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

We need to identify which lanthanoid element most easily deviates from the common +3 oxidation state.

The electronic configurations of the lanthanoids are:

Ce (Z = 58): [Xe] 4f$$^1$$ 5d$$^1$$ 6s$$^2$$

La (Z = 57): [Xe] 5d$$^1$$ 6s$$^2$$ (no 4f electrons)

Lu (Z = 71): [Xe] 4f$$^{14}$$ 5d$$^1$$ 6s$$^2$$

Gd (Z = 64): [Xe] 4f$$^7$$ 5d$$^1$$ 6s$$^2$$

Lanthanoids tend to show +3 as the most stable oxidation state. However, deviation occurs when attaining a particularly stable electronic configuration (empty, half-filled, or fully filled 4f orbitals).

Ce$$^{3+}$$: [Xe] 4f$$^1$$ — By losing one more electron, Ce$$^{4+}$$: [Xe] 4f$$^0$$ achieves a noble gas configuration (empty 4f). This makes Ce$$^{4+}$$ quite stable, so Ce easily deviates to +4 oxidation state.

La$$^{3+}$$: [Xe] 4f$$^0$$ — already has empty 4f, so +3 is very stable. No easy deviation.

Lu$$^{3+}$$: [Xe] 4f$$^{14}$$ — fully filled 4f, so +3 is very stable. No easy deviation.

Gd$$^{3+}$$: [Xe] 4f$$^7$$ — half-filled 4f, so +3 is very stable. No easy deviation.

Cerium (Ce) most easily deviates from the +3 oxidation state because Ce$$^{4+}$$ achieves the very stable empty 4f configuration [Xe] 4f$$^0$$.

Therefore, the correct answer is Option A.