Statement-1 : $[Cr {({NH}_{3})}_{6}]^{3 +}$ has three unpaired electron in ${t_{2}}_{g}$ orbital.
Statement-2 : $[Cr {({NH}_{3})}_{6}]^{3 +}$ does not show geometrical isomerism.

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

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

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

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

answer is B.

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Detailed Solution

$[Cr {({NH}_{3})}_{6}]^{3 +} C r^{3 +} \underset{}{\to} [Ar] 3 d^{4} 4 s^{0}$

3 unpaired electron in $t_{2 g}$ orbital.

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