Question

Which one of the following complex ions has geometrical isomers?
(A) ${\left[Ni{\left(N{H}_3\right)}_{5}Br\right]}^{+}$
(B) ${\left[Co{\left(N{H}_3\right)}_2{\left(en\right)}_2\right]}^{3+}$
(C) ${\left[Co{\left(N{H}_3\right)}_4\left(en\right)\right]}^{3+}$
(D) ${\left[Co{\left(en\right)}_3\right]}^{3+}$ (en is ethylenediamine)

Answer 1

Hint: Octahedral complexes are the complexes in which the ligands have six donor atoms. Ethylenediamine is a bidentate ligand and has two donor atoms. All the given complexes are octahedral complexes but the complex which has a difference in stereo or special arrangement will have geometrical isomerism.

Complete answer:
Complex compounds are also known as coordination compounds. These compounds have a special property as they do not lose their identity when dissolved in water or any other organic solvents.
Given complexes have ligands containing six donor atoms and can be considered as octahedral complexes.
The complex given is ${\left[Ni{\left(N{H}_3\right)}_{5}Br\right]}^{+}$ , it has five ammonia ligands and one bromine ligand and there is no stereo difference and does not exhibit geometrical isomerism.
The complex given is ${\left[Co{\left(N{H}_3\right)}_2{\left(en\right)}_2\right]}^{3+}$ , it has two ammonia ligands and two ethylenediamine ligands and exist in both cis and trans forms that can exhibit geometrical isomerism.
The complex given is ${\left[Co{\left(N{H}_3\right)}_4\left(en\right)\right]}^{3+}$ , it has four ammonia ligands and one ethylenediamine ligand and there is no stereo difference and does not exhibit geometrical isomerism.
The complex given is ${\left[Co{\left(en\right)}_3\right]}^{3+}$ , it has three ethylenediamine ligands and there is no stereo difference and does not exhibit geometrical isomerism.
 ${\left[Co{\left(N{H}_3\right)}_2{\left(en\right)}_2\right]}^{3+}$ only exhibit two geometrical isomers.
Option B is the correct one.

Note:
The complex which can form different structures and there should be some difference in the arrangement of ligands can only exhibit geometrical isomers. In the complex ${\left[Co{\left(N{H}_3\right)}_2{\left(en\right)}_2\right]}^{3+}$ the ligands will arrange in two different ways and can exhibit cis and trans forms which are geometrical isomers.