Question: Which of the following ligands forms a chelate complex?
Correct Option: (D) Oxalate ion
Step 1 (Definition of Chelation): [0.5 Mark]
A chelate complex is formed when a polydentate ligand (a ligand with two or more donor atoms) binds to a single central metal ion, forming a ring structure. The governing principle is the presence of multiple donor sites capable of simultaneous coordination.
Step 2 (Analysis of Ligands): [0.25 Mark]
- (A) Ammonia ($\text{NH}_3$): Monodentate (only one N donor atom).
- (B) Water ($\text{H}_2\text{O}$): Monodentate (only one O donor atom).
- (C) $\text{NO}_2^-$: Monodentate (ambidentate, but binds through only one atom at a time).
- (D) Oxalate ion ($\text{C}_2\text{O}_4^{2-}$): Bidentate ligand. It has two oxygen donor atoms that can coordinate with the central metal ion simultaneously to form a five-membered ring.
Step 3 (Final Concluded Answer): [0.25 Mark]
Since the oxalate ion ($\text{C}_2\text{O}_4^{2-}$) is a bidentate ligand, it satisfies the condition for chelation. Therefore, (D) Oxalate ion is the correct answer.
💡 EL10 Concept Insight: Think of a chelate like a pair of tweezers; because the oxalate ion has two "hands" (donor atoms) to grab the metal at the same time, it forms a stable, ring-like loop, whereas the other options only have one "hand" and can't form a loop.