The total number of possible isomers of the compound $[C u^{0} (N H_{3})_{4}] [P t^{0} (C l_{4})]$ are?

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Solution

The correct option is C 4
Possible isomers:
$[C u {(N H_{3})}_{3} C l] [P t {(C l)}_{3} {N H}_{3}], [C u {({N H}_{3})}_{2} {C l}_{2}] [P t ({C l}_{2}) ({N H}_{3})_{3}], [C u ({N H}_{3}) {C l}_{3}] [P t C l {({N H}_{3})}_{3}], [C u {C l}_{4}] [P t {({N H}_{3})}_{4}]$
Hence, $4$ possible isomers.

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