Question

The calculated spin-only magnetic moment values in BM for [FeCl4]- and [Fe(CN)6]3- are

• A. 5.9 BM,1.732 BM
• B. 4.89 BM,1.732 BM
• C. 3.87 BM,1.732 BM
• D. 1.732 BM,2.82 BM

Answer 1

Answer: (A)

5.9 BM,1.732 BM

Answer 2

The calculated spin-only magnetic moment values in Bohr magnetons ($\mu_B$) for [4]−$[\text{FeCl}_4^−]$ and [6]³−$[\text{Fe}(\text{CN})_6^{3-}]$ are:

[4]−$[\text{FeCl}_4^−]$: The oxidation state of iron ($\text{Fe}$) in this complex is +2. The electronic configuration of iron ($\text{Fe}^{2+}$) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$. As there are 6 unpaired electrons, the spin-only magnetic moment can be calculated as: $\sqrt{n(n+2)}\ \mu_B = \sqrt{6(6+2)}\ \mu_B = 4.90\ \mu_B$.

$[\text{Fe}(\text{CN})_6^{3−}]$: The oxidation state of iron ($\text{Fe}$) in this complex is +3. The electronic configuration of iron ($\text{Fe}^{3+}$) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5$. As there are 5 unpaired electrons, the spin-only magnetic moment can be calculated as: $\sqrt{n(n+2)}\ \mu_B = \sqrt{5(5+2)}\ \mu_B = 5.92\ \mu_B$.

The correct answer is option (A): $5.9\ \text{BM}, 1.732\ \text{BM}$.