Worked Solutions to 2011 A-level H2 Chemistry Paper 1
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In TiO2, Titanium exists as Ti4+.
Ti: [Ar] 4s2 3d2 = 22 electrons.
No. of electrons for each Ti4+ = 22 – 4 = 18
Answer: A
Nuclear charge of Ne > F but shielding effect about same for both as electrons are added into
the same quantum shell.
Therefore, E.N.C of Ne > F thus IE of Ne > F
Answer: D
1 mol (Na2CO3)x.y(H2O2) : x mol CO2 : y mol H2O2
0.01 mol (Na2CO3)x.y(H2O2) : 0.048/24 = 0.02 mol CO2
x=2
MnO4- : (H2O2) = 2: 5
Amt of H2O2 = 0.024x 0.05 x 5/2 = 0.03 mol
0.01 mol (Na2CO3)x.y(H2O2) : 0.03 mol H2O2
y=3
Answer: C
PV = nRT
Since V, n, R const  P α T
If P x 2  T (in units of K) x 2
Temperature = (27 + 273) x 2 = 600 K
Answer: D
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ΔHc = (x + y)/2
Answer: C
Lattice energy 
|(q+. q)/ (r+ + r )|
Charge effect is greater that size effect where M2+ and Z2- will form the ionic compound with
the most exothermic LE.
Since the radii of L+ > J+ and Y- > X-, hence the LE of JX is more exothermic than LY.
Hence LE decreases in the order: MZ, JX, LY
Answer: C
ΔHrxn = ΔHf (products) – ΔHf (reactants) = -2.09 – 0 = -2.09 kJmol-1
ΔSrxn = ΔS (products) – ΔS (reactants) = 44.1 – 51.4 = -7.3 JK-1mol-1
ΔGrxn = ΔH – TΔS = (-2.09 x 1000) – 285(-7.3) J mol-1
Answer: D
Anode: 2Cl–(aq)
 Cl2(aq) + 2e–
Cathode: 2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Cl2 and aqueous NaOH produced during electrolysis can further react to form sodium
chlorate(I).
Cl2(g) + 2NaOH(aq)  NaClO(aq) + NaCl(aq) + H2O(l)
Answer: C
Metal that will not be easily dissolved (oxidised) should have the highest EoR value
Eo(Cu+/Cu) = +0.52 ; Eo(Cu+2/Cu) = +0.34
Eo(Fe2+/Fe) = -0.44 ; Eo(Fe3+/Fe) = -0.04
Eo(Pb2+/Pb) = -0.13
Eo(Zn2+/Zn) = -0.76
Answer: A
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Worked Solutions to 2011 A-level H2 Chemistry Paper 1
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From the given half equation, all the species are ions.
Hence, X must be inert electrode – Pt
Y must be solution containing Fe2+ and Fe3+ as the solution must contain all the ions in the
half equation.
Answer: D
Definition of buffer: Buffer maintains the pH at about the same level when small amounts of
acid or base is added.
Answer: C
 When [SO2] is const, when [NO2] is increases by 1.33 times, relative rate also
increases by 1.33 times. Therefore 1st order w.r.t. NO2.
 When [SO2] increases from 0.005 to 0.006 (1.2 times), [NO2] increases by 1.2 times,
relative rate increases due to SO2 alone is 1.44 / 1.2 = 1.2 times. Therefore 1st order
w.r.t. SO2.
Answer: B
Rate α 1/time
Plot concentration of reactant vs rate. The order can be determine from the shape of graph.
Answer: B
 A: Phosphorus exists as P4 whilst sulfur exists as S8. (False)
 B: Aluminium has the highest electrical conducitivity as it has the most number of
delocalised electrons. (False).
 C: Silicon has the highest melting point (False).
 Anionic radius trend is as follows: P3- > S2- > Cl- (False)
Answer: B
Stability of H-X to decomposition depends on H-X bond.
Answer: C
2Ca(NO3)2(s)  2CaO(s) + 4NO2(g) + O2(g)
Amt of Ca(NO3)2 = 8.2/164 = 0.05 mol
Amt of O2 produced = 0.05/2 = 0.025 mol
Vol of O2 produced = 0.025 x 24 x 1000 = 600 cm3
Answer: C
To produce oxygen gas, hydrogen peroxide must undergo oxidation. Therefore condition
must be acidic.
Answer: C
Enthalpy change of fusion is α melting point
W and X has giant structure while Y and Z has simple covalent structure. Also, X has a higher
melting point than W hence X must be Si and W Al. Also, since Z has a higher melting point
than Y, the size of the electron cloud of Z (S8) must be bigger than Y (P4).
Answer: A
But-2-yn-ol
ABCDAnswer: C
True (sp hybridised – C has 2 sigma bonds)
True. (–OH gp decolourises KMnO4 due to multiple bonds.)
False. (They are sp3 hybridised as C has 3 sigma bonds)
True. (C has 4 bp, 0 lp and according to VSEPR theory, shape = tetrahedral)
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Worked Solutions to 2011 A-level H2 Chemistry Paper 1
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CxH2x+2 + (3x+1)/2 O2  xCO2 + (x+1)H2O
Number of carbon atoms in alkane = x
Moles of oxygen gas = (3x+1)/2
Answer: B
In chloroethene, the Cl is next to the C=C. The electrons on Cl can be delocalised into the pi
bond of the alkene thus making C-Cl bond stronger (exhibiting partial double bond character)
and hence less resistant to hydrolysis.
Answer: C
Intermediate in SN1 is a carbocation.
C is intermediate in Electophilic Substitution; D is intermediate in SN2 (transition state
formed); A is an intermediate in a nucleophilic addition mechanism.
Answer: B
The reactant will undergo ester hydrolysis (there is heat from steam and acid). H2O will also
add across double bond (EA).
Note in C: a diol cannot be formed through EA unless its through mild oxidation.
Answer: B
Ease of nucleophilic substitution reaction according to classes of halogeno compounds are:
Acyl Chlorides >
Alkyl chlorides >
Aryl chlorides
Ease of nucleophilic substitution reaction for reaction within the same class of compounds
are:
R-Cl <
R-Br <
R-I
Answer: B
Cl2 should not be aq. as water will cause the AlCl3 to be hydrolysed. (recall: chemical
periodicity – reaction of AlCl3 with water to form an acidic solution).
Answer: C
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Answer: A
X is
The alkyl halide undergoes NS with ammonia to form a substituted amine.
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Worked Solutions to 2011 A-level H2 Chemistry Paper 1
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The amine above then undergoes an intermolecular NS with the acid chloride to form A.
(Note: C and D are not possible as they are ionic compound which should be soluble in
water.)
Answer: A
React with Na : alcohol, phenol OR -COOH group present
React with 2,4DNPH: ketone or aldehyde
(X, Y, Z reacts with Na; only Y reacts with 2,4 DNPH)
Answer: D
When pKa (2.1 and 4.1) < pH (7)  glutamic acid exists in a basic medium. Hence, all acidic
groups are deprotonated.
When pKa (9.5) > pH (7)  glutamic acid exists in an acidic medium. Hence all basic groups
are protonated.
Answer: B
With HCl and heat,
-amide group will undergo hydrolysis, leaving –NH3+ on oseltamivir
-amine group (basic) will react with acid to form –NH3+
-ester group will undergo hydrolysis, leaving –COOH on oseltamivir
Answer: B
Only 1 (by definition or relative molecular mass)
Answer: D (only 1)
Option 1 – True. Hydrogen gas must be at standard conditions i.e. 1 atm , 298 K
Option 2 – True. The cell produces a current but when the electrodes are connected through
a high-resistance voltmeter, the current in the external circuit is virtually zero and the cell
registers its maximum potential difference, (E0).
Option 3 – False. A concentration of 1mol dm-3 solution is required. pH = 1 infers [H+] = 10-1
mol dm-3.
Answer: B ( 1 and 2)
Option 1 – True. Same empirical formula (C6H5NO)n where n= 1 for Z and n= 2 for X
Option 2 – True. X and Y have the same molecular formula C12H10N2O2 but different
arrangement of atoms hence they are isomers.
Option 3. – True. Since (C6H5NO)n n= 1 for Z and n= 2 for Y, Mr of Y is twice that of Z.
Answer: A (1, 2 and 3)
Option 1. Correct – since it is zero order, rate does not change with time
Option 2. Correct – rate constant only depends on temp
Option 3. Wrong - rate constant will change with temp, hence the graph will linearly
increasing.
Answer: B ( 1 and 2)
AlCl3 (s) + aq  [Al(H2O)6]3+ (aq) + 3 Cl- (aq)
[Al(H2O)6]3+ (aq) + H2O (l)  [Al(H2O)5OH]2+ (aq) + H3O+ (aq)
AlCl3 undergoes hydroylsis, making it acidic.
1. Wrong. Not due to the reaction of chloride and water as chloride is a weak
conjugate base that does not hydrolyse in water.
2. Correct. Al3+ has high charge/ mass (similar to charge/surface area), causing it to
undergo hydrolysis in water.
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Worked Solutions to 2011 A-level H2 Chemistry Paper 1
3. Correct. One of the O-H bonds in H2O of is broken in [Al(H2O)6]3+ to release the
H+
Answer: C ( 2 and 3)
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NaBr and conc sulphuric acid will react as follows:
KBr + H2SO4  HBr + KHSO4
2 HBr + H2SO4  SO2 + Br2 + 2H2O
Answer: B ( 1 and 2)
1. Correct.
2. Wrong. Only 1 ester group
3. Wrong. Only 3 chiral centres
Answer: D (1 only)
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Only 1 is correct. CO and NO (both pollutants) are converted to more harmless gases, CO2
and N2. Option 2 and 3 are wrong because harmful gases are produced – CO, NO2 and SO3
Answer: D (1 only)
Baeyer’s test – Mn is reduced from +7 (in MnO4-) to +4 (in MnO2)
Fehling’s test – Cu is reduced from +2 (in Cu2+) to +1 (in Cu2O)
Tollen’s test – Ag is reduced from +1 (in [Ag(NH3)]+ to 0 (in Ag)
Answer: A (1, 2 and 3)
1. Correct. The non-polar R group (–CH3 group) in alanine can interact with each other by
VDW
2. Correct. The polar R group ( –CH2OH group) in serine can interact by H-bonding with the –
COOH gp of another amino acid
3. Wrong. The R groups do not contain -NH2 and –COOH group.Hence ionic bonding is not
present.
Answer: B (1 and 2)
END
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Worked Solutions to 2011 A-level H2 Chemistry Paper 1
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