The most common oxidation state of Lanthanoid elements is + 3 ....

The most common oxidation state of Lanthanoid elements is

+ 3

. Which of the following is likely to deviate easily from

+ 3

oxidation state?

Options

A $Ce\left(At . No, 58\right)$
B $La\left(At . No, 57\right)$
C $Lu\left(At . No, 71\right)$
D $Gd\left(At . No, 64\right)$

Correct answer

A. $Ce\left(At . No, 58\right)$

Step-by-step solution

Electronic configuration of $Ce = \left[Xe\right]\mathop{\mathop{4{f}^{1}5{d}^{1}6{s}^{2}}_{⏟}}_{}\overset{- 4{e}^{-}}{\to}\mathop{\mathop{\left[Xe\right]}_{⏟}}_{}$

Cerium in $+ 4$ oxidation state due to stable xenon configuration. $La, LuandGd$ are stable in +3 oxidation state.

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