Question: 70: Match List-I with List-II :
| List – I (Complexes) | List – II (Types) |
|---|---|
| (a) \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right] \mathrm{Cl}_{2} and \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{ONO}\right] \mathrm{Cl}_{2} | (i) ionisation isomerism |
| (b) \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Co}(\mathrm{CN})_{6}\right] and \left[\mathrm{Cr}(\mathrm{CN})_{6}\right]\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] | (ii) coordination isomerism |
| (c) \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{SO}_{4}\right)\right] \mathrm{Br} and \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4} | (iii) linkage isomerism |
| (d) \left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3} and \left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O} | (iv) solvate isomerism |
Choose the correct answer from the options given below :
(1) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
(2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(3) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(4) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Answer: Option (3)
Explanation:
In pair (a), the ligand \mathrm{NO}_{2}^{-} coordinates through nitrogen in one complex and through oxygen in the other.
This change in the point of attachment of the same ligand leads to linkage isomerism.
Therefore, (a) corresponds to (iii).
In pair (b), both complexes contain a cationic and an anionic complex,
but the metal ions exchange their ligands between the two coordination spheres.
This type of isomerism is known as coordination isomerism.
Hence, (b) corresponds to (ii).
In pair (c), one complex gives bromide ions in solution while the other gives sulphate ions.
The interchange of anions between the coordination sphere and
ionisable sphere causes ionisation isomerism.
Thus, (c) corresponds to (i).
In pair (d), the difference lies in the presence of water either inside or outside the coordination sphere.
This gives rise to solvate (hydrate) isomerism.
Hence, (d) corresponds to (iv).
Therefore, the correct matching is:
(a)-(iii),\ (b)-(ii),\ (c)-(i),\ (d)-(iv)So, option (3) is correct.