Which of the following has the highest magnetic moment ?

Question

Accepted Answer

To determine which of the given complexes has the highest magnetic moment, we need to consider the electronic configuration and the number of unpaired electrons in each complex. The magnetic moment is directly related to the number of unpaired electrons, as it is calculated using the formula:

$$ \mu = \sqrt{n(n+2)} $$

where $ \mu $ is the magnetic moment and $ n $ is the number of unpaired electrons.

1. **Option 3: $[FeF_{6}]^{3-}$**

- **Electronic Configuration of Fe:** Iron (Fe) has an atomic number of 26, with an electronic configuration of $[Ar] 3d^6 4s^2$.
   - **Fe in $[FeF_{6}]^{3-}:** In this complex, iron is in the +3 oxidation state, so its configuration becomes \([Ar] 3d^5$.
   - **Ligand Field:** Fluoride (F⁻) is a weak field ligand, which does not cause pairing of electrons in the 3d orbitals.
   - **Unpaired Electrons:** The 3d^5 configuration results in 5 unpaired electrons.
   - **Magnetic Moment:** Using the formula, $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \mu_B$.

This high number of unpaired electrons results in the highest magnetic moment among the given options.

2. **Option 1: $[Mn(CN)_{6}]^{3-}$**

- **Electronic Configuration of Mn:** Manganese (Mn) has an atomic number of 25, with an electronic configuration of $[Ar] 3d^5 4s^2$.
   - **Mn in $[Mn(CN)_{6}]^{3-}:** In this complex, manganese is in the +3 oxidation state, so its configuration becomes \([Ar] 3d^4$.
   - **Ligand Field:** Cyanide (CN⁻) is a strong field ligand, which causes pairing of electrons in the 3d orbitals.
   - **Unpaired Electrons:** The 3d^4 configuration with strong field ligands results in 2 unpaired electrons.
   - **Magnetic Moment:** $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \mu_B$.

3. **Option 2: $[Fe(CN)_{6}]^{3-}$**

- **Electronic Configuration of Fe:** As previously mentioned, iron has an electronic configuration of $[Ar] 3d^6 4s^2$.
   - **Fe in $[Fe(CN)_{6}]^{3-}:** In this complex, iron is in the +3 oxidation state, so its configuration becomes \([Ar] 3d^5$.
   - **Ligand Field:** Cyanide (CN⁻) is a strong field ligand, causing pairing of electrons.
   - **Unpaired Electrons:** The 3d^5 configuration with strong field ligands results in 1 unpaired electron.
   - **Magnetic Moment:** $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \mu_B$.

4. **Option 4: $[CoF_{6}]^{3-}$**

- **Electronic Configuration of Co:** Cobalt (Co) has an atomic number of 27, with an electronic configuration of $[Ar] 3d^7 4s^2$.
   - **Co in $[CoF_{6}]^{3-}:** In this complex, cobalt is in the +3 oxidation state, so its configuration becomes \([Ar] 3d^6$.
   - **Ligand Field:** Fluoride (F⁻) is a weak field ligand, which does not cause pairing of electrons.
   - **Unpaired Electrons:** The 3d^6 configuration results in 4 unpaired electrons.
   - **Magnetic Moment:** $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \mu_B$.

In conclusion, $[FeF_{6}]^{3-}$ has the highest magnetic moment due to having the maximum number of unpaired electrons (5), which is greater than any of the other complexes listed. This supports Option 3 as the correct answer.