GATE 2025 CY – Question 40
The effective magnetic moment, $\mu_{eff}$ value for $[\mathrm{Cr(H_2O)_6}]^{3+}$ taking into account for spin-orbit coupling is closest to [Given: Atomic number of Cr = 24, spin-orbit coupling constant $\lambda=92\ \mathrm{cm^{-1}}$, and $\Delta_o=17400\ \mathrm{cm^{-1}}$]
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Correct answer: (A) $3.79\mu_B$
Explanation
For an octahedral $d^3$ ion such as $[\text{Cr(H}_2\text{O)}_6]^{3+}$ (ground term $^4A_{2g}$), the magnetic moment is slightly reduced below the spin-only value by spin-orbit coupling. The formula (for $A_{2g}$ ground terms) is
$$\mu_{eff}=\mu_{s.o.}\left(1-\frac{4\lambda}{\Delta_o}\right),\qquad\mu_{s.o.}=\sqrt{n(n+2)}=\sqrt{15}\ \mu_B\quad(n=3).$$
**Numbers.** $\lambda=92\ \text{cm}^{-1}$, $\Delta_o=17\,400\ \text{cm}^{-1}$:
$$\frac{4\lambda}{\Delta_o}=\frac{368}{17\,400}=0.02115$$
$$\mu_{eff}=3.873\times(1-0.02115)=3.873\times0.97885=\mathbf{3.79\ \mu_B}.$$