The number of geometric isomers that can exist for square planar ${\left[\mathrm{P}\mathrm{t}\left(\mathrm{C}\mathrm{l}\right)\left(\mathrm{p}\mathrm{y}\right)\left(\mathrm{N}{\mathrm{H}}_3\right)\left(\mathrm{N}{\mathrm{H}}_2\mathrm{O}\mathrm{H}\right)\right]}^{+}$ is (py = pyridine):

• A

  6
• B

  2
• Correct answer
  C

  3
• D

  4

Solution

Complex with general formula [Mabcd]⁺ square planar complex can have three isomers.

Hence, the correct answer is option: 'C'.