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Example · Example 1

Q.Classify each of the following ligands as monodentate, bidentate, or polydentate, giving the number of donor atoms in each case:

(a) NH3\text{NH}_3,
(b) Cl−\text{Cl}^-,
(c) ethylenediamine (en\text{en}),
(d) oxalate ion (C2O42−\text{C}_2\text{O}_4^{2-}),
(e) EDTA4−^{4-}.
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Denticity is the number of donor atoms a single ligand molecule or ion uses to bind simultaneously to one metal centre. (a) NH3\text{NH}_3 has one lone pair on its single nitrogen atom available for donation, so it is monodentate.

(b) Cl−\text{Cl}^- likewise donates through a single atom, so it is monodentate.

(c) Ethylenediamine, H2N-CH2-CH2-NH2\text{H}_2\text{N-CH}_2\text{-CH}_2\text{-NH}_2, has two nitrogen atoms, each with a lone pair, and both can bind the same metal ion at once, closing a five-membered chelate ring — it is bidentate.

(d) The oxalate ion, C2O42−\text{C}_2\text{O}_4^{2-}, binds through two of its four oxygen atoms (one from each carboxylate end), also forming a ring — bidentate. (e) EDTA4−^{4-} offers six donor atoms simultaneously: the two nitrogen atoms of its central ethylenediamine backbone plus the oxygen atom of each of its four carboxylate arms (4 O), giving six donor atoms in total — it is polydentate, specifically hexadentate, and can occupy all six positions of an octahedral metal ion by itself. [!ANSWER] Monodentate: NH3\text{NH}_3 (1 N), Cl−\text{Cl}^- (1 Cl). Bidentate: en (2 N), oxalate (2 O). Polydentate/hexadentate: EDTA4−^{4-} (2 N + 4 O = 6 donor atoms).

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