Question

The effective magnetic moment, μ_eff, value for [Cr(H₂O)₆]³⁺ taking into account spin–orbit coupling is closest to:

[Given: Atomic number of Cr = 24, spin–orbit coupling constant λ = 92 cm⁻¹, and Δ_o = 17400 cm⁻¹]

• A. 3.79 $\mu_B$
• B. 3.87 $\mu_B$
• C. 4.05 $\mu_B$
• D. 3.60 $\mu_B$

Hint:

To adjust for spin-orbit coupling in magnetic moment calculations, use the correction factor involving $\lambda$ and $\Delta_o$. This is especially important for transition metal ions with partially filled d-orbitals.

Answer 1

The Correct Option is A

For the complex [Cr(H₂O)₆]³⁺:

• Cr³⁺ has electronic configuration: [Ar] 3d³
• This is a d³ system in an octahedral field (high spin)
• Number of unpaired electrons = 3

The spin-only magnetic moment is given by:

\[ \mu_{\text{spin only}} = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \mu_B \]

To account for spin–orbit coupling, we use the correction:

\[ \mu_{\text{eff}} = \mu_{\text{spin only}} \left(1 - \frac{\alpha \lambda}{10 \Delta_o} \right) \]

For a d³ ion in an octahedral field, α = 4 (empirical constant)

Substitute the values:

\[ \mu_{\text{eff}} = 3.87 \left(1 - \frac{4 \times 92}{10 \times 17400} \right) = 3.87 \left(1 - \frac{368}{174000} \right) = 3.87 \left(1 - 0.0021149 \right) = 3.87 \times 0.997885 \approx 3.79\ \mu_B \]

\[ \boxed{\mu_{\text{eff}} \approx 3.79\ \mu_B} \]