YJC/ JC2/ 2008 1

GROUP II

AJC/P2/3

CJC/P3

1 (a) (i) MgO (s) + 2HNO3 (aq) →→→→ Mg(NO3)2 (aq) + H2O (l)

Amount of MgO = 0124.00.163.24

500.0=

+mol

Amount of HNO3 = 00500.0100.01000

50=× mol (limiting reagent)

Amount of Mg(NO3)2 produced = 00250.02

005.0= mol

(a) (ii) Mg(OH)2 + 2NH4NO3 →→→→ Mg(NO3)2 + 2NH3 + 2H2O

Amount of Mg(NO3)2 = 0101.00.960.283.24

5.1=

++mol

Volume of NH3 produced = 2420101.0 ×× = 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 2

HCI/P2

OR

YJC/ JC2/ 2008 3

RJC/P2/Q3

c(i) M(NO3)2(s) heat→ MO(s) + 2NO2(g) +

1

2O2(g)

(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 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 5

IJC/P3

MJC/P2/Q2b i

ii.

iii

YJC/ JC2/ 2008 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 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 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

3Cl2 + 6OH- ���� ClO3- + 5Cl- + 3H2O [1 m for each equation]

[2]

(ii) [1m] for multiplying 2 to amount of X- and SO42-

S2O32- : X- : SO4

2- : H+ 1 : 8 : 2 : 10 [1m] 4X2 + S2O3

2- + 5H2O ���� 2SO42- + 8X- + 10H+ (1m)

[3]

(iii) Y2 + 2S2O32- ���� 2Y- + S4O6

2- [1m] The change in oxidation state of S is smaller in (ii) compared to (iii). Hence X2 has stronger oxidising power. [1m]

[2]

YJC/ JC2/ 2008 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 10

TRANSITION METALS

AJC/P3/2

HCI/P3/Q3

YJC/ JC2/ 2008 11

HCI/P3/4

YJC/ JC2/ 2008 12

MJC/P2

bi

bii

ci

ii

di

dii

diii

YJC/ JC2/ 2008 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 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 S2O8

2- 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 Step 2: 2Co2+ + S2O8

2- ���� 2Co3+ + 2SO42- E(cell) = +0.19V

Overall: S2O82- + 2I- ���� 2SO4

2- + 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+ → FeO4

2- + 4H2O + 3e

•••• 2e + H2O + OCl- → Cl- + 2OH-

•••• Overall: 10OH- + 2Fe3+ + 3OCl- → 2FeO42- + 5H2O + 3Cl-

YJC/ JC2/ 2008 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 16

RJC/P3/3 (a)(i) Electronic configurations:

25Mn2+

: [Ar]3d5 ; 26Fe

2+: [Ar]3d

6 ; 27Co

2+: [Ar]3d

7

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,

S2O82−−−− + 2e−−−− 2SO4

2−−−− EO = +2.01 V

Fe3+

+ e−−−− Fe2+

EO = +0.77 V

I2 + 2e−−−− 2I−−−− EO = +0.54 V

The reaction between S2O82−−−− and I−−−− is kinetically slow as it involves reaction

between 2 species which are negatively charged.

On addition of Fe3+

, Fe3+

oxidises I−−−− to I2 and itself is reduced to Fe2+

. The reaction

is spontaneous (E����

cell = +0.77 −−−− 0.54 = +0.23 V>0) and kinetically favourable as it

involves species which are oppositely charged.

The Fe2+

formed then reduces S2O82−−−− to SO4

2−−−− and itself is oxidised back to Fe3+

.

The reaction is spontaneous (E����

cell = +2.01 −−−− 0.77 = +1.24 V>0) and kinetically

favourable as well as it also involves species which are oppositely charged.

Since Fe3+

is regenerated, it is chemically unchanged and only a small amount is needed for it to catalyse the reaction.

(iii) Both aqueous Fe2+

and Fe3+

exist as aqua complexes with formulae [Fe(H2O)6]2+

and [Fe(H2O)6]

3+ respectively.

Due to the higher charge density of Fe3+

, water molecules coordinated to Fe3+

in [Fe(H2O)6]3+

is

more polarised so that more H3O+ are produced due to greater hydrolysis:

[Fe(H2O)6]3+

(aq) + H2O (l) [Fe(H2O)5(OH)]

2+ (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 17

ci

cii