Question:easy

Name a member of the lanthanoid series: (I) which exhibits +4 oxidation state; (II) which exhibits +2 oxidation state.

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Lanthanoids most commonly show the +3 oxidation state, but a few show +2 or +4 when doing so gives an empty, half-filled, or fully-filled 4f subshell, which is extra stable.
Updated On: Jun 16, 2026
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Solution and Explanation

Step 1: Recall the usual state.
Lanthanoids almost always show the \(+3\) oxidation state.

Step 2: Know why other states appear.
A \(+2\) or \(+4\) state shows up when it gives an empty \((4f^0)\), half-filled \((4f^7)\), or fully filled \((4f^{14})\) subshell, which is extra stable.

Step 3: Find the \(+4\) member.
Cerium can lose four electrons to reach \(4f^0\) (empty subshell), so it shows the \(+4\) state.

Step 4: Find the \(+2\) member.
Europium can stop at \(+2\) because that leaves a half-filled \(4f^7\) subshell, which is stable.

Step 5: State the answers.
So Ce gives \(+4\) and Eu gives \(+2\).

Answer: (I) \(+4\) oxidation state: Cerium (Ce); (II) \(+2\) oxidation state: Europium (Eu).
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