Which of the following isomerism is shown by [Co(NH3)4(NO2)2]NO3
(A) Linkage isomerism
(B) Ionisation isomerism
(C) Hydrate isomerism
(D) Geometrical isomerism
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
Structural isomers of coordination compounds differ in which atoms are bonded to the metal or which groups are inside the coordination sphere, while stereoisomers differ only in the arrangement in space. Each type needs a particular feature in the compound:
Linkage isomerism: an ambidentate ligand, which can bind through two different atoms.
Ionisation isomerism: a counter ion that can exchange places with a ligand inside the coordination sphere.
Hydrate (solvate) isomerism: water molecules that can be either ligands or lattice water.
Geometrical isomerism: ligands that can be arranged cis or trans, for example an octahedral [MA4B2] complex.
In \( \ce{[Co(NH3)4(NO2)2]NO3} \), the complex ion has four \( \ce{NH3} \) and two \( \ce{NO2^{-}} \) ligands, and \( \ce{NO3^{-}} \) is the counter ion.
(A) Linkage isomerism — Correct. \( \ce{NO2^{-}} \) is ambidentate: it can bind through nitrogen (nitrito-N) or through oxygen (nitrito-O).
(B) Ionisation isomerism — Correct. An \( \ce{NO2^{-}} \) ligand can exchange with the \( \ce{NO3^{-}} \) counter ion, giving \( \ce{[Co(NH3)4(NO2)(NO3)]NO2} \).
(C) Hydrate isomerism — Incorrect. The compound contains no water molecules, so hydrate isomerism is not possible.
(D) Geometrical isomerism — Correct. It is an octahedral [MA4B2] type complex, so the two \( \ce{NO2^{-}} \) ligands can be next to each other (cis) or opposite each other (trans).