Question

The number of geometrical isomers found in the metal complexes $\left[Pt\right(Cl{)}_2\left(N{H}_3{)}_2\right],\left[Ni\right(CO{)}_4],[Ru\left(H_{2}O{)}_{3}C{l}_3\right]\text{and}\left[CoC{l}_2\right(N{H}_3{)}_4{]}^{+}$ respectively, are

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• A. 2, 0, 2, 2
• B. 2, 1, 2, 1
• C. 1, 1, 1, 1
• D. 2, 1, 2, 2

Answer 1

The correct option is A 2, 0, 2, 2

$\left[Pt\right(Cl{)}_2\left(N{H}_3{)}_2\right]$ forms two possible geometrical isomers, cis and trans.
In cis, same group on same side and in trans same groups on opposite side.


In $\left[Ni\right(CO{)}_4]$, the oxidation state of Ni is 0 and it forms a outer orbital complex with $s{p}^3$ hybridisation. Thus, it form tetrahedral complex and it lacks geometrical isomerism.

In $\left[M{a}_3{b}_3\right]$ type of complex, the possible geometrical isomers are two, facial and meridional.
Facial isomer have three identical ligands on one triangular face.
Meridional isomer have three identical ligands in a plane bisecting the molecule.



In $\left[M{a}_2{b}_4\right]$ type, there are two possible geomerical isomers, cis and trans.
These isomers are given below,


Hence, option (c) is correct.