Determine Spin Only Magnetic Moments of Chromium and Copper Complexes

To determine the spin-only magnetic moments of the given coordination complexes, we calculate the number of unpaired electrons ($n$) present in the central metal ion of each complex using Crystal Field Theory (CFT).

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1. Complex 1: $[\text{Cr(NH}_3)_6]^{3+}$

• Oxidation State of Chromium ($\text{Cr}$): Let the oxidation state of $\text{Cr}$ be $x$. Since $\text{NH}_3$ is a neutral ligand: $x + 6(0) = +3 \implies x = +3$ Thus, the central metal ion is $\text{Cr}^{3+}$.

• Electronic Configuration:

  • Ground state configuration of $\text{Cr}$ ($Z = 24$): $[\text{Ar}] \, 3d^5 4s^1$
  • Configuration of $\text{Cr}^{3+}$: $[\text{Ar}] \, 3d^3$

• Crystal Field Splitting in Octahedral Field: For a $d^3$ metal ion in an octahedral complex, the three electrons enter the lower energy $t_{2g}$ orbitals according to Hund's rule: $\text{Electronic configuration} = t_{2g}^3 \, e_g^0$

  • Number of unpaired electrons, $n = 3$.

• Spin-Only Magnetic Moment ($\mu_{\text{s.o.}}$): $\mu_{\text{s.o.}} = \sqrt{n(n + 2)} \text{ BM}$ $\mu_{\text{s.o.}} = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \text{ BM}$

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2. Complex 2: $[\text{CuF}_6]^{3-}$

• Oxidation State of Copper ($\text{Cu}$): Let the oxidation state of $\text{Cu}$ be $y$. Since $\text{F}^-$ has a charge of $-1$: $y + 6(-1) = -3 \implies y = +3$ Thus, the central metal ion is $\text{Cu}^{3+}$.

• Electronic Configuration:

  • Ground state configuration of $\text{Cu}$ ($Z = 29$): $[\text{Ar}] \, 3d^{10} 4s^1$
  • Configuration of $\text{Cu}^{3+}$: $[\text{Ar}] \, 3d^8$

• Crystal Field Splitting in Octahedral Field: For a $d^8$ metal ion in an octahedral complex, the distribution of 8 electrons between $t_{2g}$ and $e_g$ orbitals is fixed regardless of ligand strength: $\text{Electronic configuration} = t_{2g}^6 \, e_g^2$

  • The $t_{2g}$ set is completely filled (6 paired electrons).
  • The $e_g$ set contains 2 electrons which occupy the degenerate $d_{x^2-y^2}$ and $d_{z^2}$ orbitals singly.
  • Number of unpaired electrons, $n = 2$.

• Spin-Only Magnetic Moment ($\mu_{\text{s.o.}}$): $\mu_{\text{s.o.}} = \sqrt{n(n + 2)} \text{ BM}$ $\mu_{\text{s.o.}} = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.84 \text{ BM}$

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Conclusion:

The calculated spin-only magnetic moments of $[\text{Cr(NH}_3)_6]^{3+}$ and $[\text{CuF}_6]^{3-}$ are $3.87 \text{ BM}$ and $2.84 \text{ BM}$, respectively.

This matches Option (A).