Question

The number of geometrical isomers found in the metal complexes $[PtCl_2(NH_3)_2]$, $[Ni(CO)_4]$, $[Ru(H_2O)_3Cl_3]$ and $[CoCl_2(NH_3)_4]^+$ respectively, are :

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

Hint:

To quickly determine the number of geometrical isomers, identify the geometry (square planar, tetrahedral, octahedral) and the type of complex (e.g., $MA_2B_2$, $MA_3B_3$). Memorizing the isomer patterns for these common types is essential. Remember that tetrahedral complexes generally do not show geometrical isomerism.

Answer 1

The Correct Option is B

Let's analyze each complex individually for geometrical isomerism.
1. $[PtCl_2(NH_3)_2]$: This is a square planar complex of the type $[MA_2B_2]$. Square planar complexes of this type exhibit cis-trans isomerism. The two $Cl^-$ ligands can be adjacent (cis) or opposite (trans). Thus, there are 2 geometrical isomers.
2. $[Ni(CO)_4]$: This is tetracarbonylnickel(0). The geometry is tetrahedral. Tetrahedral complexes with four identical ligands, $[MA_4]$, do not show geometrical isomerism as all positions are equivalent relative to each other. Thus, there are 0 geometrical isomers.
3. $[Ru(H_2O)_3Cl_3]$: This is an octahedral complex of the type $[MA_3B_3]$. Octahedral complexes of this type exhibit facial (fac) and meridional (mer) isomerism. In the fac isomer, the three identical ligands occupy the corners of one face of the octahedron. In the mer isomer, they occupy positions in a plane that bisects the molecule. Thus, there are 2 geometrical isomers.
4. $[CoCl_2(NH_3)_4]^+$: This is an octahedral complex of the type $[MA_4B_2]$. Octahedral complexes of this type exhibit cis-trans isomerism. The two $Cl^-$ ligands can be on adjacent positions (90$^\circ$ apart, cis) or on opposite positions (180$^\circ$ apart, trans). Thus, there are 2 geometrical isomers.
The number of geometrical isomers for the complexes are 2, 0, 2, and 2, respectively. This corresponds to option (B).