Q.22 Match the coordination complexes given in Column I with the most appropriate properties in Column II.
(Given: Atomic numbers of Mn: 25; Co: 27; Ni: 28)

(A) E–1, F–2, G–3, H–4
(B) E–2, F–1, G–4, H–3
(C) E–4, F–2, G–1, H–3
(D) E–1, F–4, G–3, H–2

Introduction

This article explains how to match the coordination complexes [Mn(H₂O)₆]²⁺, [CoF₆]³⁻, [NiCl₄]²⁻ and [Ni(CN)₄]²⁻ with the most appropriate properties using crystal field theory, hybridisation and spin-only magnetic moments.[web:23][web:26][web:9][web:18] The discussion is designed for competitive exams, especially where understanding of high-spin and low-spin complexes is essential.

Data from the question

Column I: Coordination complexes

• E. [Mn(H₂O)₆]²⁺
• F. [CoF₆]³⁻
• G. [NiCl₄]²⁻
• H. [Ni(CN)₄]²⁻

Column II: Properties

1. 5.92 Bohr Magneton (BM)
2. CFSE = 0.4 Δ₀
3. Metal ion hybridisation is sp³
4. Diamagnetic

Detailed analysis of each complex

E. [Mn(H₂O)₆]²⁺

Mn has atomic number 25, so Mn²⁺ has configuration 3d⁵ and with weak field H₂O ligands it forms a high-spin octahedral complex with five unpaired electrons.

The spin‑only magnetic moment is given by the formula

μ = √n(n + 2) with n = 5 gives μ = √35 ≈ 5.92 BM.

Therefore, [Mn(H₂O)₆]²⁺ corresponds to property 1 (5.92 BM) → E–1.

F. [CoF₆]³⁻

• In [CoF₆]³⁻ the oxidation state of Co is +3; Co³⁺ is 3d⁶.
• F⁻ is a weak field ligand, so the complex is octahedral, high‑spin, with configuration t₂g⁴eg², giving four unpaired electrons.
• For d⁶ high‑spin octahedral: CFSE = −0.4 × 4Δ_o + 0.6 × 2Δ_o = −1.6Δ_o + 1.2Δ_o = −0.4Δ_o, so magnitude 0.4 Δ_o as given in Column II.

Thus [CoF₆]³⁻ corresponds to property 2 (CFSE = 0.4 Δ_o) → F–2.

G. [NiCl₄]²⁻

Nickel has atomic number 28, so Ni²⁺ is 3d⁸ in both [NiCl₄]²⁻ and [Ni(CN)₄]²⁻. Chloride is a weak field ligand, so no pairing occurs and the complex adopts tetrahedral geometry with sp³ hybridisation using the 4s and 4p orbitals, containing two unpaired electrons and therefore paramagnetic.

Hence [NiCl₄]²⁻ matches property 3 (metal ion hybridisation is sp³) → G–3.

H. [Ni(CN)₄]²⁻

In [Ni(CN)₄]²⁻ the metal is again Ni²⁺ (3d⁸), but CN⁻ is a strong field ligand that causes pairing of the 3d electrons. This leads to dsp² hybridisation and a square‑planar low‑spin complex in which all d‑electrons are paired, making the complex diamagnetic.

Therefore [Ni(CN)₄]²⁻ corresponds to property 4 (diamagnetic) → H–4.

Checking the options (A–D)

Since only option (A) reproduces the correct set E–1, F–2, G–3, H–4, the answer is (A).