The correct increasing order of spin-only magnetic moment values of the complex ions

Q. The correct increasing order of spin-only magnetic moment values of the complex ions

[MnBr₄]²⁻ (A), [Cu(H₂O)₆]²⁺ (B), [Ni(CN)₄]²⁻ (C) and [Ni(H₂O)₆]²⁺ (D) is :

A. A = B < D < C

B. C = D < B < A

C. C < B < D < A

D. A = B < C < D

Correct Answer: C < B < D < A

Solution (Step-by-step)

Spin-only magnetic moment depends only on the number of unpaired electrons.

Formula used:

μ = √[n(n + 2)] Bohr Magneton where n = number of unpaired electrons

(A) [MnBr₄]²⁻

Mn in +2 oxidation state → Mn²⁺ Electronic configuration of Mn²⁺ = 3d⁵

Br⁻ is a weak field ligand → high spin complex Number of unpaired electrons = 5

(B) [Cu(H₂O)₆]²⁺

Cu in +2 oxidation state → Cu²⁺ Electronic configuration of Cu²⁺ = 3d⁹

Number of unpaired electrons = 1

(C) [Ni(CN)₄]²⁻

Ni in +2 oxidation state → Ni²⁺ Electronic configuration of Ni²⁺ = 3d⁸

CN⁻ is a strong field ligand → pairing occurs Square planar complex → all electrons paired

Number of unpaired electrons = 0

(D) [Ni(H₂O)₆]²⁺

Ni²⁺ = 3d⁸

H₂O is a weak field ligand → octahedral high spin Number of unpaired electrons = 2

Now compare number of unpaired electrons:

C : 0 B : 1 D : 2 A : 5

Hence, increasing order of spin-only magnetic moment is:

C < B < D < A