Question

Which of the following complex shows ionisation isomerism:
(A) $[Cr{(N{H_3})_6}]C{l_3}$
(B) $[Cr{(en)_2}]C{l_2}$
(C) $[Cr{(en)_3}]C{l_3}$
(D) $[CoBr{(N{H_3})_5}]S{O_4}$

Answer 1

Hint: Ionisation isomerization a type of structural isomerization in which ligands exchange between the coordination sphere and the ionic sphere. The ligands inside the coordination sphere are used to decide the coordination number and the ligands outside the coordination sphere are used to decide the oxidation state.

Complete Step by Step Solution:
Generally ionisation isomerism is shown by those coordination compounds in solution that give different ions although they possess the same chemical composition. Also this type of isomerism is seen to occur in a coordination compound when its counter ion is a potential ligand and this counter ion can displace a ligand, making it behave as a counterion.

In option A, \[N{H_3}\] cannot replace $C{l^ - }$. Only charged ligands can be present outside the coordination sphere. If a neutral ligand is present outside the coordination sphere then there will be no electrostatic interaction left between the coordination sphere and ionic sphere and they will separate from each other. This will change the chemical formula of the molecule and will no longer be called isomers. Hence, option A is not the correct option.

The exchange does not depend upon the charge of the ligands but on denticity. Thus, option B and option C does not show ionisation isomerism because the ligand \[en\]meaning ethylene diamine is a bidentate ligand because there are two \[N{H_3}\] groups and thus, two coordination sites from lone pair present on each N. Whereas $C{l^ - }$ is a monodentate ligand. It has only one donating site. Thus, they both have different denticity. This signifies that in order to undergo isomerization, two $C{l^ - }$ ligands are required to exchange one \[en\] ligand. But this will distort the geometry. Hence, the second and third option does not show ionisation isomerism.

In option D, \[S{O_4}^{2 - }\]and \[B{r^ - }\] although having different charges but have the same denticity. Also counter ion, $SO_{4}^{2-}$ can easily replace $C{{l}^{-}}$ and is known to show ionisation isomerization.
Thus, the correct option is D.

Note: There are several other categories of structural isomers. For example, if the ligand undergoing ionisation isomerism is specifically water then is it termed as hydration isomerism. Other types include linkage isomerism, coordination isomerism, coordination position isomerism and ligand isomerism.