Question

In $\left[Pt\right(N{H}_3{)}_{2}C{l}_2],pt-Cl$ bond length is $2\stackrel{\circ }{A}$ and $Cl-Cl$ distance is $2.88\stackrel{\circ }{A}$ then the compound is

• A. Tetrahedral
• B. Square pyramidal
• C. Cis-square planar
• D. Trans-square planar

Answer 1

Answer: (C)

Cis-square planar

The complex $\left[Pt\right(N{H}_3{)}_{2}C{l}_2],pt-Cl$ has square planar geometry.

The isomer will be trans if the distance between two chlorine atoms is twice the C-Cl bond length.

The isomer will be cis if the distance between two chlorine atoms is less than twice the C-Cl bond length. This is the case in the given complex as $2.8\stackrel{o}{A}<2\times 2\stackrel{o}{A}$

Thus the given complex is cis square planar complex.

Hence, the correct option is $\text{C}$