Among the following species, the metal center that has the highest number of unpaired electrons is _______.(A) VCl₄
(B) Ni(CO)₄
(C) [AuCl₄]⁻
(D) [CdBr₄]²⁻

VCl₄ has the metal center with the highest number of unpaired electrons among the given options. This coordination chemistry question tests understanding of oxidation states, d-electron configurations, ligand field strengths, and geometries in transition metal complexes. The correct answer is (A) VCl₄.​

Option Analysis

(A) VCl₄

Vanadium (V, atomic number 23) has configuration [Ar] 4s² 3d³. In VCl₄, oxidation state is +4 (4 Cl⁻ contribute -4 charge). V⁴⁺ is [Ar] 3d¹ with one unpaired electron. Cl⁻ is a weak field ligand in this tetrahedral (sp³ hybridized) complex, so high-spin configuration yields 1 unpaired electron.​

(B) Ni(CO)₄

Nickel (Ni, Z=28) is [Ar] 4s² 3d⁸. Ni(CO)₄ is neutral, so Ni oxidation state is 0. CO is a strong field ligand causing pairing to [Ar] 3d¹⁰. Tetrahedral (sp³) geometry shows 0 unpaired electrons, making it diamagnetic.​

(C) [AuCl₄]⁻

Gold (Au, Z=79) is [Xe] 4f¹⁴ 5d¹⁰ 6s¹. Oxidation state is +3 (4 Cl⁻ = -4, overall -1 charge). Au³⁺ is 5d⁸. Square planar (dsp² hybridized) due to heavy metal effects and relativistic stabilization, all electrons paired (0 unpaired).​

(D) [CdBr₄]²⁻

Cadmium (Cd, Z=48) is [Kr] 5s² 4d¹⁰. Oxidation state +2 (4 Br⁻ = -4, overall -2). Cd²⁺ is d¹⁰. Tetrahedral (sp³) with fully filled d orbitals, so 0 unpaired electrons.​

VCl₄ features the metal center with the highest number of unpaired electrons (1) due to its V⁴⁺ (3d¹) configuration and tetrahedral geometry with weak field Cl⁻ ligands. Other options show 0 unpaired electrons: Ni(CO)₄ (d¹⁰, strong field CO), [AuCl₄]⁻ (d⁸ square planar), and [CdBr₄]²⁻ (d¹⁰). This comparison highlights crystal field theory applications in coordination compounds. For CSIR NET aspirants, master these by calculating oxidation states first, then assessing ligand strength and geometry impacts on pairing.​