11 Transition metals
Answers
Answers to Topic 11 Test yourself questions
1s2, 2s2 2p6, 3s2 3p6 3d1, 4s2
1s2, 2s2 2p6, 3s2 3p6
1s2, 2s2 2p6, 3s2 3p6 3d5, 4s2
1s2, 2s2 2p6, 3s2 3p6 3d5
[Ar] 3d6
[Ar] 3d5
Fe3+, because it has a half-filled 3d sub-shell in
which the charge will be distributed more evenly.
d) Cu2+, because all its shells of electrons are
completely full.
They are d-block elements because the last electron
added to form their atomic structures enters a d subshell.
They are not transition metals because neither of
them forms a stable ion with a partially filled d subshell.
Zinc atoms have the electronic structure [Ar] 3d10
4s2. When zinc reacts, both the electrons in the 4s
sub-shell are lost to form stable Zn2+ ions, in which
all the electron shells are completely full.
a) E values for the systems M2+(aq)M(s) become
less negative (more positive).
b) The metals become less reactive.
From titanium to chromium, the positive charge on
the nucelus increases from 22 to 24 while elctrons
are being added to an inner shell of electrons. The
increasing nucleus charge attracts the outer 4s
electrons more strongly and this causes a decrease in
the atomic radius.
Possible statements are:
● most transition metals can exist in more than one
oxidation state
● elements in the middle of the series have the
greatest range of oxidation states
● all the transition elements form compounds in
the +2 state
● the +2 state becomes more stable relative to the
+3 state across the series.
a) +3: CrCl3, Cr2(SO4)3; +6: Cr2O72−, CrO42−
b) +2: MnCl2, MnSO4; +7: MnO4−
c) +2: FeCl2, FeSO4;
+3: FeCl3, Fe2(SO4)3
d) +1: CuCl, CuI;
+2: CuCl2, CuSO4
(Only one compound is expected for each
oxidation state.)
1 a)
b)
c)
d)
2 a)
b)
c)
3
4
5
6
7
8
9 Oxygen in VO2+ has an oxidation state of 2, so the
two O atoms contribute 4 to the overall charge. So
the oxidation state of V in VO2+ must be +5 to give
an overall charge of +1.
10 a) i) VO2+ + 2H+ + e−  VO2+ + H2O
ii) Zn  Zn2+ + 2e−
b) From the Data sheet of standard electrode
potentials:
[VO2+(aq) + 2H+(aq)], [VO2+(aq) + H2O(l)]Pt E = +1.00 V
and
Zn2+(aq)Zn(s)
 for the reduction of
VO2+
(VO2+(aq)
E = 0.76 V
2+
to VO by Zn,
+ 2H (aq) + e  VO2+ + H2O)  2 E = +1.00 V
+
−
Zn(s)  Zn2+(aq) + 2e−
E = +0.76 V
overall, Ecell for the reaction (under standard conditions) is +1.76 V.
The relatively high positive value of Ecell suggests
that zinc will reduce VO2+(aq) to VO2+ in acid
solution.
11 a) i) Cu+(aq)  Cu2+(aq) + e−
ii) Cu+(aq) + e−  Cu(s)
b) Cu+(aq)  Cu2+(aq) + e−
E = 0.15 V
+
−
Cu (aq) + e  Cu(s)
E = +0.52 V
c) Adding the two half-equations in part b) gives
2Cu+(aq)  Cu2+(aq) + Cu(s)
Ecell = +0.52  0.15 V
= +0.37 V
The possible overall value of +0.37 V for the
above reaction indicates that Cu2+(aq) + Cu(s)
are more thermodynamically stable than
2Cu+(aq). So, under standard conditions, aqueous
Cu+(aq) ions will be oxidised and reduced
simultaneously (i.e. disproportionate) forming
Cu2+(aq) and Cu(s).
12 a) The electronic configurations are:
Zn2+ [Ar] 3d10; Cu+ [Ar] 3d10; Sc3+ [Ar]
In Zn2+ and Cu+, the 3d level is full and in Sc3+
the 3d level is empty, so electron transitions
within the d level are not possible.
b) red – orange – yellow
13 a) In a continuous process, the reaction mixture
(gas or liquid) can flow through a solid
heterogeneous catalyst in a reactor and there is
no problem in separating the products from the
catalyst.
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11 Transition metals
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Answers
b) In a batch process, the solid catalyst can be
suspended in the reaction mixture or agitated
with it and then easily recovered by filtration
for the next batch.
a) In the first stage, Fe3+ ions are reduced to Fe2+
as they oxidise iodide ions to iodine:
2Fe3+(aq) + 2e−  2Fe2+(aq)
2I−(aq)  I2(aq) + 2e−
Then in the second stage, S2O82− ions oxidise
Fe2+ ions back to Fe3+ ready to oxidise more I−:
2Fe2+(aq)  2Fe3+(aq) + 2e−
S2O82−(aq) + 2e−  2SO42−(aq)
b) Yes, because S2O82− ions will oxidise Fe2+ ions
to Fe3+, and the Fe3+ ions can then oxidise I−
ions to I2, producing Fe2+ ions which can be
oxisidised by more S2O82−.
a) Any two of:
 add some Mn2+ ions
 heat the solution
 increase the concentration of MnO4−(aq)
 increase the concentration of C2O42−(aq).
b) i) At the start, the mixture will be purple due
to unreacted but diluted KMnO4. A few
bubbles of CO2 will be produced.
ii) As the reaction gets underway, the purple
colour will fade and disappear as MnO4−
ions form almost colourless Mn2+ ions if
ethanedioate is in excess. At the same time,
an increasing number of bubbles of CO2
will be produced.
a) 4
b) 6
c) 6
a) tetraamminecopper(II)
b) tetrahydroxozincate(II)
c) tetrahydridoaluminate(III)
d) hexaaquanickel(II)
a) +2
b) +1
c) +3 d) +2
2−
AgBr(s) + 2S2O3 (aq)
 [Ag(S2O3)2]3−(aq) + Br−(aq)
a) linear
b) octahedral
c) tetrahedral
d) octahedral.
Using the lone pair of electrons on the nitrogen
atom and one of the two lone pairs of electrons on
either of the oxygen atoms, glycine can make two
co-ordinate bonds from the same molecule to a
cation.
22 a) Because one edta ion can make six co-ordinate
bonds with the same metal ion.
b) octahedral
23 a)
b) octahedral
24 a) [Ni(edta)]2− > [Ni(en)3]2+ > [Ni(NH3)6]2+
b) The stability of complex ions will depend on
the strength and number of co-ordinate bonds
from individual ligands.
So, edta, a hexadentate ligand, will form the
most stable complex ions with Ni2+, then
1,2-diaminoethane (en), a bidentate ligand, and
then ammonia, a monodentate ligand.
25 a) [Fe(CN)6]4−
b) Fe4(Fe(CN)6)3
26 a) Co(H2O)62+(aq) + 6NH3(aq)
 Co(NH3)62+(aq) + 6H2O(l)
b) Co(NH3)62+(aq) + 4Cl−(aq)
 CoCl42−(aq) + 6NH3(aq)
c) Fe(H2O)62+(aq) + 6CN−(aq)
 Fe(CN)64−(aq) + 6H2O(l)
27 a) In ammonia solution, the following equilibrium
operates:
NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)
Initially OH− ions react with aqueous copper
ions to form a pale blue precipitate of
copper(II) hydroxide:
Cu2+(aq) + 2OH−(aq)  Cu(OH)2(s)
[More precisely: Cu(H2O)62+(aq) + 2OH−(aq)
 Cu(OH)2(H2O)4(s) + 2H2O(l)]
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© G. Hill and A. Hunt 2009 Edexcel Chemistry for A2
11 Transition metals
Answers
b) The precipitate dissolves on adding more
ammonia to form a deep-blue solution
containing Cu(NH3)42+ [more precisely
Cu(NH3)4(H2O)22+] ions:
Cu(OH)2(s) + 4NH3(aq)
 Cu(NH3)42+(aq) + 2OH−(aq)
[or Cu(OH)2(H2O)4(s) + 4NH3(aq) 
Cu(NH3)4(H2O)22+(aq) + 2OH−(aq) + 2H2O(l)]
28 a) Co(H2O)62+(aq) + 4Cl−(aq)
pink
colourless
 CoCl42−(aq) + 6H2O(l)
blue
colourless
b) When the filter paper soaked in pink cobalt(II)
chloride solution is dried in an oven, all the
water is lost and the substance on the filter
paper contains blue tetrachlorocobaltate(II)
ions, CoCl42−. In the presence of water, the blue
CoCl42− ions react to form pink Co(H2O)62+
ions.
CoCl42−(aq) + 6H2O(l)
blue
 Co(H2O)62+(aq) + 4Cl−(aq)
29 a) The solution will turn frompink
pink to green.
b) The solution will turn from green to pink.
30 edta is a hexadentate ligand forming 6 co-ordinate
bonds with Co2+ ions, whereas NH3 and H2O are
only monodentate ligands.
When edta reacts with either Co(H2O)62+ or
Co(NH3)62+, there is a large increase in entropy.
Ssystem and Stotal are both positive, which means
that Co(edta)2− is thermodynamically more stable
than Co(H2O)62+ and Co(NH3)62+.
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© G. Hill and A. Hunt 2009 Edexcel Chemistry for A2