Question:
Why do Ce and Tb show +4 oxidation state?
Why do Ce and Tb show +4 oxidation state?
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Completely filled, half-filled, or noble gas configurations impart extra stability and allow higher oxidation states.
Updated On: Jan 7, 2026
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Solution and Explanation
Step 1: Lanthanoids generally show +3 oxidation state due to the loss of two 6s electrons and one 5d or 4f electron.
Step 2: Cerium (\(\text{Ce}\)) has the electronic configuration: \[ [\text{Xe}]\,4f^1 5d^1 6s^2 \] On losing four electrons, it attains: \[ \text{Ce}^{4+} : [\text{Xe}] \] which is a noble gas configuration and hence very stable.
Step 3: Terbium (\(\text{Tb}\)) has the electronic configuration: \[ [\text{Xe}]\,4f^9 6s^2 \] On losing four electrons, it forms: \[ \text{Tb}^{4+} : [\text{Xe}]\,4f^7 \] which is a half-filled \(4f^7\) configuration, known for extra stability.
Step 4: Due to these stable electronic configurations, Ce and Tb can exhibit the +4 oxidation state.
Step 2: Cerium (\(\text{Ce}\)) has the electronic configuration: \[ [\text{Xe}]\,4f^1 5d^1 6s^2 \] On losing four electrons, it attains: \[ \text{Ce}^{4+} : [\text{Xe}] \] which is a noble gas configuration and hence very stable.
Step 3: Terbium (\(\text{Tb}\)) has the electronic configuration: \[ [\text{Xe}]\,4f^9 6s^2 \] On losing four electrons, it forms: \[ \text{Tb}^{4+} : [\text{Xe}]\,4f^7 \] which is a half-filled \(4f^7\) configuration, known for extra stability.
Step 4: Due to these stable electronic configurations, Ce and Tb can exhibit the +4 oxidation state.
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