Question

Which kind of isomerism is exhibited by octahedral $\left[ Co{{\left( N{{H}_{3}} \right)}_{4}}B{{r}_{2}} \right]Cl$ ?
[A] Geometrical and ionisation
[B] Geometrical and optical
[C] Optical and ionisation
[D] Geometrical only

Answer 1

HINT: In the given compound, it shows one structural and one stereoisomerism. The stereoisomers are not optically active. Here, the ligand outside the coordination sphere can replace one of the similar ligands in the coordination sphere.

COMPLETE STEP BY STEP SOLUTION:
Before answering this question, let us discuss what isomerism is and the different types of isomerism exhibited by compounds.
Isomerism is the phenomenon where two or more chemical compounds have the same chemical formula but different chemical structures.
Isomerism can be further divided into two types- structural isomerism and stereoisomerism.
(1) Structural isomerism includes various subparts so let us go through them-
(a) Ionisation isomerism – This isomerism occurs due to interchange of ions between the inner and the outer coordination sphere. For example - $\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}Br \right]S{{O}_{4}}\text{ and }\left[ Co{{\left( N{{H}_{3}} \right)}_{5}}S{{O}_{4}} \right]Br$
(b) Hydration isomerism – This occurs due to different modes of distribution of water molecules between inner and outer coordination spheres. For example - $\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{3}}\text{ and }\left[ Cr{{\left( {{H}_{2}}O \right)}_{5}}Cl \right]C{{l}_{2}}\cdot {{H}_{2}}O$
(c) Coordination isomerism – When salts are produced by cationic and anionic compounds, metal centres between the complex entities can be interchanged to give coordination isomers. For example, ${{\left[ Pt{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2-}},{{\left[ PdC{{l}_{4}} \right]}^{2-}}$ can be inter-converted into ${{\left[ Pd{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2-}},{{\left[ PtC{{l}_{4}} \right]}^{2-}}$
(d) Ligand isomerism – This occurs due to isomeric nature of ligands, example-$N{{\left( C{{H}_{3}} \right)}_{3}}\text{ and C}{{\text{H}}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}$
(e) Linkage isomerism – This occurs due to the presence of ambidentate ligands having different donor sites. Example- $Co{{\left( SCN \right)}_{4}}^{2-}\text{ and }Co{{\left( NCS \right)}_{4}}^{2-}$
(2) Stereo isomerism is of two types-
(a) Geometrical isomerism – These are the compounds with the same number and types of atoms but have different geometries. For example- fumaric acid and maleic acid have the same number and type of atoms but they are not the same.
(b) Optical isomerism – These are two compounds with the same number of atoms with the same bond connectivity but their spatial arrangement is different.
Now, let us discuss about the compound given to us $\left[ Co{{\left( N{{H}_{3}} \right)}_{4}}B{{r}_{2}} \right]Cl$
We can clearly see that it has ionisation isomers which we can write as - $\left[ Co{{\left( N{{H}_{3}} \right)}_{4}}B{{r}_{2}} \right]Cl\text{ and }\left[ Co{{\left( N{{H}_{3}} \right)}_{4}}BrCl \right]Br$
Its geometry is octahedral and it’s a complex of type $[M{{A}_{4}}{{B}_{2}}]$. It has 2 stereo isomers but both are optically inactive and thus are geometrical isomers.

As we can see it shows geometrical as well as ionisation isomerism.
Therefore, the correct answer is option [A] geometrical and ionisation
NOTE: Presence of geometrical isomers does not mean it will have optical isomers too. Optical isomers are attained by the difference in spatial arrangements of the ligands. If the mirror image on one compound is non-superimposable, they are enantiomers and are called optical isomers.
Among the coordination number 4 structures, tetrahedron shows no geometrical isomerism as the relative position of all the ligands is the same.