Question

The number of isomers possible for square planar complex $K_2\left[Pt\right(ClB{r}_{2}SCN\left)\right]$ is:

• A. 4
• B. 4.00
• C. 4.0

Answer 1

Answer: (A), (B), (C)

The coordination number of $Pt$ here is $4$. Hence two types of hybridizations are possible:$s{p}^3\text{or}ds{p}^2$
$Pt$ is a $5d$ element. Such elements almost always show inner orbital hybridizations. Hence, in this case, $ds{p}^2$ hybridization occurs.
Now, $ds{p}^2$ hybridization can show geometrical isomerism because they are planar. So, two isomers are possible: $Br-Br$ cis or $Br-Br$ trans.
But, the $SCN$ is an ambidentate ligand. Hence its connectivity can be either from $N$ atom this $SCN\to$ or from $S$ atom like this $NCS\to$.
So, the total isomers for this compound is $4$.