Question

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

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

4

The coordination number of Pt here is 4. Hence two types of hybridizations are possible: $s{p}^3$ or $ds{p}^2$

$Pd$ is a $5d$ element. Such elements almost always show inner orbital hybridizations. Hence, in this case, $ds{p}^2$ hybridization.

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 like this SCN→ or from S atom like this NCS

So, the total isomers for this compound is 4.

Hence option D is correct.