Understanding Spin-Only Magnetic Moment

The spin-only magnetic moment (μ) is calculated using the formula:

$\mu =\sqrt{n(n+2)}$ Bohr Magnetons (B.M.), where $n$ is the number of unpaired electrons.

Step 1: Determine n from given μ

Given μ = 2.82 B.M., we solve for $n$:

${\mu }^2=n(n+2)$

$7.95\approx n(n+2)$

Testing $n=2$: $2\times 4=8$ (close to 7.95), so $n=2$ unpaired electrons.

Step 2: Analyze each complex for unpaired electrons

Ni has atomic number 28: [Ar] 3d⁸ 4s². We consider oxidation state and geometry.

Option 1: Ni(PPh₃)₄ - Tetrahedral, Ni(0): 3d¹⁰ (all paired, μ=0)

Option 2: Ni(CO)₄ - Tetrahedral, Ni(0): 3d¹⁰ (all paired, μ=0)

Option 3: [NiCl₄]^2– - Tetrahedral, Ni(II): 3d⁸. Tetrahedral field has small Δ, high spin: two unpaired electrons (μ≈2.82 B.M.)

Option 4: [Ni(CN)₄]^2– - Square planar, Ni(II): 3d⁸. Strong field, all paired (μ=0)

Step 3: Final Answer

Only [NiCl₄]^2– has two unpaired electrons, matching μ=2.82 B.M.

Related Topics & Formulae

Crystal Field Theory: Explains splitting of d-orbitals in ligands fields, affecting electron pairing.

Magnetic Moment Formula: $\mu =\sqrt{n(n+2)}$ B.M. for spin-only contribution.

Common Geometries: Tetrahedral (weak field, often high spin) vs. Square Planar (strong field, low spin).