Period 4 transition metals exhibit variable oxidation states

The d-block elements in period 4, from titanium to copper, commonly form compounds with a variety of oxidation states. This is largely because the 3d and 4s electrons in these metals have similar energies. As a result, each element can form relatively stable ions in solution or solid compounds by losing different numbers of electrons.

This behaviour contrasts with metals in groups 1 and 2, where the ionisation energy increases significantly once the outer shell electrons are removed.

The common oxidation states of each period 4 transition element is shown below. The most common oxidation states are bolded.

• The elements at the ends of the series (scandium and zinc) have only one main oxidation state.
• The broadest range of oxidation states occurs for elements near the middle of the series.
• Most of these elements form compounds in the +2 oxidation state, using both 4s electrons in bonding.

Redox reactions of vanadium

Vanadium can exist in multiple oxidation states, each displaying a distinct colour in solution:

Changes in oxidation state occur through redox reactions, where ions gain electrons (reduction) or lose electrons (oxidation).

The following table lists the half-equations and their corresponding reduction potentials for vanadium:

These reduction potentials can be used to predict the direction of redox reactions.

For example, when zinc is introduced to an acidified vanadium(V) solution:

Zn_(s)_ + 2e^-^ ⇌ Zn^2+^(aq) E^⦵^ = $-$0.76 V

The positive cell potentials suggest that zinc will sequentially reduce vanadium(V) to (IV), then to (III) and (II):

1. VO_2_^+^ + 4H^+^ + Zn ⇌ VO^2+^ + 2H_2_O + Zn^2+^ E^⦵^ = +1.76 V
2. VO^2+^ + 4H^+^ + Zn ⇌ V^3+^ + 2H_2_O + Zn^2+^ E^⦵^ = +1.10 V
3. V^3+^ + Zn ⇌ V^2+^ + V^2+^ E^⦵^ = +0.50 V
4. V^2+^ + Zn ⇌ V + V^2+^ E^⦵^ = $-$0.42 V

As the reduction proceeds, the solution changes colour from yellow to blue to green to violet. The negative potential for the final step indicates that zinc cannot reduce vanadium(II) under standard conditions.

Redox reactions of chromium

Chromium forms ions in oxidation states +2, +3 and +6, each with a distinctive colour in aqueous solution:

The colour of chromium(VI) ions varies with pH due to the following equilibrium:

2CrO_4_^2-^(aq) + 2H^+^(aq) ⇌ Cr_2_O_7_^2-^(aq) + H_2_O_(l)_

• At high pH - The solution is yellow due to the presence of CrO_4_^2-^ ions.
• At low pH - The solution is orange due to the presence of Cr_2_O_7_^2-^ ions.

The Cr^3+^ ion is generally green in solution, but it can appear violet when surrounded by six water ligands. Cr^2+^ ions are blue but are easily oxidised to Cr^3+^ unless maintained under an inert atmosphere.

The different oxidation states of chromium can be interconverted through redox reactions:

1. Reduction of Cr(VI) to Cr(III) - In acidic solution, dichromate ions (Cr_2_O_7_^2-^) can be reduced to Cr^3+^ ions using zinc metal: Cr_2_O_7_^2-^(aq) + 14H^+^(aq) + 3Zn_(s)_ ➔ 2Cr^3+^(aq) + 7H_2_O_(l)_ + 3Zn^2+^(aq)

2. Reduction of Cr(III) to Cr(II) - Under an inert atmosphere, excess zinc metal can further reduce Cr^3+^ ions to Cr^2+^: 2Cr^3+^(aq) + Zn_(s)_ ➔ 2CrCr^2+^(aq) + Zn^2+^(aq)

3. Oxidation of Cr(III) to Cr(VI) - In alkaline conditions, hydrogen peroxide can oxidise Cr^3+^ ions to chromate ions (CrO_4_^2-^): 2Cr^3+^(aq) + 3H_2_O_2(aq)_ + 10OH^-^(aq) ➔ 2CrO_4_^2-^(aq) + 8H_2_O_(l)_