Answers for Group VII Problem-Solving Questions

Hwa Chong I nstitution (Colle (College) ge)

2011 H2 Chemistry – Group VII

GROUP VII – THE HALOGENS Problem-Solving Problem-Solving Questions Question 1 (N08/III (N 08/III/1 /1 modified) Section 3.1 (a)

Describe and explain the variation in volatility of the elements in Group VII. The halogens exist as simple, non-polar diatomic molecules. Dispersion forces exist between between molecules. Going down Group VII from fluorine to iodine, the s ize of the electron cloud of X2 increases. Thus, polarisability of the molecule increases. Dispersion forces between molecules become stronger and more energy is required to overcome the intermolecular forces of attraction. Hence, volatility decreases down Group VII.

(b)

Under certain conditions the hydrogen halides decompose into their elements. Section 5.2 2H X

(i)

⇌

H2 +

( X = C l, Br or I)

X 2

How can this reaction be carried out? By inserting a hot wire or glass rod into a test-tube of hydrogen halides.

(ii) State and describe the trend of this reaction when X = C l, X = Br and X = I. The extent of reaction increases down Group VII. HCl does not decompose even on strong heating, HBr yields brown fumes of Br 2 under strong heating while violet fumes of I2 obtained when red-hot rod is plunged into jar of H I.

(iii) Use bond energy values from the Data Booklet to calculate = C l and when X = I. 2H X (g) ⇌ H2 (g) + Bond energy

X 2

(g)

H  X

HH

X  X

For X = C l:

2 (431)

(436)

(244)

For X = I:

2 (299)

(436)

(151)

H

for HC l = 2 × 431 – 436 – 244 = +182 kJ mol –1

H

for HI = 2 × 299 – 436 – 151 = +11 kJ mol –1

Tay SA/Grace Chua/Jacqueline Tano /Tham YH

ΔH

for this reaction when X

Hwa Chong Institution (College)


(iv) Use your results from (iii) to explain the trend you stated in (ii) . Results from (iii) shows that the reaction becomes less endothermic from HC l to HI, i.e. it required less energy for the thermal decomposition of H X , therefore reaction occurs to a greater extent down the group (the trend stated in (ii)).

Question 2 (N06/II/1c) Section 5.1 Fluorspar reacts with concentrated sulfuric acid to form hydrogen fluoride gas. CaF2 + H2SO4 → CaSO4 + 2HF Data about HF, HC l, HBr and H I are given below. HF

HC l

HBr

HI

boiling point / ˚C

+20

-85

-67

-35

bond energy / kJ mol ˉ1

562

431

366

299

∆Hof / kJ mol ˉ1

-269

-92

-36

+26

(a)

State and explain the trend in boiling points of HC l, HBr and H I.

(b)

Suggest why the boiling point of HF is much higher than those of the others.

(a)

From HC l to HI, boiling point increases. This is due to increasing size of electron cloud , thus increasing the strength of van der Waals ’ forces .

(b)

HF has much higher boiling points due to hydrogen bonding which is stronger than van der Waals’ forces.




Question 3 (N93/I/7 modified) (a)

How do the chloride, bromide and iodide ions differ in their reactions with

(i)

silver nitrate followed by dilute aqueous ammonia, Section 6.1

(ii)

concentrated sulfuric acid? Section 6.3

In each case, suggest the products of the reaction and write equations, where appropriate.

(i) Halide ClBr-

I

Expected observations white AgC l precipitates; ppt soluble in excess aq ammonia pale yellow AgBr precipitates; ppt insoluble in excess NH 3 yellow Ag I precipitates; ppt insoluble in excess NH 3

Equations Ag (aq) + C l (aq)  AgCl(s) + + Ag (aq) + 2NH 3(aq)  [Ag(NH3 )2] (aq) Ag+(aq) + Br(aq)  AgBr(s) +



Ag+(aq) + I(aq)  AgI(s)

(ii) Halide Cl-

Br

-

I

(b)

Expected observations

Equations

white fumes of HC l evolve

Cl(s) + H2 SO4(l)  HCl(g) + HSO4 (s)

white fumes of HBr & a little reddish brown fumes of Br 2 evolve (may also have SO2) A little bit of white fumes of HI & purple fumes of I2 evolve (may also have pungent SO 2 or H2S). Yellow solid of S is observed.

Br(s) + H2SO4(l)  HBr(g) + HSO 4(s) 2HBr(g) + H2SO4(l)  Br2(g) + SO 2(g) + 2H2O( l) 



I (s) + H2 SO4 (l)  HI(g) + HSO4 (s)

2HI(g) + H2SO4(l)  I2(g) + SO2(g) + 2H2O( l) 6HI(g) + H2SO4(l)  3I2 (g) + S(s) + 4H 2 O(l) 8HI(g) + H2SO4 (l)  4I2(g) + H2S(g) + 4H2O( l)

What explanations can you offer for the differences in their reactions in (a)?

(i)

The difference in the reactions is due to decreasing K sp from AgC l to Ag I. For AgC l, its ionic product is lower than the K sp value when aqueous NH 3 is added. Addition of ammonia results in the formation of the diamminesilver( I) complex, [Ag(NH3)2]+. Ag+ (aq) + 2NH 3 (aq)

[Ag(NH3)2]+ (aq)

This causes [Ag +] to decrease and causes position of equilibrium Ag +(aq) + X(aq) AgX(s) to shift to the left, increasing the solubility of AgX. For AgBr and Ag I, due to their lower K sp , the ionic product is greater than K sp when aqueous ammonia is used as [Ag +] is not sufficiently lowered.



(ii)


Reducing power of halides: I- > Br- > C lHI can reduce H 2SO4 to H2S and SO 2 and itself oxidized to I2. HBr can reduce H 2SO4 to SO 2 and itself oxidized to Br 2 . HCl cannot reduce H 2SO4.

Question 4 (N90/III/6) (a)

Describe the reactions that occur when chlorine is bubbled through cold aqueous sodium hydroxide and the mixture is then heated. Section 4.4.2 Write equations and explain how the oxidation number of chlorine changes during these reactions. What observation would make it clear that the chlorine had reacted?

(b)

When chlorine is bubbled through a solution of iodine in hot aqueous sodium hydroxide, the two halogens react in the C l2:I2 ratio of 7:1, forming a white precipitate A and a solution of sodium chloride. A has the following composition by mass: Na, 16.9%; H, 1.1%; I, 46.7%; O, 35.3%. Calculate the empirical formula of A and thus deduce the balanced equation for the reaction. What is the oxidation number of iodine in A?

(a)

When chlorine is bubbled through cold aqueous s odium hydroxide, chlorine will undergo disproportionation according to the equation below: Cl2(aq) + 2NaOH(aq)  NaCl(aq) + NaC lO(aq) + H 2O( l) ------ (1) When heated, a further disproportionation reaction occurs: 3NaClO(aq)  2NaCl(aq) + NaC lO3(aq) ------ (2) The oxidation number of C l changes from 0 in C l2 to -1 in NaC l and +1 in NaC lO. The oxidation number of C l changes from +1 in NaC lO to -1 in NaC l and +5 in NaC lO 3. The characteristic smell of C l2 disappears. The yellowish-green gas dissolves into a colourless solution.

(b) Na 16.9

Mass in 100g / g No. of moles Ratio

16.9 23.0



0.7348

0.7348 0.3677

I

H 1.1



2

1.1 

46.7

1.1

1.0

46.7 

0.3677

127

1.1 0.3677

O 35.3



3

16.0

0.3677 

0.3677

35.3

1



2.2063

2.2063 

0.3677

Empirical formula of A: Na2H3IO6 Balanced equation: 7Cl2(g) + I2(aq) + 18NaOH(aq) → 14NaCl(aq) + 2Na 2 H3IO6(s) + 6H2O( l)  be y, Let the oxidation number of I in H3IO 62ֿ 3 × (+1) + y + 6 × (-2) = -2 The oxidation number of I is +7.


6



Question 5 (J89/I/5 modified) State what you would observe in each of the following experiments. Explain these observations with the aid of the Data Booklet where appropriate.

(a)

Solid iodine is shaken with aqueous potassium iodide; aqueous sodium thiosulfate is then added. Shiny black solid iodine dissolves in aqueous potassium iodide to give a brown solution. This is due to the formation of I3 −. 

Upon addition of aqueous sodium thiosulfate, I2 (s) + 2e

⇌

S4O 62(aq) + 2e  o

 E cell

E  = +0.54 V

 (aq) 2Iֿ ⇌

2S2O32(aq)

E = +0.09 V

= +0.45 V > 0 (feasible)

− 2 2 I2 (s) + 2S2O3 (aq)  2I (aq) ＋ S4O6 (aq) Overall: Thus solution turns from brown to colourless.

(b)

Aqueous chlorine is added to aqueous iron( II) sulfate and aqueous sodium hydroxide is then added. Upon addition of aqueous chlorine to aqueous iron(II) sulfate, C l2(aq) + 2e

⇌

Fe3+(aq) + e 

E

o cell

 (aq) 2Clֿ

⇌

Fe2+(aq)

E = +1.36 V E = +0.77 V

= +0.59 V > 0 (feasible)

Overall: C l2 (aq) + 2Fe2+(aq)  2Cl−(aq) + 2Fe3+(aq) Thus solution turns from green to yellow. 

(c)

When aqueous sodium hydroxide is added, reddish-brown precipitate of Fe(OH) 3 formed. Aqueous iron( III) sulfate is added to aqueous potassium iodide and a few drops of starch indicator are then added. Upon addition of aqueous iron( III) sulfate to aqueous potassium iodide, I2 (s) + 2e

⇌

2Iֿ  (aq)

Fe3+(aq) + e

Fe2+(aq)

E = +0.77 V

o

E cell = +0.23 V > 0 (feasible) 3+ 2+  (aq) + 2Fe (aq)  I2 (s) + 2Fe (aq) 2Iֿ 

Overall:

⇌

E  = +0.54 V

Thus solution turns from yellow to brown. Upon addition of starch indicator, a blue-black complex will be formed.




Question 6 (N98/II/2) Bromine has been obtained f rom sea salt for a number of years. For simplification, sea salt can be considered to be sodium chloride containing some sodium bromide. One early method of obtaining bromine is outlined below.

(a)(i) What causes the yellow colour produced in A? Section 4.4.1 (ii) Write an equation for the reaction in A. (i)

Aqueous bromine

(ii) 2Br(aq) + C l2(aq)

(b)



Br2(aq) + 2C l(aq)

Ether is an organic solvent immiscible with water. Explain why the yellow substance dissolves in the ether in B but the salt does not dissolve in ether. Br2 is non-polar and so it dissolves in a non-polar solvent like ether. The salt is ionic and will not dissolve in a non-polar solvent like ether as there is no favourable solute-solvent interaction.

(c)

Suggest an equation for reaction C. Section 4.4.2 Br2(aq) + 2KOH(aq)  KBr(aq) + KBrO (aq) + H2O( l)

(d)

Suggest an equation for reaction D, the reaction of acid on the products of reaction C. Br-(aq) + BrO -(aq) + 2H+(aq)  Br2(g) + H2 O(l)


[reverse of disproportionation]



Question 7 (N06/III/5) When sodium thiosulfate is reacted separately with chlorine and iodine, different products are formed. 4Cl2 + Na2S2O 3 + 5H2O  2NaHSO 4 + 8HC l I2 + 2Na2 S2 O3  Na2 S4 O6 + 2Na I

Use the following data and data from the Data Booklet , to suggest an explanation for this difference. S4O 62− + 2e−

2S2 O32−

+

4SO2 + 4H + 6e

−

S4O6

SO42− + 4H+ + 2e− Cl2 + 2e 

E = +0.09 V 2−

+ 2H2O

SO2 + 2H2O

Cl2 + 2S2O32-  2Cl- + S4O 622-

-

 6Cl

E = +1.27V > 0 (feasible) +

+ 4SO2 + 4H

Cl2 + SO2 + 2H2O  2Cl- + SO42- + 4H+ I2 + 2e 

2-

E = +0.85V > 0 (feasible) E = +1.09V > 0 (feasible) E = +0.54 V

2Iֿ 

⇌

I2 + 2S2O 3  S4 O6

E = +0.17 V E = +1.36 V

2Clֿ 

⇌

3Cl2 + S4O6

E = +0.51 V

2-

+ 2Iˉ

E = +0.45V > 0 (feasible)

2For the reaction between I2 and S4O 6 , E = +0.03V, is very small so reaction does not go to completion.

Question 8 (N02/II/3) In 1819, the element fluorine was known to exist but had not been isolated. The mineral fluorspar, which contained calcium fluoride, CaF 2, had many properties similar to those of other calcium halides.

(a)

What type of bonding is present in calcium fluoride? Ionic bond

(b)

When calcium fluoride reacts with concentrated sulfuric acid, gaseous hydrogen fluoride is produced. A similar reaction occurs with calcium chloride but with calcium bromide, bromine is formed.

(i) (ii) (iii)

Write an equation for the reaction of calcium fluoride with concentrated sulfuric acid. Why, when reacted with concentrated sulfuric acid, does calcium bromide form bromine whereas calcium chloride forms hydrogen chloride? Suggest a possible reaction product of the concentrated sulfuric acid in the reaction (ii) above.




(i)

CaF2(s) + 2H2SO4(l)  2HF(g) + Ca(HSO 4)2(s)

(ii)

Conc. H2SO4 oxidises HBr to Br2, but not HC l because the reducing power of HC l is lower than HBr.

(iii)

Sulfur dioxide; 2HBr(g) + H2SO4(l)  Br2(g) + 2H2O( l) + SO2(g)

Question 9 (N03/III/5 modified) (a)

Describe what you would see if chlorine water is added to aqueous sodium bromide. Write an equation for the reaction. Colourless solution turns yellow-orange. C l2 (aq) + 2NaBr (aq)  2NaCl (aq) + Br2 (aq)

(b)

When chlorine is bubbled through an aqueous solution containing sodium hydroxide and sodium bromide, bromide ions are oxidized to a bromine-containing oxyanion. Careful addition of aqueous silver nitrate to the resulting solution precipitates 4.31 g of AgC l. When the precipitate is filtered off and more silver nitrate solution added, 1.18g of cream solid is produced. The solid had the following composition by mass: Ag, 45.8%; Br, 33.9%; O, 20.3%.

(i)

Calculate the formula of the cream solid. Ag 45.8

Mass in 100g / g No. of moles

45.8 108



Br 33.9

0.4241

Ratio

33.9

0.4241





O 20.3 20.3

0.4243

79.9

16.0

0.4243

1



0.4241



1.269

1.269

1



0.4241

3

0.4241

Formula of cream solid is AgBrO 3.

(ii)

Calculate the number of moles of cream solid and of silver chloride formed. Number of mol of cream solid =

Number of mol of AgC l =

1.18 108  79.9  (3  16.0)

4.31 108  35.5





5.00  10

3

mol

0.0300 mol

(iii) Construct a balanced equation for the reaction between chlorine, sodium bromide and sodium hydroxide. 3Cl2(g) + 6NaOH(aq) + NaBr(aq) → 6NaC l(aq) + NaBrO 3(aq) + 3H2O( l)

Question 10 (N97/II/3)




Potassium chlorate( V), KC lO 3, is widely used in fireworks and match heads. It decomposes when heated with a catalyst to form potassium chloride and oxygen.

(a)

Construct an equation for this decomposition of potassium chlorate( V). 2KClO 3(s)

(b)

(c)

2KCl(s) + 3O2(g)



(i)

Suggest why potassium chlorate( V) is used in fireworks and match heads.

(ii)

Suggest one substance which could be used as the combustible material in fireworks or match heads.

(i)

Potassium ion gives a lilac flame, OR KClO3 produces O 2 which supports combustion.

(ii)

Carbon, or any other combustible materials, e.g. P, S, sugar.

How can potassium chlorate( V) be obtained from chlorine? Illustrate your answer with an equation. Pass chlorine through potassium hydroxide at 70°C: 3Cl2(aq) + 6KOH(aq)  5KCl(aq) + KC lO3(aq) + 3H2O( l)

(d)

The standard redox potential of chlorate( V) ions is given below. ClO 3 + 6H+ + 6e

⇌

Cl + 3H2O

E = +1.45 V

Use the Data Booklet to predict what you would expect to observe when acidified potassium chlorate( V) is added separately to each of the following reagents. Write a balanced equation for any reaction that occurs.

(i)

aqueous iron(II) sulfate 3+

6Fe

+ 6e

⇌

+

2+

ClO3 + 6H + 6e 

E

o cell

E = +0.77 V

6Fe

⇌

Cl + 3H2O E = +1.45 V

= +0.68 V > 0 (feasible)

6Fe2+ + C lO3 + 6H+  6Fe3+ + C l + 3H2O Thus solution turns from green to yellow.

(ii)

acidified potassium manganate( VII) 2+

E (MnO4, Mn ) = +1.52 V No reaction will occur, since both KMnO 4 and KClO 3 are oxidising agents. The purple colour of KMnO 4 remains.


Answers for Group VII Problem-Solving Questions

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