Question

Which of the following complexes will show geometrical isomerism?
A.Potassium amminetrichloroplatinate (II)
B.Potassium tris(oxalate) chromate (III)
C.Aquachlorobis (ethylenediamine) cobalt (II) chloride
D.Pentaaquachlorochromium (III) chloride.

Answer 1

For a square planar geometry, the geometrical isomerism is not shown by the complexes of the type ${\text{M}}{{\text{A}}_3}{\text{B}}$. In case of octahedral geometry the complexes with the type ${\text{M}}{{\text{A}}_5}{\text{B}}$ does not show geometrical isomerism. All the other types of complexes show geometrical isomerism.

Complete step by step answer:
There are two types of geometrical isomers that are, Cis and Trans isomers. Cis isomer is that isomer in which the same groups are places at $90^\circ$ with each other and in Trans isomers the similar groups are place at an angle of $180^\circ$ with each other.
In a square planar geometry all groups are placed at an angle of $90^\circ$ with each other. If three groups come out to be similar then no other arrangement of groups are possible. Like in the option A there is trichloro which means 3 chlorine ligand and one amine group. There is only one way these can be put in square planar geometry and hence no isomerism exists in this.
In option B, since oxalate is bi dentate ligand it forms an octahedral complex. All the ligands are same so there is no possibility for cis or Trans isomerism. In option D there are five chlorine and 1 water as a ligand. Here also similar groups are at $90^\circ$ and $180^\circ$ in the same molecule, so no geometrical isomerism exists.
Option C has two ethylenediamine, one chlorine and one water as ligand. It forms an octahedral complex because one ethylenediamine attaches with two sites with metal ions. Since the other two ligands are different, both the ethylenediamine groups can be placed at $90^\circ$ and $180^\circ$ with each other and hence will form geometrical isomers.
Option D has 5 water molecules and 1 chlorine atom surrounding the chromium metal. Thus, due to the presence of only 1 ligand of chlorine, it cannot show any geometrical isomerism. Thus, option D is incorrect.

Option C is the correct option.

Note:
Octahedral complexes are formed when the coordination number of metal is 6 and tetrahedral or square planar complexes are formed when the metal has coordination number 4. Coordination number is the number of ligands that are directly attached with metal in a coordination sphere.