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Question

The species with the lowest Valence Electron (VE) count is :

This question was previously asked in
MH SET 2023 Paper 2 Computer Science and Applications Question Paper (26-Mar-2023)
The correct answer is

[AuCl4]–

Using the ionic model, the valence electron count = d electrons of the metal ion + 2 electrons from each donor atom (each neutral 2-electron ligand, each halide and each end of a chelate).

[V(CO)6]–: V(–I) is d6; six CO give 12 electrons, so VE = 6 + 12 = 18.

[Cu(MA3)6]2+: Cu(II) is d9; six neutral monodentate donors give 12 electrons, so VE = 9 + 12 = 21.

[AuCl4]–: Au(III) is d8; four chloride donors give 8 electrons, so VE = 8 + 8 = 16, the usual count for a square-planar d8 complex.

[Zn(en)3]2+: Zn(II) is d10; three en ligands give six N donors and 12 electrons, so VE = 10 + 12 = 22.

Check: by the neutral (covalent) method for the gold complex, Au contributes 11, four Cl radicals 4 and the negative charge 1, giving 11 + 4 + 1 = 16, the same as the ionic count. So the order is 16 < 18 < 21 < 22.

Hence, [AuCl4]–.

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