Group VI Chemistry - Suggested Solutions
Topic 9.3: Group VI Chemistry
ACJC/2009/P2/Q2d,e
1
ACJC/2009/P3/Q2f
2
AJC/2009/P3/Q5a,b
3
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Group VI Chemistry - Suggested Solutions
CJC/2009/P3/Q2b
4
(i)
Pass chlorine gas into hot aq. sodium hydroxide
OR Heat NaClO.
3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O
3ClO- 2Cl- + ClO3(ii)
ClO3- + H2O ClO4- + 2H+ 2eOverall: 4ClO3- 3ClO4- + Cl-
(iii)
pH independent.
H+ or OH- is not involved in the overall equation.
HCI/2009/P2/Q5a-b
5 (a) (i) • dehydration of alcohol to form alkene:
CH3CH2OH
conc. H2SO4
H2C CH2 +
OR esterification: RCO2H + R'OH
conc. H2SO4
• nitration of benzene:
+ conc. HNO3
H 2O
RCO2R' + H2O
NO2
conc. H2SO4
OR esterification: RCO2H + R'OH
+ H 2O
conc. H2SO4
OR hydration of alkene:CH2=CH2 + H2O
RCO2R' + H2O
conc. H2SO4
CH3CH2OH
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Group VI Chemistry - Suggested Solutions
(ii)
• H2SO4 (l) + NaBr (s) → HBr (g) + NaHSO4 (aq)
• H2SO4 (l) + 2HBr (g) → Br2 (g) + SO2 (g) + 2H2O (l)
(b)
(i)
Tetrachloromethane / hexane (or any other non-polar organic solvent)
(ii)
The halide is iodide.
2I–(aq) + Cl2(g) → 2Cl–(aq) + I2(aq)
HCI/2009/P3/Q1d
6 (i) Both Cl2 and I2 exist as diatomic molecules and have dispersion forces existing
between molecules.
I, being lower down in the group than Cl, has a bigger electron cloud size and hence,
stronger dispersion forces. Therefore I2 is less volatile than Cl2.
(ii) 4Cl2 (aq) + S2O32– (aq) + 5H2O (l) → 8Cl – (aq)
+ 2SO42– (aq) + 10H+ (aq)
I 2 (aq) + 2S2O32– (aq)
(aq)
→
2I – (aq) + S4O62–
Cl2 + 2e = 2Cl –
I2 + 2e = 2I –
E, / V
+1.36
+0.54
E,Cl2/Cl– is more positive than E,I2/I–, indicating that
Cl2 is a stronger oxidizing agent than I2. Hence, Cl2
can oxidize S2O32– to SO42– while I2 can only oxidize
S2O32– to S4O62–.
JJC/2009/P3/Q3b,c
7
(b)
A white precipitate of AgCl is obtained when sodium chloride is mixed
with silver nitrate. Precipitate dissolves in aqueous ammonia to give a
colourless solution of Ag(NH3)2+. [1]
Ag+(aq) + Cl–(aq) → AgCl(s)
Ag+(aq) + 2NH3(aq) → Ag(NH3)2+(aq) [1]
When NH3 is added, the formation of the complex ion decreases the
concentration of Ag+, causing equilibrium position of the equilibrium AgCl
Ag+ + Cl– to shift to the right such that ionic product < Ksp(AgCl).
[1]
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Group VI Chemistry - Suggested Solutions
(c)
NaAt(s) + H2SO4(l) → HAt(g) + NaHSO4(s)
[1]
2HAt(g) + H2SO4(l) → At2(g) + SO2(g) + 2H2O(l)
8HAt(g) + H2SO4(l) → 4At2(g) + H2S(g) + 4H2O(l) [1]
MI/2009/P3/Q5d
8
2Cl– → Cl2 + 2e
2I– → I2 + 2e
Eθ(Cl–/Cl2) = –1.36 V
Eθ(I–/I2) = –0.54 V
[½]
[½]
I– is a stronger reducing agent than Cl–, i.e. more likely to be oxidised by conc.
H2SO4, to form I2 instead, than Cl–.
I– + H2SO4 → HI + HSO4–
;
8HI + H2SO4 → 4I2 + H2S + 4H2O
;
Pungent smell detected with the formation of a black solid.
;
MJC/2009/P3/Q2
9
a(i)
Gas liberated - Chlorine gas
ClO- (aq) + Cl- (aq) + H2O (l)
Cl2 (g) + 2OH- (aq)
(ii)
Chlorine is only slightly/ moderately soluble in water as the VDW of attraction is not
compatible to the hydrogen bonding between the water molecules
(iii)
Concentrated H2SO4 is only able to oxidize HI to I2 or Br- to Br2
NaCl + H2SO4 → HCl + NaHSO4
NaBr + H2SO4 → HBr + NaHSO4
2 HBr + H2SO4 → Br2 + SO2 + 2 H2O
or
NaI + H2SO4 → HI + NaHSO4
8 HI + H2SO4 → 4 I2 + H2S + 4 H2O
(iv) Thermal stability decreases from HBr to HAt as the covalent bond strength decreases
down the group resulting in the decrease of the bond dissociation energy
BE of HI (+299 kJ mol–1) < BE of HBr (+366 kJ mol–1)
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Group VI Chemistry - Suggested Solutions
RI/2009/P3/Q4a,b
10
(a)
Iodine oxidises thiosulfate to tetrathionate resulting in a change of average
oxidation number from +2 to +2.5:
I2 + 2S2O32− → 2I− + S4O62−
Chlorine oxidises thiosulfate to sulfate(VI) resulting in a change of average
oxidation number from +2 to +6:
4Cl2 + S2O32− + 5H2O → 8Cl− + 2SO42− +10H+
Since chlorine is able to increase the oxidation state of S in thiosulfate more greatly
than iodine, chlorine has higher oxidising power than iodine.
(b)(i) NaCl(s) + H2SO4(l) → NaHSO4(s) + HCl(g)
(ii) Hydrogen iodide formed is oxidised by conc H2SO4 to iodine since HI is easily
oxidised.
(iii) Concentrated phosphoric(V) acid.
SAJC/2009/P2/Q7a,b
11
(a)
From data booklet, atomic size: F 0.072 nm ;
Cl: 0.099 nm ; Br : 0.114 nm ; I : 0.133 nm
As size of halogen atom increases down the group, the bond length of H – X
increases/effectiveness of overlap of orbitals decreases. Bond energy of H – X
decreases and ease of decomposition of HX increases.
Or
From data booklet: Bond energy values: H – F : 562
kJ mol-1; H – Cl : 431 kJ mol-1 ; Br : 366 kJ mol-1 and H – I : 299 kJ mol-1
As the bond energy of H – X decreases down the group, the ease of decomposition
of HX increases. The bond energy decreases down the group as the atomic size of
halogen atoms increases and H – X bond length increases.
(b)
(i)
2 HBr + H2SO4 SO2 + Br2 + 2H2O
(ii)
HI is a stronger reducing agent than HBr
It is able to reduce S from oxidation state +6 (in H2SO4) to -2 (in H2S).
HBr is only able to reduce S to +4 oxidation state (in SO2)
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Group VI Chemistry - Suggested Solutions
SRJC/2009/P3/Q3a
12
(i) Chlorine is used in water purification to kill bacteria or disinfect water
due to its strong and effective oxidising property.
(ii) 2OH- (aq) + Cl2 (aq) → Cl- (aq) + ClO- (aq) + H2O (l)
OR
2 NaOH (aq) + Cl2 (aq) → NaCl (aq) + NaClO (aq) + H2O (l)
TJC/2009/P2/Q1a
13
Number of moles of NaOH used = 0.250 x 0.100 – 0.250 x 10−(14 – 12.0)
= 0.0225 mol
Number of moles of chloride = 2.69 / (107.8 + 35.5)
= 0.0188 mol
Number of moles of chloride : number of moles of NaOH
= 0.0188 : 0.0225
= 5:6
Which is in line with the equation:
3Cl2 + 6OH- → 5Cl- + ClO3- + 3H2O
Number of moles of chlorine gas used = 0.0188 x 3 / 5
= 0.0113 mol
Cl2 + 2OH−
→
Cl− + ClO− + H2O
● Number of moles of chloride formed (cold) = 0.0113 mol
● Mass of silver chloride formed = 0.0113 x (107.8 + 35.5)
= 1.62 g
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Group VI Chemistry - Suggested Solutions
TPJC/2009/P3/Q2a
14
2(a) Reducing power of halides (or ease of oxidation of halides) increases down the group
and this is evident from their decreasing Evalues
OR
Evalues becomes less positive as shown:
E= + 1.36 V
Cl2 (g) + 2e <-----> 2Cl – (aq)
–
E= + 1.07 V
Br2 (g) + 2e <-----> 2Br (aq)
–
I2 (g) + 2e <-----> 2I (aq)
E= + 0.54 V
Sodium chloride
*Sodium chloride reacts with conc. H2SO4 to produce white fumes of hydrogen chloride.
OR
2 NaCl (s) + H2SO4 (l)
NaCl (s) + H2SO4 (l)
2 HCl (g) + Na2SO4 (aq)
NaHSO4 (aq) + HCl (g)
Sodium bromide
*Sodium bromide reacts with conc. H2SO4 to produce white fumes of hydrogen bromide.
Some HBr gas can be further oxidised by concentrated H2SO4 to form reddish brown Br2 gas.
Conc. H2SO4 is reduced to form pungent SO2.
OR
2 NaBr (g) + H2SO4 (l)
NaBr (s) + H2SO4 (l)
2 HBr (g) + H2SO4 (l)
2 HBr (g) + Na2SO4 (aq)
NaHSO4 (aq) + HBr (g)
Br2 (g) + SO2 (g) + 2 H2O (l)
Sodium iodide
*Sodium iodide reacts with conc. H2SO4 to produce white fumes of hydrogen iodide.
Most of the HI can be further oxidised by concentrated H2SO4 to form purple fumes of I2 gas
Conc. H2SO4 is reduced to form pungent H2S.
OR
2 NaI (g) + H2SO4 (l)
NaI (s) + H2SO4 (l)
8 HI (g) + H2SO4 (l)
2 HI (g) + Na2SO4 (aq)
NaHSO4 (aq) + HI (g)
4 I2 (s) + H2S (g) + 4 H2O (l)
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Group VI Chemistry - Suggested Solutions
YJC/2009/P2/Q1b
15
YJC/2009/P3/Q4a
16
166