Question

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<C<D
• B. B<D<C
• C. C<B<A
• D. C<B<D<A

Hint:

For magnetic moment problems: 1. Determine oxidation state and \(d\)-electron count. 2. Identify ligand strength (weak field → high spin, strong field → low spin). 3. Count unpaired electrons and apply \(\mu = \sqrt{n(n+2)}\).

Answer 1

The Correct Option is D

Concept:
Spin-only magnetic moment is given by:
μ = √[ n(n + 2) ] BM
where n is the number of unpaired electrons.

Step 1: Analyze each complex.

[MnBr4]2− (A):
Mn is in +2 oxidation state with d5 configuration.
Br− is a weak field ligand, so the complex is high spin.
Number of unpaired electrons = 5.
Magnetic moment = √35 ≈ 5.92 BM.

[Cu(H2O)6]2+ (B):
Cu is in +2 oxidation state with d9 configuration.
Number of unpaired electrons = 1.
Magnetic moment = √3 ≈ 1.73 BM.

[Ni(CN)4]2− (C):
Ni is in +2 oxidation state with d8 configuration.
CN− is a strong field ligand, so the complex is low spin.
Number of unpaired electrons = 0.
Magnetic moment = 0 BM.

[Ni(H2O)6]2+ (D):
Ni is in +2 oxidation state with d8 configuration.
H2O is a weak field ligand, so the complex is high spin.
Number of unpaired electrons = 2.
Magnetic moment = √8 ≈ 2.83 BM.

Step 2: Arrange in increasing order.

[Ni(CN)4]2− < [Cu(H2O)6]2+ < [Ni(H2O)6]2+ < [MnBr4]2−

Final Order:
C < B < D < A