# Transition Metals and their Chemistry

> Edexcel International A-Level Chemistry · IAL
> Source: https://www.owlsprep.com/study/edexcel-ial-chemistry-u5-transition-metals-and-their-chemistry/

This guide covers all Edexcel IAL Chemistry Unit 5 content for transition metals, including electron configurations, complex ion properties, redox reactions, characteristic observations, and catalysis for your WCH15 exam.

**Prerequisites:** [Redox reactions and standard electrode potentials (Edexcel IAL U2)](https://www.owlsprep.com/study/edexcel-ial-chemistry-u2-redox-electrode-potentials/); [Bonding and structure (Edexcel IAL U1)](https://www.owlsprep.com/study/edexcel-ial-chemistry-u1-bonding-structure/)

## Learning objectives

- Define transition metals and write electron configurations for Period 4 d-block atoms and ions
- Explain complex ion formation, d-orbital splitting, and the origin of colour in transition metal complexes
- Recall and predict observations for reactions of transition metal ions with NaOH, NH₃, and ligand exchange reagents
- Interpret redox interconversions of vanadium and chromium compounds using standard electrode potentials
- Distinguish between heterogeneous, homogeneous and autocatalysis, with named industrial and laboratory examples
- Write balanced ionic equations for deprotonation, ligand exchange and redox reactions of transition metals

## 1. Introduction to Transition Metals

**Transition Metal** — A d-block element that forms at least one stable ion with an incompletely filled d-subshell.

For Period 4 d-block elements (Sc to Zn), the 4s subshell fills before the 3d subshell, but 4s electrons are always lost first when forming positive ions. Note two electron configuration exceptions: Cr = $[Ar]4s^13d^5$ and Cu = $[Ar]4s^13d^{10}$, as half-full and full d-subshells have extra stability. Sc and Zn are d-block elements but not transition metals, as explained in the FAQ.

**Worked example:** Write the full electron configuration of the Fe³⁺ ion.

1. 1. Write the electron configuration of neutral Fe: $[Ar]4s^23d^6$
2. 2. Remove 4s electrons first when forming ions: lose 2 electrons from 4s, then 1 electron from 3d to give a +3 charge
3. 3. Final configuration: $[Ar]3d^5$

> **tip**
>
> Always remove 4s electrons before 3d when writing transition metal ion configurations. This is one of the most frequently lost marking points in transition metal questions.

**Check your understanding**

1. Which of the following is a transition metal?

   - A) Zn
   - B) Sc
   - C) Cu
   - D) Ca

   *Why:* Cu forms the Cu²⁺ ion which has an incomplete $3d^9$ subshell, so it is a transition metal. Zn only forms Zn²⁺ ($3d^{10}$, full), Sc only forms Sc³⁺ ($3d^0$, empty), and Ca is an s-block element.

## 2. Complex Ions and Origin of Colour

**Complex Ion** — A central metal ion surrounded by ligands bonded via dative covalent bonds. The overall charge equals the oxidation state of the central metal plus the sum of the charges of all ligands.

Ligands are classified by denticity (number of lone pairs donated per ligand): monodentate (e.g. $H_2O$, $OH^-$, $NH_3$, $Cl^-$), bidentate (e.g. ethylenediamine, en), hexadentate (e.g. $EDTA^{4-}$). Coordination number determines shape: 6-coordinate complexes are octahedral (small ligands like $H_2O$), 4-coordinate complexes with large ligands like $Cl^-$ are tetrahedral, and square planar complexes include the cancer treatment drug cis-platin.

Colour in transition metal complexes arises when ligands split the 3d orbitals into two different energy levels. Visible light energy is absorbed to promote an electron from the lower energy d-orbitals to the higher energy d-orbitals. The wavelength of light not absorbed is transmitted, which is the colour we see. Ions with $d^0$ or $d^{10}$ configurations are colourless, as there are no electrons to promote or no empty orbitals to accept electrons.

**Worked example:** Explain why $[Cu(H_2O)_6]^{2+}$ is blue, but $[Zn(H_2O)_6]^{2+}$ is colourless.

1. 1. $Cu^{2+}$ has an electron configuration of $[Ar]3d^9$, so it has an incomplete d-subshell.
2. 2. Water ligands split the Cu 3d orbitals into two energy levels. Red/orange visible light is absorbed to promote a d electron, so blue light is transmitted, making the solution appear blue.
3. 3. $Zn^{2+}$ has an electron configuration of $[Ar]3d^{10}$, with a full d-subshell. There are no available empty d-orbitals for electron promotion, so no visible light is absorbed, and the solution is colourless.

> **warning**
>
> Complex ion colours are not provided in the data booklet, so you must memorise the exact colours for all required complexes for the exam.

## 3. Reactions of Transition Metal Ions

You are required to recall observations (colour, precipitate formation, solubility in excess reagent) for reactions of $Cr^{3+}$, $Mn^{2+}$, $Fe^{2+}$, $Fe^{3+}$, $Co^{2+}$, $Ni^{2+}$, $Cu^{2+}$, $Zn^{2+}$ with $NaOH(aq)$ and $NH_3(aq)$. Reactions are either deprotonation (hydroxide ions remove $H^+$ from water ligands to form insoluble hydroxide precipitates) or ligand exchange (excess ammonia replaces water ligands to form soluble complexes).

**Worked example:** Write balanced ionic equations and state observations for the reaction of $[Cu(H_2O)_6]^{2+}$ with (a) limited $NaOH(aq)$, (b) excess $NH_3(aq)$.

1. (a) Limited $NaOH$: Deprotonation reaction
2. $$[Cu(H_2O)_6]^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(H_2O)_4(OH)_2(s) + 2H_2O(l)$$
3. Observation: Pale blue precipitate forms
4. (b) Excess $NH_3$: Initial deprotonation followed by ligand exchange
5. $$Cu(H_2O)_4(OH)_2(s) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 2H_2O(l) + 2OH^-(aq)$$
6. Observation: Pale blue precipitate dissolves to form a deep blue solution

> **mnemonic**
>
> For common copper complex colours: *B*lue hexaqua, *Y*ellow tetrachloro, *D*eep blue tetraammine → *B*oys *Y*ell *D*uring copper practicals.

Chromate and dichromate ions exist in equilibrium, dependent on pH: $Cr_2O_7^{2-}$ (orange, acidic conditions) ⇌ $2CrO_4^{2-}$ (yellow, alkaline conditions). The full equilibrium equation is: $Cr_2O_7^{2-}(aq) + H_2O(l) \rightleftharpoons 2CrO_4^{2-}(aq) + 2H^+(aq)$.

## 4. Redox Reactions of Transition Metals

Transition metals have variable oxidation states, so they participate in a wide range of redox reactions. Standard electrode potentials ($E^o$), provided in the data booklet, are used to predict if a redox reaction is feasible (positive $E^o_{cell}$ = feasible under standard conditions).

Vanadium has four common oxidation states: +5 ($VO_2^+$, yellow), +4 ($VO^{2+}$, blue), +3 ($V^{3+}$, green), +2 ($V^{2+}$, violet). Zinc in acidic conditions reduces vanadium from +5 all the way to +2. For chromium: orange $Cr_2O_7^{2-}$ can be reduced to green $Cr^{3+}$ then pale blue $Cr^{2+}$. $Cr^{3+}$ can be oxidised to yellow $CrO_4^{2-}$ using $H_2O_2$ in alkaline conditions.

**Worked example:** Use $E^o$ values to explain if $Fe^{3+}$ can oxidise $V^{2+}$ to $V^{3+}$. Given: $E^o(V^{3+}/V^{2+}) = -0.26 V$, $E^o(Fe^{3+}/Fe^{2+}) = +0.77 V$.

1. 1. Calculate overall $E^o_{cell}$: $E^o_{cell} = E^o(oxidation) + E^o(reduction) = +0.26 V + 0.77 V = +1.03 V$
2. 2. A positive $E^o_{cell}$ means the reaction is feasible under standard conditions.
3. 3. $Fe^{3+}$ is reduced to $Fe^{2+}$, while $V^{2+}$ is oxidised to $V^{3+}$.

> **Exam tip**
>
> Always include the sign when stating $E^o_{cell}$ values, as a negative sign confirms the reaction is not feasible under standard conditions.

## 5. Catalysis by Transition Metals

**Catalysis** — A process where a catalyst increases reaction rate without being consumed, by providing an alternative reaction pathway with a lower activation energy.

- **Heterogeneous catalysis**: Catalyst is in a different phase to reactants, e.g. $V_2O_5$ (solid) in the Contact process, platinum/rhodium (solid) in catalytic converters.
- **Homogeneous catalysis**: Catalyst is in the same phase as reactants, e.g. $Fe^{2+}$ (aqueous) catalysing the reaction between $I^-$ and $S_2O_8^{2-}$.
- **Autocatalysis**: A product of the reaction acts as the catalyst, e.g. $Mn^{2+}$ catalysing the reaction between acidified $MnO_4^-$ and $C_2O_4^{2-}$.

**Worked example:** Explain why $Mn^{2+}$ acts as an autocatalyst in the reaction between acidified manganate(VII) and ethanedioate ions.

1. 1. The overall reaction is: $2MnO_4^-(aq) + 16H^+(aq) + 5C_2O_4^{2-}(aq) \rightarrow 2Mn^{2+}(aq) + 10CO_2(g) + 8H_2O(l)$
2. 2. Initial reaction rate is slow, as two negative reactant ions repel each other, leading to high activation energy.
3. 3. As $Mn^{2+}$ product forms, it acts as a homogeneous catalyst, providing an alternative lower activation energy pathway, so reaction rate increases as more $Mn^{2+}$ is produced.
4. 4. This is autocatalysis because the catalyst is a product of the reaction.

## Common pitfalls

- **Wrong:** Writing $Fe^{2+}$ electron configuration as $[Ar]4s^23d^4$
  - Why it fails: You must remove 4s electrons before 3d when forming transition metal ions.
  - Correct: $Fe^{2+} = [Ar]3d^6$, as 2 electrons are lost from the 4s subshell first.
- **Wrong:** Classifying Sc and Zn as transition metals
  - Why it fails: Sc only forms $Sc^{3+}$ ($3d^0$, empty d-orbitals), Zn only forms $Zn^{2+}$ ($3d^{10}$, full d-orbitals), so they do not meet the transition metal definition.
  - Correct: Explicitly state that Sc and Zn are d-block elements but not transition metals.
- **Wrong:** Assigning incorrect charge to complex ions
  - Why it fails: Complex charge = oxidation state of central metal + sum of charges of all ligands.
  - Correct: Always calculate total charge after ligand exchange, e.g. $[CuCl_4]^{2-}$ has charge = $+2 + 4(-1) = -2$.
- **Wrong:** Explaining colour as emitted light when electrons fall back to lower orbitals
  - Why it fails: Colour of transition metal complexes comes from transmitted light that is not absorbed during d-electron promotion.
  - Correct: State that absorbed light matches the energy difference between split d-orbitals, and the unabsorbed transmitted light is the colour observed.
- **Wrong:** Confusing heterogeneous and homogeneous catalysis
  - Why it fails: Heterogeneous catalysts are in a different physical state to reactants, homogeneous catalysts are in the same state.
  - Correct: For the Contact process, solid $V_2O_5$ and gaseous reactants = heterogeneous; $Fe^{2+}$ catalysing aqueous reactants = homogeneous.
- **Wrong:** Stating cis-platin is tetrahedral
  - Why it fails: Cis-platin has a square planar structure that allows it to bind to DNA and stop cancer cell replication.
  - Correct: Identify cis-platin as a square planar 4-coordinate complex used in cancer treatment.

## Cheatsheet

| Key Complex/Ion | Metal Oxidation State | Colour | Notes |
| --- | --- | --- | --- |
| $[Cu(H_2O)_6]^{2+}$ | +2 | Pale blue | Hexaaqua copper(II) |
| $[Cu(NH_3)_4(H_2O)_2]^{2+}$ | +2 | Deep blue | Excess $NH_3$ product |
| $[CuCl_4]^{2-}$ | +2 | Yellow | Conc HCl ligand exchange |
| $VO_2^+$ | +5 | Yellow | Vanadium(V) |
| $VO^{2+}$ | +4 | Blue | Vanadium(IV) |
| $V^{3+}$ | +3 | Green | Vanadium(III) |
| $V^{2+}$ | +2 | Violet | Vanadium(II) |
| $Cr_2O_7^{2-}$ | +6 | Orange | Acidic chromium(VI) |
| $CrO_4^{2-}$ | +6 | Yellow | Alkaline chromium(VI) |
| $[Fe(H_2O)_6]^{3+}$ | +3 | Yellow/pale violet | Iron(III) hexaqua |

## What's next

Now that you have mastered transition metal chemistry for Edexcel IAL Chemistry Unit 5, you can move on to the second part of Unit 5: Organic Nitrogen Chemistry, which covers amines, amides, amino acids, and polymers, and makes up the remaining 50% of the WCH15 exam content. You should also practice past paper questions on transition metals to reinforce your recall of complex colours, reaction observations, and balanced equations, as these are frequent high-mark question topics. Make sure you are confident with standard electrode potential calculations, as these are often combined with transition metal redox questions. Finally, review the CP14 practical on preparing a transition metal complex, as practical-based questions make up ~15% of the Unit 5 paper.

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