2008 H2 Chemistry Preliminary Examinations Paper 2 Mark Scheme

1 (a) On heating, Group I metal nitrates such as sodium nitrate(V) decompose

giving the metal nitrate(III) and oxygen, while Group II metal nitrates, for
example magnesium nitrate(V), decompose giving different products.

(i) Write balanced equations for the decomposition of sodium nitrate(V)

and magnesium nitrate(V) respectively.

NaNO3 NaNO2 + ½O2

Mg(NO3)2 MgO + 2NO2 + ½O2

15.35 g of a mixture of sodium nitrate(V) and magnesium nitrate(V) was

heated in a fume cupboard until no more gases were evolved.

The water soluble part of the residue was dissolved in water to prepare
1.00 dm3 of solution.

10.00 cm3 of this solution was reacted with 20.00 cm3 (in excess) of
0.0200 mol dm-3 potassium manganate(VII) solution, acidified with dilute
sulphuric acid.

(ii) The nitrate(III) half equation is

NO2- + H2O NO3- + 2H+ + 2e-.

Write a balanced equation for the reaction between nitrate(III) ions

and manganate(VII) ions.

2MnO4- + 5NO2- + 6H+ 2Mn2+ + 5NO3- + 3H2O

The excess potassium manganate(VII) required 12.00 cm3 of

0.0500 mol dm-3 ethanedioic acid solution for complete reaction.

[ 2MnO4- + 5C2O42- + 16H+ 2Mn2+ + 10CO2 + 8H2O]

(iii) Calculate the amount in moles of the nitrate(III) ions in the 10.00 cm3
solution.

no of mols of C2O42- ions in 12.00 cm3 = 0.012 x 0.05 = 6.00 x10-4

no of mols of excess MnO4- = (2/5) x 0.0006 = 2.40 x 10-4
no of mols of MnO4- in 20.00 cm3 = 0.02 x 0.02 = 4.00 x 10-4

no of mols of MnO4- reacted with NO2- in 10.00 cm3

= 4 x 10-4-2.40 x 10-4
= 1.60 x 10-4 l
no of mols of NO2- in 10.00 cm3
= (5/2) x 1.60 x 10-4 = 4.00x 10-4

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 (iv) Hence, calculate the mass of each nitrate in the mixture.

NaNO3≡ NaNO2

mass of NaNO3 in 1 dm3

= (1000/10) x 4.00 10-4 x (23.0+14.0+16.0x3)
= 0.0400 x 85 = 3.40g

mass of Mg(NO3)2 = 15.35-3.40 =12.0g ( 3s.f.) (or 11.95 g)

(b) Magnesium nitrate(V) and strontium nitrate(V) decompose similarly on

heating. Magnesium nitrate decomposes at a lower temperature than
strontium nitrate.
Explain why these two nitrates decompose at different temperatures.

• Mg2+ has a smaller ionic radius than Sr2+

• higher charge density and greater polarising power
• hence Mg2+ distorts electron cloud around NO3- to a greater
extent, thus decomposing at a lower temperature.

(c) Ammonium nitrate(V) decompose to produce nitrous oxide, N2O.

Nitrous oxide is relatively inert at room temperature but at 500oC, it
decomposes to oxygen, nitrogen and nitric oxide, NO. In the spaces
provided, draw the dot and cross diagrams of these two oxides of nitrogen.

Formula Nitrogen Dot and Cross Diagram

oxidation
state
N2O +1
x
N xx N O x
x x
Nx N O x
x x
x x
OR

NO +2
x xx
x x
x N O x
OR
x
x
x
N O

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2 The first ionisation energies of nine elements from sodium to potassium are
shown in the sketch below.

Ar
(a)
Cl
•

First IE / kJ mol-1
P •
•
Si •S
Mg •
•
•
• Al
Na
•K
11 12 13 14 15 16 17 18 19
number of protons

Give reasons for:

(i) the general trend across the period from Na to Ar.
First IE increases across a period.
Across a period, number of protons increases.
Shielding effect remains the same as number of inner shell electrons
is the same. Hence effective nuclear charge on valence electrons is
stronger across a period. Atomic size is also smaller. More energy is
required to remove an outermost electron due to increasing
electrostatic attraction.
(ii) the discontinuity between Mg and Al.

Mg 1s2 2s2 2p6 3s2

Al 1s2 2s2 2p6 3s2 3p1

Al has a lower first IE. Less energy is required to remove a 3p

electron in Al than a 3s electron in Mg since 3p electron is further
away from the nucleus.

(iii) the discontinuity between P and S.

P 1s2 2s2 2p6 3s2 3px1 3py1 3pz1 OR 1s2 2s2 2p6 3s2 3p3
S 1s2 2s2 2p6 3s2 3px2 3py1 3pz1 1s2 2s2 2p6 3s2 3p4

S has a lower first IE. Less energy is required to remove an electron

from paired 3p electrons in S since inter-electron repulsion is
experienced between the paired electrons.

(iv) the difference between the first ionisation energies of Na and K.

Na 1s2 2s2 2p6 3s1

K 1s2 2s2 2p6 3s2 3p6 4s1

K has a lower first IE than Na. Less energy is required to remove a 4s

electron in K than a 3s electron in Na since 4s electron is further away
from the nucleus.

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(b) (i) Sketch the melting point trend of elements in period 3.
Si
•

Melting point / ºC
Mg •Al
• S
•
•
Na •P

Cl • Ar•
11 12 13 14 15 16 17 18
number of protons
(ii) Explain the difference in melting points for sodium and silicon in terms of
their structures and bondings.
Silicon has higher melting point than sodium.
Silicon has covalent bond and giant molecular structure.
Sodium has metallic bond and giant lattice structure .
More energy is required to break the relatively stronger bonds in
silicon than in sodium.
atomic radius of silicon is smaller than atomic radius of sodium;
hence silicon atoms are more closely packed together.

(c) Sodium was first produced commercially in 1855 by thermal reduction of sodium
carbonate with carbon in what is known as the Deville process.

Na2CO3(l) + 2C(s) → 2Na(g) + 3CO(g)

The standard entropy change of reaction, ∆Srθ, is +549 J K-1 mol-1

(i) Explain why the entropy change of above reaction is positive.

Change in phase: Sgas >> Sliquid >> Ssolid
There is a change in phase from solid or liquid reactants to gaseous
products. Entropy of a gas is much greater than that of a liquid or
solid as its particles are free to move and system becomes more
disorderly.
Entropy increases as the number of gaseous particles increases and
system becomes more disorderly.

(ii) Determine the range of temperatures for the above reaction to be feasible.

Na2CO3(l) CO(g) Na(g)

ΔHfӨ / kJ mol-1 -1103 -111 +107

ΔHrӨ = ∑nΔHfӨ(products) − ∑mΔHfӨ(reactants)

= 2 ΔHfӨ(Na) + 3 ΔHfӨ(CO) - ΔHfӨ(Na2CO3) - 2 ΔHfӨ(C)
= (2)(107) + (3)(-111) – (-1103) – 0
= +984 kJ mol-1

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 Assume that ΔH and ΔS remain constant.
For reaction to be feasible, ΔGr < 0
ΔHr - TΔSr < 0
+984,000 - (T)(549) < 0
T > 1792 K

3 (a) (i) Calcium is a fairly soft, silvery-grey metal which quickly tarnishes in
air; hence metallic calcium has no commercial uses. However
titanium is a commercially important engineering metal.

State two physical properties which make titanium a very useful

material in the aircraft industry and suggest another property that
allows titanium to be used in artificial hip joints.

(Low density/ light) ; (strong/high tensile strength)

Does not corrode

(ii) Calcium can only exist as Ca2+ ions in its compounds but titanium
forms ions with different charges (+2,+3 and +4) in its solid
compounds.
TiCl3 is coloured while TiF4 ( an ionic compound) is a white powder.

Unlike calcium, explain why titanium can exhibit several oxidation

states in its compounds and suggest why TiF4 is not coloured.

Ti : [Ar]3d24s2. As both 4s and 3d electrons are close in energy,

it can use both 4s and 3d electrons for bond formation leading
to several possible oxidation states.

In the +4 oxidation state, Ti has no 3d electron and hence d-d

transition is not possible.

(b) Vanadium was named after Vanadis, the Scandinavian goddess of beauty
and it forms many coloured compounds. Using the following data, choose
a reagent which will convert vanadium(V) to vanadium(IV) but not to
vanadium(III). Write a balanced equation for the conversion.
Eθ/V
VO2+ + 2H+ + e- V3+ H2O +0.34
+ + - 2+
VO2 + 2H + e VO + H2O +1.00
2- + -
SO4 + 4H + 2e SO2 + 2H2O +0.17
Fe3+ + e- Fe2+ +0.77

VO2+ + 2H+ + Fe2+ VO2+ + Fe3+ + H2O

Ecell = +1.00 - 0.77 = + 0.23V > 0

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(c) Some data for iron and ruthenium are given below.
Proton number Electronic configuration
Fe 26 1s22s22p63s23p63d64s2
Ru 44 1s22s22p63s23p63d104s24p64d75s1

[Ru(H2O)6]3+ + e- [Ru(H2O)6]2+ Eθ = + 0.23V

[Fe(H2O)6]3+ + e- [Fe(H2O)6]2+ Eθ = + 0.77V
2+
(i) Write the electronic configuration of Ru (aq) and hence explain why
Ru2+(aq) is less stable than Ru3+(aq).

1s22s22p63s23p63d104s24p64d6

In +2 oxidation state, there is a pair of electrons in one of the 4d

orbitals which results in inter-electronic repulsion.

(ii) [Ru(H2O)6]3+ can be made from the complex [Cl5Ru-O-RuCl5]4-.

What is the oxidation state of ruthenium in [Cl5Ru-O-RuCl5]4-?
+4

(iii) Cyanide ligands can form a very stable complex with Fe2+ ions. How
can the formation of this very stable complex explain the highly
poisonous nature of the cyanide ion?

Fe2+ is important as an O2 carrier in haemoglobin. CN- binds to

it preventing O2 from being carried, causing suffocation.

(d) (i) 2,3-dihydroxybutanedioate ions are oxidised by hydrogen peroxide to

carbon dioxide and water. The reaction is catalysed by Co2+(aq). The
solution is pink at the beginning and end of the reaction, but green
during it.
The redox reaction between 2,3-dihydroxybutanedioate ions and
hydrogen peroxide is represented by the following half-equations:
(CHOHCO2-)2(aq) + 2H2O(l) 4CO2(g) + 8H+(aq) + 10e-
2H+(aq) + H2O2(aq) + 2e- 2H2O(l)
Suggest the mechanism by which Co2+(aq) ions catalyse this
reaction.

Step 1
2Co2+ + 2H+ + H2O2 2Co3+ + 2H2O

Step 2
(CHOHCO2-)2 + 2H2O + 10Co3+ 4CO2 + 8H+ + 10Co2+

(ii) The typical blue colour of the famous Delft pottery is due to the Co2+
ions which are incorporated in the thin layer of glaze on the pottery.
Explain why cobalt compounds are usually coloured.
• In cobalt compounds, the five 3d orbitals/ 3d subshell are split
into two energy levels with an energy difference ΔE
corresponding to that of visible light energy
• When 3d e- from the lower 3d orbital is promoted to a vacancy
in the upper 3d orbital, visible light energy ΔE is absorbed.
• The colour seen is complementary to the colour absorbed.

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4 (a) 1.00 g of magnesium ammonium phosphate, MgNH4PO4, was added to 50 cm3
of a 1.00 mol dm-3 aqueous solution of sodium hydroxide. The mixture was
then boiled.
(The values of Ksp of magnesium hydroxide = 1.0 x 10-11 and of
Kb of ammonia = 1.0 x 10-5)

(i) Predict whether a precipitate of magnesium hydroxide would form in the

mixture above.

Mr of MgNH4PO4, = 137.3
Amt of MgNH4PO4, = 1/137.3 = 0.00728 mol
[Mg2+] = 0.00728 / 0.05 mol dm-3 = 0.146 mol dm-3

[Mg2+] [OH-]2 = 0.146 x (1.00)2

= 0.146 >> Ksp of Mg(OH)2 .
A ppt would form
(Taking into account the amount of NaOH that reacts with NH4+ , the
[OH-]remaining works out to be 0.85 mol dm-3. Using this value to work out
the IP of Mg(OH)2 is absolutely correct albeit tedious. The approximate
value of [OH-]= 1.00 mol dm-3 in the above calculation does not detract
much from the actual value)

(ii) If all the ammonia liberated from the mixture were completely dissolved in
water to give a 50.0 cm3 solution, what volume of 0.20 mol dm-3 aqueous
hydrochloric acid would be needed for complete reaction with the aqueous
ammonia solution?
NH3 ≡ HCl ≡ Mg2+
Amt of HCl needed = 0.00728
Vol of acid needed = 0.00728 / 0.20 = 0.0364 dm3 (or 36.4 cm3)

(iii) Hence, calculate the pH value of the mixture in (ii) after hydrochloric acid
has completely reacted with it.

Total vol = 50 +36.4 = 86.4 cm3

[NH4+(aq)] = 0.00728 mol / 0.0864

NH4+ (aq) + H2O(l) NH3(aq) + H3O+(aq)

+
Ka = [H3O (aq)] [NH3 aq)]
[NH4+(aq)]

= [H3O+(aq)]2
[NH4+(aq)]

[H3O+(aq)]2 = (10-14 / 10-5) ( 0.00728 mol / 0.0864)

[H3O+(aq)] = 9.18 x 10-6 pH = 5.04

(b) The conversion of A2 is as follows:

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 A2(g) 2A(g)

The conversion was studied using a fixed amount of A2 in a reaction vessel. At

different times during the experiment, changes were made to the conditions in the
reaction vessel. The change in the concentrations in the equilibrium mixture with
time is given by the graph below:
Concentration

A2

Time
T1 T2 T3

Suggest the change in condition that caused the change at time

(i) T1.
The volume of the mixture was reduced/ the mixture was compressed

(ii) T2

A2 was added to the mixture at equilibrium

(iii) Explain whether you expect the conversion of A2 to A to be exothermic or

endothermic. Sketch on the graph above the changes in the concentrations
of A2 and A when the mixture was cooled at time T3.

Endothermic. Involves bond breaking

5 (a) An organic compound X (spirit of amber) plays an important biochemical role in the
Krebs cycle. It is also produced in the fermentation of sugar and gives wine a
characteristic flavour.
(i) Compound X has the following composition by mass.

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 C, 40.7%; H, 5.1%; O, 54.2%

Calculate the empirical formula of X.

Assume 100 gram of sample

C H O
Mass / g 40.7 5.1 54.2
Amount / mol 3.39 5.1 3.39
Simplest Mole Ratio 1 1.5 1
2 3 2

Empirical Formula of X: C2H3O2

(Incomplete presentation of table: -1m)

(ii) When a 0.204g sample of compound X was vapourised in a suitable
apparatus, the vapour occupied 74.2 cm3 at 250 ºC and 101 kPa.

Calculate the Mr of compound X.

Assume X is an ideal gas

PV = nRT

(101,000 ) ( 74.2 × 10−6 ) = ⎛⎜

0.204 ⎞
⎟ ( 8.31)( 250 + 273 )
⎝ Mr ⎠
Mr = 118.3
(iii) Determine the molecular formula of X.
Let molecular formula of X be (C2H3O2)n

[(2 x 12) + (3 x 1) + (2 x 16)] n = 118.3

n=2
Molecular formula of X: C4H6O4

(b) The following tests are carried out on compound X. State all deductions about
compound X in each case.

(i) When aqueous sodium carbonate is added to compound X, effervescence

occurs. Colourless gas forms white precipitate in lime water.

X is a carboxylic acid.

(ii) Compound X gives white fumes when treated with thionyl chloride, SOCl2.

X has –OH or –COOH groups.

(iii) On heating compound X under reflux with potassium manganate (VII),

purple colour of solution remains.

X is neither an alkene, an aldehyde nor (primary or secondary alcohol).

(iv) Compound X does not give yellow precipitate with aqueous alkaline iodine.

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 X does not have this structure: CH3CO─ or CH3CH(OH)─

(v) Compound X does not give orange precipitate with

2,4-dinitrophenylhydrazine.

X is neither a ketone nor aldehyde (or not a carbonyl compound).

(c) (i) Draw two possible displayed formulae of compound X.

H H O H O
O O

C C C C H O C C C O H
H O O H
H H H C H

(ii) Given that compound X can be obtained from an alkene, suggest reagents
and conditions for the reaction.

Reagent: KMnO4, dilute H2SO4

Condition: Heat under reflux

(d) Compound Y has the same molecular formula as compound X. If compound Y

undergoes positive tests in (b)(i) to (b)(v), suggest a structural formula for it.

CH3CH(OH)COCOOH or CH3COCH(OH)COOH

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