Download A2 Module 2814: Chains, Rings and Spectroscopy

EEE2 14-15 transition elements p.1
A2 Module F325: Equilibria, Energetics and Elements
EEE 2: Transition elements
For the elements up to Ca the 3d orbitals are higher in energy than the 4s orbital. Therefore, after
argon (element 18), the 4s orbital is filled: Ca has electron configuration [Ar] 4s2. From scandium on,
the 3d orbitals are filled, until they have ten electrons at zinc. The term “d-block elements” refers to
those elements in which this d-subshell is filling (Sc–Zn), but the term “transition elements” is used
for d-block elements that form one or more stable ions with a partially filled d-subshell. This excludes
Sc and Zn, since their only common oxidation states are Sc3+ (3d0) and Zn2+ (3d10). This distinction is
made because the main features of the chemistry of the transition elements depend largely on this
partially filled d-subshell.
You will be expected to use your Periodic Table to deduce the electronic configurations of atoms and
ions. Remember that: (i) the stability of the half-filled sub-shell means that d5 and d10 configurations
are particularly stable e.g. Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1 (not 3d4 4s2); (ii) the 3d is written before the
4s; (iii) in ions the 4s electrons are always lost first: Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6.
Remember: the 4s fills before the 3d and is emptied before the 3d.
Properties of transition elements
(i) Variable oxidation state
A Group 2 element like Mg shows only one oxidation state (+2); transition elements, however,
normally show a range of different oxidation states e.g. Fe2+ and Fe3+; Cu+ and Cu2+.
Generally transition metals show a fairly steady increase in successive ionisation energies until all the
4s and 3d electrons have been removed, when there is a sharp jump as the closed [Ar] shell is broken
into.
Generally the +2 oxidation will be common as this corresponds to the removal of the element’s two
4s electrons. Higher oxidation states than +2 mean that 3d electrons (close in energy to the 4s) have
also been removed. Copper is the only transition metal to form a significant number of compounds in
the +1 state because the one 4s electron can be removed to leave a stable 3d 10 configuration.
However, even here the +1 state is very easy to oxidise, with the result that copper(I) compounds are
only stable if they are in a complex (see below), or if they are insoluble.
(ii) Coloured ions
Unlike the elements of, for example, Groups 1 and 2, which exhibit white solids and colourless
solutions, transition elements show a wide variety of colours in their compounds. When ligands (see
below) pack around a metal ion, the d-orbitals no longer have exactly the same energies. If they are
partially filled, it is possible for an electron to jump from a lower-energy d-orbital to an unoccupied
higher-energy d-orbital. These “d–d transitions” are of an energy corresponding to absorption in the
visible region, and so the compound appears to be coloured. The colour is that of the light which is
not absorbed: e.g. copper(II) ions look blue because they absorb red light. The energies of d-orbitals,
and so the colour of the complex, are very sensitive to the ligands present.
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Colours of some common complexes of iron and copper:
Fe(H2O)62+ very pale green;
Fe(OH)2 dirty grey/green
Fe(H2O)63+ red/brown
Fe(OH)3 red/brown (rust)
Fe(H2O)5SCN2+ blood red
CuI, CuCl, Cu2SO4 all white (d10 is full, so no d–d transitions possible)
Cu(H2O)62+ blue
CuCl42– yellow/green
Cu(NH3)4(H2O)22+ deep blue
(iii) Catalytic behaviour
A wide range of industrial catalysts consist of transition elements or their compounds. They work
through many different mechanisms, of which two will be mentioned here.
In the Haber process, Fe2O3 is reduced to finely divided iron metal, and this provides a surface for the
nitrogen and hydrogen to react (heterogeneous catalysis – different phase). It is likely that the low
energy empty orbitals on the iron can be used to accept co-ordinate bonds, and so absorb the gases
temporarily and allow short-lived intermediates to be formed.
Homogeneous catalysis – in which the catalyst is in the same phase as the reactants – often involves
a reversible change in oxidation states by a d-block metal. Here Fe(OH)3 is insoluble, but the
decomposition probably takes place via the small amounts which remain in solution. The Fe3+ can
oxidise the hydrogen peroxide, changing to Fe2+, and then another hydrogen peroxide molecule can
oxidise it back to Fe3+ again:
2 Fe3+ +
H2O2
2 Fe2+ + 2 H+ + H2O2
 2 Fe2+ + 2H+ + O2
 2 Fe3+ + 2 H2O
Other examples of catalysts in industry (all heterogeneous) include Ni in hydrogenation of oils to
make margarine, V2O5 in the contact process (oxidation of SO2 to SO3 to make sulfuric acid), and
platinum + rhodium in catalytic converters in cars.
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(iv) Complexes
Ions of transition elements are able to accept the donation of several lone pairs into their outer shells.
The lone pairs will come from species able to donate lone pairs of electrons into coordinate bonds –
we call such species ligands in this context. Examples of ligands include H2O, CN-, NH3 and Cl-. A
complex ion is one in which a central positive ion is surrounded by ligands, which are co-ordinately
bonded to it; e.g. Cr(H2O)63+, Fe(CN)64–.
The transition elements are not unique in forming complexes (there are small numbers formed by
metals in groups 2, 3 and 4), but they form a much wider range than other elements. This is because
the transition metal ions are small and polarising, since their nuclei are poorly shielded, and so they
attract ligands strongly.
The most common type of complex ion by far is the aqua-ion, where the transition element ion is
surrounded by water ligands (usually six, because this is the number able to fit around the ion).
Ligands and Complexes
The co-ordination number of a metal ion in a complex is the number of atoms that are coordinately
bonded to the metal ion. This can be 2 (rarely), e.g. CuCl2–, Ag(CN)2–; or 4, e.g. CuCl42–; or most
commonly 6, e.g. Fe(H2O)63+.
A chelating ligand can co-ordinate from more than one atom at once, creating a ring structure.
A bidentate ligand is one which can co-ordinate from two positions at once, in the same molecule or
ion.
e.g.
O
H2N
NH2
H2C
CH2
ethane-1,2-diamine
O
C
C
O
O
ethanedioate
"en"
"ox"
(old name: ethylenediamine)
(old name: oxalate)
You are expected to be able to predict the formula and charge of a complex. For this you need to
know : (a) the charge on the metal ion; (b) the charge on the ligand; (c) the likely co-ordination
number [C.N.]. Examples:
Fe2+, with CN–, C.N. 6: gives Fe(CN)64–
Cr3+, with C2O42– (bidentate), C.N. 6: gives Cr(C2O4)33–
Conversely, in Fe(C2O4)34–, the oxidation state of Fe is given by x – 6 = –4; x = +2.
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Shapes of 4- and 6-co-ordinate complexes
(a) 4-co-ordinate
Complexes with four ligand positions can be either tetrahedral or square planar. Simple repulsion of
ligands would give the former, but in some complexes (especially d8 complexes of Ni-Pd-Pt)
interaction with the d-electrons means that the square-planar arrangement is more stable. We shall
treat the shape as something to be determined experimentally.
A square-planar shape can give rise to cis-trans isomers in complexes like ML2Q2 (where L and Q are
different ligands). Comparison with carbon compounds like CH2Br2 reminds us that no cis-trans
possibilities occur in a tetrahedral shape: it follows that, if a metal complex of formula ML2Q2 shows
cis-trans isomerism, this is proof that it must be square planar.
2-
Cl
H3N
Co
Ni
Cl
Cl
Cl
H3N
Cl
tetrahedral
e.g. CoCl42-
Cl
square-planar
e.g. cis-Ni(NH3)2Cl2
The cis-isomer of PtCl2(NH3)2, called cis-platin, is effective at treating certain types of cancer, by
binding to DNA (in place of the two Cl– ions), thus distorting the shape of DNA and inhibiting
replication and transcription in cell production. Trans-platin has no effect, confirming that the shape is
critical in enabling the cis-isomer to bind to the DNA. Cis-platin also inhibits repair processes by
blocking proteins. The drug is highly toxic, and must be administered with much water and diuretic
drugs to limit damage to the kidneys.
(b) 6-co-ordinate
Complexes with six ligand positions are octahedral [though occasionally distorted, as in some
copper(II) complexes, such as Cu(NH3)4(H2O)22+ ]. Octahedral complexes of the type ML4Q2 can show
cis-trans isomerism (the two Q ligands can be axial, or opposite, each other, or can be adjacent. If
they have three bidentate ligands, as in Ni(NH2CH2CH2NH2)32+, they also show optical isomerism. A
complex of this type is chiral, with the three bidentate ligands acting like propeller blades — the
propeller can be left-handed or right-handed, like a corkscrew.
NH2
NH2
=
CH2
CH2
NH2
NH2
NH2
Ni
NH2
NH2
Ni
NH2
NH2
NH2
NH2
NH2
mirror
EEE2 14-15 transition elements p.5
When prepared from symmetrical material (e.g. by adding 1,2-diaminoethane to Ni(H2O)62+) a 50:50
racemic mixture of the two enantiomers results. If the enantiomers are separated (e.g. by crystal
picking), each rotates the plane of polarised light in the opposite direction to the other.
Reactions of Complex Ions
One set of ligands may be displaced by others that form stronger co-ordinate bonds, or which are
present in very high concentration – this is called ligand substitution:
When copper(II) sulfate solution is treated with dilute aqueous ammonia, the solution starts blue
because of the Cu(H2O)62+ ion. It first forms a pale blue precipitate of Cu(OH)2, and then this dissolves
to give a deep blue coloured solution, containing the Cu(NH3)4(H2O)22+ ion.
[N.B. the hydroxide is formed first because ammonia solution is alkaline – see below for the effect of
alkaline solutions on transition element ions]
When copper(II) sulfate solution is treated with concentrated hydrochloric acid (or sodium chloride
solution), the solution starts blue because of the Cu(H2O)62+ ion. As Cl– ions are added they displace
water molecules, forming CuCl42–, which is yellow. The colour changes from blue through lime-green
to yellow-green, and becomes more intensely coloured despite the dilution:
Cu(H2O)62+ + 4Cl– ⇌ CuCl42– + 6H2O
blue
yellow
Another example: in solution Co(H2O)62+ ions are pink. When concentrated hydrochloric acid is added
to this pink solution, the solution eventually becomes deep blue, as a ligand substitution reaction
takes place, and the co-ordination number changes from 6 (octahedral) to 4 (tetrahedral):
Co(H2O)62+ + 4Cl– ⇌ CoCl42– + 6H2O
pink
deep blue
The reason the CN changes from 6 to 4 is that the Cl- ligands are larger and negatively charged, so only
4 of them can fit around the Co2+ ion, as opposed to 6 water ligands.
Ligand substitution is also involved in iron’s role in haemoglobin. Each haemoglobin molecule contains
four haem ligands, which are tetradentate – i.e. they can make four coordinate bonds to an ion in
their centre. The ion in question is Fe2+. Oxygen molecules can form (weak) coordinate bonds to the
Fe2+ ions, and this is how they are transported around the blood stream. However, carbon monoxide
molecules can also act as ligands and they form stronger coordinate bonds to the Fe2+ than O2, so as
CO gets into the blood stream, it stops the haemoglobin molecules from transporting oxygen and the
person may die of suffocation.
EEE2 14-15 transition elements p.6
The Stability Constant, Kstab
A ligand substitution reaction can be viewed as an equilibrium and as such can have an equilibrium
constant assigned to it – in this case called Kstab, as it shows the stability of the new complex formed
relative to the original complex.
The stability constant of a complex ion is the equilibrium constant for its formation from its
constituent ions in a particular solvent.
e.g. in a generalized substitution reaction:
M(H2O)62+ + 6X- ⇌ MX64- + 6H2O
The equilibrium constant will be represented by the expression
Kstab = [MX64-] / [M(H2O)62+] [X-]6
[H2O] is omitted because in aqueous solution the concentration of water is essentially constant.
Clearly the more the equilibrium is to the right, the more stable the new complex in question and the
larger the Kstab value. Thus complexes with the largest Kstab values are considered the most stable.
Reaction with Alkalis (NaOH or dilute NH3)
Reacting a transition element ion with an alkali such as NaOH or dilute NH3 (which can also effectively
provide OH- ions) will result in the formation of a coloured precipitate:
(a) Cu2+(aq), a blue solution, gives a mid/light-blue precipitate:
Cu2+(aq) + 2OH–(aq)
 Cu(OH)2(s)
2+
(b) Fe (aq), a very pale green solution, gives a dirty grey/green precipitate:
Fe2+(aq) + 2OH–(aq)
 Fe(OH)2(s)
3+
(c) Fe (aq), a yellow solution, gives a brown (rust-colour) precipitate:
Fe3+(aq) + 3OH–(aq)
 Fe(OH)3(s)
2+
(d) Co (aq), a pink solution, gives a blue precipitate:
Co2+(aq) + 2OH–(aq)
 Co(OH)2(s)
Redox systems
Because transition elements have variable oxidation states, they will frequently be involved in redox
processes in which they switch between these oxidation states.
Example 1:
Observation: an Fe2+ solution starts very pale green. Dilute sulfuric acid is added, then purple
potassium manganate(VII) solution dropwise. The yellow colour of the Fe 3+ begins to appear, and at
EEE2 14-15 transition elements p.7
the end point the whole solution turns pink, due to the excess of KMnO4.
The half equations:
MnO4-
Fe2+  Fe3+ + e+ 8H+ + 5e-  Mn2+ 4H2O
Overall:
MnO4- + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O
This reaction could be used in a redox titration, e.g.:
Calculation: 5.00g of impure FeSO4.7H2O (Mr = 278) is weighed out and made up to 250 cm3 in a
standard flask. 25.0cm3 aliquots of this solution are titrated with 0.0200 mol dm –3 KMnO4 solution,
and the mean titre is 14.8cm3. Calculate the % purity of the iron(II) sulfate.
Steps:
(a) Find moles of MnO4–:
amount of MnO4– in 14.8cm3 = 0.02 x 14.8/1000 = 0.000296 mol
(b) Use chemical equation (above) to find no. of moles of Fe2+
from equation, 1 mol of MnO4– reacts with 5 mole Fe2+
therefore 0.000296 mol of MnO4– reacts with 5×0.000296 = 0.00148 mol Fe2+
(c) Convert to mass of pure FeSO4.7H2O in 250cm3
amount of Fe2+ in 250cm3 solution = 10 × 0.00148 = 0.0148mol
mass of pure FeSO4.7H2O in 250cm3 = 0.0148 × 278 = 4.11g
(d) Find percentage purity
percentage purity = ( 4.11 / 5.00 ) x 100 = 82.2%
Example 2:
A copper coin weighing 0.99g is reacted with concentrated nitric acid, producing a solution of Cu 2+
ions which is made up to 250cm3 using distilled water. A 25cm3 portion of the solution is reacted with
an excess of potassium iodide solution, producing a white precipitate of CuI and a brown solution of
I2. This I2 solution is titrated against 0.1 mol/dm3 sodium thiosulfate solution, requiring 15.00cm3 to
reach the end point (noted when starch indicator turns colourless from black/blue). The relevant
equations are shown below. What is the percentage by mass of copper in the coin?
2Cu2+(aq) + 4I–(aq)  2CuI(s) + I2(s)
I2(aq) + 2S2O32–(aq)  S4O62–(aq) + 2I–(aq)
(a) Find moles of S2O32-:
amount of S2O32- in 15.00cm3 = 0.1 x 15/1000 = 0.0015
(b) Use second equation above to find moles of I2:
from equation, 2 moles S2O32- require 1 mole of I2; therefore, 0.0015 moles S2O32- required
0.00075 moles of I2
(c) Use first equation above to find moles of Cu2+:
EEE2 14-15 transition elements p.8
from equation, 1 mole of I2 is produced by 2 moles of Cu2+; therefore, 0.00075 moles of I2 is
produced by 0.0015 moles of Cu2+
(d) Find original number of moles of Cu2+:
0.0015 moles of Cu2+ in 25cm3 portion; therefore, in 250cm3 solution there are 0.015 moles
of Cu2+
(e) Convert to mass of copper:
mass of 0.015 moles of copper = 0.015 x 63.5 = 0.9525g
(f) Find percentage purity:
percentage purity by mass of copper in 0.99g coin = (0.9525 / 0.99) x 100 = 96.2%