Statement-I : Complex ion $[Co {({NH}_{3})}_{6}]^{2 +}$ is readily oxidized to $[Co {({NH}_{3})}_{6}]^{3 +} .$
Statement-II : Unpaired electron in complex ion $[Co {({NH}_{3})}_{6}]^{2 +}$ is present in 4p orbital.

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Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for
statement-1.

Statement-1 is false, statement-2 is true.

Statement-1 is true, statement-2 is false.

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

answer is C.

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Detailed Solution

$2 [C o {(N H_{3})}_{6}]^{2 +} + 2 H^{+} + \frac{1}{2} O_{2} \underset{}{\to} 2 [C o {(N H_{3})}_{6}]^{3 +} + H_{2} O$

The unpaired electron in $[C o {(N H_{3})}_{6}]^{+ +}$ is present in $4 d$ orbital. As $4 p$ orbitals are to be involved in Hybridization, hence migration of $e^{-}$ occurs to $4 d .$

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