Question

Match the Column-I (Ligands) with Column-II (Ligand type):
Column-I: (I) \(NO_{2}^{-}\), (II) \(OH^{-}\), (III) \(C_{2}O_{4}^{2-}\), (IV) EDTA.
Column-II: (A) Hexadentate, (B) Bidentate, (C) Monodentate, (D) Ambidentate.

• A. I-D, II-C, III-B, IV-A
• B. I-C, II-D, III-B, IV-A
• C. I-C, II-D, III-A, IV-B
• D. I-A, II-C, III-B, IV-D

Hint:

Remember these important ligands: \[ \boxed{ \begin{aligned} OH^- &\rightarrow \text{Monodentate}\\ H_2O &\rightarrow \text{Monodentate}\\ NH_3 &\rightarrow \text{Monodentate}\\ NO_2^- &\rightarrow \text{Ambidentate}\\ SCN^- &\rightarrow \text{Ambidentate}\\ C_2O_4^{2-} &\rightarrow \text{Bidentate}\\ EDTA &\rightarrow \text{Hexadentate} \end{aligned} } \] These ligands are frequently asked in competitive examinations.

Answer 1

The Correct Option is A

$NO_2^-$ is ambidentate because it can coordinate through N (nitro) or O (nitrito); similarly $CN^-$ can bind through C or N. Ethylenediamine (en) is a bidentate chelating ligand that binds a metal ion simultaneously through both its $NH_2$ groups, forming a stable five-membered chelate ring.