The number of isomers possible for square planar complex $K_{2} [P d C l B r_{2} S C N]$ is:

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Solution

The correct option is A 2
The given planar complex can have upto two optical isomes 1. Cis- $[P d C l B r_{2} S C N]^{2 -}$ and 2.trans- $[P d C l B r_{2} S C N]^{2 -}$ .In first complex the Br-Br will be cis to each other and in second isomer they are trans to one another. Hence the correct option is A.

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