Transition Metals and their Chemistry

Q: How can I remember the oxidation state colours of vanadium?

Use the mnemonic: Yellow (+5: ), Blue (+4: ), Green (+3: ), Violet (+2: ) → You Better Get Vanadium.

Edexcel International A-Level Chemistry· Unit 5 (WCH15) Topic 17· 70 min read

1. 1. Introduction to Transition Metals★★☆☆☆⏱ 10 min

📘 Definition

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 = $[A r] 4 s^{1} 3 d^{5}$ and Cu = $[A r] 4 s^{1} 3 d^{10}$ , 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: $[A r] 4 s^{2} 3 d^{6}$
2
1. Remove 4s electrons first when forming ions: lose 2 electrons from 4s, then 1 electron from 3d to give a +3 charge
3
1. Final configuration: $[A r] 3 d^{5}$

✓ Quick check

Which of the following is a transition metal?
- A) Zn
- B) Sc
- C) Cu
- D) Ca
Reveal answer
C —
Cu forms the Cu²⁺ ion which has an incomplete $3 d^{9}$ subshell, so it is a transition metal. Zn only forms Zn²⁺ ( $3 d^{10}$ , full), Sc only forms Sc³⁺ ( $3 d^{0}$ , empty), and Ca is an s-block element.

2. 2. Complex Ions and Origin of Colour★★★☆☆⏱ 15 min

📘 Definition

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. $H_{2} O$ , $O H^{-}$ , $N H_{3}$ , $C l^{-}$ ), bidentate (e.g. ethylenediamine, en), hexadentate (e.g. $E D T A^{4 -}$ ). Coordination number determines shape: 6-coordinate complexes are octahedral (small ligands like $H_{2} O$ ), 4-coordinate complexes with large ligands like $C l^{-}$ 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 $[C u (H_{2} O)_{6}]^{2 +}$ is blue, but $[Z n (H_{2} O)_{6}]^{2 +}$ is colourless.

1
1. $C u^{2 +}$ has an electron configuration of $[A r] 3 d^{9}$ , so it has an incomplete d-subshell.
2
1. 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
1. $Z n^{2 +}$ has an electron configuration of $[A r] 3 d^{10}$ , 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.

3. 3. Reactions of Transition Metal Ions★★★★☆⏱ 20 min

You are required to recall observations (colour, precipitate formation, solubility in excess reagent) for reactions of $C r^{3 +}$ , $M n^{2 +}$ , $F e^{2 +}$ , $F e^{3 +}$ , $C o^{2 +}$ , $N i^{2 +}$ , $C u^{2 +}$ , $Z n^{2 +}$ with $N a O H (a q)$ and $N H_{3} (a q)$ . 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 $[C u (H_{2} O)_{6}]^{2 +}$ with (a) limited $N a O H (a q)$ , (b) excess $N H_{3} (a q)$ .

1
(a) Limited $N a O H$ : 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 $N H_{3}$ : 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

Chromate and dichromate ions exist in equilibrium, dependent on pH: $C r_{2} O_{7}^{2 -}$ (orange, acidic conditions) ⇌ $2 C r O_{4}^{2 -}$ (yellow, alkaline conditions). The full equilibrium equation is: $C r_{2} O_{7}^{2 -} (a q) + H_{2} O (l) ⇌ 2 C r O_{4}^{2 -} (a q) + 2 H^{+} (a q)$ .

4. 4. Redox Reactions of Transition Metals★★★★☆⏱ 15 min

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 $E_{ce l l}^{o}$ = feasible under standard conditions).

Vanadium has four common oxidation states: +5 ( $V O_{2}^{+}$ , yellow), +4 ( $V O^{2 +}$ , 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 $C r_{2} O_{7}^{2 -}$ can be reduced to green $C r^{3 +}$ then pale blue $C r^{2 +}$ . $C r^{3 +}$ can be oxidised to yellow $C r O_{4}^{2 -}$ using $H_{2} O_{2}$ in alkaline conditions.

📐 Worked Example

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

1
1. Calculate overall $E_{ce l l}^{o}$ : $E_{ce l l}^{o} = E^{o} (o x i d a t i o n) + E^{o} (r e d u c t i o n) = + 0.26 V + 0.77 V = + 1.03 V$
2
1. A positive $E_{ce l l}^{o}$ means the reaction is feasible under standard conditions.
3
1. $F e^{3 +}$ is reduced to $F e^{2 +}$ , while $V^{2 +}$ is oxidised to $V^{3 +}$ .

5. 5. Catalysis by Transition Metals★★★☆☆⏱ 10 min

📘 Definition

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. $V_{2} O_{5}$ (solid) in the Contact process, platinum/rhodium (solid) in catalytic converters.
Homogeneous catalysis: Catalyst is in the same phase as reactants, e.g. $F e^{2 +}$ (aqueous) catalysing the reaction between $I^{-}$ and $S_{2} O_{8}^{2 -}$ .
Autocatalysis: A product of the reaction acts as the catalyst, e.g. $M n^{2 +}$ catalysing the reaction between acidified $M n O_{4}^{-}$ and $C_{2} O_{4}^{2 -}$ .

📐 Worked Example

Explain why $M n^{2 +}$ acts as an autocatalyst in the reaction between acidified manganate(VII) and ethanedioate ions.

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

6. Common Pitfalls

Wrong move:

Writing $F e^{2 +}$ electron configuration as $[A r] 4 s^{2} 3 d^{4}$

Why:

You must remove 4s electrons before 3d when forming transition metal ions.

Correct move:

$F e^{2 +} = [A r] 3 d^{6}$ , as 2 electrons are lost from the 4s subshell first.

Wrong move:

Classifying Sc and Zn as transition metals

Why:

Sc only forms $S c^{3 +}$ ( $3 d^{0}$ , empty d-orbitals), Zn only forms $Z n^{2 +}$ ( $3 d^{10}$ , full d-orbitals), so they do not meet the transition metal definition.

Correct move:

Explicitly state that Sc and Zn are d-block elements but not transition metals.

Wrong move:

Assigning incorrect charge to complex ions

Why:

Complex charge = oxidation state of central metal + sum of charges of all ligands.

Correct move:

Always calculate total charge after ligand exchange, e.g. $[C u C l_{4}]^{2 -}$ has charge = $+ 2 + 4 (- 1) = - 2$ .

Wrong move:

Explaining colour as emitted light when electrons fall back to lower orbitals

Why:

Colour of transition metal complexes comes from transmitted light that is not absorbed during d-electron promotion.

Correct move:

State that absorbed light matches the energy difference between split d-orbitals, and the unabsorbed transmitted light is the colour observed.

Wrong move:

Confusing heterogeneous and homogeneous catalysis

Why:

Heterogeneous catalysts are in a different physical state to reactants, homogeneous catalysts are in the same state.

Correct move:

For the Contact process, solid $V_{2} O_{5}$ and gaseous reactants = heterogeneous; $F e^{2 +}$ catalysing aqueous reactants = homogeneous.

Wrong move:

Stating cis-platin is tetrahedral

Why:

Cis-platin has a square planar structure that allows it to bind to DNA and stop cancer cell replication.

Correct move:

Identify cis-platin as a square planar 4-coordinate complex used in cancer treatment.

7. Quick Reference Cheatsheet

Key Complex/Ion	Metal Oxidation State	Colour	Notes
$[C u (H_{2} O)_{6}]^{2 +}$	+2	Pale blue	Hexaaqua copper(II)
$[C u (N H_{3})_{4} (H_{2} O)_{2}]^{2 +}$	+2	Deep blue	Excess $N H_{3}$ product
$[C u C l_{4}]^{2 -}$	+2	Yellow	Conc HCl ligand exchange
$V O_{2}^{+}$	+5	Yellow	Vanadium(V)
$V O^{2 +}$	+4	Blue	Vanadium(IV)
$V^{3 +}$	+3	Green	Vanadium(III)
$V^{2 +}$	+2	Violet	Vanadium(II)
$C r_{2} O_{7}^{2 -}$	+6	Orange	Acidic chromium(VI)
$C r O_{4}^{2 -}$	+6	Yellow	Alkaline chromium(VI)
$[F e (H_{2} O)_{6}]^{3 +}$	+3	Yellow/pale violet	Iron(III) hexaqua

8. Frequently Asked

Why are Sc and Zn not classified as transition metals?

Sc only forms the Sc³⁺ ion, which has an empty 3d subshell ( $3 d^{0}$ ). Zn only forms the Zn²⁺ ion, which has a full 3d subshell ( $3 d^{10}$ ). Transition metals require at least one stable ion with an incomplete d-orbital, so both are excluded from the classification.

How can I remember the oxidation state colours of vanadium?

Use the mnemonic: Yellow (+5: $V O_{2}^{+}$ ), Blue (+4: $V O^{2 +}$ ), Green (+3: $V^{3 +}$ ), Violet (+2: $V^{2 +}$ ) → You Better Get Vanadium.

What causes the chelate effect?

The chelate effect is driven by a positive change in system entropy ( $Δ S_{sy s t e m}$ ). When one multidentate ligand replaces multiple monodentate ligands, the total number of particles in solution increases, raising entropy and making the complex more thermodynamically stable.

Going deeper

official_specEdexcel IAL Chemistry Unit 5 SpecificationRefer to section 17 for full topic requirements
practical_guideCP14 Preparation of a Transition Metal Complex Guide

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.