Question

Total no. of possible isomers for complex compound $\left[Cu\right(N{H}_3{)}_4\left]\right[PtC{l}_4]$ are:

• A. 3
• B. 2
• C. 4
• D. none of these

Answer 1

The correct option is C 4

$\left[Cu\right(N{H}_3{)}_4\left]\right[PtC{l}_4],$ $\left[Cu\right(N{H}_3{)}_{3}Cl\left]\right[PtC{l}_{3}N{H}_3],$ $\left[Cu\right(N{H}_3{)}_{2}C{l}_2\left]\right[PtC{l}_2\left(N{H}_3{)}_2\right],$ $\left[Cu\right(N{H}_3\left)C{l}_3\right]\left[PtCl\right(N{H}_3{)}_3],$ $\left[Cu\right(Cl{)}_4\left]\right[Pt\left(N{H}_3{)}_4\right]$

So five isomers are possible ,but third one do not exist due to unsatisfie charge on complex so only $4$ exist.