Copolymers : (Refer Ans. 7). 27. (i) Nylon6 : Caprolactum NH CH 2

C O

H N (CH ) NH + HO C (CH ) C OH 2 2 6 2 2 4

CH 2 CH 2 CH2 CH 2

(ii) Propene : CH 3 – CH

O

CH 2

O

HN (CH ) NH C (CH ) C 2 6 2 4

28. Refer Ans. 9.

Nylon 6,6

29. (i) Elastomers are the polymers in which polymer chains are held by weakest intermolecular force e.g. BunaS, BunaN. (ii) Condensation polymers : The polymers which are formed by repeated condensation reaction between bifunctional monomer units are called condensation polymers.

JJJ

n

(iii) Polymers formed by addition reaction of monomers having double or triple bond e.g. n CH CH ( CH CH ) 2 2 2 2 n Ethene

Polythene


98

Chapter16 Chemistry in Everyday Life

16. What are antibiotics ? Distinguish between narrow spectrum and broad spectrum antibiotics. Classify the following into bactericidal and bacteriostatic antibiotics. [AI 2008] (i) tetracycline (ii) penicillin [AI 2008] [AI 2008] 17. How do antiseptics differ from disinfectant? Give one


1.

Describe the following with examples : (i) Analgesics

2.

Hormones are chemical messengers. Explain.

3.

example of each type. [AI 2009] Name a substance that can be used as an antiseptic as well as a disinfectant. [Delhi 2008 C] 18. What are the following substances? Give one example of each type. Explain these substances with suitable example. (i) Antacid (ii) Nonionic detergents. Food preservatives. [AI 2009, AI SetII 2012] (iii) Antiseptic. [AI 2009] Describe the terms : Antifertility drugs. [AI 2010] 19. Explain the following terms with one example in each case: What is meant by a ‘broad spectrum antibiotic’? (i) Food preservatives (ii) Enzymes [Foreign SetI 2011] (iii) Detergents. [AI 2010] What are antiseptics? Give an example. 20. Write the preparation of anionic detergents. Write their two [Foreign SetII 2011] main uses. [AI 2010] Differentiate between disinfectants and antiseptics. 21. Explain the following terms with an example for each: [AI SetI 2012] (i) Antibiotics (ii) Antiseptics What is the cause of a feeling of depression in human beings? (iii) Analgesics. [AI 2010] Name a drug which can be useful in treating this depression. 22. What are analgesic medicines? How are they classified and [AI SetIII 2012] when are they commonly recommended for use?

4. 5. 6. 7. 8. 9.


[AI 2010]

10. What are antibiotics? Name any two antibiotics which are 23. Determine the following giving one example for each: (i) Detergents specific for certain diseases. [Delhi 2008] (ii) Food preservatives 11. Define the following and give one example of each. (iii) Antacids [Delhi SetI 2011] (a) Antipyretics 24. Explain the following terms with one suitable example for (b) Antibiotics [Delhi 2008] each : 12. Give the structures of (i) A sweetening agent for diabetic patients (i) Aspirin (ii) Paracetamol (ii) Enzymes (iii) Chloroxylenol [AI 2008] (iii) Analgesics [Delhi SetII 2011] 13. What are biodegradable and nonbiodegradable detergents? 25. Answer the following questions : Gives one example of each class. [Delhi 2008] (i) Why do soaps not work in hard water? (ii) What are the main constituents of dettol? 14. Explain the following terms with suitable example. (iii) How do antiseptics differ from disinfectants? (i) Cationic detergents. [Delhi SetIII 2011] (ii) Tranquillizers. [AI 2008, 2009] 15. (a) What are antihistamines and how do they act? Give an 26. What are following substances? Give one example of each. (i) Food preservatives (ii) Synthetic detergents example. (iii) Antacids [AI SetI 2011] (b) Mention the name of a substance which can be used both as an antiseptic as well as a disinfectant. 27. (a) What are disinfectants and antiseptics? Give one [Delhi 2008, 2009] example of each.

Chemistry in Everyday Life

(b) What is the tincture of iodine?

99

[AI SetII 2011]

(i) Tranquillizers (iii) Synthetic detergents

(ii) Food preservatives [Delhi SetI 2012]

28. State the reason in each of the following cases: 30. Explain the following terms giving one example of each type. (i) Soaps do not work well in hard water. (ii) Synthetic detergents are better than soaps. (i) Antacids, (ii) Disinfectants, (iii) Enzymes. [Delhi SetIII 2012] [Foreign SetI 2011] 29. What are the following substances? Give one example of each 31. Explain the cleaning action of soap. Why do soaps not work in hard water? [AI SetII 2012] one of them?

a n s w e r s 1. Analgesics : The medicines which are used to relieve pains 11. (a) Antipyretics : The medicines taken to lower the body are known as analgesics. temperature in high fever are called antipyretics. e.g. Aspirin, novalgin, butazolidine, etc. e.g. Crocin, aspirin, paracetamol, etc. (b) Refer Ans. 10. 2. Hormones are molecules that transfer information from one group of cells to distant tissue or organ and thus 12. (i) Aspirin : control the metabolism. So they act as chemical messengers. 3. Phenol : 0.2% solution of phenol acts as antiseptic where as 1% solution acts as disinfectant. 4. Preservatives are the substances used to prevent spoilage of food due to microbial growth during storage. The most common preservatives used are sodium benzoate, sodium metabisulphite etc. Salt, sugar and oil also act as food preservatives and have been used for many years. 5. Drugs which are used to control birth rate are called antifertility drugs. e.g. Norethindrove, novestrol.

OCOCH 3 COOH OH

(ii) Paracetamol : NHCOCH 3

CH 3 Cl

(iii) Chloroxylenol : 6. The antibiotic which is effective against a wide range of H3 C OH gram positive and gram negative becteria is known as broad spectrum antibiotic. For example, chloramphenicol. 13. Biodegradable detergents : Detergents having straight chain hydrocarbons are easily decomposed by micro organisms. 7. Chemical substances which prevent the growth of These are called biodegradable detergents. microorganisms or kill them but are not harmful to living tissues are called antiseptics. For example, Dettol, Savlon, etc.

Example : Sodium 4 (1, 3, 5, 7 Tetramethylocyl) benzene sulphonate. Nonbiodegradable detergents : Detergents having highly branched hydrocarbon chains are not degraded by bacteria easily. These are called nonbiodegradable detergents. Example : Trimethyl stearyl ammonium chloride.

8. Antiseptics are applied to the living tissues such as wounds, cuts etc. Disinfectants are applied to inanimate things like floors, drainage systems etc. (i) Furacine, soframicine are antiseptics. (ii) Chlorine in aqueous solution and sulphur dioxide are 14. (i) Cationic detergents which are mostly acetates or chlorides of quaternary amines e.g., Cetyltrimethyl disinfectants. ammonium chloride. 9. Low level of noradrenaline is the cause of depression of (ii) Tranquillizers : They act on the central nervous system human being. and help in reducing anxiety. They help in treatment Equanil is used for treatment. of mental diseases and in the removal of emotional 10. Antibiotics : The medicines which are prepared from one distress e.g Equanil. type of microorganisms and are used to kill or prevent the 15. (a) The drug which prevents the interaction of histamine growth of some other type of microorganisms are called with the receptors present in the stomach wall are antibiotics. These are fungi, bacteria etc. e.g., called antihistamines. They control release of acid in (i) Dysidazirine Toxic towards cancer cells. stomach. e.g. Synthetic drug brompheniramine (ii) Ampicillin is used to cure typhoid, pneumonia, soar (Dimetaple) and terfenadine (Seldane) acts as throat.


100

antihistamines. (b) Refer Ans. 3.

(ii) Refer Ans. 7. (iii) Refer Ans. 1.

16. Antibiotics : These are the chemical substances, which are 22. Analgesic medicines : The chemical substances which are partly produced as metabolic products by some specific used to relieve pain are called analgesic medicines. organisms (bacteria, fungi) and can destroy some other There are two types of analgesic medicines. microorganisms. e.g., Penicillin. (i) Non narcotic drugs : These are effective in relieving skeletal pain, preventing heart attack and viral Broad spectrum antibiotics are those antibiotics which kill inflammation etc. or inhibit a wide range of gram positive and gram negative (ii) Narcotic drugs : These are recommended for the relief bacteria. On the other hand, narrow spectrum antibiotics in postoperative pain, cardiac pain, terminal cancer, are those antibiotics which kill or stop the growth of gram and child birth. These are morphine derivatives and positive or negative bacteria. habit forming. Penicillin is a narrow spectrum bactericidal and tetracycline 23. (i) Refer Ans. 19 (iii). is a broad spectrum bacteriostatic antibiotic. Example : Sodium dodecylbenzene sulphonate. 17. Antiseptics: These are the chemical substances used to (ii) Refer Ans. 4. prevent the growth of microorganisms or to kill them and (iii) Refer Ans. 18 (i). are safe to be applied to living tissues. These are applied to wounds and cuts. For examples (i) iodoform (ii) tincture of iodine. Disinfectants : These are chemical substances used to kill microorganisms in inanimate materials and are not safe to be applied to the living tissues. Examples (i) 1% solution of phenol. (ii) Solution of o–, m– and pcresols in soapy water is called lysol and is used as a disinfectant. 18. (i) Antacids are those substance which neutralise the excess acid and bring the pH to appropriate level in stomach. e.g., Magnesium hydroxide and sodium bicarbonate. (ii) Nonionic detergents are esters of long chain fatty acids with long chain alcohols. e.g., CH 3 (CH 2 ) 16 COO(CH 2 CH 2 O) n CH 2 CH 2 OH Ester of polyethylene glycol with stearic acid (iii) Refer Ans. 17.

24. (i) Artificial sweetening agents are chemical substances which are sweet in taste but do not add calories to our body. (ii) Refer Ans. 19 (ii). (iii) Refer Ans. 1. 25. (i) Hard water contains Mg 2+ and Ca 2+ ions which form insoluble salt with soap. That is why soap does not work in hard water. (ii) The main constituents of dettol are terpineol and chloroxylenol. (iii) Refer Ans. 8. 26. (i) Refer Ans. 4. (ii) Refer Ans. 19 (iii). (iii) Refer Ans. 18 (i). 27. (i) Refer Ans. 8. (ii) A 23% solution of iodine in alcohol water mixture is known as tincture of iodine. It is used as an antiseptic.

19. (i) Refer Ans. 4. (ii) Enzymes : The proteins which perform the role of biological catalysts in the body are called enzymes e.g., 28. (i) Refer Ans. 25 (i). (ii) Synthetic detergents can be used in hard water as well Trypsin, pepsin. as in acidic solutions. Hence those are better than soaps. (iii) Detergents : Chemical substances which help in removal of fats which bind dirty materials to the fabric or skin 29. (i) Refer Ans. 14 (ii). (ii) Refer Ans. 4. and improve cleansing properties of water are called (iii) Refer Ans. 19 (iii). detergents. 20. Preparation : Alkyl hydrogen sulphate formed by treating 30. (i) Refer Ans. 18 (i). (ii) Disinfectants are the chemicals which either kill or long chain alcohols with concentrated sulphuric acid are prevent the growth of microorganisms but are not safe neutralised with alkali to form anionic detergents. for the living tissues. Disinfectants are applied to H 2SO 4 CH3(CH2 )10 CH2 OH CH3(CH2)10CH2OSO3 H inanimate objects like floors, drains etc. Alkyl hydrogen sulphate e.g. (a)Aqueous solution of chlorine in the concentration 0.2 to 0.4 ppm is disinfectant. NaOH (aq ) – + (b) Sulphur dioxide in very low concentration is CH 3(CH 2)10CH 2OSO 3 Na disinfectant. Some substances can act as an antiseptics Anionic detergents at low concentration and as disinfectant at high Uses : These are mostly used (i) for household work (ii) concentration. in tooth paste. 21. (i) Refer Ans. 10.

Chemistry in Everyday Life

101

e.g. 0.2 percent solution of phenol is an antiseptic while its one percent solution is disinfectant. OH

CH 3

CH 3

H3 C

H3 C

Cl Chloroxylenol

Cl

OH CH 3

Terpineol

OH OH Cl S

Cl

Cl Bithional

Stearate ion Water

C

O CO

OO

C OO

(iii) Refer Ans. 19 (ii). 31. Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOO – Na + (e.g., sodium stearate CH 3 (CH 2 ) 16 COO – Na + . When dissolved in water, it dissociates into RCOO – and Na + ions. The RCOO – ions, however, consist of two parts a long hydrocarbon chain R(also called nonpolar ‘tail’) which is hydrophobic (water repelling), and a polar group COO – (also called polar ionic ‘head’), which is hydrophilic (water loving).

OOC

OO C

COO CO O

O

(a)

CO

OO

C

Ionic micelle

O CO

(b)

JJJ

The RCOO – ions are, present on the surface with their COO – groups in water and the hydrocarbon chains R staying away from it. At critical micelle concentration, the anions are pulled into the bulk of the solution of aggregate to form a spherical shape. An aggregate thus formed is known as ‘ionic micelle’. These micelles may contain as many as 100 such ions. The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part is in the oil droplet and hydrophilic part interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats. The negatively charged sheath around the globules prevents them from coming together and forming aggregates.

(a)

(b)

(c) Hard water contains calcium and magnesium ions. These ions form insoluble calcium and magnesium salts when sodium or potassium soaps are dissolved in hard water. These insoluble soaps separate as scum in water and are useless as cleansing agent.

Chemistry 5 Years Class12

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