The document describes several chemical reactions involving group VII elements and transition metals. It provides reaction equations to show how concentrated sulfuric acid reacts with potassium halides to produce hydrogen halides. It also describes how transition metal ions like iron and copper form colored complexes in aqueous solution due to d-d electron transitions. The reactions of cobalt ion as a catalyst in the oxidation of iodide by persulfate are also outlined.

1. GROUP II
AJC/P2/3
CJC/P3
1 (a) (i) MgO (s) + 2HNO3 (aq) → Mg(NO3)2 (aq) + H2O (l)
0.500
Amount of MgO = = 0.0124 mol
24.3 + 16.0
50
Amount of HNO3 = × 0.100 = 0.00500 mol (limiting reagent)
1000
0.005
Amount of Mg(NO3)2 produced = = 0.00250 mol
2
(a) (ii) Mg(OH)2 + 2NH4NO3 → Mg(NO3)2 + 2NH3 + 2H2O
1.5
Amount of Mg(NO3)2 = = 0.0101 mol
24.3 + 28.0 + 96.0
Volume of NH3 produced = 0.0101 × 2 × 24 = 0.485 dm3
(iii) 2Mg(NO3)2 → 2MgO + 4NO2 + O2
Brown gas of NO2 will be observed.
(iv) Ba(NO3)2 is more thermally stable than Mg(NO3)2, hence it will not
decompose at 330°C.
Ba2+ is less polarising than Mg2+ as it has a larger size, hence the
NO3- is polarised to a smaller extent in Ba(NO3)2.
YJC/ JC2/ 2008 1

2. HCI/P2
OR
YJC/ JC2/ 2008 2

3. RJC/P2/Q3
heat 1
c(i) M(NO3)2(s)  MO(s) + 2NO2(g) +
→ O2(g)
2
(ii) Amt of nitrate = 1/2 x Amt of NO2 = ½ x 1.84/(14.0 + 2(16.0)) = 0.0200 mol
Mr of nitrate = 2.96/0.0200 = 148.0 g mol─1
Mr of M = 148.0 – 2(14.0 + 3 x 16.0) = 24.0 g mol─1
Hence, M is magnesium.
(d) Lattice energy cannot be used to determine relative thermal stability of Group II
nitrates.
Barium nitrate has a higher decomposition temperature because
Ba2+ is a larger ion or Ba2+ has a lower charge density, therefore Ba2+ has lower
polarising power.
Hence, Ba2+ polarises the electron cloud of the nitrate ion less strongly and
weakens the N–O bonds to a lesser extent in barium nitrate.
IJC/P3/Q3
YJC/P2/3
ai Magnesium
Mg(s) + H2O(l) MgO(s) + H2(g)
Mg reacts with slowly with cold water to give H2(g).
Y, Y(s) + H2O(l) Y(OH)2(s) + H2(g), Y reacts vigorously with cold water to give
hydrogen gas
ii. The more negative the reduction potential, the greater the reducing power as
the valence electrons are more readily lost.
b. Down the gp, size of cation increases and the charge density decreases,
polarising power of Y2+ decreases [1m] and C-O bond is less polarised [1m] and
require more energy to break, therefore it has a high decomposition
temperature.
ci. By losing 2 electrons, Be forms a stable duplet structure
ii Be2+ has a highest charge density (high charge and smallest size). It polarises
the oxide ion significantly which leads to covalency . Hence, the oxide is ionic
with covalent character and is amphoteric.
YJC/ JC2/ 2008 3

4. TJC/P3
2a) • Down Group II, the thermal stability of the nitrates increases.
• Down the group, since the charge density of the M2+ decreases, the
M2+ becomes less polarising. Thus the covalent bonds within the
NO3− anion is the least polarised with Ba2+ and hence would require
greater heat energy to decompose into simpler substances.
• 2 M(NO3)2 → 2 MO + 4 NO2 + O2
b) • Mass of MO residue obtained = (100 − 65.8) × 1.00
100
= 0.342 g
Number of moles of MO obtained = 0.342 ÷ (Ar M + 16.0)
• Since M(NO3)2 ≡ MO,
1.00 ÷ (Ar M + 124.0) = 0.342 ÷ (Ar M + 16.0)
• Ar M = 40.1
Hence the metal is calcium.
ci) • BaO + H2SO4 → BaSO4 + H2O
• BaO2 + H2SO4 → BaSO4 + H2O2
ii) Using the Data Booklet, H2O2 ≡ 2I− ≡ I2
• Number of moles of H2O2 produced = 1.18 × 10−3 mol
Hence, number of moles of BaSO4 produced from BaO2 = 1.18 × 10−3
mol
• Mass of BaO2 present in 1.00 g solid = 1.18 × 10−3 × 169.0
= 0.199 g
[Mr BaO2 = 169.0]
• Mass of BaO present in 1.00 g solid = 1.00 − 0.199
= 0.801 g
VJC/P2/Q2/b
YJC/ JC2/ 2008 4

5. IJC/P3
MJC/P2/Q2b
i
ii.
iii
YJC/ JC2/ 2008 5

6. GROUP VII
CJC/P2
2 (a)(i) HCl is a weaker acid than HI because the H-Cl bond is stronger than H-I
bond .
Hence HCl dissociate to give H+ to a smaller extent than HI. (Or vice versa)
HX + H2O H3O+ + X- (or) HX H+ + X-
(ii) Cl- + H2SO4 HCl + HSO4-
I- + H2SO4 HI + HSO4-
8 HI + H2SO4 4I2 + 4H2O + H2S
Cl- is a weaker reducing agent than I-. con. sulphuric acid is strong enough
to oxidise HI to I2
AJC/P3
NJC/P2/
YJC/ JC2/ 2008 6

7. SRJC/P3/3
ai
White / steamy fumes of HCl will be observed
HCl is not oxidized by concentrated H2SO4, no Cl2 obtained.
KCl (s) + H2SO4(l) KHSO4 (aq) + HCl (g)
ii
Concentrated H2SO4 reacts with KI to form HI.
KI(s) + H2SO4 (aq) KHSO4 (aq) + HI (g)
HI is then oxidized by concentrated H2SO4 to give the violet fumes, I2
and the pungent gas, SO2.
2HI (g) + H2SO4 (aq) I2 (g) + SO2 (g) + 2H2O (l)
iii Test the fumes with moist starch paper.Starch paper should turn
black/blue/blue-black.
bi
Bottle C is Cl2 (aq) which oxidized / displaced Br- (aq) to Br2 (aq) and I-
(aq) to I2 (aq) which account for the brown mixture obtained.
Cl2 (aq) + 2Br- (aq) 2Cl- (aq) + Br2 (aq)
Cl2 (aq) + 2I- (aq) 2Cl- (aq) + I2 (aq)
ii. Add hexane to Bottle A and B after reaction with Cl2 (aq). Shake and
allow to stand
The bottle with reddish brown organic layer contains KBr (aq) initially.
The bottle with violet / purple organic layer contains NaI (aq) initially.
YJC/ JC2/ 2008 7

8. IJC/P3/4
RJC/P2/3a
i. Add hot, aqueous sodium hydroxide to aqueous chlorine, which will then
disproportionate to form aqueous sodium chloride and aqueous sodium
chlorate(V).
ii. Ethanedioic acid is functioning as the reducing agent because the oxidation state
of chlorine decreased from +5 in ClO3– to +4 in ClO2.
iii. 2ClO2(g) + H2O(l)  ClO2–(aq) + ClO3–(aq) + 2H+(aq) E
→ ££¢¡
= –0.20 V
Since E ££¢¡
= E R – E L +0.95 V – (+1.15 V) 0 V, the disproportionation of
ClO2(aq) to ClO2–(aq) and ClO3–(aq) is energetically not feasible and would not
occur under standard conditions.
YJC/P3/3
3 (a) (i) Cl2 + 2OH- ClO- + Cl- + H2O [2]
3Cl2 + 6OH- ClO3- + 5Cl- + 3H2O [1 m for each equation]
(ii) [1m] for multiplying 2 to amount of X- and SO42- [3]
S2O32- : X- : SO42- : H+
1 : 8 : 2 : 10 [1m]
4X2 + S2O32- + 5H2O 2SO42- + 8X- + 10H+ (1m)
(iii) Y2 + 2S2O32- 2Y- + S4O62- [1m] [2]
The change in oxidation state of S is smaller in (ii) compared to (iii).
Hence X2 has stronger oxidising power. [1m]
YJC/ JC2/ 2008 8

9. TPJ/P2
6(a) Boiling points increase as you go down group
Electron cloud / size of molecule / atomic radius increases going down
group
Id-id VDW intermolecular forces increases going down group
More energy required to overcome intermolecular forces
(b) (i) H2 + Cl2 2HCl
white fumes are evolved
(ii) Going down group,
Electron affinity decreases
Strength of H-X bond decreases
Tendency of halogen to be reduced decreases
OR Oxidation strength of halogen decreases
(c) (i) NaCl + H2SO4 NaHSO4 + HCl
(ii) NaI + H2SO4 NaHSO4 + HI
8HI + H2SO4 4I2 + H2S + 4H2O
Going down group,
Reducing power of halogen increases
HI is oxidized to iodine, hence pure HI cannot be obtained.
HCI/P3/Q2
YJC/ JC2/ 2008 9

10. TRANSITION METALS
AJC/P3/2
HCI/P3/Q3
YJC/ JC2/ 2008 10

11. HCI/P3/4
YJC/ JC2/ 2008 11

12. MJC/P2
bi
bii
ci
ii
di
dii
diii
YJC/ JC2/ 2008 12

13. PJC/P3
a
bi
SAJC/P3
2(d) The five 3d orbitals of d-block free metal ions are degenerate (at same energy
level). When they are bonded to ligands, the electronic repulsion between the
ligands and the metal ion causes the 3d orbitals to undergo splitting, results in 2
groups of non-degenerate orbitals. called the d-d splitting. By absorption of
energy, electrons can be promoted from a lower energy d-orbital to a higher
energy d-orbital. This is called the d-d transition. This energy is related to the
wavelength of the light absorbed, and the light not absorbed is thus seen as the
colour of the complex
Sc and Zn do not form coloured complexes because Sc3+ [Ar]3d0, does not have
3d electrons and Zn2+ [Ar]3d10, has completely- filled 3d-subshell . Hence d-d
transition is not possible.
2(e)(i) Copper(II) sulphate salts dissolve in water to give a blue colour solution,
which is attributed to [Cu(H2O)6]2+ (aq)
CuSO4 (s) + 6H2O(l) Cu(H2O)6]2+ (aq) + SO42- (aq)
Blue solution
When dilute aqueous NH3 is gradually added, a blue precipitate of Cu(OH)2 is first
formed.
NH3 + H2O ⇔ NH4+ + OH-
[Cu(H2O)6]2+(aq) + 2OH-(aq) Cu(OH)2(s) + 6H2O
YJC/ JC2/ 2008 13

14. blue ppt.
When excess aqueous NH3 is added, the precipitate dissolves to give a deep blue
solution, due to the formation of the soluble complex ion, [Cu(NH3)]2+.
Cu(OH)2 + 4NH3 + 2H2O ⇔ [Cu(NH3)4(H2O)2]2+ + 2OH-
deep blue soln.
NH3 is a stronger ligand than H2O and so, displaces water from [Cu(H2O)6]2+ to
form [Cu(NH3)4(H2O)2]2+.
NB: The identification of the coloured compounds is important.
2(f) The uncatalysed reaction is slow due to high activation energy since 2
negatively ions S2O82- and I- are involved and they repel each other.
The Co3+ acts as a homogenous catalyst. The catalysed pathway involves two
steps:
Step 1: 2Co3+ + 2I- 2Co2+ + I2 E(cell)= +1.28V
2+ 2-
Step 2: 2Co + S2O8 2Co3+ + 2SO42- E(cell) = +0.19V
Overall: S2O82- + 2I- 2SO42- + I2
Both steps in the catalysed reaction involve the interaction of oppositely charged
ions, which attract one another strongly; hence the activation energy is lower and
enhances the rate of reaction.
TJC/P3
1ai) • Colour of Fe2+ (aq): green
Colour of Fe3+ (aq): yellow
ii) • For transition metals ions e.g. Fe3+ (aq), In the presence of water
ligands, the degeneracy of the 5 3d orbitals is raised and split into
two energy levels due to the repulsion of the electrons in the metal
ion and the lone pairs on the water ligands.
• When an electron is promoted from the d-orbital of lower energy to
one of higher energy (d-d transition), an amount of energy, , which
happens to be in the visible region of the electromagnetic spectrum,
has to be absorbed.
• The colour observed is complement to the light energy absorbed.
bi) • Oxidation state of iron in Na2FeO4 = +6
8OH- + Fe3+ → FeO42- + 4H2O + 3e
•
2e + H2O + OCl- → Cl- + 2OH-
• Overall: 10OH- + 2Fe3+ + 3OCl- → 2FeO42- + 5H2O + 3Cl-
YJC/ JC2/ 2008 14

15. VJC/P3
2ai.
ii
b
c
TPJC/P3/1
(d) Cu2+ ion has the electronic configuration of [Ar] 3d9. When CuSO4(s) is
dissolved in water, Cu2+ ions are hydrated by water ligands. This causes d-
orbital splitting and enables electrons from the orbital of lower energy to
absorb light energy and jump to the partially-filled orbital of higher energy.
Light is absorbed form red spectrum. Hence, a blue solution is seen.
YJC/ JC2/ 2008 15

16. RJC/P3/3
(a)(i) Electronic configurations:
2+ 5 2+
25Mn : [Ar]3d ; 26Fe : [Ar]3d6 ; 27Co
2+
: [Ar]3d7
Third ionisation energy of iron is lower than 3rd ionisation energy of Co since Co2+
has one more proton than Fe2+ and hence greater effective nuclear charge than Fe2+
(shielding being approximately the same).
Despite its greater effective nuclear charge, Fe has lower 3rd IE than Mn. This is
because the electron to be removed in Fe2+ is from a paired orbital and is aided by
electron−electron repulsion.
(ii) With reference to the Data Booklet,
2−
− 2− O
S2O8 + 2e− 2SO4 −
E = +2.01 V
3+ 2+ O
Fe + e− Fe E = +0.77 V
O
I2 + 2e− 2I− E = +0.54 V
2−
The reaction between S2O8 − and I− is kinetically slow as it involves reaction
between 2 species which are negatively charged.
3+ 3+ 2+
On addition of Fe , Fe oxidises I− to I2 and itself is reduced to Fe . The reaction
is spontaneous (E cell = +0.77 − 0.54 = +0.23 V0) and kinetically favourable as it
involves species which are oppositely charged.
2+ 2− 2− 3+
The Fe formed then reduces S2O8 − to SO4 − and itself is oxidised back to Fe .
The reaction is spontaneous (E cell = +2.01 − 0.77 = +1.24 V0) and kinetically
favourable as well as it also involves species which are oppositely charged.
3+
Since Fe is regenerated, it is chemically unchanged and only a small amount is
needed for it to catalyse the reaction.
2+ 3+ 2+
(iii) Both aqueous Fe and Fe exist as aqua complexes with formulae [Fe(H2O)6]
3+
and [Fe(H2O)6] respectively.
3+ 3+ 3+
Due to the higher charge density of Fe , water molecules coordinated to Fe in [Fe(H2O)6] is
+
more polarised so that more H3O are produced due to greater hydrolysis:
3+ 2+ +
[Fe(H2O)6] (aq) + H2O (l) [Fe(H2O)5(OH)] (aq) + H3O (aq)
Hence, for the same concentration, pH of aq FeCl3 is lower than pH of aq FeCl2 as
+
concentration of H3O in aq FeCl3 is higher.
NJC/P3/4a.
b.
YJC/ JC2/ 2008 16

17. ci
cii
YJC/ JC2/ 2008 17