JEE Main · 2024easyCORD-094

The coordination environment of Ca²⁺ ion in its complex with EDTA⁴⁻ is:

Coordination Compounds · Class 12 · JEE Main Previous Year Question

Question

The coordination environment of Ca²⁺ ion in its complex with EDTA⁴⁻ is:

Options
  1. a

    tetrahedral

  2. b

    trigonal prismatic

  3. c

    octahedral

  4. d

    square planar

Correct Answerc

octahedral

Detailed Solution

🧠 Ca²⁺ + EDTA = 6-Coordinate

EDTA (ethylenediaminetetraacetate) is hexadentate: 2 amine N + 4 carboxylate O. When EDTA wraps a single metal ion, the metal is necessarily 6-coordinate octahedral.

Ca²⁺ is large enough to accept all 6 donors → octahedral [Ca(EDTA)]2\mathrm{[Ca(EDTA)]^{2-}}.

🗺️ Eliminate the Wrong Geometries

  • (1) Tetrahedral — only 4 sites, but EDTA needs 6. ✗
  • (2) Trigonal prismatic — 6-coordinate but only seen for very specific d⁰/d¹⁰ ions (W, Mo); not Ca²⁺. ✗
  • (3) Octahedral — 6 sites, fits EDTA perfectly. ✓
  • (4) Square-planar — only 4 sites. ✗

The "EDTA → Octahedral" Default

For any small-to-medium-sized metal, EDTA gives an octahedral chelate. Only for very large metals (lanthanides) does the geometry expand beyond 6.

⚠️ Don't Confuse Ca with Pt or Pd

Square-planar geometries belong to d⁸ ions (Pt²⁺, Pd²⁺, Ni²⁺ + strong field). Ca²⁺ is d0\mathrm{d^0} (a main-group ion in the s-block) — never square-planar.

Answer: (3) octahedral\boxed{\text{Answer: (3) octahedral}}

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