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KINGSTON COLLEGE
GRADE 12
CHEMISTRY
First Row Transition Elements
What is a Transition Metal?
The term transition metal generally refers to an element that has an atom, or forms at least one
ion that has partially filled d orbitals. These are the so-called d-block metals. Table 1 below
gives the electronic configurations of the first row transition elements.
Table 1 The electronic configurations of the first row transition elements
You will notice the unexpected electronic configurations for chromium and copper. You would
have expected an electronic configuration of [Ar]3d44s2 for chromium and [Ar]3d94s2 for copper.
Instead we have [Ar]3d54s1 and [Ar]3d104s1 respectively. This is explained on the basis of the
fact that filled (10 electrons) and half-filled (5 electrons) d orbitals have a greater stability, so an
electron is promoted from the 4s orbital to achieve this.
Properties of the First Row Transition Elements
The properties of the transition elements do not vary greatly across a period. Like other metals
they are good conductors of heat and electricity, strong, hard, shiny, malleable and ductile. They
differ from a typical s block metal such as calcium in the following ways:
(i) they have higher densities
(ii) they have higher melting and boiling points
(iii) they form coloured compounds and ions
(iv) they form complex ions
(v) the compounds are often paramagnetic (they contain unpaired electrons and are attracted to a
magnetic field)
(vi) they often show catalytic activity
(vii) they show variable oxidation states in their compounds
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Properties of First Row Transition Elements
Z and Symbol
Melting Point/oC
Density/gcm–3
Atomic radius/pm
First Ionization
Energy/kJ mol-1
M2+ ionic
radius/pm
21 Sc
1541
2.99
161
22 Ti
1668
4.54
145
23 V
1910
6.11
132
24 Cr
1857
7.19
125
25 Mn
1246
7.33
124
26 Fe
1538
7.87
124
27 Co
1495
8.90
125
28 Ni
1455
8.90
125
29 Cu
1083
8.92
128
30 Zn
420
7.13
133
631
658
650
653
717
759
758
737
746
906
na
90
88
84
80
76
74
72
69
74
M3+ ionic
radius/pm
81
76
74
69
66
64
63
62
na
na
Common
Oxidation States
+3 only
+2,3,4
+2,3,4,
5
+2,3,6
+2,3,4,
6,7
+2,3,6
+2,3
+1,2
+2
only
Outer Electron
Configuration
[Ar]...
3d14s2
3d24s2
3d34s2
3d54s1
3d54s2
3d64s2
3d74s2
3d84s2
3d104s1
3d1042
Electrode potential
M(s)/M2+(aq)
na
-1.63V
–1.18V
–0.90V
–1.18V
-0.44V
-0.28V
–0.26V
+0.3V
-0.76V
Electrode potential
M(s)/M3+(aq)
–2.03V
-1.21V
–0.85V
–0.74V
–0.28V
-0.04V
+0.40
na
na
na
Electrode potential
M2+(aq)/M3+(aq)
na
-0.37V
–0.26V
–0.42V
+1.52V
+0.7V
+1.8V
na
na
+2,+3
Atomic Radius
The atomic radius may be defined as the distance between the centre of the nucleus and the
outermost electron (valence) shell. The two main factors that affect the atomic radius are:
(i) the size of the charge on the nucleus;
(ii) the number of shells between the nucleus and the outermost electron (inner shells).
The decrease in atomic radii from left to right across the first row of transition elements is small
and irregular. The decrease is much less than that seen on moving across a short period such as
sodium to argon. In the first row of transition elements, as the nuclear charge increases across the
period, each electron enters an inner 3d orbital and this increases the shielding experienced by the
4s electrons. This results in a relatively small difference in the effective nuclear charge felt by the
outer 4s electrons and so the contraction in atomic radius across the row is small.
Density
As the atomic radius decreases with increasing relative atomic mass, the density increases.
Ionic Radius
Ionic radii for transition metals are quite complicated due to the variable oxidation states that they
exhibit. However, there is a small decrease in ionic radii from left to right across the first row of
transition elements. This arises in the same way as the small changes in atomic radius.
Ionization Energy
Because the 3d electrons efficiently shield the 4s electrons from the nucleus, there is only a
minimal increase in effective nuclear charge felt by the outer 4s electrons. This results in
relatively small changes in the energy required to remove an outer 4s electron.
Variable oxidation states
Transition metals have electrons in both the 3d and 4s orbitals and given the similar energies of
these orbitals, one would expect that the electrons in these orbitals would also have similar
energies. As a result, it is easy for a transition metal to form ions of roughly the same stability by
losing different numbers of electrons from the 4s and 3d orbitals. In other words the transition
metal will exhibit variable oxidation states. Note that when d-block elements form ions, it is the
4s electrons that are lost first.
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Complex formation
A complex is formed when a transition metal ion is surrounded by other molecules or ions that
use lone pairs to form dative covalent bonds with the d-block metal. The molecules or ions,
which form the dative bonds, are called ligands. Ligands are negative ions or polarized
molecules, the negative end of which forms the dative bond. Ligands include: :Cl–, :NH3, :OH–,
:OH2. The number of ligands that surround the transition metal ion is usually four or six. Typical
complex ions include [Co(NH3)6]3+ and [CoCl4]2– shown overleaf.
In [Co(NH3)]3+, since the metal ion has a charge of +3 and the ligands are all neutral, the complex
ion has an overall charge of +3. The geometry of the ion is octahedral. In [CoCl4]2–, the metal
ion has a charge of +2 and each of the four ligands has a charge of –1. The complex ion has an
overall charge of –2. The geometry is tetrahedral.
The number of ligands surrounding the metal ion is called the coordination number.
Coordination numbers of four and six are the most common. The usual geometries are tetrahedral
and octahedral respectively. A few complexes of coordination number four adopt a square planar
geometry. The following nickel complex is an example:
The Formation of Coloured Ions by Transition Metals
Transition metal ions show a wide range of colours, both in the solide state and in solution. This
is due to the fact that their ions contain unfilled d orbitals. Before bonding with a ligand, the
isolated transition metal ion has five d orbitals of equal energy – five degenerate orbitals.
However, bonding with ligands causes the d orbitals to split into two groups having different
energies. In octahedral complexes, two of the orbitals experience greater repulsion and are
pushed to a higher energy level.
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Diagram showing the splitting of d orbitals in an octahedral complex
In tetrahedral complexes, the reverse occurs and three of the orbitals end up at higher energy and
two at lower energy. The magnitude of the energy difference ΔE depends on the strength of the
electrostatic field caused by the lone pairs of the ligand. Therefore different ligands give rise to
diferent ΔE values.
A substance appears coloured if it absorbs light in the visible region of the electromagnetic
spectrum. The energy gap ΔE between the split orbitals for most transition metal complexes
corresponds to energies in the visible region of the spectrum. Electrons in the lower d energy
level can absorb visible light of particular frequencies and jump up to the higher energy level.
The colour of the complex is a complement of the colours absorbed from the visible light. If a
substance absorbs green light, for example, it will let through red and blue and thus appear
purple. Electrons can only move from one d orbital to another if there are spaces in the d orbitals,
so ions which have d0 and d10 electronic arrangements, have no possible d-d transitions and hence
appear colourless. So ions like Ti4+ and Sc3+ (both d0) and Cu+ and Zn2+ (both d10) generally do
not give rise to coloured compounds.
Ligand exchange reactions
Ligand exchange is a reaction in which one ligand in a complex ion is replaced by another
ligand. This replacement occurs since some ligands form stronger bonds with a particular metal
ion than other ligands do. Thus, stronger ligands may displace weaker ligands from a complex.
Ligand exchange is a reversible process and even though each ligand is replaced one at a time, it
is more convenient write an equilibrium expression for the overall ligand exchange reaction.
The aqueous chemistry of the cobalt(II) ion (Co2+) demonstrates the principle of ligand
exchange.
•
In aqueous solution, in the absence of complexing agents, the cobalt(II) ion forms the
stable pink (or magenta) hexaaquacobalt(II) complex, [Co(H2O)6]2+(aq). This is an
octahedral complex.
•
The addition of aqueous sodium hydroxide or aqueous ammonia to a solution of
cobalt(II) ions produces the hydrated cobalt(II) hydroxide as a blue-green precipitate,
which turns pink on standing.
Co2+(aq) + 2OH– → Co(OH)2(s)
This blue precipitate is usually represented as an ordinary metal hydroxide but is better
written as [Co(H2O)4(OH)2], showing the aqua ligands. In fact, the hexaaaquacobalt (II)
ion and hexaaqua ions of the general type [M(H2O)6]n+, are acidic in nature and will
donate protons from the ligands to a base as follows:
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[Co(H2O)6]2+ + 2OH– → [Co(H2O)4(OH)2] + 2H2O
•
When sodium chloride or hydrochloric acid is added to an aqueous solution of a
cobalt(II) salt, the deep blue, tetrachlorocobaltate complex ion, [CoCl4]2– is formed.
This is a tetrahedral complex. There are some key points to note about this particular
ligand exchange reaction:
(i)
(ii)
the larger chloride ion ligand leads to a change in coordination number from 6 to
4;
the complex ion shape changes from octahedral to tetrahedral
(iii)
the complex changes from a cationic to an anionic ion
(iv)
there is no change in oxidation state
This is quite a good reaction to demonstrate the Le Châtelier principle concerning
chemical equilibria:
[Co(H2O)6]2+(aq) + 4Cl– ⇌ [Co(Cl)4]2–(aq) + 6H2O
(i)
dilution shifts the equilibrium to the left, towards an increase in pink colour;
(ii)
increasing the chloride ion concentration shifts the equilibrium to the right,
towards an increase in blue colour;
(iii)
increasing the solution temperature shifts the equilibrium to the right, towards an
increase in blue colour;
(iv)
if prepared at higher temperature, with just enough chloride to turn the solution
blue, on cooling the solution will become pink; this shows that left to right is
endothermic and right to left is exothermic.
Stability Constant Kstab
Like any other equilibrium reaction, ligand exchange has an associated equilibrium constant
called the stability constant, Kstab. For the ligand exchange,
[Co(H2O)6]2+(aq) + 4Cl– ⇌ [Co(Cl)4]2–(aq) + 6H2O
Kstab = [Co(Cl)4]2–]______
[Co(H2O)6]2+] [Cl–]4
The stability constant measures the stability of the formed [Co(Cl)4]2–(aq) ion and is indicative of
the ease with which the chloride ion can replace the aqua ligand. The higher the value of the
stability onstant, the more stable is the complex formed. Since Kstab values occur over a very
wide range, they are often expressed as the logarithm, log10Kstab so that numbers are simplified
and patterns can be more easily recognized.
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When excess ammonia is added to a cobalt(II) salt solution, the brown,
hexaamminecobalt(II) ion, [Co(NH3)6]2+ is formed but this is unstable in the presence
of dissolved oxygen and is oxidized to the cobalt(III) complex.
[Co(H2O)6]2+(aq) + 6NH3(aq) ⇌ [Co(NH3)6]2+(aq) + 6H2O(l)
The NH3 and H2O ligand molecules are similar in size and ligand exchange occurs
without a change in coordination number, which remains at 6. Dissolved oxygen then
oxidizes Cr2+ to Cr3+.
4[Co(NH3)6]2+(aq) + O2(g/aq) + 4H+ → 4[Co(NH3)6]3+(aq) + 2H2O(l)
This change in cobalt’s oxidation state from +2 to +3 via an oxidizing agent is quite
common if a complexing agent is also present.
The aqueous chemistry of the copper(II) ion (Cu2+)
• Copper (II) sulfate forms the blue hexaaqua copper (II), [Cu(H2O)6]2+ ion in solution.
On adding aqueous ammonia, a pale blue precipitate is forms which is soluble in excess
ammonia to give a deep blue solution. The ammonia acts as both a base and a ligand.
With a small amount of ammonia, the ammonia ions accepts hydrogen ions from the
water ligands attached to the copper (II) ion to form a pale blue precipitate of copper (II)
hydroxide.
[Cu(H2O)6]2+(aq) + 2NH3(aq) ⇌ [Cu(H2O)4(OH)2](s) + 2NH4+(aq)
This precipitate is often written as Cu(OH)2, without including the water ligands. In
excess ammonia, the water ligands are replaced to form tetraamminediaquacopper(II)
ions.
[Cu(H2O)4(OH)2](s) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 2OH–(aq) + 4H2O(l)
•
If concentrated hydrochloric acid is added to a blue solution of copper(II) sulfate, the
solution turns green and then yellow. The chloride ions gradually replaces the water ligands in
the blue [Cu(H2O)6]2+ ions to form a yellow wolution of [CuCl4]2– ions. The equilibrium mixture
of the blue [Cu(H2O)6]2+ ions and the yellow [CuCl4]2– ions give the green colour. Dilution of the
solution reverses the colour change.
[Cu(H2O)6]2+(aq) + 4Cl–(aq) ⇌ [CuCl4]2–(aq) + 6H2O(l)
Note
The larger chloride ion ligand leads to a change in co–ordination number from 6 to 4.
The complex ion shape changes from octahedral to tetrahedral.
The colour of the complex changes from blue to yellow–brown (green due to residual
blue),
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The complex changes from a cationic complex ion to an anionic complex ion.
There is no oxidation state change at all, copper is in the +2 state throughout the
reaction.
Le Chatelier’s Principle can be applied to equilibria involving the hexaaquacopper(II)
ion a similar way as was done for the hexaaquacobalt (II) ion.
Ligand Exchange Reactions of Haemoglobin
Haemoglobin is the iron-containing pigment of the blood and is responsible for transporting
oxygen from the lungs to the other cells and tissue of the body. It consists of four globular
proteins. At the centre of each protein is a Fe2+ ion complex known as haem. The Fe2+ ion has a
co-ordination number of 6. Five of the co-ordination sites are occupied by nitrogen, four from a
planar ring structure called a porphyrin and the fifth from a nitrogen atom of the protein. Water
or a molecular oxygen ligand can reversibly attach itself to the sixth site.
Water and oxygen ligands are reversibly bonded to the Fe2+ complex.
In the lungs, oxygen attaches to the Fe2+ ion forming oxyhaemoglobin. In the other parts of the
body, the oxygen is replaced by a water ligand forming deoxyhaemoglobin. This allows
haemoglobin to transport oxygen in the blood from the lungs to the other parts of the body.
Haemoglobin is capable of transporting 4 molecules of oxygen at any one time. Stronger ligands
such as cyanide and carbon monoxide can also bond to this site. However, they bond irreversibly
to the site and prevent haemoglobin from transporting oxygen in the body. This is the poisonous
action of these substances.
The Oxidation States of Vanadium
The ground state electronic configuration of vanadium is [Ar]3d3 4s2. Vanadium has five
electrons outside of the [Ar] core and these can be lost in such a way that vanadium is able to
exhibit the four common oxidation states +5, +4, +3 and +2. Each of these oxidation states can
be distinguished by its colour. These colour changes can be shown by shaking a solution
containing vanadium(V) with zinc and dilute acid.
The solution of vanadium (V) is made by dissolving about 3 g of ammonium vanadate (V),
NH4VO3, in 40 cm3 of 2 mol dm–3 H2SO4. This ammonium vanadate (V) solution is yellow in
colour and contains dioxovanadium (V) ions, VO2+, in acid solution.
VO3–(aq) + 2H+(aq) → VO2+(aq) + H2O(l)
yellow solution; oxidation state +5
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Shaking the dioxovanadium (V) solution with granulated zinc will cause a gradual transition of
vanadium from the +5 to the +2 oxidation state. The colour of the solution changes from yellow
to green (mixture of VO2+ and VO2+) then to bue (VO2+) then to green (V3+) and finally to violet
(V2+).
VO2+(aq) + 2H+(aq) + e– → VO2+(aq) + H2O(l)
blue solution; oxidation state +4
VO2+(aq) + 2H+(aq) + e– → V3+(aq) + H2O(l)
green solution; oxidation state +3
V3+(aq) + e– → V2+(aq) + H2O(l)
violet solution; oxidation state +2
The Feasibility of Vanadium Reactions
θ
E values can be used to show that the vanadium reactions showing different colours of vanadium
are all feasible. The scheme below shows that positive E
θ
cell
(emf) are obtained for each reaction.