Question

The ionisation isomer of \[\left[ {{\rm{Cr(}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{)}}_{\rm{4}}}{\rm{Cl}}\left( {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right)} \right]{\rm{Cl}}\] is
A. \[\left[ {{\rm{Cr(}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{)}}_{\rm{4}}}\left( {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right)} \right]{\rm{C}}{{\rm{l}}_2}\]
B. \[\left[ {{\rm{Cr(}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{)}}_{\rm{4}}}{\rm{C}}{{\rm{l}}_2}} \right]{\rm{N}}{{\rm{O}}_{\rm{2}}}\]
C. \[\left[ {{\rm{Cr(}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{)}}_{\rm{4}}}{\rm{Cl(ONO)}}} \right]{\rm{Cl}}\]
D. \[\left[ {{\rm{Cr(}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{)}}_3}{\rm{C}}{{\rm{l}}_2}{\rm{(N}}{{\rm{O}}_2}{\rm{)}}} \right]{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{O}})\]

Answer 1

Hint: A set of molecules possessing the same molecular formula but their structures are different is termed isomers. And this phenomenon is termed isomerism. The different isomerism types are linkage, ionisation, etc.

Complete Step by Step Answer:
Let’s understand what ionisation isomerism is. In this isomerism, the production of different ions by the coordination compounds takes place.
\[{\rm{[Co(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{5}}}{\rm{Br]S}}{{\rm{O}}_{\rm{4}}} \to {[{\rm{Co(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{5}}}{\rm{Br}}]^{2 + }} - {\rm{Cobalt(III)ion}} + {\rm{S}}{{\rm{O}}_4}^{2 - }\]

In the solution of \[{\rm{BaC}}{{\rm{l}}_{\rm{2}}}\], the above isomer gives a white coloured precipitate of \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\].
\[{\rm{[Co(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{5}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{]Br}} \to {[{\rm{Co(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{5}}}{\rm{S}}{{\rm{O}}_4}]^ + } - {\rm{Cobalt(III)ion}} + {\rm{B}}{{\rm{r}}^ - }\]
The above isomer produces a precipitate of light yellow colour.

Let’s observe the options one by one.
In option A, the complex has the coordination number of 5 but the given compounds have the coordination number of 6. So, A is not an ionisation isomer of the given complex.
In option C, the ONO acts as ambidentate ligand. So, it is a linkage isomer. Therefore, C is wrong.
In option D, the molecule of water cannot come out of the coordination sphere as the complex is formed as a water of crystallisation. Therefore, D is incorrect.
And the correct ionization isomer of \[\left[ {{\rm{Cr(}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{)}}_{\rm{4}}}{\rm{Cl}}\left( {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right)} \right]{\rm{Cl}}\]is \[\left[ {{\rm{Cr(}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{)}}_{\rm{4}}}{\rm{C}}{{\rm{l}}_2}} \right]{\rm{N}}{{\rm{O}}_{\rm{2}}}\].
Therefore, option B is right.

Additional Information: Geometrical isomerism is a phenomenon where coordination compounds have differed on the basis of the arrangement of the ligands in around the central metal atom in a coordination sphere. This isomerism is of two types, cis and trans. Generally, this isomerism is observed in unsaturated compounds having a double bond.

Note: The phenomenon of ionisation isomerism is also defined as the ion exchange of the inside and outside of the coordination sphere. For example, pentaammine sulphato cobalt (III) bromide and pentaamminenitrocobalt (III) sulphate are ionisation isomers.