The most common oxidation state of Lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?

Solution

Chemistry ma’am · 3 steps

Write the 4f count of each M³⁺ ion. Only Ce can lose one more electron and reach an empty 4f.

1. What to check. A lanthanoid leaves +3 easily when the new state gives a stable 4f shell: empty (4f⁰), half-filled (4f⁷) or full (4f¹⁴). If the M³⁺ ion already has one of these, +3 is the stable state.

   Element	Configuration	M³⁺ ion	Other state?
   La (57)	[Xe] 5d¹ 6s²	4f⁰, already empty	No, +3 only
   Ce (58)	[Xe] 4f¹ 5d¹ 6s²	4f¹, one extra electron	Yes, +4 gives 4f⁰
   Gd (64)	[Xe] 4f⁷ 5d¹ 6s²	4f⁷, already half-filled	No, +3 only
   Lu (71)	[Xe] 4f¹⁴ 5d¹ 6s²	4f¹⁴, already full	No, +3 only

2. Why cerium. Ce³⁺ keeps a single 4f electron. Losing it gives Ce⁴⁺ with the noble gas configuration of xenon, so Ce(IV) is stable. This is why Ce⁴⁺ salts are common oxidising agents.
3. So. Cerium is the element that moves away from +3 easily, to +4. The answer is (a).

The idea to keep. A lanthanoid shows a state other than +3 when that state gives an empty, half-filled or full 4f shell: Ce(IV) is 4f⁰, Eu(II) and Tb(IV) are 4f⁷, Yb(II) is 4f¹⁴.

Answer from the NTA final answer key. Solution by PrepWiser.