Question

The correct increasing order of spin-only magnetic moment values of the complex ions \([\ce{MnBr4}]^{2-}\) (A), \([\ce{Cu(H2O)6]^{2+}\) (B), \([\ce{Ni(CN)4}]^{2-}\) (C) and \([\ce{Ni(H2O)6}]^{2+}\) (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: \[ \mu = \sqrt{n(n+2)} \, \text{BM} \] where \(n\) = number of unpaired electrons.
Step 1: Analyze each complex. - \([\ce{MnBr4}]^{2-}\): Mn(II), \(d^5\), weak field ligand (Br\(^-\)), high spin → 5 unpaired electrons. \[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \,\text{BM} \] - \([\ce{Cu(H2O)6}]^{2+}\): Cu(II), \(d^9\), one unpaired electron. \[ \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \,\text{BM} \] - \([\ce{Ni(CN)4}]^{2-}\): Ni(II), \(d^8\), strong field ligand (CN\(^-\)), low spin → no unpaired electrons. \[ \mu = 0 \,\text{BM} \] - \([\ce{Ni(H2O)6}]^{2+}\): Ni(II), \(d^8\), weak field ligand (H\(_2\)O), high spin → 2 unpaired electrons. \[ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \,\text{BM} \]
Step 2: Arrange in increasing order. \[ [\ce{Ni(CN)4}]^{2-} (0)<[\ce{Cu(H2O)6}]^{2+} (1.73)<[\ce{Ni(H2O)6}]^{2+} (2.83)<[\ce{MnBr4}]^{2-} (5.92) \] \[ \text{Order: C<B<D<A} \]