Question

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

• A. $16$
• B. $12$
• C. $8$
• D. $24$

Answer 1

The correct option is B $12$

The square-planar complex of type $[Mabcd{]}^{n\pm }$, 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 $\left[Pt\right(Cl\left)\right(N{O}_2\left)\right(N{O}_3\left)\right(SCN){]}^{2-}$ is $12$.