JEE MAIN - Chemistry (2024 - 9th April Evening Shift - No. 19)

The coordination environment of $$\mathrm{Ca}^{2+}$$ ion in its complex with $$\mathrm{EDTA}^{4-}$$ is :
trigonal prismatic
octahedral
square planar
tetrahedral

Explanation

The coordination environment of the $$\mathrm{Ca}^{2+}$$ ion when it forms a complex with $$\mathrm{EDTA}^{4-}$$ (ethylenediaminetetraacetic acid) is octahedral. Ethylenediaminetetraacetic acid (EDTA) is a hexadentate ligand, which means it has six donor atoms that can bind to a central metal ion. In the case of EDTA, these donor atoms are four oxygen atoms from its four carboxyl groups and two nitrogen atoms from its two amine groups.

When $$\mathrm{Ca}^{2+}$$ forms a complex with $$\mathrm{EDTA}^{4-}$$, all six donor atoms from EDTA coordinate with the calcium ion. As a result, the coordination number of the calcium ion is 6, leading to an octahedral geometry. This is because the octahedral geometry is the most common and energetically favorable arrangement for a coordination number of 6. In such a geometry, the six ligands are placed at equal distances from the central ion and at 90° angles relative to adjacent ligands, maximizing the distance between all ligands to minimize repulsion.

The correct answer is Option B, octahedral.

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