Question

The complex $\left[Pt\right(Py\left)\right(N{H}_3\left)BrCl\right]$ will have how many geometrical isomers?

• A. 2
• B. 0
• C. 4
• D. 3

Answer 1

The correct option is D

3

The complex is square planar and shows two cis and one trans isomers.

There are 4 ligands. This reduces to a problem of circular permutation such that we count only the sequence in which atoms are connected. Meaning, we disregard the direction - clockwise or counter-clockwise. The answer should be $\frac{(n-1)!}{2}$ = $\frac{6}{2}$ = 3. So we get 3 geometric isomers.