3.1.9 (HL)—Variable oxidation states
- Syllabus
- First assessment 2025
- Objective
- 3.1.9
- Level
- HL
Variable Oxidation States
HL onlyFirst-row transition elements can show variable oxidation states because successive ionization energies for outer s and d electrons are relatively close. Forming an ion removes 4s electrons before 3d electrons.
Write the neutral configuration, remove the required 4s electrons first, then remove 3d electrons and count the remaining configuration.
Start Fe as [Ar]4s²3d⁶. Fe²⁺ loses the two 4s electrons to give [Ar]3d⁶; Fe³⁺ loses one more 3d electron to give [Ar]3d⁵. The written filling order of the neutral atom does not change the rule that 4s electrons are removed first.
Deducing Transition-Ion Configurations
HL onlyAssessment in practice
Representative question
Question 1
[Maximum number: 3]
The first four ionization energies of beryllium and iron are shown.
One common property of transition elements is that they have variable oxidation states. Discuss, referring to the graph, why iron, but not beryllium, displays this characteristic.
IE values of Fe gradually increase
AND
IE values of Be show a sudden rise
first and second ionization energies close together therefore do not form a +1 oxidation state / singly charged ion
further IEs of Fe are close to second IE, so the oxidation state/number of electrons Fe loses can vary «according to the oxidizing agents present»
Marking guidance:
Accept Be always loses 2 electrons /
forms Be2+ / only has +2 oxidation state
for M2 .
The Periodic Table Summary
Retrieve the route: locate an element from configuration, explain periodic and group trends, write oxide/reaction and oxidation-state answers, then connect incomplete d-sublevels to transition properties, ion configurations, and colours.
Check that every trend explanation names its particle-level cause, every equation is balanced, every oxidation state is a formal charge convention, and every transition colour uses absorbed/observed complementarity.