CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: COURSEBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Coursebook answers
Chapter 24
Science in context
b
For scandium the only observed
oxidation state is +3, so the electronic
configuration of Sc3+ is 1s2 2s2 2p6 3s2
3p6 4s0. This ion has no d electrons,
so does not satisfy the definition of
a transition element. The only ion
of zinc is Zn2+, with the electronic
configuration 1s2 2s2 2p6 3s2 3p6 3d10
4s0. This ion has a completely filled,
not a partially filled, d subshell − so
zinc is not a transition element.
c
The + 7 oxidation state involves all of
the 3d and 4s electrons in manganese.
d
Oxidation state of vanadium in a =
(VO2+) = +5; b (VO2+) = +4;
c (V3+) = +3; d (V2+) = +2.
e
i
An internet search engine will produce results that
learners can use to explain how cis-platin interacts
with DNA when treating cancerous tumours.
Working together with a biology student could be
beneficial when discussing the effect of cis-platin
on DNA. Here is the basic mechanism:
• The cis-platin can be taken by patients in solution
and can pass through cell membranes and into
the nucleus of the cell. One of the chloride ion
ligands is first replaced by a water molecule,
forming the complex [PtCl(H2O)(NH3)2]+.
• On contact with a DNA molecule, the water
ligand is itself replaced by a guanine base as
one of its nitrogen atoms forms a dative bond
with the platinum ion.
• Then the remaining chloride ion is ideally
positioned to interact with another adjacent
guanine base. The chloride ion is replaced and
the new platinum complex has formed a bridge
in a DNA strand.
• This ‘bridging’ can also take place to a lesser
extent between the two strands in DNA, as well
as with the base adenine.
Discuss the use of drugs that can have harmful
side-effects in small groups so that learners get the
chance to voice their opinions. Ask each group
to write four bullet points to summarise their
discussion. Pin the summaries up to share with the
whole class.
Self-assessment questions
1
1
a
i
Ti 1s2 2s2 2p6 3s2 3p6 3d2 4s2
ii
Cr 1s2 2s2 2p6 3s2 3p63d5 4s1
iii
Co 1s2 2s2 2p6 3s2 3p6 3d7 4s2
iv
Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5 4s0
v
Ni2+ 1s2 2s2 2p6 3s2 3p6 3d8 4s0
vi
Cu+ 1s22s2 2p6 3s2 3p6 3d10 4s0
+4 as this involves all the 4d and
5s electrons, leaving the noble gas
electronic configuration of krypton.
ii
2
a
ZrO2
Fe (aq) → Fe3+(aq) + e−
2+
Cr2O72−(aq) + 14H+(aq) + 6e− →
2Cr3+(aq) + 7H2O(l)
b
6Fe2+(aq) → 6Fe3+(aq) + 6e−
Cr2O72−(aq) + 14H+(aq) + 6e− →
2Cr3+(aq) + 7H2O(l)
Cr2O72−(aq) + 6Fe2+(aq) + 14H+(aq) →
2Cr3+(aq) + 6Fe3+ + 7H2O(l)
c
E⦵ = +1.33 V + (−0.77 V) = +0.56 V
The positive value indicates that the
reaction as written is feasible and its
relatively large value suggests that the
reaction is likely to occur (although
values of E⦵ tell us nothing about the
rate of a reaction).
d
6
Cambridge International AS & A Level Chemistry © Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: COURSEBOOK
i
0.0153 × 0.001 = 0.000 015 3 mol
ii
0.000 015 3 × 6 = 0.000 091 8 mol
e
iii
0.000 0918
0.025
i
+3
ii
+2
iii
+3
iv
+3
v
+2
3
4
a
= 0.003 67 mol dm−3
i
H2N
Cl
NH2
SCN− has a higher value of Kstab
than H2O. So the position of
equilibrium is shifted to the right.
iii
[Fe(H2O)5SCN]2+(aq)
iv
Yes; a colour change is likely /
possible. F− has a higher value
of Kstab than SCN−. So F
substitutes for SCN (and for
water) because the position of
equilibrium is shifted to the right.
NH2
Cl
H2N
Cl
Cl
The ligands in a complex cause
the d orbitals to split, forming two
sets of non-degenerate orbitals.
The difference in the energy (ΔE)
between the non-degenerate d orbitals
corresponds to the energy of part
of the visible spectrum of light. So
when light travels through a solution
or a solid containing the complex, an
electron from one of the three lower
non-degenerate orbitals absorbs that
amount of energy (ΔE) and jumps into
one of the two higher non-degenerate
orbitals. This leaves the transmitted
light coloured.
CN
trans-isomer
Non-polar, as the charge
is distributed perfectly
symmetrically around the central
nickel (with both cyanide ligands
diagonally opposite each other in
the square planar structure, and
similarly with the two chloride ions).
a
+2
b
[CoCl4] (aq) + 6H2O(l) →
[Co(H2O)6]2+(aq) + 4Cl−(aq)
c
A
c
2−
i
ii
3d
Ni2+ ...3d8
[ PtCl (NH ) (aq)] Cl (aq) 
[ PtCl ] (aq)  [ NH (aq) ]


−
2
b
Cl
NC
cis-isomer
a
orbitals at the same energy level
Ni
NC
6
a
2–
2–
Cl
Ni
5
7
They are mirror images, which are
not superimposable.
NC
OH2
ii
Co
ii
OH2
H2O
6
ii
i
H2O
H2O
ethanedioate ion (ox) and
ethane-1,2-diamine (en)
b
[Fe(H2O)6]3+(aq)]:
Fe
c
iii
2+
H2O
[Ni(EDTA)]2−
i
b
 Ni ( NH3 )4 ( H2O )2  (aq)
 Ni ( H2O ) 2+ (aq)   NH3 ( aq )  4
6


b
a
iii
2
2
3 2
2−
4
2
3
 Cr ( H O ) Cl + (aq) 
2
2
4
 

 Cr ( H O ) 3+ (aq)  Cl – ( aq ) 2
2

6
 
 
Cambridge International AS & A Level Chemistry © Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: COURSEBOOK
d
e
4
Sc3+ ions have electronic configuration
[Ar]3d04s0. If d-orbital splitting were
to occur in a complex ion containing
Sc3+, there would be no electrons in
the three 3d orbitals of lower energy,
so visible light would not be absorbed
in promoting an electron from a lower
energy 3d orbital to a higher energy
3d orbital.
a
b
c
b
Zn2+ ions have electronic configuration
[Ar]3d104s0. If d-orbital splitting were
to occur in a complex ion containing
Zn2+, each of the 3d orbitals would
contain two electrons, and would
therefore be fully occupied. Visible
light could not be absorbed in
promoting an electron from a lower
energy 3d orbital to a higher energy 3d
orbital.
2
[1]
[1]
a
[Ar] 3d 4s or 1s 2s 2p 3s 3p 3d6 4s2[1]
b
[Ar] 3d7 or 1s2 2s2 2p6 3s2 3p6 3d7 4s0
[1]
c
[Ar] 3d1 or 1s2 2s2 2p6 3s2 3p6 3d1 4s0
[1]
2
2
2
6
2
6
[Total: 3]
e.g. FeCl3
[1]
oxidation state +3
[1]
e.g. FeCl2
[1]
oxidation state +2
[1]
3
a
b
d orbitals split / form two sets of nondegenerate orbitals;
[1]
an electron from one of the lower
orbitals absorbs energy from visible
light;
[1]
and is promoted to one of the higher
orbitals.
[1]
[Total: 7]
3
Cu(OH)2(H2O)4(s) + 4NH3(aq) →
Cu(H2O)2(NH3)4]2+(aq) + 2H2O(l) +
2OH−(aq) [1]
[1]
giving a deep blue solution
[1]
a
The electrode potential for the Cl2 / Cl−
redox system is more negative than the
one for MnO4−/Mn2+;
[1]
this means that
MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O can
gain electrons, proceeding to the right,
and the reaction Cl2 + 2e− ⇌ 2Cl−
can proceed to the left, forming Cl2.
Chlorine is toxic, and this also gives
an inaccurate titration result as the
MnO4− reacts with the Cl− as well
as the Fe2+.
[1]
The electrode potential for the SO42− /
SO2 redox system is more negative
than the one for MnO4−/Mn2+;
[1]
this means that the reaction
SO42− + 4H+ + 2e− → SO2 + 2H2O can
proceed to the left but not to the right
and the acid is unchanged.
[1]
b
[Total: 4]
6
[1]
the precipitate dissolves
5
a molecule or ion capable of bonding
to a positive ion by donating a
lone-pair of electrons and forming a
co-ordinate bond
[1]
joined to one or more ligands
[1]
[Total: 5]
an element forming one or more
compounds that contain an ion which
has a partly filled 3d subshell
[1]
a positive ion
[Cu(H2O)6]2+(aq) + 2OH−(aq) →
Cu(OH)2(H2O)4(s) + 2H2O(l)
pale blue precipitate
Exam-style questions
1
a
Mr of FeSO4·7H2O = 55.8 + 32.1 + 64.0
+ 126.0 = 277·9
[1]
amount (in mol) of FeSO4·7H2O
=
5.56
277.9
= 0.0200 mol
[FeSO4] =
c
i
0.02
0.250
[1]
= 0.0800 mol dm−3
[1]
5Fe + 8H + MnO4 →
5Fe3+ + 4H2O + Mn2+ [2]
2+
+
−
[1 mark for formulae; 1 mark for
balancing]
ii
d
hen a permanent pink colour is
w
obtained
[1]
amount in mol of Fe2+ (FeSO4) in
25 cm3 = V × C = 0.025 × 0.0800
= 0.00200 mol
1
amount in mol of MnO4− = × amount
5
in mol of Fe2+
= 4.00 × 10−4 mol
[1]
Cambridge International AS & A Level Chemistry © Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: COURSEBOOK
[MnO4−] =
=
e
iii
n
V (in dm3 )
4.00 × 10 −4
0.0212
correct drawing of mirror image; [1]
= 0.0189 mol dm−3
[1]
5SO2 + 2H2O + 2MnO4 →
2Mn2+ + 5SO42− + 4H+
[ignore charge]
en
en
−
[2]
Ni
[1 mark for formulae; 1 mark for balancing]
amount in mol of MnO4− = 0.0189 ×
0.025 = 4.73 × 10−4 mol
[1]
amount in mol of SO2
=
5
2
× 4.73 × 10−4 mol = 1.18 × 10−3 mol [1]
volume of SO2 at r.t.p.
= 1.18 × 10−3 × 24 dm3 = 0.0283 dm3
[1]
[Total: 17]
To make question 5e easier to answer,
break it up:
1 How many moles of MnO4− are we given?
6
[1]
correct drawing of one isomer;
2
How many moles of SO2 will this react
with? (the reaction equation is needed
here)
3
What is the volume of this number of
moles of SO2?
a
i
4Cl−(aq) in gap on left
[1]
6H2O(l) in gap on right
[1]
ii
ellow-green colour turns to
Y
light blue;
[1]
excess water shifts position of
equilibrium to the left;
[1]
some of the Cl in the complex
replaced by water molecules.
[1]
en
en
Ni
en
en
recognition of a single ligand
attached to two points in the
structure in either of the isomers; [1]
octahedral structure in either of
the isomers
[1]
[Total: 14]
−
iii
−
ne or more Cl replaced by
O
ammonia
because ammonia has greater
value of Kstab (than Cl−).
b
i
4
[1]
[1]
bidentate ligands
Ligands have two N lone pairs
per ligand molecule available for
complex formation.
ii
[1]
iaminoethane because it has a
d
higher value of Kstab
[1]
[1]
Cambridge International AS & A Level Chemistry © Cambridge University Press 2020