C4 Periodic Properties-1

INDEX S.No

Name of the sub-topic

Do you know

?

Page Number

1.

Periodic properties

Fluorine Borazole

7-8

2.

Atomic Radii

Chlorine Caesium

9 - 10

3.

Periodicity of Atomic Radii

Noble Gases Vant Hoff

11 - 12

4.

Ionization Energy

Oxygen H1 1 Diamond

13 - 15

Ru & Os Silver D 2O

16 - 18

5.

Electron Affinity

6.

Electronegativity

Phosphorus Gold & Platinum Lucas Reagent

7.

Factors affecting Electronegativity

Antichlor Alumina

22 - 23

8.

Electropositivity

Anti-freezing agent Haber's Process

24 - 25

9.

Oxidising and Reducing property

Arsenic glass Caro's acid

26 - 27

10.

Valency

Asbestos Aspirin

28 - 29

11.

Oxidation numbers

Haematite Bitter salt

30 - 31

5

19 - 21

INDEX S.No

Name of the sub-topic

Do you know

?

Page Number

12

Diagonal relationship and chemical reactivity

Black ash Blue stone

32 -33

13.

General tendency of periodic properties

Metals

34

14.

Orange worksheets 1 to 5

Bone ash Marble Blue lead Burnt lime

35 -38

15.

16.

17.

Yellow worksheets 1 to 5

Diaspore Mohr's Salt Contact process

Blue worksheet 1 to 3

Polymorphism Hydrolith Sodium

Previous EAMCET, AIEEE and IIT Questions

6

Fluorspar Ferro prussiate

39 - 41

42 - 44 45 - 46

GOAL– 1

SYNOPSIS

PERIODIC PROPERTIES INTRODUCTION The properties which are directly or indirectly dependent on the electronic configuration of the elements are called atomic properties or periodic properties.The meaning of the word periodic is that a particular property is repeated in a system at regular intervals.In modern periodic table,these intervals are 2,8,8,18,18 and 32 i.e.,similar properties are observed with elements belonging to the same group which have been arranged in groups after the difference of either 2 or 8 or 18 or 32 in atomic numbers as similar valence shell electronic configurations recur after certain regular intervals of atomic number.This is the cause of periodicity in atomic properties. Modern periodic law : According to Modern periodic law “ the physical and chemical properties of the elements are the periodic functions of their atomic numbers or the electronic configuration”. Properties such as Atomic radii, Ionization energy, Electron gain enthalpy, Electronegativity, Electropositivity, Valency, Oxidation number etc.,shows periodicity. To understand periodic properties in a better manner we should learn the following terms. 1)

The screening effect or shielding effect : In a multi electron atom, the electrons of the valence shell are attracted towards the nucleus and also

these electrons are repelled by the electrons present in the inner shells. On account of this, the actual force of attraction between the nucleus and the valence electrons is some what decreased by the repulsive force acting in opposite direction. This decrease in the force of attraction exerted by the nucleus on the valence electrons due to the presence of electrons in the inner shells, is called Screening effect or Shielding effect. The magnitude of the screening effect depends upon the number of inner electrons, i.e., higher the number of inner electrons, greater shall be the value of screening effect. 2)

Effective nuclear charge : Due to screening effect the valence electron experiences less attraction towards nucleus. This brings

decrease in the nuclear charge. The resultant nuclear charge is termed as effective nuclear charge. It is represented by Z*.

Do you know ? Highest electronegative element is Fluorine (F) 7

GOAL– 1

ASSIGNMENT

Multiple choice questions : 1. Elements belonging to same group of periodic table have generally the same a) Electronic configuration b) Number of electrons in outermost shell c) Chemical properties d) Physical properties

[

]

[

]

2.

Among the following, the element with greater value of screening effect is a) Li b) Na c) Cs d) K

3.

According to modern periodic law, the physical and chemical properties of the elements are the periodic functions of their [ ] a) group number b) period number c) mass number d) electronic configuration

4.

The group number of element in periodic table indicates: a) Valency with respect to hydrogen b) The atomicity d) Oxidation number c) The number of electrons in the outermost shell

[

]

5.

According to modern periodic law, variations in the properties of elements is related to their [ a) Atomic weights b) Nuclear weights c) Atomic numbers d) Neutron to proton ratio

]

Test your self (Matrix Matching) __________________________________________________________________________________________ This question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B– q and r; C – p and q; and D – s and ; then the correct darkening of bubbles will look like the following. p q r s t

Column-I A) Na B) Cl C) Al D) O

A B C D

q

r

p r p

q

s

t

s

t

r

p) q) r) s) t)

Column-II IA Group 3rd period P – block Non–metal 2nd Period

p A B C D

Do you know ? Formula of Borazole is B3N3H6 8

q

r

s

t

GOAL– 2

SYNOPSIS

ATOMIC RADII Atomic radii defined as the distance between the centre of the nucleus and the outer most shell where electrons are present. It is expressed in nano metre (1nm = 10-9m) or in Angstrom units (1A0 = 10-10m ) or in pico metre (1pm = 10 -12m) Various names have been proposed for atomic radii depending upon their nature of the atoms. They are : a) Covalent Radii b) van der Waals’ radii c) Crystal Radii (or) Metallic Radii a) Covalent Radii : In a homo nuclear di - atomic molecule it is one half of the distance between the centre of the two nucleii bonded by a single covalent bond. Bond length dA - A = rA + rA = 2 rA A

A

↔ covalent radii Homo nuclear di - atomic molecule

In a hetero nuclear diatomic molecule of AB type, if the bonding is purely covalent, then the bond length dA-B is equal to the sum of the covalent radii of A and B. dA-B = rA + r B A

B

↔ covalent radii Hetero nuclear diatomic molecule

In case of molecules with one element with greater electronegativity than other, then the bond length dA-B is calculated as dA-B = rA + rB - 0.09( c A- c B) Where c A and c B are the electronegativities of the elements A and B. *Covalent radius is used for non metals. b)

A

A

B

B

van der Waals’ radii : It is one half of the inter neuclear distance between the two atoms of A and B, in the two neighbouring molecules of the substance in the solid state.

NOTE : van der Waals’ radii of an element is always larger than its covalent radii.

↔ van der Waals’ radii

Do you know ? Highest electron affinity element is chlorine (Cl) 9

c)

Crystal radii : It is also called metallic radii. It is defined as “half of the distance between the nucleii of two adjacent metal atoms in the metallic closed packed crystal lattice”.It is used for metal atoms which are assumed to be closely packed spheres in the metallic crystal.

NOTE : Metallic radii is always larger than the covalent radii. Comparision of three radii of an element : van der Waals’ radii > Metallic (crystal) radii > Covalent radii

↔

Crystal radii

GOAL– 2

ASSIGNMENT

Multiple choice questions : 1.

Atomic radii expressed in a) Nano metre b) Angstrom units

c) Both a and b

[

]

d) Dalton

2.

Among the following, the correct order of radii is a) van der Waals’ radii < metallic radii b) Metallic radii < van der Waals’ radii c) Covalent radii > van der Waals’ radii d) Crystal radius < Covalent radius

[

]

3.

Covalent radii is used for a) Metals b) Non - metals

[

]

[

]

[

]

4.

5.

c) Both a and b

d) Metalloids

A and B are two elements whose electronegativities are c A and c B where c A > c B . then the bond length of dA-B is given by a) dA-B = rA+ rB - 0.09( c A - c B) b) dA-B = rA- rB + 0.99( c A - c B) c) dA-B = rA+ rB - 0.99( c A + c B) d) dA-B = rA+ rB - 0.09( c A + c B) Statement type question : Statement A : van der Waals’ radii of an element is always larger than its covalent radii. Statement B : Metallic radii is always larger than the covalent radii. Statement C : Crystal radius is used for metals a) All the statements are correct. b) Statements A, B are correct and C is incorrect. c) Statements B, C are correct and A is incorrect. d) Statements A, C are correct and B is incorrect.

Do you know ? Highest electropositive element is caesium (Cs) 10

GOAL– 3

SYNOPSIS

PERIODICITY OF ATOMIC RADII Variation of atomic radii in period : The atomic radii decreases from left to right along a period in the periodic table. In a period the atomic number increases and distinguishing electrons enters the same outer shell, hence nuclear charge increases. This increases the attractions between the nucleus and the extra nuclear electrons as the number of orbitals remain same. Due to this all electrons in orbitals are pulled closer to the nucleus. This goes on from atom to atom in a period. Example : 2nd period Atomic radii (in A°)

Li > Be > B > C 1.23 0.89 0.82 0.77

>

N 0.75

>

O > F 0.73 0.72

Variation in a group : In a group from top to bottom, atomic number increases, the valence shell increases, hence the atomic radii increases. The increase in size is due to the presence of extra energy shells in the elements as we go down the group. In vertical columns of transition elements, there is an increase in the size from the first member to the second member. But from the second member to the third member, there is a small change in the size and some times sizes are same. This is due to Lanthanide contraction. Lanthanide contraction: In lanthanides, the differentiating electron enter in to the 4f level. Since these electrons do not effectively screen the valence electron from the increased nuclear charge, the size gradually decreases. This decrease is termed as Lanthanide contraction. Ionic Radii : Definition : It is a distance of outermost shell of an anion or cation from its nucleus. The radius of the cation is always smaller than the atomic radius of parent atom. Eg: Na>Na+ The radius of the anion is always larger than the atomic radius of the parent atom. Eg: Cl->Cl Iso electronic species : Some atoms and ions which contain the same number of electrons are known as iso electronic species. Eg: C-4, N-3, O-2 The size of iso electronic ions decreases with increase in the atomic number. C-4>N-3>O-2>F->Ne>Na+>Mg+2>Al+3>Si+4

Do you know ? Electron affinity of Noble gases is zero. 11

GOAL– 3

ASSIGNMENT


The correct order of atomic radii is a) Na < Be < B

2.

b) Atomic radii

c) Both a and b

b) B

c) C

]

d) Fe+3 < Fe+2 < Fe+4 [

]

d) Atomic number

Among the following,the element with smallest covalent radii is a) Be

4.

c) Na < Li < K

If we go from lithium to fluorine in the second period,there would be a decrease in a) Atomic mass

3.

b) F-< O-2 < N-3

[

[

]

[

]

[

]

d) N

The ions O-2, F-, Na+, Mg+2,Al+3 are iso electronic. Their ionic radii show a) An increase from O-2 to F- and then decrease from Na+ to Al+3 b) A decrease from O-2 to F- and then increase from Na+ to Al+3 c) A significant increase from O-2 to Al+3 d) A significant decrease from O-2 to Al+3

5.

Among the following,the group which represents a collection of iso electronic species is (atomic number of Cs = 55 and Br = 35) a) N-3, F-, Na+

6.

8.

9.

b) Cl-

d) Na+, Ca+2, Mg+2

c) Br-

[

]

[

]

[

]

d) I-

The ionic radius of a cation is always a) less than atomic radius

b) more than atomic radius

c) equal to atomic radius

d) cannot be predicted

The lanthanide contraction is responsible for the fact that a) Zr and Y have about the same radius

b) Zr and Nb have similar oxidation state

c) Zr and Hf have about the same radius

d) Zr and Zn have the same oxidation state

The ion with largest radius is a) O-2

10.

c) Ca+2, Cs+1, Br

The ion having biggest size is a) F-

7.

b) Be, Al+3, Cl-

b) Mg+2

c) Na+

b) P+3 < P+5

c) K+ > Cl-

d) P+3 > P+5

Do you know ? First Nobel prize of chemistry was given to Vant Hoff. 12

]

[

]

d) F-

Ionic radius of a) Ti+4 < Mn+7

[

GOAL– 4 & 5

SYNOPSIS

IONIZATION ENERGY Definition of Ionization Energy : It is the minimum amount of energy required to remove an electron from the valence shell of an isolated neutral gaseous atom. It is represented by I.E The ionization energy required to remove first electron from neutral gaseous atom is termed as first ionization energy (I.E1). Similarly energy required to remove second and third electrons is referred as second ionization energy(I.E2) and third ionization energy (I.E3) respectively. ∆ H = IE1

M(g)

M+(g)+e-

M+(g)

M2+(g)+e- ∆ H = IE2

M2+(g)

M3+(g)+e- ∆ H = IE3 Generally,

IE3>IE2>IE1

Ionization energy values are often expressed in K.cal/g.atom or KJ/g.atom or KJ/mol or eV/atom 1 eV/atom = 23.06kcal/mol = 96.39KJ/mol Factors affecting Ionization Energy : 1)

Atomic size : As the size of atom increases the nuclear attraction on valence electron decreases. Therefore, in large atoms less energy is required to remove valence electron i.e., lower IE or IP Value.

2)

Nuclear Charge : More the nuclear charge, the attraction between nucleii and the electron of outer shell will be more, therefore more energy will be required for its removal.

3)

Shielding Effect or Screening Effect : The phenomenon in which inner electrons acts as a screen or shield in between nucleus and valence electrons and there by reducing nuclear charge is known as shielding effect. The increase in the inner electrons will tend to decrease the nuclear charge on valence electrons and hence ionization energy will decrease.

Do you know ? Most abundant element on earth is Oxygen. 13

4)

Penetration effect : The ionization potential also depends on the type of the subshell in which the electron is present. As s-electron penetrates nearer to the nucleus, so it is more tightly held than p-electrons and similarly d-electrons are more tightly held than f-electrons. Therefore ionization energy decreases in the order of s p d f subshells. Hence ionization energy of s-electron is greater than p-electron. Which inturn is greater than that of d-electron and so on. Penetration power of s,p,d and f sub shells is s > p > d > f

5)

Completely or half filled electronic configuration : Half filled or completely filled electronic configurations are more stable and it is difficult to remove electron and thus ionization energy is more i.e.,removal of electron from completely filled configuration requires much more IE than in a normal case. Eg: IE1 of N > IE1 of O but IE2 of O > IE2 of N

6)

Net charge on the atom : The energy required to remove the electron from an atom or ion is directly proportional to the charge on the residual ion. The successive IE of various electrons are in the following order : 1st < 2nd < 3rd < 4th < 5th and so on. Eg: Al+ < Al+2 < Al+3

Variation of Ionization energy in a period : Generally ionization energy increases as we move from left to right along a period, because with increase in atomic number in a period, nuclear charge increases and hence electrons are more tightly held. 2nd Period

Li

Be

B

C

N

O

F

IE1(in eV) 5.4

9.3

8.2

11.2

14.54

13.64

17.42

Ne 21.56

Note: Nitrogen has half filled p orbital. So it requires more ionization energy to remove an electron. Variation of Ionization energy in a group : On moving down the group valence shell become far away from the nucleus and thus nuclear attraction towards valence electrons decreases which results decrease in ionization energy. IA group Li IE1 (eV) 5.4

Na 5.1

K 4.3

Rb 4.2

Cs 3.9

Do you know ? Element containing no neutron is 1H1 14

GOAL– 4 & 5

ASSIGNMENT

Multiple choice questions : 1. Among the Alkali metals, the metal with highest ionization enthalpy is a) Li b) Na c) Cs d) Rb

[

]

2.

[

]

[

]

Among the following elements(whose electronic configurations are given below) the one having the highest ionization energy is [ 2 1 2 3 2 2 10 2 3 a) [Ne]3s 3p b) [Ne]3s 3p c) [Ne]3s 3p d) [Ar] 3d 4s 4p

]

3. 4.

The element with the highest ionization energy is a) Oxygen b) Nitrogen c) Helium

d) Carbon

The set representing the correct order of first ionization enthalpy is a) K > Na > Li b) Be > Mg > Ca c) B > C > N

d) Ge > Si > C

5.

Ionization energy of nitrogen is more than oxygen because [ ] a) Nucleus has more attraction for electrons. b) Half filled p orbitals are more stable. c) Nitrogen atom is small. d) More penetration effect. Statement type questions : [ ] 6. Statement A : The first ionization energy of Al is less than the first ionization energy of Mg. Statement B : The first ionization energy of Na is less than the first ionization energy of Mg. Statement C : The second ionization of Mg is greater than the second ionization energy of Na. a) All the statements are correct b) All the statements are incorrect c) A, B are correct and C is incorrect d) A, B are incorrect and C is correct 7.

Statement A : Penetration power of s,p,d and f subshells are s > p > d > f. Statement B : IE1 of N > IE1 of O but IE2 of O > IE2 of N. Statement C : IE3 > IE2 > IE1 a) All the statements are correct b) All the statements are incorrect c) A, B are correct and C is incorrect d) A, B are incorrect and C is correct

[

]

Statement A : In large atoms less energy is required to remove valence electrons. [ Statement B : The more the effective nuclear charge,the more will be the ionization energy. Statement C : If net charge increases ionization energy decreases. a) All the statements are correct b) All the statements are incorrect c) A, B are correct and C is incorrect d) A, B are incorrect and C is correct Assertion and Reason type : a) Both A and R are correct and R is the correct explanation of A. b) Both A and R are correct but R is not the correct explanation of A. c) A is correct, R is incorrect. d) A is incorrect but R is correct

]

9.

Assertion (A): The first ionization energy of Be is greater than that of B. Reason (R) : 2p electrons are more loosely held than 2s electrons.

[

]

10.

Assertion (A): First ionization energy of nitrogen is lower than that of oxygen. Reason (R) : Across the period effective nuclear charge increases.

[

]

8.

Do you know ? Diamond is the Crystal form of carbon. 15

GOAL– 6

SYNOPSIS

ELECTRON AFFINITY (ELECTRON GAIN ENTHALPY) Definition : Electron Affinity of an element is equal to “The energy released when an electron is added to valence shell of an isolated neutral gaseous atom”. (or) When an electron is added to a neutral gaseous atom to convert it into a negative ion, the enthalpy change accompanying the process is defined as electron gain enthalpy. ( ∆ egH ) A(g) + e-

A-1(g) + EA1

∆ egH1 = - EA1

The addition of second electron in an anion is more difficult and energy is needed to over power the repulsion forces between negatively charged atomic sphere and the valence electron. This energy being greater than energy released during gaining up of electron, i.e., why higher EA Values (EA2, EA3) are Endoergic nature. +e A2-(g) → A3-(g) ∆ egH = +EA3 i.e., EA3 > EA2 > EA1

A-(g) + e∆ egH = +EA2

EA2 , EA3 are always Endoergic EA1 Values are exoergic (however for some elements endoergic e.g. Noble gases, and alkali earth metals) while rest are endoergic. Factors affecting Electron Affinity : The magnitude of electron affinity depends on number of factors such as : 1)

Atomic size : The smaller is the size of the atom the stronger is the attraction for the added electron towards the nucleus.Thus EA increases. Example : EA of Be > EA of Mg

2)

Nuclear charge : The greater the nuclear charge, the stronger the attraction for the added electron towards the nucleus. Thus EA increases as nuclear charge increases. Example : EA of O < EA of F

Do you know ? Metals showing highest oxidation number is Ru & Os. 16

3)

Electronic configuration : The elements with stable electronic configuration have very little or no tendency to accept additional electron. Hence, EA is low or zero in these cases.

Variation of Electron gain enthalpy or Electron Affinity in a group : The electron affinity decreases from top to bottom as the atomic size increases and added electron would be farther from the nucleus. But generally the second element in a group i.e., third period element (IIA to VIIA groups) has greater electron enthalpy than the first element i.e., second period element. Order of electron affinity of various groups is

Si > C > Ge > Sn > Pb ( IVA group)

P > N > As > Sb > Bi ( VA group)

S > O > Se > Te ( VIA group)

Cl > F > Br > I ( VIIA group)

Variation of Electron gain enthalpy or Electron Affinity in a period : Electron gain enthalpy increases across a period from left to right. The nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom.

Do you know ? Among metals, best electricity conductor is silver (Ag) 17

GOAL– 6

ASSIGNMENT


2.

3.

The element with the highest electron affinity is a) O b) F c) Cl

[

]

d) N

Increasing order of electron affinity is a) N < O < Cl < Al b) O < N < Al < Cl

[ d) Cl < N < O < Al

]

c) N < Al < O < Cl

The electronic configuration of four elements is given below. Arrange these elements in the correct order of the magnitude(without sign) of their electron affinity. ii) 3s23p5 iii) 2s22p4 iv) 3s23p4 i) 2s22p5 Select the correct answer using the codes given below. a) i < ii < iii < iv b) ii < i < iv < iii c) iii < iv < ii < i

[

]

d) iii < iv < i < ii

Assertion and Reason type : 4.

Assertion (A) : Electron affinity of oxygen is less than that of fluorine but greater than that of nitrogen. Reason (R) : Fluorine has small atomic size. [ ] a) Both A and R are correct and R is the correct explanation of A. b) Both A and R are correct but R is not the correct explanation of A. c) A is correct, R is incorrect. d) A is incorrect but R is correct

Statement type questions : 5.

[

]

Statement A : Electron affinity increases with increase in the atomic number across a period. Statement B : The decreasing order of electron affinity of fluorine, chlorine and bromine is F > Cl > Br. Statement C : The element with stable electronic configuration will have low electron affinity. a) All the statements are correct b) A and C are correct and B is incorrect c) A, B are correct and C is incorrect d) A, B are incorrect and C is correct

Do you know ? Coolant in nuclear reactors is D2O 18

GOAL– 7 & 8

SYNOPSIS

ELECTRONEGATIVITY Definition : The tendency of an atom of an element to attract the shared pair of electrons in a covalent bond towards its own direction is called Electronegativity. Electronegativity is the property of bonded atom and it has no units. Expressions of Electronegativity The following approaches were used to express the electronegativity: 1)

Pauling’s Approach

2)

Mulliken’s Approach

3)

Allred - Rochow scale

1)

Pauling’s Approach : In 1931 Linus Pauling proposed a scale of measurement of electronegativity (EN). As per this scale, EN of an element is calculated from the values of bond energies. The bond energy of a compound (a diatomic molecule) A-B is taken as the average value of the bond energies of A-A and B-B molecules. 1 (E + E ) or EA–B = (EA–A ´ EB–B)1/2 2 A–A B–B But the experimentally determined values of E(A–B) are found to be more than calculated values. The extra bond energy is called ∆ (Delta). EA–B =

∆ (A-B) = [(EA-B)exp – (EA-B)cal] 1/2 ∆ (A-B) = [(EA-B) – (EA-A ´ EB-B) ]

If c A and c B are EN values of A and B, then they are related to ∆ (A-B) according to Pauling by the following relation. ( c A- c B ) = 0.208

∆ ( A− B )

Where ∆ (A-B) is expressed in kcal/mol and In SI units ( c A- c B ) = 0.1017

∆ ( A− B )

Where ∆ (A-B) is expressed in kJ/mol On Pauling scale, fluorine has the highest EN value i.e., 4.0

Do you know ? Phosphorus stored in water. 19

2)

Mulliken’s approach : In1937 Mulliken suggested that the electronegativity of the element is the average of ionization energy and electron affinity. Ionization Enthalpy + Electron Affinity 2 Relation between Pauling scale and Mulliken scale of EN Electronegativity =

(EN)M = (2.8) (EN)P Where (EN)M = EN on Mulliken scale, (EN)P = EN on Pauling scale NOTE : Pauling’s EN values are commonly used values. 3)

Allred - Rochow scale : Allred - Rochow defined EN as the electrostatic force exerted by the nucleus on the valence electrons at a distance equal to the covalent radius. This force F is given by e 2 .Zeff F= r2 Where e = charge of an electron, r = Covalent radius, Zeff = Effective nuclear charge

NOTE : The zero value of electronegativity for noble gases is due to their inert nature.

Do you know ? Gold and Platinum are noble metals. 20

GOAL– 7 & 8

ASSIGNMENT

Multiple choice questions : 1. 2.

The most electronegative element is a) N b) F

IE - EA 2

c) EN =

[

]

IE +EA 2

d)

EN =

EA +IE 6

Relation between electronegativity on Mulliken scale((EN)M) and Pauling scale((EN)P) is [ M a) (EN) 2.8 = (EN)P

4.

b) EN =

]

d) Cl

According to Mulliken, electronegativity of the element is a) EN = IE +4EA

3.

c) O

[

b)

(EN)M = (EN)P 2

c)

(EN)M = (EN) +1 P

d)

(EN)P = (EN)M 2.8

The electronegativity value of neon is a) 4.0

]

[

b) 3.0

c) 2.1

]

d) 0

Statement type question : 5. Statement A : Electronegativity is the property of isolated atom. [ Statement B : EN has no units. Statement C : Noble gases have highest electronegative value i.e.,4.0 a) All the statements are correct b) All the statements are incorrect c) A, B are correct and C is incorrect d) A, C are incorrect and B is correct

]

Test your self (Matrix Matching) __________________________________________________________________________________________ This question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B– q and r; C – p and q; and D – s and ; then the correct darkening of bubbles p q r s t will look like the following. q r A p s t B r s t C q p r D

Column-I A) Na B) Hg C) Br D) Au

Column-II p) liquid at room temperature q) Metal r) Valency is one s) Halogen t) +1 Oxidation state

p

q

A B C D

Do you know ? Lucas reagent : Conc.HCl + anhydrous ZnCl2 21

r

s

t

GOAL– 9

SYNOPSIS

FACTORS AFFECTING ELECTRONEGATIVITY a)

Nuclear Charge:

If the nuclear charge of an atom increases towards a shared pair of electrons then the electronegativity also increases. b)

Screening Effect :

More screening effect is produced by inner electrons, more repulsive force is experienced by valence electrons. Therefore possibility of entry of electron will be less.Therefore EN decreases. c)

Atomic size : As the atomic size increases, the nuclear attractions on valence electrons decreases, thus electronegativity decreases. VII A group E.N.values

d)

4.0

F

Cl

Br

3.0

2.8

2.5

I

Oxidation state : An atom with higher oxidation state is more electronegative, as it has more effective nuclear charge. Eg: Electronegativity of Fe3+> Electronegativity of Fe2+

Variation of Electronegativity in a period : On moving from left to right in a period, the electronegativity value increases because atomic size decreases and nuclear charge increases. 2nd Period

Li

Be

B

C

N

O

F

Ne

E.N Value

1.0

1.5

2.0

2.5

3.0

3.5

4.0

0

Variation of Electronegativity in a group : On moving down the group electronegativity decreases because atomic size increases. NOTE : The highest electronegative element is fluorine (F), while the least electronegative element is Cs. However for Noble gases, EN value is zero (0)

Do you know ? Sodium thiosulphate acts as an Antichlor. 22

GOAL– 9

ASSIGNMENT


The electronegativity of the following elements increases in the order of a) F, Cl, Br, I b) I, Br, F, Cl c) I, Br, Cl, F

[ d) F, I, Br, Cl

2.

In the series C,N,O and F electronegativity a) decreases from carbon to fluorine b) remains same c) decreases carbon to oxygen and then increases d) increases from carbon to fluorine

]

[

]

[

]

[

]

[

]


Assertion : Ferrous ion has more electronegativity than ferric ion. Reason : An atom with higher oxidation state has more effective nuclear charge.

a) Both A and R are correct and R is the correct explanation of A. b) Both A and R are correct but R is not the correct explanation of A. c) A is correct, R is incorrect. d) A is incorrect but R is correct Statement type questions : 4.

Statement A : On moving down the group electronegativity decreases. Statement B : Cs has least electronegativity Statement C : Cl has more electronegativity than F. a) All the statements are correct c) A, B are correct and C is incorrect

5.

b) All the statements are incorrect d) A, B are incorrect and C is correct

Statement A : If atomic size increases, electronegativity decreases. Statement B : With increase in oxidation state, electronegativity increases. Statement C : As nuclear charge increases, electronegativity decreases. a) All the statements are correct c) A, B are correct and C is incorrect


Do you know ? Common name of Aluminium oxide is Alumina. 23

GOAL– 10

SYNOPSIS

ELECTRO POSITIVITY Definition : Tendency of atom of an element to lose one or more electrons and form positive ion is known as electropositivity. Metals are electropositive in nature and have a tendency to lose electrons. M → M+ + eAs the electropositivity increases metallic nature also increases. Factors affecting Electropositivity : 1)

Atomic size : Larger the size of the atom, the element can lose the valence electron easily, hence EP increases.

2)

Nuclear charge : The greater the nuclear charge, the stronger will be the attraction between the nucleus and the valence electrons. Thus EP decreases as nuclear charge increases.

3)

Shielding effect : Due to the increase in the shielding effect the valence electrons have less nuclear attraction and thus can be easily removed. Therefore the greater the shielding effect, the greater will be the EP.

Variation of electropositivity in a period : In a period from left to right electropositivity decreases, this is due to increase in ionization energy along a period which makes loss of electrons difficult. Variation of electropositivity in a group : In a group from top to bottom electropositivity increases. This is due to decrease in ionization energy on going down a group.

Do you know ? Ethylene glycol is used as a Antifreezing agent. 24

GOAL– 10

ASSIGNMENT


The most electropositive element in the periodic table is a) Cs

2.

b) Ca

c) Li

b) Metals

c) Non-metals

]

[

]

d) Pb

Electropositivity is seen in a) Metalloids

[

d) zero group elements.


Statement A : Metals are electropositive in nature.

[

]

[

]

[

]

Statement B : With increase in atomic size, electropositivity increases. Statement C : As the shielding effect increases, EP decreases.

4.

a) All the statements are correct

b) All the statements are incorrect

c) A, B are correct and C is incorrect

d) A, B are incorrect and C is correct

Statement A : With increase in nuclear charge, electropositivity decreases. Statement B : As the EP increases metallic nature also increases. Statement C : In a period from right to left EP decreases. a) All the statements are correct





Assertion (A) : Cs has highest EP. Reason (R) : Atomic size of Cs is large. a) Both A and R are correct and R is the correct explanation of A. b) Both A and R are correct but R is not the correct explanation of A. c) A is correct, R is incorrect.

d) A is incorrect but R is correct

Do you know ? Haber’s process is used to prepare NH3. 25

GOAL– 11

SYNOPSIS

OXIDISING AND REDUCING PROPERTY Oxidation : Removal of electrons or loss of electrons from an atom is called oxidation. Oxidising property : Elements with high EN have the property of pulling the shared pair of electrons towards itself from the neighbouring bonded atom. This property is called oxidising property. Example : F>O>Cl Reduction: Gain of electrons is reduction. Reducing property : Elements with high EP and low IP have the property of losing an electron. This property is called Reducing property. Example : Na, Mg, Al. etc., Metallic and non-metallic property : 1) Metallic property : Metals have the tendency to form cations by losing electrons. Metals are electropositive in nature. Metals are good reducing agents. Metals have low IP and low EN values. Example : Na, K, Al etc., 2)

Non-Metallic property : Non-Metals have the tendency of pulling the shared pair of electrons towards themselves. Non-Metals are highly EN and possess high IP. Non-Metals are good oxidising agents. Example : C, N, O, S, P......etc.,

Variation in a group : The metallic property increases down the group and non metallic property decreases down the group, due to the following reasons: i) IP decreases down the group ii) Shielding effect increases. iii) Nuclear charge decreases. Example : In IVA group, Carbon is a non-metal where as Tin and Lead are metals. Variation in a period : The Metallic property decreases along the period from left to right and non metallic property increases along the period, due to the following reasons: i) Nuclear charge increases in a period ii) IP increases, EN increases. Example : IA and IIA group elements are metals where as VIIA,VIIIA group elements are non-metals.

Do you know ? Common name of Arsenic trioxide is Arsenic glass. 26

GOAL– 11

ASSIGNMENT


2.

Across the period the reducing nature of elements

]

a) increases

b) remains constant

c) decreases

d) increases upto VA group and then decreases

Among the following, the element with metallic property is a) P

3.

[

b) N

c) K

[

]

[

]

[

]

[

]

d) S

Choose the incorrect statement a) Loss of electrons is called oxidation. b) Gain of electrons is called reduction. c) Non-metals are highly electronegative and possess high IP. d) Metals have low IP and high EN values.


Statement A : Metals are good reducing agents. Statement B : Metals lose electrons easily. Statement C : Metals have low IP and high EN.

5.

a) All the statements are correct




Statement A : Down the group shielding effect increases. Statement B : Across a period nuclear charge increases. Statement C : Non-metals are highly EN and possess high IP. a) All the statements are correct




Do you know ? Formula of Caro’s acid is H2SO5 27

GOAL– 12

SYNOPSIS

VALENCY Definition : Valency is the combining capacity of an element. (OR) According to the new concept, valency may be defined as “The number of electrons that are lost or gained or shared with one atom of that element, to acquire the stable configuration of the nearest noble gas element. The valency of metals is given by the number of valence electrons present in an atom. The valency of non metals is given by subtracting the number of valence electrons present in an atom from 8, i.e., (8 - number of valence electrons) Valency of s and p - block elements : In case of representative elements the number of valence electrons increases from 1 to 8 from left to right in a period so the valency increases from left to right and then decreases to zero. General Valence No. of bonds Group Electronic Valency formed configuration electrons IA IIA IIIA IVA VA VIA VIIA VIIIA

ns1 ns2 ns2np1 ns2np2 ns2np3 ns2np4 ns2np5 ns2np6

1 2 3 4 5 6 7 8

1 2 3 4 3 2 1 0

1 2 3 4 3 2 1 0

Valency of d - block elements : These elements have the outer most electronic configuration ns 1-2(n-1)d1-10 These elements show variable valency involving valence electrons and d electrons of the penultimate shell. The common valency of the transition elements is either 2 or 3 . Example : Iron has a variable valency of 2 & 3.

Do you know ? Asbestos’s chemical name is Magnesium silicate. 28

GOAL– 12

ASSIGNMENT


‘X’ has 8 valence electrons, then its valency is a) 2

2.

b) 6

c) 0

b) 6 oxygen atoms

]

[

]

d) 3

‘X’ has 6 valence electrons, then it combines with a) 6 hydrogen atoms

[

c) 2 Cl atoms

d) both b and c

Statement type question : 3.

Statement A : Valency of s-block elements is equal to number of electrons present in ultimate shell. Statement B : The common valency of d-block elements is 2. Statement C : General electronic configuration of transition elements is ns 1-2(n-1) d0-10. a) All the statements are correct




[

]

[

]

[

]

Match the following : 4.

Valency

Group

1) 3 2) 0 3) 4

A) Carbon family B) Boron family C) Chalcogens D) Inert gases

a) c) 5.

1

2

3

A C

B D

D A

b) d)

Element 1) B 2) Ar 3) Si 4) Ba

a) c)

1

2

3

A B

C D

B A

Number of bonds formed A) 4 B) 3 C) 2 D) 1 E) 0

1

2

3

4

B B

C E

E A

A C

b) d)

1

2

3

4

B E

E A

C C

D B

Do you know ? Aspirin’s chemical name is Acetylsalicylic acid. 29

GOAL– 13

SYNOPSIS

OXIDATION NUMBERS The oxidation number or oxidation state of an element is defined as the possible electronic charge which the atom of the element appears to have acquired in the given chemical form of the element. The common oxidation states of the elements are related to their valence electronic configuration. Variation in a group : The oxidation number of the elements is mostly equal to the group number.It varies from group to group. For s -block elements, the maximum positive oxidation state shown by any element is equal to the total number of valence electrons. Example :

Valence Group

electronic configuration

Valence Oxidation Electrons number

IA Group

ns1

1

+I

IIA Group

ns2

2

+II

p -block elements exhibits various oxidation states. These elements often gain or lose sufficient number of electrons in order to attain noble gas electronic configuration.

Group

General Electronic configuration

Valence electrons

Oxidation number

IIIA

ns2np1

3

+III

IVA

ns 2np2

4

+IV

VA

ns 2np3

5

-III, +III, +V

VIA

ns 2np4

6

-II, +II, +IV, +VI

VIIA

ns 2np5

7

-I, +I, +III, +V, +VII

VIIIA

ns 2np6

8

0

Do you know ? Formula of Haematite is Fe2O3 30

d-block elements exhibits more than one oxidation states. The outer most electronic configuration of d-block elements is ns1-2(n-1)d1-10.They show the common oxidation state of +II due to the loss of ns2 electrons. Ru and Os exhibits a maximum Oxidation state of +VIII in RuO4 and Os O4 respectively. Example : Outer most electronic Configuration

Element

Oxidation states

Sc

[Ar] 3d1 4s2

+ II, +III

Ti

[Ar] 3d2 4s2

+ II, +III, +IV

V

[Ar] 3d3 4s2

+ II, +III, +IV, +V

Cr

[Ar] 3d5 4s1

+ II, +III, +IV, +V, +VI

Mn

[Ar] 3d5 4s2

+ II, +III, +IV, +V, +VI, +VII

Fe

[Ar] 3d6 4s2

+ II, +III, +IV, +V, +VI

Co

[Ar] 3d7 4s2

+ II, +III, +IV

Ni

[Ar] 3d8 4s2

+ II, +III, +IV

Cu

[Ar] 3d10 4s1

+ I, +II

Zn

[Ar] 3d10 4s2

+ II

GOAL– 13

ASSIGNMENT


The set which shows highest oxidation state is a) Os and Rh b) Ru and Fe

[ c) Os and Ru

]

d) As and Ru

2.

Oxidation number of the elements is mostly equal to a) Period number b) Group number c) Both a and b

3.

‘X’ is an element which belongs to VA group and 3 rd period. The oxidation state and name of the element are [ ] a) +5 and P b) -3 and N c) +5 and N d) -3 and S

4.

Number of valence electrons in argon is a) 0 b) 8

5.

c) 3

Valence electronic configuration of VIII group elements is a) ns2(n-1)d6 b) ns2(n-1)d8 c) Both a and b

[ ] d) Atomic number

2

d) ns (n-1)d

3

Do you know ? Bitter Salt’s chemical name is Magnesium Sulphate. 31

[

]

[

]

d) 6

GOAL– 14 & 15

SYNOPSIS

DIAGONAL RELATIONSHIP AND CHEMICAL REACTIVITY Diagonal Relationship : There exists a characteristic analogy in elements of 2nd and 3rd period. The first element of a group resembles closely with second element of next successive group. This phenomenon is known as diagonal relationship. Therefore the first three elements of second period( Li, Be and B) not only show resemblance with the elements of their own groups but show resemblance with the elements diagonally placed in the higher groups. Period Groups I II III IV 2nd Period Li Be B C 3rd Period Na Mg Al Si Lithium shows close resemblance with magnesium, beryllium with aluminium and boron with silicon. The resemblance, however, disappears after these pairs. On moving far away in period or in group the relationship disappears since the elements located diagonally show large differences in their atomic radii because the increasing or decreasing trend in atomic radii does not occur in same order. The diagonally similar elements possess, the same polarising power. Polarising power of species is the ratio of its ionic charge to the square of the ionic radius. i.e., Polarising Power =

Ionic ch arg e

( Ionic radius)

2

Chemical Reactivity : Reactivity of elements depends on two properties of elements. They are a) Electropositivity b) Electronegativity If the elements are highly electropositive in nature, they can lose the electrons easily and easily react with other elements. If the elements are highly electronegative in nature, they can easily pull the electrons towards themselves and easily react with other elements. In a group : The Electropositivity character increases down the group and Electronegativity decreases. Therefore the reactivity of elements increases down the group. Example : Caesium is the last element in IA group, hence it is highly reactive. But in the case of p-block elements, as the electronegativity nature decreases down the group, the reactivity decreases down the group. Example : Fluorine is highly reactive as electronegativity of Fluorine is 4. The reactivity decreases down the group. In a period : The Electropositivity decreases from left to right, but electronegativity increases from left to right. From this, electropositive nature is more to the extreme left in the period. So reactivity is high in IA and IIA group elements, then it decreases to the right. Therefore halogen group elements are highly reactive as electronegativity is high to the extreme right of the period. Example : Fluorine is the first element in VIIA group which has highest electronegativity.

Do you know ? Black ash is the crude form of sodium carbonate. 32

GOAL– 14 & 15

ASSIGNMENT


2.

Aluminium is diagonally related to a) Li b) Si Reactivity of elements depends on a) EA b) EP

c) Be

[

]

[

]

[

]

[

]

[

]

d) B

c) IP

d) none


Assertion (A) : Caesium is highly reactive Reason (R) : Caesium is highly electronegative in nature a) Both A and R are correct and R is the correct explanation of A. b) Both A and R are correct but R is not the correct explanation of A. c) A is correct, R is incorrect. d) A is incorrect but R is correct

Statement type question : 4.

Statement A : IA group elements are highly reactive in nature. Statement B : VIIA group elements are highly reactive in nature. Statement C : Chemical reactivity depends on EP and EN a) All the statements are correct c) A, B are correct and C is incorrect

5.


Match the following as per their diagonal relationship : Set A

Set B

1) Li

A) Na

2) Be 3) B

B) Mg C) Si D) P E) Al

a) c)

1 A B

2 C C

3 E D

b) d)

1 B E

2 E B

3 C D

Do you know ? Blue stone is the crystal form of copper sulphate. 33

GENERAL TENDENCY OF PERIODIC PROPERTIES PERIODIC PROPERTY

ALONG A PERIOD (From left to right)

DOWN THE GROUP

ATOMIC RADIUS

Decreases

Increases

IONIZATION ENERGY

Increases

Decreases

ELECTRON AFFINITY

Increases

Decreases

ELECTRONEGATIVITY

Increases

Decreases

ELECTROPOSITIVITY

Decreases

Increases

METALLIC PROPERTY

Decreases

Increases

NON-METALLIC PROPERTY

Increases

Decreases

NOTE : This table represents general tendency, but in some cases there will be exceptions. Refer theory for exceptions.

Electron gain enthalpy

Ionization energy

Atomic radius

Electronegativity

Electron gain enthalpy

Ionisation energy

ter arac h C ic etall -M cter n o N hara C c alli Met

Atomic radius Electronegativity

Minimum

Maximum

The periodic trends of elements in the periodic table.

Do you know ? Metals comprise more than 78% of all known elements. 34

ORANGE WORK SHEET - 1 Multiple Choice Questions : 1.

Among the following, the iso electronic speices are a) O-1, F-

2. 3.

b) Mg+2, Ne

Atomic radius of a) Li > Be > B

b) C < N < O

[

]

c) O-2, Ne

d) Na+, Cl]

c) Li < Be < B

[ d) N < O < F

The distance between the centre of the nucleus and the outer most shell where electrons are present is [ ] a) Covalent radius b) van der Waals’ radius c) Atomic radius d) Metallic radii

Assertion & Reason type question : 4.

Assertion (A) :Elements belonging to same group have similar properties.

[

]

Reason (R): Same groups after the difference of either 2 or 8 or 18 or 32 in the atomic number and having similar valence shell electronic configuration have similar properties. a) Assertion and reason are correct and reason is the correct explanation of assertion b) Assertion and reason are correct but reason is not the correct explanation of assertion c) Assertion is correct but reason is incorrect d) Assertion is incorrect but reason is correct. 5.

Match the following : SET - A

SET - B

1) Atomic radii

A) Non-metals

2) Radius of cation

B) Metals

3) Crystal radius

C) Z*

4) Covalent radius

D) Nano metre

5) Effective nuclear charge

E) Smaller than atomic radius of parent atom

1

2

3

4

5

a)

A

D

C

B

E

c)

D

C

A

E

B

[

1

2

3

4

5

b)

D

E

B

A

C

d)

D

A

E

B

C

Do you know ? Bone ash is the crude form of calcium phosphate. 35

]

ORANGE WORK SHEET - 2 Multiple Choice Questions : 1. 2.

High ionization potential is exhibited by a) Metals b) Non-metals

[ ] d) Radioactive elements

c) Metalloids

The energy required to remove the second electron from an atom is a) Ionization energy c) Second ionization energy

[

]

b) First ionization energy d) Third ionization energy

Statement based question : 3.

Statement A : The magnitude of E.A. depends on atomic size and electronic configuration. Statement B : The energy released is when an electron is added to a valence shell of an [ ] isolatedneutral gaseous atom. Statement C : Electron gain enthalpy decreases with increase in the atomic number across a period. a) All the statements are correct b) A & B statements are correct & C is incorrect c) A & C statements are correct & B is incorrect d) B & C statements are correct & A is incorrect

Assertion & Reason type question : 4.

Assertion (A) : Electron affinity of chlorine is more than that of fluorine.

[

]

[

]

Reason (R) : Fluorine has small atomic size. a) Assertion and reason are correct and reason is the correct explanation of assertion b) Assertion and reason are correct but reason is not the correct explanation of assertion c) Assertion is correct but reason is incorrect d) Assertion is incorrect but reason is correct. 5.

Match the following : SET - A

SET - B

1) Electron affinity

A) s > p > d > f

2) Second IE

B) 23.06 k.cal/mole

3) Penetration power

C) A( g) + e− → A(−g) + EA1

4) 1 ev/atom

D) M(+g) → M(2g)+ + e −

a) c)

1

2

3

4

A C

D D

C A

B B

b) d)

1

2

3

4

C B

B C

D D

A A

Do you know ? Marble contains calcium carbonate. 36

ORANGE WORK SHEET - 3 Multiple Choice Questions : 1. The units of electronegativity are a) KJ/mole b) eV/atom 2.

c) Kcal / mole

I .E. + E.A 2

]

[

]

[

]

[

]

[

]

d) no units

According to Mulliken electronegativity is a) E.N . =

[

b) (E.N)M = (2.8) . (EN)P

e 2 .Z eff I .E . d) E.N. = F = 2 r2 Statement based question : 3. Statement A: Electronegativity of Fe 3+ is greater than electronegativity of Fe2+. Statement B: Electronegativity increases as the screening effect increases. Statement C : The highest electronegative element is F. a) All the statements are correct b) A & B statements are correct & C is incorrect c) A & C statements are correct & B is incorrect d) B & C statements are correct & A is incorrect c) E . N . =

Assertion & Reason type question : 4. Assertion (A) : On Pauling scale, fluorine has the highest EN value i.e, 4.0 Reason (R) : Fluorine has small atomic size. a) Assertion and reason are correct and reason is the correct explanation of assertion b) Assertion and reason are correct but reason is not the correct explanation of assertion c) Assertion is correct but reason is incorrect d) Assertion is incorrect but reason is correct. 5.

Match the following : SET - A 1) e 2) r 3) Zeff

SET - B A) Covalent radius B) Charge of an electron C) Effective nuclear charge D) Atomic number

a) c)

1 A C

2 C A

3 B B

b) d)

1 B B

2 C A

3 A C

Do you know ? Blue lead is the common name of Lead sulphate. 37

ORANGE WORK SHEET - 4 & 5 Multiple Choice Questions : 1. Cause of reducing property is a) Gaining electrons b) lose of electrons

[

]

[

]

[

]

[

]

[

]

6.

Statement A : Elements often gain or lose sufficient number of electrons to attain noble gas [ electronic configuration. Statement B : Oxidation number of the element is mostly equal to its group number. Statement C : The valence electronic configuration of III B group in ns2(n - 1)d1. a) All the statements are correct b) A & B statements are correct & C is incorrect c) A & C statements are correct & B is incorrect d) B & C statements are correct & A is incorrect

]

7.

Statement A : The metallic property decreases along the period from left to right. Statement B : The non-metallic property decreases along the period. Statement C : Metals have the tendency to form cations by gaining electrons. a) All the statements are correct b) A & B statements are incorrect & C is correct c) A & C statements are incorrect & B is correct d) B & C statements are incorrect & A is correct

[

]

[

]

2.

4. 5.

d) Both a and c

As the electropositivity increases metalllic nature a) increases

3.

c) Sharing electrons

b) decreases

c) remains constant

Valence electronic configuration of carbon family is a) ns2 np1 b) ns2 np2 c) ns2 np3 The least stable ion amongst the following is b) Bea) Li-

c) B

Li is diagonally related to a) Mg b) Al

c) S

-

d) no change d) ns2 np4 d) C

-

d) Na

Statement based question :

Assertion & Reason type question : 8. Assertion (A) : Valency is the combining capacity of an element. Reason (R) : The valency of oxygen is 2. a) Assertion and reason are correct and reason is the correct explanation of assertion b) Assertion and reason are correct but reason is not the correct explanation of assertion c) Assertion is correct but reason is incorrect d) Assertion is incorrect but reason is correct.

Do you know ? Burnt lime is the common name of calcium oxide. 38

9.

Assertion (A) : 2nd and 3rd period elements have diagonal relationship. [ Reason (R) : The first element of a group resembles closely with second element of next successive group. a) Assertion and reason are correct and reason is the correct explanation of assertion b) Assertion and reason are correct but reason is not the correct explanation of assertion c) Assertion is correct but reason is incorrect d) Assertion is incorrect but reason is correct.

10.

Match the following : Element 1) Fluorine 2) Osmium 3) Phosphorus 4) Carbon 5) ‘S’ in H2S a) c)

1 E E

2 A A

3 B D

Oxidation states A) +8 B) +3, +5 C) +4, - 4, 0 D) - 2 E) - 1 4 C C

5 D B

b) d)

1 E E

2 C B

[

3 B C

4 A D

]

]

5 D A

YELLOW WORK SHEET - 1 Multiple Choice Questions : 1.

Among the following, the process which requires maximum amount of energy is [ + + a) Na(g) → Na (g) + e b) Al(g) → Al (g) + e 2+ c) Na+(g) → Na (g) + e d) Al+(g) → Al2+(g) + e Among the following ions, the set with isoelectronic ions is [ b) Mg2+, Ba2+, Ca2+ c) Na+, Mg+, Al3+ d) Ca2+, S2-, Cla) N3-, Na+, Al3+

]

3.

Correct order of first ionization potentials of elements of the second period [ a) Li < Be > B < C < N > O < F < Ne b) Li < Be < B < C < N < O < F < Ne c) Li < Be > B < C < N < O < F < Ne d) Li < Be < B < C < N > O < F < Ne

]

4.

Some statements related to ionization potential are given. Among them the correct statements are [ ] a) IE2 of sodium is greater than that of magnesium b) IE2 of neon is greater than that of oxygen c) IE2 of lithium is greater than IE2 of helium d) IE2 of lithium is greater than IE2 of beryllium

5.

Chemical similarity between B and Al is due to a) Similar outermost electronic configuration b) Diagonal relationship c) Both belongs to same period

2.

[

d) Both b and c

Do you know ? Formula of Diaspore is Al2O3.H2O 39

]

]

YELLOW WORK SHEET - 2 & 3 Multiple Choice Questions : 1. 2.

Among the following, the speices with the highest electron affinity is a) Fb) O c) OAmong the following, the smallest in size is a) Clb) Na+

c) Mg

2+

[

]

[

]

+

d) Na 2-

d) S

3.

Among the following, the metal whch requires radiation of highest frequency to cause emission of electrons is [ ] a) Na b) Mg c) K d) Ca

4.

Atoms with high value of IE1 always have a) Large atomic radiius c) strongly bonded valence electrons

5.

[ b) Small atomic radius d) loosely bonded valence electrons

Among the following rearrangements which is not truely represents the property indicated against it [ ] b) Br2 < F2 < Cl2 : electron affinity a) Br2 < Cl2 < F2 : electronegativity c) Br2 < Cl2 < F2 : electropositivity

6.

7.

d) Br2 < Cl2 < F2 : oxidising power

Among the following, the correct order of the size is a) Na < Mg < K b) P < Cl < Br d) I- > I > I +

[ ] 2+ c) O < F < Mg 2-

-

The second ionization energies of Li, Be, B and C are in the order of a) B > C > Li > Be

8.

]

b) B > C > Be > Li

c) Li > B > C > Be

[

d) Li > C > B > Be

Among the following, trends of ionic radii are correct is -

+

a) F > Na > Mg

2+

3+

2-

b) Al > O > N

3-

3-

]

[ 3+

-

c) P > Al < Cl

-

+

] +

d) H > H > He

9.

The electorn affinity of K+(g) ion is numerically equal to a) Electron affinity of Cl atom b) Electronegativity of Cl atom c) Ionization energy of Ar atom d) Ionization energy of K atom

[

]

10.

The ions O2-, F-, Na+, Mg2+ and Al3+ are isoelectronic, their ionic radii show a) A significant increase from O2- to Al3+ b) A Significant decrease from O2- to Al3+ c) An increase from O2- to F- and then decrease from Na+ to Al3+ d) A decrease from O2- to F- an then increase from Na+ to Al3+

[

]

Do you know ? Formula of Mohr's Salt is FeSO4.(NH4)2SO4.6H2O 40

YELLOW WORK SHEET - 4 & 5 Multiple Choice Questions : 1. The second ionization energy can be represented as a) M(s) → M+(g) + eb) M+(g) → M2+(g) + e- c) M(g) + e- → M-(g)

[ +2

d) M

(g)

→ M

2.

The first IE of magnesium is greater than that of aluminium because a) Al atom is very large when compared to Mg b) Al atom has an electron more when compared to Mg c) Mg has a stable electronic configuration d) Al has a stable electronic configuration

3.

The outermost electronic configuration of the most electronegative element is a) ns2 np3 b) ns2 np4 c) ns2 np5 d) ns2 np6

3+ (g)

] -

+e

[

[

]

]

4.

The electronegativity of the following elements increases in the order [ ] a) C, N, Si, P b) N, Si, C, P c) Si, P, C, N d) P, Si, N, C Assertion and Reason type questions : In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as a) If both assertion and reason are correct and reason is the correct explanation of assertion b) If both assertion and reason are correct but reason is not the correct explanation of assertion c) If assertion is correct but reason is incorrect d) If both assertion and reason are incorrect e) If assertion is incorrect but reason is correct. 5. Assertion (A) : Smaller the size of an atom greater is the electronegativity. [ ] Reason (R) : Electronegativity refers to the tendency of atom to share electrons with other atom. 6. Assertion (A) : Outermost electronic configuration of most electropositive elements is ns 2 np3. Reason (R) : ns2 np3 is stable due to half filled subshell. [ ] 7. Assertion (A) : The highest I.E. in a period is shown by noble gas. [ ] Reason (R) : Noble gases are at the extreme right of the period. Match the following : 8. SET A 1) High electron affinity 2) 2s2 2p5 3) EA1

SET B A) F B) Exoergic C) Cl D) Endoergic

4) EA2 & EA3 a) c)

1 C A

2 A C

3 B B

4 D D

b) d)

1 A B

[

2 B D

3 C C

4 D A

Do you know ? Contact process is used to prepare H2SO4 41

]

9.

SET A

SET B

[

I .E + E. A A) E.N = 2 ( E.N ) M B) (E.N)P = 2.8

1) Pauling scale 2) Mullikens’ scale

]

3) Relation between Pauling and Mullikens’ scale

C) F =

4) Allred & Rochow scale 1 A D

a) c) 10.

2 D A

3 B B

e2 .Z eff

D) (XA- XB) = 0.208

4 C C

b) d)

1 B C

General Electronic Configuration

Valency

1) ns2 np3

A) 0

2) ns1

B) 2

3) ns2

2 A D

3 C B

∆

(A− B )

KCal / mol

4 D A

[

]

C) 1

2

4) ns np a) c)

r2

1 D A

6

D) 3 2 C C

3 A B

4 B D

b) d)

1 C D

2 D C

3 B B

4 A A

BLUE WORK SHEET - 1 In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as a) If both assertion and reason are correct and reason is the correct explanation of assertion b) If both assertion and reason are correct but reason is not the correct explanation of assertion c) If assertion is correct but reason is incorrect d) If both assertion and reason are incorrect e) If assertion is incorrect but reason is correct. 1.

Assertion (A): In case of isoeletronic ions the ionic size increases with the increase in atomic number Reason(R) : The greater the attraction of nucleus, greater is the ionic radius [ ]

Do you know ? Oxygen, Sulphur & Phosphorus exhibits polymorphism. 42

2.

Assertion (A) : Second period consists of 8 elements. [ Reason (R) : Number of elements in each period is four times the number of atomic orbitals available in the energy level that is being filled. Assertion (A) : Na+ is smaller in size than Na. [ + Reason (R) : Na has same nuclear charge as that of Na.

]

4.

Atomic radii of fluorine and neon in Angstorm units are respectively given by [ a) 0.72, 1.60 b) 1.60, 1.60 c) 0.72, 0.72 d) both b and c

]

5.

Which one of the following is the smallest in size ? a) N3b) O2c) F-

]

3.

[

]

d) Na+

BLUE WORK SHEET - 2 In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as a) If both assertion and reason are correct and reason is the correct explanation of assertion b) If both assertion and reason are correct but reason is not the correct explanation of assertion c) If assertion is correct but reason is incorrect d) If both assertion and reason are incorrect e) If assertion is incorrect but reason is correct. 1.

Assertion (A) : Ionization enthalpy is the energy released to remove an electron from an isolated gaseous atom in its ground state. [

]

Reason (R) : Element has a tendency to lose the electron to attain the stable configuration. 2.

Assertion (A) : Second ionization enthalpy will be higher than the first ionization enthalpy.

[

]

Reason (R) : Ionization enthalpy is a quantitative measure of the tendency of an element to lose electron. 3.

The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is [ a) C > N > O > F b) O > N > F > C c) O > F > N > C d) F > O > N > C

]

4.

The element with the highest first ionization potential is a) boron b) carbon c) nitrogen

]

5.

[ d) oxygen

The first ionization potential in electron volts of nitrogen and oxygen atoms are respectively given by [ ] a) 14.6, 13.6

b) 13.6, 14.6

c) 13.6, 13.6

d) 14.6, 14.6

Do you know ? Formula of Hydrolith is CaH2 43

BLUE WORK SHEET - 3 In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as a) If both assertion and reason are correct and reason is the correct explanation of assertion b) If both assertion and reason are correct but reason is not the correct explanation of assertion c) If assertion is correct but reason is incorrect d) If both assertion and reason are incorrect

e) If assertion is incorrect but reason is correct.

1.

Assertion (A) : First ionization enthalpy of oxygen is greater than nitrogen. [ ] Reason (R) : Nitrogen atom has three 2p electrons residing in different atomic orbitals whereas electronic configuration of oxygen is 1s 2 2s2 2p4

2.

Assertion (A) : Nuclear charge does not affect the ionization potential of the atom. Reason (R) : Nuclear charge tells about the stability of configuration.

[

]

3.

Assertion (A) : Shielding effect increases as we go down the group. Reason (R) : More is the number of electrons in the penultimate shell, more is shielding.

[

]

4.

Assertion (A) : F- ion is larger in size compared to F. [ ] Reason (R) : Electron repulsion increases because of addition of electron which results in decrease in effective nuclear charge.

5.

Assertion (A) : Ionization potential across the period is Na < Al < Mg < Si Reason (R) : Ionization potential decreases with decreases in atomic size.

6.

Assertion (A) : Electron gain enthalpy can be exothermic or endothermic. [ ] Reason (R) : Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion.

7.

Assertion (A) : Electron gain enthalpy of O or F is less than that of the succeeding element. [ Reason (R) : Electron gain enthalpy increases as we go down the group.

]

8.

The first ionization potential of Na, Mg, Al and Si are in the order a) Na < Mg > Al < Si b) Na > Mg > Al > Si c) Na < Mg < Al < Si d) Na > Mg > Al < Si

[

]

9.

The incorrect statement among the following is a) I.E1 of Al < I.E1 of Mg c) I.E1 of Na < I.E1of Mg

[

]

[

]

10.

[

]

b) I.E2 of Mg > I.E2 of Na d) I.E3 of Mg > I.E3 of Al

The set representing the correct order of first ionization potential is a) K > Na > Li b) Be > Mg > Ca c) B > C > N

Do you know ? Sodium stored in Kerosene. 44

d) Ge > Si > C

PREVIOUS EAMCET, AIEEE AND IIT QUESTIONS In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as a) If both assertion and reason are correct and reason is the correct explanation of assertion b) If both assertion and reason are correct but reason is not the correct explanation of assertion c) If assertion is correct but reason is incorrect d) If both assertion and reason are incorrect e) If assertion is incorrect but reason is correct. 1.

Assertion (A) : Isoelectronic species have same radii. Reason (R) : They contain same number of electrons.

2.

Assertion (A) : Electron affinity of oxygen is less than the nitrogen but larger than fluorine. Reason (R) : N, O, F belong to same period of the periodic table.

3.

Assertion (A) : I.E. of 7N is more than that of 8O as well as 6C. Reason (R) : This is due to difference in reactivity towards oxygen.

4.

Assertion (A) : P has high I.E. than sulphur. (IIT) Reason (R) : Electronegativity refers to the tendency of atom to share electrons with other atom.

5.

Assertion (A) : Properties of atom and its corresponding ions remain same. (IIT) Reason (R) : Electronic configurations of both atom and ion remain same.

6.

The correct order of atomic radii is a) N < Be < B b) F- < O2- < N3-

c) Na < Li < K

(IIT)

(IIT)

(IIT)

(IIT) d) Fe3+ < Fe2+ < Fe4+

7.

According to the Periodic Law of elements, the variation in properties of elements is related to their (AIEEE) a) nuclear neutron - proton number ratios b) atomic masses c) nuclear masses d) atomic number

8.

The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest I.E 2? (AIEEE) a) Fe b) V c) Cr d) Mn

9.

Which one of the following groupings represents a collection of isoelectronic species ? ( At. nos. : Cs = 55, Br = 35) (AIEEE) 2+ + + 2+ 2+ 3+ a) Ca , Cs , Br b) Na , Ca , Mg c) N , F , Na d) Be, Al3+, Cl-

10.

Which of the following ions has the highest value of ionic radius ? a) Li+ b) B3+ c) O2-

(AIEEE) d) F-

Do you know ? Fluorspar is the natural form of calcium fluoride. 45

11.

Which one of the following sets of ions represents the collection of isoelectronic species ? (AIEEE) + 2+ 3+ a) K , Ca , Sc , C1 b) Na+, Ca2+, Sc3+, Fc) K+, C1-, Mg2+, Al3+, Sc3+ d) Na+, Mg2+, Al+3, Cl-

12.

In which of the following arrangements the order is not according to the property indicated against it ? a) Li < Na < K < Rb (Increasing metallic radius) (AIEEE) b) I < Br < F < Cl ( Increasing electron gain enthalpy (with negative sign)) c) B < C < N < O ( Increasing first ionization enthalply) d) Al3+ < Mg2+ < Na+ < F- ( Increasing ionic size)

13.

Of the following sets which one does not contain isoelectronic species a) BO33- , CO32-, NO3b) SO32- , CO32-, NO3c) CN- , N2, C22d) PO43- , SO42-, ClO4-

14.

The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first ) is (AIEEE) a) F < S < P < B b) P < S < B < F c) B < P < S < F d) B < S < P < F

15.

Which one of the following sets of ions represents a collection of isoelectronic species ? (AIEEE) a) K+, Cl-, Ca2+, Sc3+ b) Ba2+, Sr2+, K+, Ca2+ c) N3-, O2-, F-, S2d) Li+, Na+, Mg2+, Ca2+

16.

The ion that is isoelectronic with CO is b) O2+ a) CN-

c) O2-

Element with atomic number 38, belong to a) IIA group and 5th period c) VA group and 2nd period

(EAMCET) b) IIA group and 2 period d) IIIA group and 5th period

17.

18.

(AIEEE)

(IIT) d) N2+ nd

Observe the following statement :

(EAMCET)

I) The physical and chemical properties of elements are periodic functions of their electronic configuration. II) Electronegativity of fluorine is less than the electronegativity of chlorine. III) Electropositive nature decreases from top to bottom in a group. The correct answer is: a) I, II and III are correct c) only I and II are correct

b) only I is correct d) only II and III are correct

19.

Identify the correct order in which the covalent radius of the following elements increases I) Ti II) Ca III) Sc (EAMCET) a) I, II, III b) III, II, I c) II, II, III d) I, III, II

20.

Identify the correct order in which the ionic radius of the following ions increases I) FII) Na+ III) N-3 (EAMCET) a) III, I, II b) I, II, III c) II, III, I d) II, I, III

Do you know ? Ferro prussiate’s chemical name is Potassium ferricyanide. 46

C4 Periodic Properties-1

Recommend Documents