Introduction to Transition Metal Reactions

Welcome to one of the most colourful and exciting chapters in A Level Chemistry! In the previous chapter (Topic 15A), you learned the basics of transition metals—like how they form complex ions and why they have such vibrant colours. Now, we are going to look at how these metals actually react.

Transition metals are like the "chameleons" of the periodic table. They can change oxidation states, swap ligands, and speed up reactions as catalysts. Don't worry if the number of colours and equations seems overwhelming at first; we will break them down into simple patterns that are easy to remember!

1. Vanadium: The Rainbow Metal

Vanadium is famous for having four distinct oxidation states, each with a different colour. You can observe this by reducing ammonium vanadate(V) using zinc in acidic conditions.

The Four Oxidation States of Vanadium

  • Oxidation State \(+5\): Yellow colour. The ion is the dioxovanadium(V) ion, \(VO_2^+\).
  • Oxidation State \(+4\): Blue colour. The ion is the oxovanadium(IV) ion, \(VO^{2+}\).
  • Oxidation State \(+3\): Green colour. The ion is \(V^{3+}\).
  • Oxidation State \(+2\): Violet colour. The ion is \(V^{2+}\).

Mnemonic Hint: To remember the order of colours from \(+5\) to \(+2\), try: You Better Get Violet (Yellow, Blue, Green, Violet).

Using Electrode Potentials (\(E^{\theta}\))

We use standard electrode potential values to predict if a reaction will happen. For zinc to reduce vanadium, the \(E^{\theta}\) for the zinc half-cell must be more negative than the vanadium half-cell. Since zinc has a very negative \(E^{\theta}\) (\(-0.76 V\)), it is a strong enough reducing agent to take vanadium all the way from \(+5\) down to \(+2\).

Quick Takeaway: Changing the oxidation state of a transition metal usually results in a distinct colour change.


2. Chromium Chemistry

Chromium exhibits some fascinating "reversibility" and reacts differently depending on whether the solution is acidic or alkaline.

The Chromate(VI) / Dichromate(VI) Equilibrium

In solution, chromium(VI) exists in an equilibrium between the yellow chromate ion and the orange dichromate ion:

\(2CrO_4^{2-} (aq) + 2H^+ (aq) \rightleftharpoons Cr_2O_7^{2-} (aq) + H_2O (l)\)

  • Add Acid (\(H^+\)): The equilibrium shifts to the right, turning the solution orange.
  • Add Alkali (\(OH^-\)): The \(OH^-\) reacts with \(H^+\), removing it. The equilibrium shifts left, turning the solution yellow.

Note: This is not a redox reaction because the oxidation state of chromium remains \(+6\) in both ions!

Reduction and Oxidation of Chromium

  • Reduction: Zinc in acidic conditions can reduce orange dichromate(\(VI\)) (\(Cr_2O_7^{2-}\)) to green chromium(\(III\)) (\(Cr^{3+}\)), and then further to blue chromium(\(II\)) (\(Cr^{2+}\)).
  • Oxidation: In alkaline conditions, hydrogen peroxide (\(H_2O_2\)) can oxidise green \(Cr^{3+}\) (specifically the \([Cr(OH)_6]^{3-}\) ion) back up to yellow chromate(\(VI\)) (\(CrO_4^{2-}\)).

3. Reactions with Sodium Hydroxide and Ammonia

This is a core part of the syllabus. You need to know what happens when you add aqueous sodium hydroxide (\(NaOH\)) and aqueous ammonia (\(NH_3\)) to five specific metal ions. Usually, a precipitate forms first, and then it might redissolve if you add excess reagent.

The Five Key Ions

1. Chromium(III) - \(Cr^{3+}\)
  • With \(NaOH\): Forms a green precipitate of \(Cr(OH)_3(H_2O)_3\).
  • In Excess \(NaOH\): The precipitate redissolves to form a dark green solution of \([Cr(OH)_6]^{3-}\). This is because chromium(III) hydroxide is amphoteric (it can react as an acid).
  • With \(NH_3\): Forms the same green precipitate.
  • In Excess \(NH_3\): The precipitate redissolves to form a purple solution of \([Cr(NH_3)_6]^{3+}\). This is ligand exchange.
2. Iron(II) - \(Fe^{2+}\)
  • With \(NaOH\) or \(NH_3\): Forms a green precipitate of \(Fe(OH)_2(H_2O)_4\).
  • In Excess: No further change (precipitate does not redissolve).
  • Common Observation: The top of the green precipitate turns brown on standing as it is oxidised by air to Iron(III).
3. Iron(III) - \(Fe^{3+}\)
  • With \(NaOH\) or \(NH_3\): Forms a brown precipitate of \(Fe(OH)_3(H_2O)_3\).
  • In Excess: No further change.
4. Cobalt(II) - \(Co^{2+}\)
  • With \(NaOH\): Forms a blue precipitate of \(Co(OH)_2(H_2O)_4\) (often turns pinkish-brown over time).
  • In Excess \(NaOH\): No change.
  • With \(NH_3\): Forms the blue precipitate.
  • In Excess \(NH_3\): The precipitate redissolves to form a yellow/brown solution of \([Co(NH_3)_6]^{2+}\).
5. Copper(II) - \(Cu^{2+}\)
  • With \(NaOH\): Forms a pale blue precipitate of \(Cu(OH)_2(H_2O)_4\).
  • In Excess \(NaOH\): No change.
  • With \(NH_3\): Forms the pale blue precipitate.
  • In Excess \(NH_3\): The precipitate redissolves to form a deep blue solution of \([Cu(NH_3)_4(H_2O)_2]^{2+}\).

Important Distinction: If a precipitate dissolves in excess \(NaOH\), it is amphoteric behaviour. If it dissolves in excess \(NH_3\), it is usually ligand exchange.


4. Ligand Exchange and the Chelate Effect

Ligand exchange is when one ligand in a complex is replaced by another. Sometimes the coordination number (the number of dative bonds) changes. For example, if you add concentrated \(HCl\) to a copper solution, the large \(Cl^-\) ligands replace the \(H_2O\) ligands, changing the shape from octahedral to tetrahedral:

\([Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O\)

The Chelate Effect

Multidentate ligands (like EDTA\(^{4-}\) or en) form much more stable complexes than monodentate ligands (like \(H_2O\) or \(NH_3\)). This is called the chelate effect.

Why does it happen? It’s all about entropy (\(\Delta S\)). When one multidentate ligand replaces several monodentate ligands, the total number of particles in the solution increases. For example:

\([Cu(H_2O)_6]^{2+} + EDTA^{4-} \rightarrow [Cu(EDTA)]^{2-} + 6H_2O\)

Here, we start with 2 particles on the left and end with 7 particles on the right. This leads to a huge increase in entropy (\(\Delta S_{system} > 0\)). Since \(\Delta G = \Delta H - T\Delta S\), a large positive \(\Delta S\) makes \(\Delta G\) more negative, meaning the reaction is very feasible and the new complex is very stable.


5. Catalysis

Transition metals are excellent catalysts because they can change oxidation states easily and provide a surface for reactions to occur.

Heterogeneous Catalysis

The catalyst is in a different phase (usually a solid) than the reactants (usually gases or liquids).

  • Vanadium(V) oxide (\(V_2O_5\)): Used in the Contact Process to make sulfuric acid. It works by changing oxidation state from \(+5\) to \(+4\) and back again.
  • The Catalytic Converter: Uses metals like Platinum or Rhodium to convert harmful \(CO\) and \(NO\) into \(CO_2\) and \(N_2\). The gases adsorb onto the surface, react, and then the products desorb.

Homogeneous Catalysis

The catalyst is in the same phase as the reactants. This usually involves the formation of an intermediate species.

  • Iron (\(Fe^{2+}\)) catalysis: Used in the reaction between \(S_2O_8^{2-}\) and \(I^-\). Both reactants are negative, so they repel each other, making the reaction slow. \(Fe^{2+}\) acts as an intermediate:
    1. \(S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2SO_4^{2-} + 2Fe^{3+}\)
    2. \(2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2\)
  • Autocatalysis: This is when a product of the reaction acts as the catalyst. A famous example is the reaction between \(MnO_4^-\) and \(C_2O_4^{2-}\) (ethanedioate), where the \(Mn^{2+}\) produced speeds up the reaction.

Summary Checklist

  • Can you list the colours of Vanadium from \(+5\) to \(+2\)?
  • Do you know which metal hydroxide is amphoteric (\(Cr^{3+}\))?
  • Can you explain why the chelate effect happens using entropy?
  • Do you know the difference between adsorption and desorption in heterogeneous catalysis?
  • Can you identify the deep blue solution formed by excess ammonia with \(Cu^{2+}\)?

Keep practicing those equations and colours—you've got this!