Chelation — Why ring-forming ligands grip metals tighter
A polydentate ligand that forms a ring structure with the central metal ion produces a chelate. The word comes from the Greek chela (claw) — the ligand wraps around the metal through two or more donor atoms within the same molecule. Ethylenediamine (en), EDTA, and oxalate (C₂O₄²⁻) are standard examples from NCERT Class 12 Chemistry Chapter 5.
What makes chelation thermodynamically favourable? The chelate effect: replacing two monodentate ligands with one bidentate ligand increases the number of free species in solution (entropy rises). For the reaction [Cu(H₂O)₄]²⁺ + 2 en → [Cu(en)₂]²⁺ + 4 H₂O, you start with 3 species and end with 5. This positive ΔS drives the equilibrium forward even when ΔH is comparable to the monodentate case.
Chelate ring size matters. Five-membered chelate rings (formed by en, glycinate) are the most stable. Four-membered rings (formed by carbonate binding through both oxygens) are strained. Six-membered rings (acetylacetonate) are stable but less common at NEET level.
Denticity vs. coordination number — a frequent confusion. Denticity is how many donor atoms one ligand provides. Coordination number is the total number of donor atoms bonded to the metal from all ligands. EDTA is hexadentate (6 donor atoms from one molecule), so one EDTA occupies all 6 coordination sites of an octahedral metal ion.
NEET relevance. Questions test whether you can identify chelating ligands, count denticity correctly, explain why chelates are more stable, and distinguish chelation from simple coordination. The definition of chelate itself (NCERT Class 12 Chemistry Chapter 5, page 121) is directly evaluable.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following ligands can act as a chelating agent?
Answer: A. Ethylenediamine has two nitrogen donor atoms in the same molecule that can simultaneously bind to a metal ion, forming a chelate ring. This is the definition of a chelating (polydentate) ligand (NCERT Class 12 Chemistry Chapter 5, page 121).
Why B is wrong: B is wrong because NH₃ is monodentate — its single nitrogen lone pair provides only one point of attachment to the metal.
Why C is wrong: C is wrong because Cl⁻ is a monodentate ligand — it has only one donor atom and cannot form a ring with the metal ion.
Why D is wrong: D is wrong because H₂O is monodentate — it coordinates through one oxygen lone pair only.
The term 'chelate' is derived from the Greek word chela, which means:
Answer: A. 'Chelate' comes from the Greek chela meaning claw, reflecting how a polydentate ligand grips the metal ion from multiple points, like a claw (NCERT Class 12 Chemistry Chapter 5).
Why B is wrong: B is wrong because although chelates form ring structures, the etymological origin is 'claw,' not 'ring.'
Why C is wrong: C is wrong because 'bond' does not correspond to the Greek root; the imagery is of a gripping claw, not a generic bond.
Why D is wrong: D is wrong because 'chain' does not match the Greek etymology; chelation refers to claw-like encirclement, not a linear chain.
The denticity of EDTA as a ligand is:
Answer: C. EDTA (ethylenediaminetetraacetic acid) has 6 donor atoms — 2 nitrogen and 4 oxygen — making it a hexadentate ligand (NCERT Class 12 Chemistry Chapter 5).
Why A is wrong: A is wrong because 2 would make EDTA bidentate, counting only the two nitrogen donors while ignoring the four carboxylate oxygen donors.
Why B is wrong: B is wrong because 4 counts only the carboxylate oxygens, missing the two nitrogen lone pairs that also coordinate to the metal.
Why D is wrong: D is wrong because EDTA has only 6 donor atoms (2N + 4O); no eighth donor site exists in the molecule.
In the complex [Co(en)₃]³⁺, the coordination number of cobalt is:
Answer: C. Each ethylenediamine (en) ligand is bidentate (2 donor atoms). Three en ligands contribute 3 × 2 = 6 donor atoms bonded to Co, so the coordination number is 6. The key is not to confuse the number of ligands (3) with the coordination number (total donor atoms).
Why A is wrong: A is wrong because 3 counts the number of en ligand molecules, not the total donor atoms. Coordination number counts individual donor atom–metal bonds, and each en provides 2.
Why B is wrong: B is wrong because 12 appears to double-count the donor atoms or count total atoms in the ligands rather than donor atoms directly bonded to the metal.
Why D is wrong: D is wrong because 2 is the denticity of a single en ligand, not the coordination number of the metal in this complex.
Why are chelate complexes generally more stable than analogous complexes with monodentate ligands?
Answer: B. Replacing multiple monodentate ligands with one polydentate ligand increases the total number of free particles in solution (e.g., 3 reactant species → 5 product species), giving a positive ΔS that drives the equilibrium toward the chelate complex. This thermodynamic entropy advantage is the chelate effect (NCERT Class 12 Chemistry Chapter 5).
Why A is wrong: A is wrong because individual M–N or M–O bond strengths in chelates are often comparable to those in monodentate analogues. The extra stability comes from entropy, not from inherently stronger single bonds.
Why C is wrong: C is wrong because electronegativity of donor atoms is not systematically higher in chelating ligands; en and NH₃ both donate through nitrogen. Chelate stability is an entropy-driven effect.
Why D is wrong: D is wrong because chelate complexes can still dissociate — they are more stable, not infinitely stable. The equilibrium constant is larger due to the entropy contribution, but dissociation is not 'prevented.'
Which of the following chelate ring sizes is generally the most stable?
Answer: B. Five-membered chelate rings (as formed by ethylenediamine with a metal ion: M–N–C–C–N) have optimal bond angles with minimal ring strain, making them the most stable chelate ring size. This is a standard result discussed in coordination chemistry (NCERT Class 12 Chemistry Chapter 5).
Why A is wrong: A is wrong because 3-membered rings would have extreme angular strain — the bond angles would be far from the natural tetrahedral/trigonal angles of the donor atoms and carbon bridges.
Why C is wrong: C is wrong because 4-membered chelate rings (e.g., carbonate bridging through both oxygens to one metal) suffer significant ring strain due to compressed bond angles.
Why D is wrong: D is wrong because 8-membered rings are too large — the chelate ring loses the entropic and strain advantages; the ligand arms are too flexible to maintain a tight grip on the metal.
Glycinate ion (H₂N–CH₂–COO⁻) acts as a bidentate ligand. Through which donor atoms does it coordinate to a metal ion?
Answer: D. Glycinate coordinates through the amino nitrogen (lone pair on N) and one carboxylate oxygen, forming a stable 5-membered chelate ring: M–N–C–C–O(–M). This is its established coordination mode as a bidentate chelating ligand.
Why A is wrong: A is wrong because if both carboxylate oxygens coordinated to the same metal, the ring formed would be a strained 4-membered ring (M–O–C–O), and the nitrogen lone pair — a strong donor — would be wasted. Glycinate's chelation uses N and O.
Why B is wrong: B is wrong because the carbon atoms in glycinate are saturated and do not have available lone pairs for coordination. Metal–ligand bonds form through atoms with lone pairs (N, O, S, etc.).
Why C is wrong: C is wrong because glycinate has only one nitrogen atom. A ligand would need two nitrogen donors (like en) to coordinate through two nitrogens.
Consider the reaction: [Ni(H₂O)₆]²⁺ + 3 en → [Ni(en)₃]²⁺ + 6 H₂O. How does the total number of species change from reactants to products, and what is the thermodynamic consequence?
Answer: D. Reactant side: 1 (complex) + 3 (en molecules) = 4 independent species. Product side: 1 (chelate complex) + 6 (water molecules) = 7 independent species. Going from 4 to 7 free particles means ΔS > 0 (positive entropy change), which contributes a favourable −TΔS term to ΔG, driving chelate formation. This is the quantitative basis of the chelate effect.
Why A is wrong: A is wrong because although it correctly counts 4 → 7 species, it draws the wrong conclusion. More free particles in solution means HIGHER translational entropy (positive ΔS), not lower. Molecular size is not the relevant factor — the number of independent particles is.
Why B is wrong: B is wrong because it reverses the species count. The product side has MORE free species (7) than the reactant side (4), not fewer. The entropy change is positive, not negative.
Why C is wrong: C is wrong because it ignores the individual ligand and solvent molecules as separate species. You must count all independent species on each side: 1 + 3 = 4 reactant species vs. 1 + 6 = 7 product species.
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Given
• Reaction: [Cu(H₂O)₄]²⁺ + 2 en → [Cu(en)₂]²⁺ + 4 H₂O• Individual Cu–N bond energies (in en complex) are comparable to Cu–O bond energies (in aqua complex).
Required
Explain why the chelate complex is more stable. Count independent species on each side and identify the thermodynamic driving force.
Concept
The chelate effect: when polydentate ligands replace monodentate ligands, the total number of free species in solution increases, causing a positive entropy change (ΔS > 0). Since ΔG = ΔH − TΔS and ΔH is approximately similar, the positive ΔS makes ΔG more negative, favouring the chelate.
Formula / Principle
ΔG = ΔH − TΔS
If ΔH(chelate) ≈ ΔH(monodentate), the sign of ΔG is controlled by the −TΔS term. Positive ΔS → negative contribution → more negative ΔG → more stable product.
Substitution (species counting)
Reactant side: [Cu(H₂O)₄]²⁺ (1) + en (2) = 3 independent species
Product side: [Cu(en)₂]²⁺ (1) + H₂O (4) = 5 independent species
Change: 3 → 5 species. Net increase = +2 free particles.
Calculation / Reasoning
More free particles in solution → greater translational entropy → ΔS > 0.
Since ΔH ≈ 0 (comparable bond strengths), ΔG ≈ −TΔS < 0.
The equilibrium lies strongly toward the chelate product.
Final answer
The chelate complex [Cu(en)₂]²⁺ is more stable because the replacement of 4 monodentate H₂O ligands by 2 bidentate en ligands increases the number of free species from 3 to 5, giving ΔS > 0 and thus a more negative ΔG. This entropy-driven stability advantage is the chelate effect.
Common trap
Assuming chelate stability comes from stronger individual bonds. It does not — the Cu–N bond in [Cu(en)₂]²⁺ is not dramatically stronger than the Cu–O bond in [Cu(H₂O)₄]²⁺. The driving force is entropy (more free particles), not enthalpy.
Similar NEET-style question
"Explain why [Ni(en)₃]²⁺ has a much larger formation constant than [Ni(NH₃)₆]²⁺, even though both involve Ni–N bonds." (Answer: same chelate effect reasoning — 4 → 7 species for en substitution vs. 7 → 7 for NH₃, so the en pathway gains entropy.)
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Ligand: ion or neutral molecule donating electron pair to metal (Lewis base). Monodentate (1 site, e.g. NH₃, Cl⁻). Bidentate (2 sites, e.g. en, ox²⁻). Polydentate (chelate, e.g. EDTA⁴⁻ has 6 sites). Coordination number = number of donor atoms attached.
-- NCERT Class 12 Chemistry, Ch. 5, p. 121More in Coordination Compounds: 2 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
1 question from NEET 2023. Answers verified against NTA official keys.
Which complex compound is most stable?
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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