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 answer this question, we must identify the coordination complex exhibiting both a crystal field splitting energy ($\Delta_{0}$) of zero and a magnetic moment ($\mu$) of 5.96 Bohr Magnetons (B.M.).

Let's examine each option:

• The magnetic moment is determined by the formula: \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons, where \(n\) is the count of unpaired electrons.
• A magnetic moment of 5.96 B.M. implies the presence of 5 unpaired electrons, as \(\sqrt{5(5+2)} = \sqrt{35} \approx 5.92\), which closely approximates 5.96.

We will now evaluate each option based on its number of unpaired electrons:

1. \([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\): This complex features Fe in the +2 oxidation state (d⁶ configuration). Due to cyanide (CN^-) being a strong field ligand, this is a low-spin complex with zero unpaired electrons, thus not matching the required magnetic moment.
2. \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\): Here, Co is in the +3 oxidation state (d⁶ configuration). Ammonia (NH₃) is a neutral ligand, resulting in a low-spin complex with no unpaired electrons, and therefore cannot achieve the 5.96 B.M. magnetic moment.
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 computed magnetic moment is approximately 4.90 B.M., which deviates from 5.96 B.M.
4. \([\mathrm{Mn}(\mathrm{SCN})_{6}]^{4-}\): Mn is in the +2 oxidation state (d⁵ configuration). Each SCN^- ligand exhibits weak field strength, resulting in a high-spin complex with 5 unpaired electrons. The calculated magnetic moment is approximately 5.92 B.M., closely matching the expected 5.96 B.M.

Consequently, the complex that fulfills both the zero $\Delta_{0}$ condition (signifying a high-spin configuration) and a magnetic moment of 5.96 B.M. is:

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

Conclusion: The appropriate answer is \([\mathrm{Mn}(\mathrm{SCN})_{6}]^{4-}\), as it accurately reflects the anticipated magnetic properties.