Much of the chemistry of transition metal ions is a consequence of the fact that the 3d and 4s orbitals are close in energy. 4s electrons are lost before 3d electrons when transition metal ions are formed and a number of different electronic configurations are stable – transition metals form ions with variable oxidation states e.g. Ti2+ and Ti3+, Mn2+ and MnO4– (ions with a charge of more than +3 are rare, higher oxidation states are seen in oxides and oxo-anions).
Trends to note:
- from titanium to manganese it is common to see the maximum oxidation state in compounds which is the result of the loss of all 4s and 3d electrons
- from manganese to copper the +2 oxidation state is common where only the 4s electrons have been lost
The stability of a transition metal ion depends on partly on the anion which makes up the compound. For example, fluoride ions are small and not easily polarised so they are able to stabilise high oxidation states compared with iodide ions.
Compounds in which the transition metal ion has a high oxidation state, e.g. manganese has an oxidation state of +7 in KMnO4 and chromium has an oxidation state of +6 in Na2Cr2O7, are strong oxidising agents because the transition metal ion readily gains electrons reducing it to a lower oxidation state. Redox reactions are very common!
Case study: chromium
The trend for variable oxidation states in the transition metals can be explained if we look at the trend in ionisation energies. Let’s take chromium as an example.
| Cr(g) ⇾ Cr+(g) + e– | 653 kJ mol-1 |
| Cr+(g) ⇾ Cr2+(g) + e– | 1591 kJ mol-1 |
| Cr2+(g) ⇾ Cr3+(g) + e– | 2987 kJ mol-1 |
| Cr3+(g) ⇾ Cr4+(g) + e– | 4743 kJ mol-1 |
| Cr4+(g) ⇾ Cr5+(g) + e– | 6702 kJ mol-1 |
| Cr5+(g) ⇾ Cr6+(g) + e– | 8745 kJ mol-1 |
| Cr6+(g) ⇾ Cr7+(g) + e– | 15455 kJ mol-1 |
Notice that the first 6 ionisation energies (which correspond to losing the 4s and 3d electrons successively) are very close together, and then there is a big jump. Oxidation states +1 to +6 are easily achieved.
Losing the 7th electron means taking one from the 3p sub shell – the effective nuclear charge experienced by this electron is significantly greater. An oxidation state of +7 for chromium is exceptionally unfavourable.
We should also be aware that as the charge on the Cr ion increases the lattice enthalpy for compounds formed with the ion will become more exothermic (increased electrostatic attraction between the ions) as will the hydration enthalpy. Both of these enthalpy changes compensate for increasingly endothermic ionisation energies as we remove electrons from a positively charged ion.
Case study: vanadium
We can illustrate how the oxidation state of vanadium can be reduced from +5 to +2 through a series of redox reactions. Each ion has a distinct colour (and some exam syllabi require you to know them).
| Ion | VO2+ | VO2+ | V3+ | V2+ |
| Colour in aqueous solution | yellow | blue | green | violet |
| Oxidation state | +5 | +4 | +3 | +2 |
| Name | dioxovanadium(V) | oxovanadium(IV) | vanadium(III) | vanadium(II) |
When powdered zinc is added to an acidified solution of dioxovanadium(V) ions we see effervescence and the solution changes from yellow to green to blue to green to violet.
To explain what’s going on here, we need to consult the relevant electrode potentials for these redox reactions.
| Half equation | E⦵ / V |
|---|---|
| Zn2+(aq) + 2e– ⇌ Zn(s) | – 0.76 |
| V3+(aq) + e– ⇌ V2+(aq) | – 0.26 |
| VO2+(aq) + 2H+(aq) + e– ⇌ V3+(aq) + H2O(l) | + 0.34 |
| VO2+(aq) + 2H+(aq) + e– ⇌ VO2+(aq) + H2O(l) | + 1.00 |
The Zn2+ / Zn half cell has a less positive standard electrode potential than the VO2+ / VO2+ half cell so Zn will be oxidised to Zn2+ and VO2+ will be reduced to VO2+. The colour change is yellow to blue but there is an intermediate stage in which we have both vanadium ions present, hence the green colour seen.
Zinc then goes on to reduce each of the vanadium ions in turn since whichever vanadium half cell we consider, the electrode potential for the Zn2+ / Zn half cell is always less positive.
The solution must be acidified because the reduction of both VO2+ and VO2+ requires H+ ions (see half equations above) and this also explains the observed effervescence as powdered zinc reacts with acid to give hydrogen gas.
In addition to the practice questions below you can find more of this style of exam question that tackles the feasibility of redox reactions here. Exam technique and terminology is VERY important!
Let’s take a look at the standard electrode potential of the highest oxidation state for titanium, vanadium, chromium and manganese …
| Half equation | E⦵ / V |
|---|---|
| TiO2+(aq) + 2H+(aq) + e– ⇌ Ti3+(aq) + H2O(l) | + 0.10 |
| VO2+(aq) + 2H+(aq) + e– ⇌ VO2+(aq) + H2O(l) | + 1.00 |
| Cr2O72-(aq) + 14H+(aq) + 6e– ⇌ 2Cr3+(aq) + 7H2O(l) | +1.33 |
| MnO4–(aq) + 8H+(aq) + 5e– ⇌ Mn2+(aq) + 4H2O(l) | +1.51 |
You’ll notice that as the highest oxidation state increases from +4 for titanium in TiO2+ to +7 for manganese in MnO4–, the standard electrode potential becomes more positive i.e. the strength of the oxidising agent increases down the table.
However, the value of the standard electrode potential is pH dependent because transition metal elements form different ions in acid or alkaline conditions (but in which they have the same oxidation state).
Chromium forms the orange dichromate ion, Cr2O72-, in acidic solution which is a strong oxidising agent but in an alkaline solution it exists as a yellow chromate(VI) ion, CrO42- . In both ions chromium has an oxidation state of +6.
The chromate(VI) ion is not an oxidising agent because the CrO42- / Cr(OH)3 half cell has a negative standard electrode potential.
CrO42-(aq) + 4H2O(l) + 3e– ⇌ Cr(OH)3(s) + 5OH–(aq) E⦵ = – 0.11V
- transition metal ions in high oxidation states are reduced in acidic solutions
- transition metal ions in low oxidation states are oxidised in alkaline solutions (a high pH favours the formation of negative ions which are more likely to be oxidised because it is easier for them to lose electrons compared with positive ions)
Case study: cobalt
Co2+ ions in alkaline solution are rapidly oxidised to Co3+.
If we add ammonia solution to [Co(H2O)6]2+, a precipitate of Co(H2O)4(OH)2 is formed (you can find out more about this reaction here) which dissolves in excess ammonia.
[Co(H2O)6]2+(aq) + 2OH–(aq) ⇾ Co(H2O)4(OH)2(s) + 2H2O(l)
Co(H2O)4(OH)2(s) + 6NH3(aq) ⇾ [Co(NH3)6]2+(aq) + 2OH–(aq) + 4H2O(l)
And [Co(NH3)6]2+(aq) is oxidised to [Co(NH3)6]3+(aq) by oxygen in the air with a change in colour from a pale brownish solution to a yellow solution.
Redox titrations
This redox behaviour of transition metal ions can turned into a quantitative analytical technique in the form of redox titrations. You can find an explanation of how to tackle these popular exam questions and some exam style Q&A here.
Practice questions
- Aqueous potassium iodide was added to a solution of dioxovanadium(V) ions and the solution turned from clear yellow to muddy brown. Following the addition of aqueous sodium thiosulfate to the test tube, a clear blue solution was observed.
| Half equation | E⦵ / V | Colour of oxidised species |
|---|---|---|
| I2(aq) + 2e– ⇌ 2I–(aq) | + 0.54 | brown |
| S4O62−(aq) + 2 e− ⇌ 2S2O32−(aq) | + 0.08 | colourless |
| V3+(aq) + e– ⇌ V2+(aq) | – 0.26 | green |
| VO2+(aq) + 2H+(aq) + e– ⇌ V3+(aq) + H2O(l) | + 0.34 | blue |
| VO2+(aq) + 2H+(aq) + e– ⇌ VO2+(aq) + H2O(l) | + 1.00 | yellow |
(a) Use the standard electrode potential data to explain the initial observation.
(b) Why was aqueous sodium thiosulfate added to the test tube?
(c) Explain why no further colour changes were observed.
2. In a redox reaction between vanadium(II) chloride and chlorine gas, the reactants react in a 2:3 molar ratio.
(a) Write a half equation with state symbols for the reduction of chlorine.
(b) Deduce the final oxidation state of vanadium in this redox reaction.
(c) State the initial and final colours of the solution.
Answers
- (a) VO2+ is reduced to VO2+ by iodide ions because the I2 / 2I– half cell has a less positive standard electrode potential. This produces iodine which explains the muddy brown colour observed which masks the blue of the VO2+ ions.
(b) The thiosulphate ions react with the iodine in the test tube, reducing it back to colourless iodide ions. The blue VO2+ ions are now visible.
Exam tip: you should be familiar with all the required practicals you have done – redox titrations between I2 and S2O32- ion are very common.
(c) Iodide ions are not able to reduce VO2+ to V3+ because the I2 / 2I– half cell has a more positive standard electrode potential than the VO2+ / V3+half cell.
2. (a) Cl2(g) + 2e– ⇌ 2Cl–(aq)
(b) 2 moles of V2+ : 3 moles of Cl2 ; 3 moles of Cl2 gain 6e– which are lost from 2 moles of V2+ ; each V2+ must lose 3e– so the final oxidation state must be +5.
(c) initial colour is violet, final colour is yellow