Transition metals form compounds in which they display more than one
valency. This is due to the variable oxidation states attainable by
losing different numbers of '3d' electrons.
Syllabus reference S3.1.9
Structure 3.1.9 - The formation of variable oxidation states in transition elements can be explained by the fact that their successive ionization energies are close in value. (HL)
- Deduce the electron configurations of ions of the first-row transition elements.
Oxidation state
The oxidation state is defined as the apparent charge on an atom within
a compound.
The oxidation state of atoms within elements is always taken to be zero.
An atom increases its oxidation state (or number) by losing electrons to
become more positive.
Even though in many cases the systems are not ionic, it is possible to
designate oxidation states to atoms in covalent systems as if they were
ionic.
Multiple oxidation states of the d-block (transition metal) elements are
due to the proximity of the 4s and 3d sub shells (in terms of energy).
All transition metals exhibit a +2 oxidation state (the first electrons
are removed from the 4s sub-shell) and all have other oxidation states.
The common transition metal oxidation
states (Sc and Zn included for comparison)
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Sc
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Ti
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V
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Cr
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Mn
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Fe
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Co
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Ni
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Cu
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Zn
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+1
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+2
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+2
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+2
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+2
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+2
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+2
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+2
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+2
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+2
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+3
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+3
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+3
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+3
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+3
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+3
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+3
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+3
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+4
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+4
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+4
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+5
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+6
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+6
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+6
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+7
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Electronic configuration
Variable oxidation states may be understood rather better by a consideration
of the electronic configurations of the states formed and the sum of the ionization energies needed to remove the electrons.
Transition metals all have similar atomic radii, so the force of attraction of the nucleus for the outer electrons is compensated for by the force of inter-electron repulsion. This means that the ionizatio energies are similar in magnitude.
Iron, for example has two common oxidation states, +2 and +3.
The element has the configuration [Ar]4s2 3d6. Clearly,
the +2 oxidation state arises from the loss of the 4s electrons. However,
loss of a further electron from the 'd' shell leaves a configuration of
[Ar]4s0 3d5. This half-full set of 'd' orbitals is
spherically symmetrical and has an extra degree of stability. Consequently
the iron(III) state is also stable and common.
| iron |
iron(II) |
iron(III) |
| [Ar]4s2 3d6 |
[Ar]4s0 3d6 |
[Ar]4s0 3d5 |
This is not quite as simple as stated, as the nature of the environment in
which the transition metal atom finds itself is also of great importance
as regards stability.
An oxidation state that is stable in a solid compound may not be stable
in aqueous solution and vice versa.
This is due to the crystal, or ligand field effect and depends on the molecules
or ions surrounding the transition metal atom.
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Worked examples
Q543-01 The electron configuration
for Mn
2+ is which of the following?
- [Ar] 4s2 3d3
- [Ar] 3d5
- [Ar] 4s1 3d5
- [Ar] 4s1 3d4
Answer
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Manganese is element number 25 with an electronic configuration [Ar]
4s2 3d5. The manganese 2+ ion has lost its pair
of 4s electrons and is left with the configuration [Ar]
3d5
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Q543-02 Vanadium, electronic
configuration [Ar] 3d
3 4s
2, exists in which of the following
oxidation states?
- I. 0
- II. +2
- III. +4
- I, II and III are correct.
- I and II are correct.
- II and III are correct.
- I is the only correct response.
- III is the only correct response.
Answer
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Vanadium has all of the possible oxidation states as well as +3 and
+5. The correct response is I, II and III.
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Q543-03 "Zn
2+ complexes
are generally atypical of d-block complexes in general." Which answer below
is correct and supports this statement?
- Zn2+ complexes are paramagnetic.
- Zn2+ complexes tend to be colourless.
- Zn2+ complexes are always octahedral.
- Zn2+ is one of several oxidation states of Zn.
Answer
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Paramagnetism occurs when an atom or ion has unpaired electrons.
Zn2+ has no unpaired electrons. Zn2+
complexes are colourless and this is atypical of transition
metals. The other two possibilities are false.
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Q543-04 Which properties are
typical of d block elements?
- I. Complex ion formation
- II. Catalytic behaviour
- III. Colourless compounds
- I, II and III
- II and III only
- I and III only
- I and II only
Answer
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Transition metal complexes tend to be coloured, so III is false.
The other two are correct, I and II only
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Q543-05 The cyanide ion, CN
-,
can form two complexes with iron ions. The formulas of these ions are [Fe(CN)
6]
4-
and [Fe(CN)
6]
3- . What is the oxidation state of the iron
in each of the complexes?
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[Fe(CN)6]4-
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[Fe(CN)6]3-
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| A. |
-4
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-3
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| B. |
+2
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+3
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| C. |
+3
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+2
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| D. |
-3
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-4
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Answer
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In the first complex the oxidation state is Fe + (6-) = 4-.
∴ Fe = +2.
In the second complex the oxidation state of Fe is given by Fe +
(6-) = 3-.
∴ Fe = +3
The correct answer is then, +2 and +3
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Q543-06 A transition metal
ion X
3+ has the electronic configuration [Ar] 3d
4. What
is the atomic number of element X.
- 22
- 24
- 25
- 27
Answer
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If the ion has lost 3 electrons to form a 3+ ion, you must add three
electrons to [Ar] 3d4 to get the configuration of the element.
The first two electrons go into the 4s orbital and the remaining electron
adds into the 3d shell.
∴ [Ar] 3d5 4s2
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Q543-07 Consider the following
coordination compounds
- I. [Pt(NH3)4]Cl2
- II. [Pt(NH3)3Cl]Cl
- III. [Pt(NH3)2Cl2]
What are the charges on the complex ions?
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I
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II
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III
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A.
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+2
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+1
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0
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B.
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-2
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-1
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0
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C.
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0
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+1
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+2
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D.
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0
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-1
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-2
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Answer
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In compound I, the complex ion has a charge that is balanced by two
chloride ions. Thus the the charge on the complex ion is 2+. Pt is
with four NH3 ligands that are neutral.
∴ Pt = 2+
In compound II, the complex ion has a charge that is balanced by
one chloride ion. Thus the the charge on the complex ion is 1+. Pt
is with three neutral NH3 ligands and one chloride ion
with a charge of -1. Thus: Pt + (-1) = +1
∴ Pt = 2+
In compound III, there are no balancing chloride ions, but there
are two chloride ligands. Thus the the charge on the compound is zero
(as with all compounds). Pt is with two neutral NH3 ligands
and two chloride ion with a charge of -1. Thus: Pt + (-2) = 0
∴ Pt = 2+
The charges on the species are +2, +1,
0
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Q543-08 A certain element
has the electronic configuration 1s
2 2s
2 2p
6
3s
2 3p
6 4s
2 3d
3. Which oxidation
states would this element most likely show?
- +2 only
- +3 only
- +2 and +5 only
- +2, +3, +4, +5
Answer
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Common oxidation states are given by first losing the 4s electrons,
and then the species can lose up to the next five 'd' electrons. In
this case the common oxidadation states are +2,
+3, +4, +5
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Q543-09 What is the oxidation
state of the platinum ion in the compound [Pt(NH3)4]Cl2?
Answer
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In the compound, the complex ion has a charge that is balanced by
two chloride ions. Thus the the charge on the complex ion is 2+. Pt
is with four NH3 ligands that are neutral.
∴ Pt = +2
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Q543-10 What is the oxidation
state of the manganese ion in the compound [Mn(NH3)4Cl2]SO4?
Answer
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In the compound, the complex ion has a charge that is balanced by
one sulfate ion (2-). Thus the the charge on the complex ion is 2+.
Mn is with four NH3 neutral ligands that are neutral and
two single negative chloride ligands. Thus: Mn + (2-) = 2+
∴ Mn = +4
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