AQA · A-Level · Chemistry
Course version: 7405 · Version from Sept 2016
Transition elements show variable oxidation states.
Full specification wording
Variable oxidation states (A-level only): Transition elements show variable oxidation states.
The 4s and 3d sub-levels lie close in energy, so once 4s electrons are removed, 3d electrons can be lost without a large jump in ionisation energy.
Checked against the AQA 7405 specification (opens in a new tab) on .
- Point 1 of 3
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
The 4s and 3d sub-levels lie close in energy, so once 4s electrons are removed, 3d electrons can be lost without a large jump in ionisation energy. This gives a ladder of accessible oxidation states. Iron shows +2 and +3, copper +1 and +2, chromium +2, +3 and +6, and manganese runs from +2 to +7. The highest states survive only when bonded to oxygen or fluorine, which is why manganese(VII) appears as MnO₄⁻ rather than a simple cation. Deducing an oxidation state from a formula is a standard task: oxygen counts as −2 and the sum equals the overall charge. Crucially, positive oxidation states must always include an explicit positive sign (e.g., +7, not 7).
What earns the marks
- Correct oxidation state of the metal in a given species, ensuring explicit signs are used (e.g., +7 in MnO₄⁻).
- Explaining variable oxidation states by the similar energies of the 4s and 3d sub-levels.
- Identifying the change in oxidation state during a named redox reaction.
- Using oxidation state changes to balance a half-equation with electrons.
Common mistakes and how to put them right
Reading the overall charge on MnO₄⁻ as the oxidation state of the manganese.
Omitting the positive sign on a positive oxidation state (writing 7 instead of +7).
Assuming every transition metal reaches +7, when only manganese does.
Forgetting that oxygen is −2 in an oxoanion when working out the metal's oxidation state.
Exam tips
- Set the arithmetic out: the sum of the oxidation states equals the overall charge on the species.
- Always include the sign for oxidation states, explicitly writing '+' for positive values.
Try the lesson
The start of the Variable oxidation states (A-level only): Transition elements show variable oxidation states. lesson, made from checked teaching notes for this specification point.
Have a go first
Which statement explains why transition elements can form ions with different positive charges?
Listen
The 4s and 3d sub-levels lie close in energy, so once 4s electrons are removed, 3d electrons can be lost without a large jump in ionisation energy. This gives a ladder of accessible oxidation states. Iron shows +2 and +3, copper +1 and +2, chromium +2, +3 and +6, and manganese runs from +2 to +7. The highest states survive only when bonded to oxygen or fluorine, which is why manganese(VII) appears as MnO₄⁻ rather than a simple cation. Deducing an oxidation state from a formula is a standard task: oxygen counts as −2 and the sum equals the overall charge. Crucially, positive oxidation states must always include an explicit positive sign (e.g., +7, not 7). A strong answer earns marks for these points. Correct oxidation state of the metal in a given species, ensuring explicit signs are used (e.g., +7 in MnO₄⁻). Explaining variable oxidation states by the similar energies of the 4s and 3d sub-levels. Identifying the change in oxidation state during a named redox reaction. Using oxidation state changes to balance a half-equation with electrons. Exam tip: Set the arithmetic out: the sum of the oxidation states equals the overall charge on the species.
See it
Key words
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variable oxidation states
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4s and 3d energies
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deducing oxidation states
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See every definition
- variable oxidation states
- Transition elements show this behaviour, in which more than one oxidation state is accessible.
- 4s and 3d energies
- These sub-levels lie close in energy, so once 4s electrons are removed, 3d electrons can be lost without a large jump in ionisation energy.
- deducing oxidation states
- A standard task in which oxygen counts as −2 and the sum equals the overall charge, with positive values always given an explicit positive sign.
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