Variable Oxidation State of Transition Elements

Variable oxidation state definition

The number of electrons available for bonding in an atom determines its oxidation state. In transition metals, the 4s and 3d electrons are available for bonding. This gives transition metals a unique characteristic: variable oxidation states.

To understand variable oxidation states, you must imagine that only ionic bonding is possible. If an atom could only lose up to three electrons to make bonds with another atom, you can say that it has 3 variable oxidation states, depending on the atom it bonds to.

• We will explore why transition metals have variable oxidation states.
• We will learn why the redox potential of a transition metal when going from a high oxidation state to a lower one depends on the pH and ligand.
• We will discover the five oxidation states of vanadium through a reaction between vanadate(V) and zinc.
• Finally, we will carry out the ‘silver mirror’ test, or the Tollens' test.

Variable oxidation states of transition metals

As we've already mentioned, the 4s and 3d electrons in transition metals are available for bonding (visit Transition Metals for lots more information). The reason is that there is not a considerable energy gap between the 3d orbital and the 4s orbital. The 4s electrons are lost first. In transition metals, the ionisation energy required to remove the third electron (on the 3d sub-shell) is not much bigger than the ionisation energies required to remove the first two (on the 4s sub-shell).

The graph below shows the first ionisation energies of vanadium (a transition metal), calcium, and scandium (non-transition metals). Notice how the first ionisation energies of vanadium are closer together than in scandium and calcium? As you can see, the small difference between the ionisation energies makes it easy for transition metals like vanadium to have variable oxidation states. Not much energy is required to remove a small number of electrons from transition metals.

Fig. 1 - Ionisation energies of vanadium, calcium, and scandium

The table below shows examples of transition metals and their common oxidation states.

Redox potential

When we talk about redox reactions, we are usually only interested in the transfer of electrons, so we only write what we call the half-reaction. In the example of the half-reaction below, the zinc 2+ ion accepts two electrons and gets reduced to zinc with an oxidation number of zero.

Zn²⁺_(aq) + 2e^- ⇌ Zn_(s) Eº = -76 V

You will also notice we have given this half-reaction an Eº value of -76 V. Voltage (V) is the unit of measure for the electrical potential of a half-reaction. Essentially, it is the difference between the demand for electrons in one half of a half-reaction and the tendency to lose electrons in the other half. E is what we call the electrode potential (or reduction potential). It shows how easily a substance gets reduced. (You can learn more about standard electrode potentials in Electrochemical Cells.)

The redox potential of transition metal ions when going from a high oxidation state to a lower one depends on two things:

• The pH

• and ligand.

pH

What do we mean by pH? When transition metal ions in an aqueous solution go through a redox reaction, it usually involves hydrogen ions. In other words, it requires acidic conditions. You can see this clearly in the example of manganese(VII).

The half-reaction below shows that the manganate(VII) ion easily readily reduces in an acidic solution. As you can see from the positive Eº value, the process moves towards the right. You can also see that manganese gets reduced from the +8 to +2 oxidation state.

MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- ⇌ Mn²⁺(aq) + 4H₂O_(l) Eº = +1.51 V

On the other hand, if you use a neutral solution like water, manganese ions only get reduced to the +4 oxidation state. Notice how the Eº value of the following reaction is much lower? The reason is that the manganese ions are less willing to accept electrons in a neutral solution.

MnO₄^- _(aq) + 2H₂O _(l) + 3e^- ⇌ MnO_{2 (s)} + 4OH^- _(aq) Eº = +0.59 V

Ligand

How do ligands affect the redox potential of transition ions? Notice the Eº values of the half-reactions below. They separately show the reduction of the nickel(II) ion and the hexaaminenickel(II) ion. What can you conclude about the ligands by looking at the Eº values?

[Ni(H₂O)₆]²⁺ _(aq) + 2e^- ⇌ Ni _(s) + 6H₂O _(l) Eº = -0.26 V

[Ni(NH₃)6]²⁺ _(aq) + 2e^- ⇌ Ni _(s) + 6NH_{3 (aq)} Eº = -0.49 V

By comparing the above two equations, we can state the following:

• Eº becomes increasingly negative when ammonia replaces the water ligands.

• The ammonia ligands are more firmly attached to the nickel ions than the water ligands.

• The process with nickel(II) is more positive than with ammonia. So the equilibrium lies slightly more to the right.

Variable oxidation state of vanadium

You can observe the variable oxidation states of vanadium in a reaction between vanadate(V) ions and zinc.

Vanadium has four oxidation states: vanadium (II), (III), (IV), and (V). Vanadium(IV) is the most stable oxidation state. We can form the vanadium species in various oxidation states through a redox reaction between vanadate(V) ions and zinc. In that reaction, the vanadate ions get reduced by zinc in an acidic solution.

For the reaction, we use ammonium vanadate, a white powder, that dissolves in hydrochloric acid to produce a yellow-coloured solution. When we add zinc to the solution, vanadium gets reduced from +5 to +2. You can tell this is happening because the solution changes colour from yellow to blue to green to violet.

The image below shows the colours of the solution as the vanadium gets reduced from the +5 to +2 oxidation state.

• You can see the species of vanadium present in the solution as it changes colour from yellow to blue to green to violet.

• You will also notice that the higher oxidation states of vanadium do not exist as simple ions such as V⁵⁺_(aq) or V⁴⁺_(aq).

Fig. 2 - Variable oxidation states of vanadium

The 'silver mirror' test is another reaction that uses the variable oxidation states of a complex ion. Read on to discover why we call it that!

Variable oxidation state of dichromate ions

As mentioned previously, transition metal ions can have a variety of oxidation states. We have gone through this for vanadium, so now we will be exploring dichromate ions.

Firstly, let's explore how the dichromate(VI) ion, Cr₂O₇²⁻ can be reduced to Cr³⁺ and Cr²⁺ ions. This can be done using zinc and a dilute acid such as sulphuric acid or hydrochloric acid. Cr₂O₇²⁻ is orange and when it is reduced using zinc and a dilute acid, it can form Cr³⁺ (green), which can be further reduced to Cr²⁺ (blue). This is represented with the following equations.

This equation shows the reduction from +6 to +3.

Cr₂O₇^2- + 14H⁺ + 3Zn → 2Cr³⁺ + 7H₂O + 3 Zn²⁺

This shows the reduction from +3 to +2.

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺

Now we shall explore how dichromate ions can be produced from the oxidation of Cr³⁺. This is done using hydrogen peroxide in alkaline conditions which is then followed by acidification. When a transition metal in a low oxidation state is in an alkaline solution, it is more easily oxidised than when it is in an acidic solution. This can be seen in the following equation:

[Cr(H₂O)₆]³⁺ (aq) → [Cr(OH)₆]^3- (aq) in excess sodium hydroxide (NaOH)

[Cr(OH)₆]^3- can then be oxidised by warming it with a hydrogen peroxide solution to produce a yellow solution of of chromate ions.

The reduction can be seen here:H₂O₂ + 2e- → 2OH^-The oxidation can be seen here: [Cr(OH)₆]^3- + 2OH^- → CrO₄^2-+ 3e- + 4H₂O

This then leads to the following equation:

2[Cr(OH)₆]^3- + 3H₂O₂ → 2CrO₄^2- + 2OH^- + 8H₂O

Finally, let us explore how dichromate(VI) ion Cr₂O₇²⁻, can be converted into chromate(VI). Chromate can be converted to dichromate using this equilibrium equation:

2CrO₄^2- + 2H⁺ ⇌ Cr₂O₇^2- + H₂O

It is important to note that this reaction is not a redox reaction. This is because the oxidation number is alway +6. is instead an acid base reaction.

CrO₄^2- is a yellow solution and can be turned into Cr₂O₇^2-, an orange solution by adding dilute sulphuric acid. To change from the orange solution to the yellow solution, we need the addition of sodium hydroxide.

Reduction of Tollens' reagent

We carry out a test to distinguish between an aldehyde or ketone by using the complex ion diamminesilver(I). This test, also called Tollens' test, is one of the ways we use to identify the functional group in an unknown organic compound. Diamminesilver(I) [Ag(NH₃)₂]⁺ is also known as Tollens' reagent after the German chemist Bernard Tollens.

The test involves reducing Tollens' reagent, which contains silver(I) nitrate, to metallic silver. In order to carry out the test, you must first prepare Tollens' reagent. We prepare it for each test since Tollens' reagent is unstable in solution.

To prepare Tollens' reagent:

1. Add some sodium hydroxide to silver nitrate to produce silver(I) oxide, a brown precipitate.

2. Add concentrated ammonia solution to redissolve the silver(I) oxide back into diamminesilver(I).

Now the Tollens' reagent is ready for Tollens' test. To carry out the test:

1. Add a few drops of the unknown organic compound to Tollens' reagent.

2. Then gently warm in a water bath.

3. If the unknown compound is a ketone, you will observe no change in the colourless solution.

4. If the unknown compound is an aldehyde, you will get a grey silver precipitate. The Ag⁺ ions have reduced to Ag while the aldehyde oxidised to carboxylic acid.

We can observe the silver precipitate when the substance contains an aldehyde because aldehydes are reducing agents and reduce the silver(I) nitrate to metallic silver. The Tollens' test is also called the 'silver mirror' test because of the silver coating formed inside the test tube.

You can find the half-equations for the reduction of Tollens' reagent by an aldehyde below, as well as the net ionic equation.

Reduction of Tollens' reagent

Ag(NH₃)₂⁺ + e^- ⟶ Ag + 2NH₃

Oxidation of aldehyde

RCHO + 3OH^- ⟶ RCOO^- + 2H₂O + 2e^-

Net ionic equation

2Ag(NH₃)₂⁺ + RCHO + 3OH^- ⟶ 2Ag + 4NH₃ + RCOO^- + 2H₂O

Redox titrations with manganese

Potassium manganate (VII) is a violently strong oxidiser in acidic conditions - it can even break carbon-carbon bonds! Today, we use it as an oxidising agent to synthesise organic compounds such as aldehydes, ketones, and carboxylic acids.

Potassium manganate (VII) is also frequently used to calculate the concentration of reducing agents in redox titrations. Essentially, potassium manganate (VII) oxidises any redox couple with an Eº value less than +1.33V. We will briefly examine two such processes with MnO4^- ions:

1. With iron (II) and iron (III) ions.
2. With ethanedioate ions.

In these redox reactions, manganese gets reduced from +7 to +2.

MnO₄^- + 8H⁺ + 5e^- ➔ Mn²⁺ + 4H₂O

We carry out these redox reactions as titrations. Always set up the experiment with potassium permanganate in the burette. We place the other solution in a flask with dilute sulphuric acid. You must use dilute sulphuric acid in these titrations because other acids can act as oxidising agents in side reactions that will give inaccurate readings.

Manganese(II) appear colourless in solution. As soon as one drop of excess MnO₄^- is added, the solution turns a visible pale pink, indicating the endpoint of the titration.