AQA A-Level Chemistry (7405)

## Learning Outcomes

- Explain why transition metals can show variable oxidation states
- Apply the exact AQA definition of a transition metal and distinguish copper from zinc
- Work out oxidation states in transition metal ions and oxyanions
- Identify oxidation and reduction in reactions involving Fe, Mn, Cr and V species
- Write and combine half-equations in acidic conditions
- Recall the colour changes linked to common oxidation states in aqueous solution
- Avoid common exam errors involving charges, names and oxidation-state calculations

## AQA A-Level Chemistry (7405) Syllabus

For the AQA A-Level Chemistry (7405), you are expected to understand variable oxidation states in transition metals, with a focus on the following syllabus points:

- Know the AQA definition of a transition metal as an element that forms at least one stable ion with an incomplete d subshell
- Explain why transition metals show variable oxidation states in terms of the similar energies of the 3d and 4s subshells
- Identify common oxidation states for Fe, Cr, Mn, Cu and V
- Calculate oxidation states in ions such as MnO4− and Cr2O7^2−
- Explain oxidation and reduction using both electron transfer and oxidation-state changes
- Write reduction half-equations for acidified manganate(VII) and acidified dichromate(VI)
- Write combined ionic equations with Fe2+ in acidic conditions
- Recall the colour changes for Fe2+/Fe3+, MnO4−/Mn2+, Cr2O7^2−/Cr3+ and the vanadium sequence

## Test Your Knowledge

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision._

1. Why can transition metals form ions with different oxidation states more easily than Group 1 or Group 2 metals?
2. Why is copper a transition metal under the AQA definition, but zinc is not?
3. What is the oxidation state of manganese in MnO4− and chromium in Cr2O7^2−?
4. What colour change is seen when acidified manganate(VII) reacts with Fe2+?
5. How do you combine half-equations to produce a full ionic equation?

## Introduction

Variable oxidation states are one of the defining chemical features of transition metals. They explain why these elements take part in many redox reactions, why they form different ions, and why several useful oxidising agents in inorganic chemistry contain transition metals.

In exams, this topic is rarely tested as isolated fact recall. It is usually linked to redox equations, colour changes, ion identification, titrations, and precise use of oxidation-state language. That means you need more than a list of numbers. You need to know what oxidation states mean, why they vary, and how to use them in equations.

> **Key Term: oxidation state**
> The apparent charge of an atom in an element, ion or compound, used to track oxidation and reduction.

### Why transition metals show variable oxidation states

A transition metal is not simply any element in the d block. The exact AQA definition matters.

> **Key Term: transition metal**
> An element that forms at least one stable ion with an incomplete d subshell.

This definition explains one of the most common exam traps. **Copper counts as a transition metal**, because **Cu2+** has an incomplete d subshell. **Zinc does not**, because **Zn2+** has a full **3d10** subshell.

Why do transition metals show variable oxidation states? The main reason is that the **3d and 4s subshells are close in energy**. Because of this, different numbers of electrons can be removed with similar amounts of energy. In contrast, for many main-group elements, removing an extra electron would require a much larger jump in energy, so fewer oxidation states are common.

Take iron as the clearest example:

- Fe atom: **[Ar] 3d6 4s2**
- Fe2+: **[Ar] 3d6**
- Fe3+: **[Ar] 3d5**

The first two electrons are lost from the **4s subshell**, not the 3d subshell. That point is essential. Students often write the electron configuration correctly for the atom but remove electrons from 3d first when forming ions. That is wrong.

Iron can therefore exist commonly in the **+2** and **+3** oxidation states. The small energy gap between 4s and 3d makes both ions possible.

Other useful examples:

- **Cr** commonly shows **+2, +3, +6**
- **Mn** commonly shows **+2, +4, +7**
- **Cu** commonly shows **+1, +2**
- **V** commonly shows **+2, +3, +4, +5**

These are not random values. They arise because transition metals can lose different numbers of electrons or share them differently in bonding.

> **Key Term: incomplete d subshell**
> A d subshell that is neither empty nor full, so it contains between 1 and 9 electrons.

A useful contrast helps fix the idea:

- **Scandium** is not usually treated as a transition metal under the AQA definition because its stable ion **Sc3+** has no d electrons left
- **Zinc** is not a transition metal because **Zn2+** is **3d10**, a full d subshell

So being in the d block is not enough. The stable ion must have an incomplete d subshell.

> **Test Tip:** If a question asks why transition metals show variable oxidation states, do not say only “because they are in the d block”. State that the **3d and 4s subshells are close in energy**, so different numbers of electrons can be removed or involved in bonding.

### How to calculate oxidation states and recognise redox changes

To use oxidation states properly, you need a reliable method. In simple ions, the oxidation state equals the ion charge:

- Fe2+ is **+2**
- Fe3+ is **+3**
- Cu+ is **+1**
- Mn2+ is **+2**

In molecular ions, calculate the unknown value from the total charge.

Worked example 1: oxidation state of Mn in MnO4−

Step 1: oxygen is usually **−2**  
Step 2: four oxygens give total **−8**  
Step 3: overall ion charge is **−1**

Let manganese oxidation state = x

x + (4 × −2) = −1  
x − 8 = −1  
x = **+7**

So manganese in **MnO4−** is **+7**.

Worked example 2: oxidation state of Cr in Cr2O7^2−

Step 1: oxygen is **−2**  
Step 2: seven oxygens give total **−14**  
Step 3: overall ion charge is **−2**

Let oxidation state of each chromium = x

2x + (7 × −2) = −2  
2x − 14 = −2  
2x = 12  
x = **+6**

So chromium in **Cr2O7^2−** is **+6**.

> **Key Term: oxidising agent**
> A species that causes oxidation by accepting electrons and is itself reduced.

Now connect oxidation states to redox:

- **Oxidation** = increase in oxidation state, or loss of electrons
- **Reduction** = decrease in oxidation state, or gain of electrons

Example:

Fe2+ → Fe3+ + e−

Iron goes from **+2** to **+3**, so it is oxidised.

MnO4− → Mn2+

Manganese goes from **+7** to **+2**, so it is reduced.

Cr2O7^2− → Cr3+

Chromium goes from **+6** to **+3**, so it is reduced.

This is why acidified manganate(VII) and acidified dichromate(VI) act as oxidising agents. Their metal ions start in high oxidation states and are easily reduced to lower ones.

Common student errors here:

- confusing oxidation state with the overall ion charge of a polyatomic ion
- saying the oxidising agent is oxidised rather than reduced
- forgetting that oxygen is usually **−2** in these ions
- missing charges in half-equations
  > **Test Tip:** In redox questions, always track both ideas: the **oxidation-state change** and the **electron movement**. If both agree, your answer is more likely to be correct.

### Half-equations, ionic equations and colour changes in acidic solution

This is where variable oxidation states becomes practical chemistry. The exam often combines oxidation-state reasoning with half-equations and observations.

> **Key Term: reduction half-equation**
> An equation showing the gain of electrons by a species during reduction.

The key half-equations you should know are:

For iron:

Fe2+(aq) → Fe3+(aq) + e−

Fe3+(aq) + e− → Fe2+(aq)

For manganate(VII) in acidic solution:

MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l)

For dichromate(VI) in acidic solution:

Cr2O7^2−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l)

The acidic condition matters because hydrogen ions are needed to balance oxygen as water. If you forget the acidic condition, you often produce an incomplete or impossible equation.

Now combine half-equations.

Worked example 3: acidified manganate(VII) with Fe2+

Step 1: write the half-equations

MnO4− + 8H+ + 5e− → Mn2+ + 4H2O

Fe2+ → Fe3+ + e−

Step 2: equalise electrons

Multiply the iron half-equation by 5:

5Fe2+ → 5Fe3+ + 5e−

Step 3: add the equations and cancel electrons

MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

Step 4: check charges

Left side: −1 + 8 + 10 = +17  
Right side: +2 + 15 = +17

Balanced.

Worked example 4: acidified dichromate(VI) with Fe2+

Step 1: write half-equations

Cr2O7^2− + 14H+ + 6e− → 2Cr3+ + 7H2O

Fe2+ → Fe3+ + e−

Step 2: equalise electrons

Multiply iron half-equation by 6:

6Fe2+ → 6Fe3+ + 6e−

Step 3: add and cancel electrons

Cr2O7^2− + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+

Again, check both atoms and charge.

The linked colour changes are very high yield:

- **MnO4−** is **purple**
- **Mn2+** is **very pale pink** or nearly colourless
- **Cr2O7^2−** is **orange**
- **Cr3+** is **green**
- **Fe2+** is **pale green**
- **Fe3+** is **yellow** to **orange-brown** in aqueous solution

These colours are not just memory facts. They show changing oxidation state. So when acidified manganate(VII) oxidises Fe2+, you should expect:

- purple manganate(VII) disappears
- very pale pink Mn2+ forms
- Fe2+ pale green changes to Fe3+ yellow/orange-brown

In practice, the dominant observed change may be described simply as **purple to colourless** or **purple decolourised**, because Mn2+ is so pale.

For dichromate(VI) with Fe2+, the key visual change is usually **orange to green**.

A very common synoptic link is vanadium. In acidic solution, reducing vanadium(V) with zinc gives a sequence of colours as oxidation state falls step by step:

- **V(V)** yellow
- **V(IV)** blue
- **V(III)** green
- **V(II)** purple

This sequence is useful because it shows that variable oxidation states are not just two-state systems like Fe2+/Fe3+. Some transition metals can pass through several stable oxidation states in solution.

### Common exam traps and how to answer with precision

This topic rewards exact wording. Small slips can lose marks even when the chemistry idea is mostly right.

First trap: definition of transition metal.

Wrong: “an element in the d block”  
Better: “an element that forms at least one stable ion with an incomplete d subshell”

Second trap: oxidation state versus ion charge.

For **Cr2O7^2−**, the ion charge is **−2**, but the oxidation state of each chromium is **+6**. These are not the same thing.

Third trap: electron removal.

When transition metal atoms form ions, **4s electrons are removed before 3d electrons**. This is especially tested in iron and copper questions.

Fourth trap: naming compounds.

Use Roman numerals correctly for variable oxidation states:

- iron(II) sulfate
- iron(III) sulfate
- copper(I) oxide
- copper(II) oxide

The Roman numeral refers to the oxidation state of the metal, not the number of that atom in the formula.

Fifth trap: balancing half-equations.

In acidic solution:

1. balance atoms other than O and H
2. balance O with H2O
3. balance H with H+
4. balance charge with e−

If you add electrons too early, mistakes are more likely.

A final distinction that students often miss: the colour quoted is usually the colour of the **aqueous ion** or complex in water, not the solid compound in every case.

> **Test Tip:** For 3- or 4-mark explanation questions, use a chain: “The 3d and 4s subshells are close in energy, so different numbers of electrons can be removed using similar amounts of energy, so the metal forms ions with different oxidation states.”

## Key Point Checklist

_This article has covered the following key knowledge points:_

- A transition metal is an element that forms at least one stable ion with an incomplete d subshell
- Transition metals show variable oxidation states because the 3d and 4s subshells are close in energy
- When transition metal ions form, 4s electrons are removed before 3d electrons
- Iron commonly forms Fe2+ and Fe3+, showing +2 and +3 oxidation states
- Zinc is not a transition metal under the AQA definition because Zn2+ has a full 3d10 subshell
- Manganese in MnO4− has oxidation state +7, and chromium in Cr2O7^2− has oxidation state +6
- Oxidation is an increase in oxidation state or loss of electrons; reduction is a decrease in oxidation state or gain of electrons
- The key acidic half-equations are for MnO4− to Mn2+ and Cr2O7^2− to Cr3+
- The ionic equation for acidified manganate(VII) with Fe2+ is MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
- The ionic equation for acidified dichromate(VI) with Fe2+ is Cr2O7^2− + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
- Key colour changes are purple to very pale pink for manganate(VII), orange to green for dichromate(VI), and pale green to yellow/orange-brown for Fe2+ to Fe3+
- Vanadium in acidic solution shows the sequence yellow to blue to green to purple as the oxidation state falls from +5 to +2

## Key Terms and Concepts

- oxidation state
- transition metal
- incomplete d subshell
- oxidising agent
- reduction half-equation


Transition metals - Variable oxidation states