ANSWERS U6 CHEMISTRY (TRIAL EXAM 2009)
Section A
1
(a)
(i) The atomic radius decrease.(1)
(ii) The nuclear charge increases(1) while the screening effect does not change much.(1)
(iii) Carbon.(1)
(b)
(i)
Ion
No. of electrons in the 3d
orbitals.
Colour of aqueous ion
Each row correct- (1)
Mn2+
5
Fe2+
6
Cu2+
9
Pink
Green
Blue
(ii) As heterogeneous catalyst.(1)
The transition elements have empty d-orbitals that can be used to form temporary
bonds (adsorption) with the reacting molecules.(1)
These adsorptions weaken the intramolecular bonds in the reacting molecules (1)
thus decreasing the activation energy of the reaction.(1)
2
(a)
(i) [OH-] = [(1.8 x 10-5)(0.80)]1/2 = 3.8 x 10-3
pOH = -log (3.8 x 10-3)
= 2.4…………………………………………………………………………….(1)
(ii) Volume = 0.80 x 30.0
1.00
= 24 cm3………………………………………………………………………….…(1)
(iii) correct shape….(1), correct axis with unit & volume……..(1)
pH
(1)……(iv)
Buffer
Volume of HNO3
(cm3)
0
24
1
(v) The concentrations of the weak base and its conjugate acid must be kept large and
approximately equal. (1)
(b)
(i)
pOH = pKb + log10 ([NH4+] / [NH3] )
= - (log10 1.8 x 10-5) + log10 ( 0.08/0.12)………………………………....(1)
= 4.6
pH = 14.0 – 4.6
= 9.4………………………………………………………………………..(1)
(ii)
Number of moles of H+ added = 2 x 0.050 x 20.0 = 0.0020 mole
1000
Number of moles of NH3 = 0.12 - 0.0020 = 0.118 moles
Number of moles of NH4+ = 0.080 + 0.0020 = 0.082 moles
pOH = pKb + log10 ([NH4+] / [NH3] )
= - (log10 1.8 x 10-5) + log10 ( 0.082/0.118)……………………………….(1)
= 4.6
pH = 14.0 – 4.6
= 9.4………………………………………..………………………………..(1)
3
(a)
(i) I: Reflux with methanol(1), concentrated H2SO4(1)
(i) II: React with ethanoyl chloride(1), at room temperature(1)
(b) Add aqueous sodium carbonate to both compounds………………………….(1)
There is no change with X. With Y, effervescence is observed and the gas liberated turns
lime water chalky……………………………….(1)
2
(c) (i)
(ii)
4
(a) Each row correct…..(1)
A
B
Substance
Titanium
Steel
Terminal
Positive
negative
(b) C is a asbestos diaphragm(1). It prevents the gases formed from mixing/ It prevents
reaction of chlorine and the hydroxide ion(1).
(c) At A: 2Cl- (aq) 2e + Cl2(g)…………………………………………………………..(1)
At B: 2 H+(aq) + 2e H2(g)…………………………………………………………....(1)
(d) The brine solution will flow under pressure through the asbestos diaphragm towards the
cathode compartment/ Prevents flow of hydroxide ions from right to left.(1)
(e) Balanced equation: Cl2 + 2NaOH  NaCl + NaClO + H2O…..….(correct-1, balanced-1)
Type of reaction: Disproportionation reaction…………………………………………(1)
Section B
5
(a)
(i) Isotopes are atoms of the same element having the same number of protons and
electrons but different number of neutrons. (1)
E.g. Carbon-12 and carbon-14 are isotopes of carbon.(1)
(i) Allotropes are atoms of the same element having the same number of protons,
electrons and neutrons but differ in the way the atoms are bonded and arranged in the
substance.(1)
E.g. Graphite and diamond are allotropes of carbon.(1)
(b)
(i) A phase diagram is a graphical representation of the conditions of temperature and
pressure(1) at which solids, liquids and gases exist, either as single phases or as two
or more phases in equilibrium with each other.(1)
3
(ii) Phase diagram for water: (1)
Phase diagram for oxygen: (1)
(iii) To the left of the triple point, water vapour forms ice which then liquefies when
higher pressure is applied.(1)
To the right of the triple point, water vapour liquefies when higher pressure is
applied.(1)
To the left of the triple point, oxygen vapour solidifies when higher pressure is
applied.(1)
To the right of the triple point, oxygen vapour liquefies and then solidifies when
higher pressure is applied.(1)
(iv) Liquid carbon dioxide has a lower density than solid carbon dioxide.
When pressure increase, the molecules are pushed more closer and the system favors
the formation of solid.
A higher temperature is needed to melt solid carbon dioxide.
(correct explaination-2 marks)
Hence, melting point of carbon dioxide increase when the pressure increase.(1)
6
(a)
(i) Solubility product is the product of the equilibrium concentrations of all the aqueous
zinc and hydroxide ions, with each concentration raised to the power of its
stoichiometric coefficient in the solubility equilibrium.(2)
Ksp = [Zn2+] [OH-]2…………………………………………………..(1)
(ii) Ksp = [Zn2+] [OH-]2
(iii)
Number of moles of Zn(OH)2 = ½ x 0.0050 = 0.0025 mole……………………(1)
Mass of Zn(OH)2 = 0.0025 x 99.4 = 0.25 g……………………………………..(1)
Mass of dissolved Zn(OH)2 is ignored due to small Ksp value………………….(1)
4
(b) Cr  Cr3+ + 3e
Quantity of charged used = 12.0 x 35.0 x 60 = 25 200 C……………………………….(1)
Number of Faradays used = 25 200/ 96 500
Mass of chromium = 52.0 x 25 200 F
3F
96 500
= 4.53 g……………………………………………………………..(1)
Purify of chromium = 4.53/4.92 x 100% = 92.1%...........................................................(1)
(c)
(i)
(ii)
7
(a)
(i)
 When carbon atom in the ground state get the sufficient energy, 1 electron
from the 2s orbital will promote to 2p orbital………………………..(1)
 During sp hybridisation, 2 equivalent sp hybrid orbitals are formed by the
mixing of one 2s and one 2p orbitals………………………………….(1)
 Each carbon atom in ethyne molecule has two sp hybrid orbitals which is
linear shape…………………………………………………………….(1)
 The two sp hybrid orbitals in a carbon atom is used to undergoes head-on
overlapping with one 1s orbital of hydrogen atom and one sp hybrid
orbital of another carbon atom respectively to form two sigma covalent
bonds…………………………………………………………………..(1)
5
 Each of the carbon atoms has two unhybridised 2p orbitals which undergo
sideways overlapping to form two pi covalent bonds between carbon
atoms…………………………………………………………………..(1)
 Thus, in a ethyne molecule, it contains two C-H sigma covalent bonds, one
C-C sigma covalent bond and two C-C pi covalent bonds……………(1)
----------------------(1)
(ii)
 The atomic size of nitrogen is smaller than carbon…………….…………..(1)
 When ethyne reacts, two pi bonds are broken, but when nitrogen reacts,
three covalent bonds (two pi bonds and one sigma bond) must be
broken…………………………………………………………………….(1)
 Hence, much more energy is required to break a stronger nitrogen-nitrogen
triple bond. This explains why nitrogen is unreactive……………………(1)
6
8
(a)
 All the Group 2 nitrates decompose when strongly heated to form oxides,
nitrogen dioxide and oxygen gas / 2Be(NO3)2  2BeO + 4NO2 +
O2 ……………………………………………………………………...…(1)
 When going down Group 2, the thermal stability of the nitrates increase.....(1)
 This is due to the increasing ionic character of the nitrates…………..……(1)
 The lower nitrates such as beryllium nitrate and magnesium nitrate easily
decompose because the high charge density of the Be2+ and Mg2+ ions
effectively polarises the electron cloud of the nitrate ion…………..……(1)
 This weakens the oxygen-nitrogen bond in the nitrate……………….…....(1)
(b)
(i) In the tetrachlorides, the M-Cl bond is covalent and the molecules are held
together by weak van der Waals forces(1). Since stronger van der Waals
forces exixts between the tetrachloride molecules due to the increasing size of
the MCl4 molecules(1), the melting point increases(1).
(ii)
 Carbon, silicon and germanium form stable tetrachlorides that do not
decompose when heated while tin and lead tetrachloride decompose when
heated………………………………………………………………….(1+1)
 The thermal stability decreases when going down the group.
(1)
 This is because the M-Cl bond that becomes increasingly longer, becomes
weaker due to the smaller overlap of the orbitals………………………...(1)
(c)
 Lead(II) chloride is an ionic compound where the ions are held together by
strong electrostatic forces of attraction………………………………..…(1)
 Lead(IV) chloride is a covalent compound and the molecules are held
together by weak van der Waals forces…………………………………..(1)
 This is because the high charge density of Pb4+ effectively polarises the
electron cloud of the chloride ion………………………………...………(1)
9
(a)
(i)
Element
% by mass
Mole ratio
Simplest
ratio
C
66.4
66.4/12 = 5.5
5.5/0.79 = 7
Cl
28.1
28.1/35.5 = 0.79
0.79/0.79 = 1
H
5.5
5.5/1 = 5.5
5.5/0.79 = 7
(1 mark)
Empirical formula of P is C7H7Cl
Let the molecular formula of P be (C7H7Cl)n
Hence, [(12 x 7) + 7 + 35.5]n = 126.5……………………………………...(1)
n= 1
Hence, molecular formula of P is C7H7Cl………………………………….(1)
7
(ii) The ratio of C:H = 1 and the number of carbon atoms in one molecule of P
exceeds 6. Hence, P is an aromatic compound(1). P can be hydrolysed by
aqueous potassium hydroxide. This show that the chlorine atom is not
attached directly to the benzene ring, but is attached to a side chain(1).
8
(b)
(i)
(ii) The rate equation for the hydrolysis is given by:
Rate = k [2-bromo-2-methylpropane] or
The rate is independent of the concentration of the OH- or KOH………….(1)
Hence, the concentration of KOH has no effect on the rate of hydrolysis….(1)
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(a)
---------- (1+1+1 marks)
Benzene reacts with bromine in the presence of the Lewis acid, aluminium
bromide, as catalyst to form bromobenzene(1). Electrophilic substitution
occurs(1).
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Cyclohexene reacts with bromine in the presence of the organic solvent
tetrachloromethane to form 1,2-dibromocyclohexane, bromine is decolourised(1).
Electrophilic addition occurs(1).
Cyclohexane reacts with bromine gas in the presence of ultraviolet light, one of
the products is bromocyclohexane(1). Free radical substitution reaction occurs(1).
(b) In the phenol molecule, the lone pair electrons of the oxygen atom can interact
with the pi orbitals on the aromatic carbon(1). This activates the ring towards
aromatic electrophilic substitution(1). So, phenol reacts with aqueous bromine
at room temperature(1) whereas benzene reacts with bromine gas in the
presence of a Lewis acid as a catalyst(1).
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