Question

The total number of possible isomers for squareplanar ${\left[Pt(Cl)\left(N{O}_2\right)\left(N{O}_3\right)(SCN)\right]}^{2-}$ is:

• A. 16
• B. 12
• C. 8
• D. 24

Answer 1

Answer: (B)

The square-planar complex of type $[Mabcd{]}^{nt}$, where all four ligands are different, has 3 geometrical isomers. But if one of the ligands is ambidentate, then $2\times 3=6$ geometrical isomers are possible.

Here, two ligands are ambidentate, then $4\times 3=12$ geometrical isomers are possible.
In our example, both $N{O}_2^{-}$and $SC{N}^{-}$are ambidentate ligands.
Hence option B is correct.
The total number of possible isomers for square-planar ${\left[Pt(Cl)\left(N{O}_2\right)\left(N{O}_3\right)(SCN)\right]}^{2-}$ is $12.$