Question:easy

Which of the following ligands forms a chelate complex ?

Show Hint

Common chelating ligands: \[ C_2O_4^{2-},\; en,\; EDTA^{4-} \] Oxalate ion is a bidentate ligand and readily forms chelate complexes.
Updated On: Jun 29, 2026
  • Ammonia
  • Water
  • \(NO_2^{-}\)
  • Oxalate ion
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Define chelation.
A chelate complex forms when one ligand bonds to the central metal through two or more donor atoms simultaneously, closing a ring. Only multidentate ligands (bidentate or higher) can chelate; monodentate ligands cannot form rings.
Step 2: Check the monodentate ligands.
$NH_3$ coordinates via one $N$ atom only. $H_2O$ coordinates via one $O$ atom only. $NO_2^-$ is ambidentate (bonds through $N$ or $O$, never both at once to the same metal). None of these close a ring, so none are chelating.
Step 3: Evaluate oxalate ion.
Oxalate $C_2O_4^{2-}$ carries two equivalent $-COO^-$ ends, each providing an oxygen donor. Both oxygens coordinate to the same metal simultaneously, forming a stable five-membered ring. This makes oxalate a classic bidentate chelating ligand.
\[ \boxed{\text{Oxalate ion}} \]
Was this answer helpful?
0