Write the ground‑state electron configuration of the lanthanide element terbium (Tb, Z = 65) and state the oxidation state that is most commonly exhibited by lanthanide ions in aqueous solution.

medium 2 marks Class 12 · Chemistry · d- and f-Block Elements

Answer

Ground‑state configuration:

[X e] 4 f^{9} 6 s^{2}

. The most common oxidation state is +3.

The correct answer is Ground‑state configuration: $[X e] 4 f^{9} 6 s^{2}$ . The most common oxidation state is +3. — Hence +3 is the oxidation state most frequently observed in aqueous lanthanide compounds.

Step-by-step solution

Start from the nearest noble gas core: Xe (Z=54).
The remaining electrons are 65‑54 = 11.
The 4f subshell is filled after Xe; it can hold up to 14 electrons. Place 9 electrons in 4f and the remaining 2 in the 6s orbital, giving $[X e] 4 f^{9} 6 s^{2}$ .
Lanthanide ions typically lose the two 6s electrons and one 4f electron, resulting in a +3 charge.
Hence +3 is the oxidation state most frequently observed in aqueous lanthanide compounds.

Common mistake: Writing

5 d^{1}

instead of

4 f^{9}

or assuming +2 oxidation state as common

Concept tested: Electron configuration and common oxidation state of lanthanides

Published 6 September 2026 · How this solution was prepared and checked

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