The total number of possible isomers of the compound $[C u^{I I} {(N H_{3})}_{4}] [P t^{I I} C l_{4}]$ are :

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Solution

The correct option is C 4
$[C u^{I I} {(N H_{3})}_{4}] [P t^{I I} C l_{4}]$ will show 4 isomers (2 coordinate isomers and 2 geometrical isomer of coordinate isomer).

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[{C u}^{I I} {({N H}_{3})}_{4}) [{P t}^{I I} {C l}_{4}]

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