Question

Which one of the following complexes will have $\Delta_{0}=0$ and $\mu=5.96$ B.M.?

• A. $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}$
• B. $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$
• C. $\left[\mathrm{FeF}_{6}\right]^{4}$
• D. $\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4}$

Hint:

The magnetic moment and crystal field stabilization energy depend on the ligand field strength.

Answer 1

The Correct Option is D

To solve this question, we need to determine which coordination complex has a crystal field splitting energy ($\Delta_{0}$) of zero and a magnetic moment ($\mu$) of 5.96 Bohr Magnetons (B.M.).

Let's analyze each option:

• The magnetic moment can be calculated using the formula: \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons, where \(n\) is the number of unpaired electrons.
• A magnetic moment of 5.96 B.M. corresponds to a system with 5 unpaired electrons because \(\sqrt{5(5+2)} = \sqrt{35} \approx 5.92\), which is very close to 5.96.

Let's evaluate each option in terms of the number of unpaired electrons:

1. \([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\): In this complex, Fe is in the +2 oxidation state (d⁶ configuration). With cyanide (CN^-) being a strong field ligand, this complex is low-spin with no unpaired electrons. Thus, it cannot have a magnetic moment of 5.96 B.M.
2. \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\): Here, Co is in the +3 oxidation state (d⁶ configuration). Ammonia (NH₃) is a neutral field ligand leading to a low-spin with no unpaired electrons. Therefore, it cannot have a magnetic moment of 5.96 B.M.
3. \([\mathrm{FeF}_{6}]^{4-}\): In this complex, Fe is in the +2 oxidation state (d⁶ configuration). Fluoride (F^-) is a weak field ligand, leading to a high-spin complex with 4 unpaired electrons. The calculated magnetic moment is approximately 4.90 B.M., which does not match 5.96 B.M.
4. \([\mathrm{Mn}(\mathrm{SCN})_{6}]^{4-}\): Mn is in the +2 oxidation state (d⁵ configuration), with each SCN^- ligand contributing weak field strength, leading to a high-spin complex with 5 unpaired electrons. The calculated magnetic moment is about 5.92 B.M., which is very close to the expected 5.96 B.M.

Thus, the complex that satisfies both conditions of zero $\Delta_{0}$ (indicating a high-spin configuration) and a magnetic moment of 5.96 B.M. is:

$[\mathrm{Mn}(\mathrm{SCN})_{6}]^{4-}$

Conclusion: The correct answer is \([\mathrm{Mn}(\mathrm{SCN})_{6}]^{4-}\), as it correctly demonstrates the expected magnetic properties.

Answer 2

1. $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}$: - $\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}$ - $\mathrm{CN}^{-}$ is a strong field ligand. - $\mu = 0$ 
2. $\left[\mathrm{CO}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$: - $\mathrm{Co}^{3+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}$ - $\mathrm{NH}_{3}$ is a strong field ligand. - $\mu = 0$ 
3. $\left[\mathrm{FeF}_{6}\right]^{4}$: - $\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0}$ - $\mathrm{F}^{-}$ is a weak field ligand. - $\mu = 0$ 
4. $\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4}$: - $\mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0}$ - $\mathrm{SCN}^{-}$ is a weak field ligand. - $\mu = \sqrt{35} \mathrm{~BM} = 5.96 \mathrm{~BM}$ - $\Delta_{0} = 0$ 
Therefore, the correct answer is (4) $\left[\mathrm{Mn}(\mathrm{SCN})_{6}\right]^{4}$.