Question

Total number of possible isomers for $\left[Cu\right(N{H}_3{)}_4\left]\right[PtC{l}_4]$ is:

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is B $4$

(i) $\left[Cu\right(N{H}_3{)}_4{]}^{2+}[PtC{l}_4{]}^{2-}$
(ii) $\left[CuCl\right(N{H}_3{)}_3{]}^{+}\left[PtC{l}_3\right(N{H}_3){]}^{-}$
(iii) $\left[CuC{l}_2\right(N{H}_3{)}_2{]}^0\left[PtC{l}_2\right(N{H}_3{)}_2{]}^0$
(iv) $\left[PtCl\right(N{H}_3{)}_3{]}^{+}\left[CuC{l}_3\right(N{H}_3){]}^{-}$
(v) $\left[Pt\right(N{H}_3{)}_4{]}^{2+}[CuC{l}_4{]}^{2-}$

(iii) is not possible because the coordination spheres acquire zero charge and the cationic and anionic spheres no longer exists.

Therefore a total of 4 isomers are possible.