The complex compound havimg maximum magnetic moment is

Options

A. `[CoF_(6)]^(3-)`
B. `[Cr(NH_(3))_(6)]^(3+)`
C. `[FeF_(6)]^(3-)`
D. `[Mn(CN)_(6)]^(4-)`

Answer

To determine the complex compound with the maximum magnetic moment, we need to analyze the number of unpaired electrons in each complex. The magnetic moment (μ) is calculated using the formula:

\[ \mu = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons. The more unpaired electrons there are, the higher the magnetic moment.

### Step-by-Step Solution:

1. **Identify the Complexes**: The complexes we need to analyze are:
- \( \text{CoF}_6^{3-} \)
- \( \text{Cr(NH}_3)_6^{3+} \)
- \( \text{FeF}_6^{3-} \)
- \( \text{Mn(CN)}_6^{4-} \)

2. **Calculate the Oxidation State**:
- For \( \text{CoF}_6^{3-} \):
- \( \text{F} = -1 \) (6 F gives -6)
- Let \( x \) be the oxidation state of Co: \( x - 6 = -3 \) → \( x = +3 \)
- For \( \text{Cr(NH}_3)_6^{3+} \):
- \( \text{NH}_3 \) is neutral, so \( x = +3 \) for Cr.
- For \( \text{FeF}_6^{3-} \):
- \( x - 6 = -3 \) → \( x = +3 \) for Fe.
- For \( \text{Mn(CN)}_6^{4-} \):
- \( \text{CN} = -1 \) (6 CN gives -6)
- Let \( x \) be the oxidation state of Mn: \( x - 6 = -4 \) → \( x = +2 \)

3. **Determine Electron Configuration**:
- \( \text{Co}^{3+} \): Argon configuration → \( [Ar] 3d^6 \)
- \( \text{Cr}^{3+} \): Argon configuration → \( [Ar] 3d^3 \)
- \( \text{Fe}^{3+} \): Argon configuration → \( [Ar] 3d^5 \)
- \( \text{Mn}^{2+} \): Argon configuration → \( [Ar] 3d^5 \)

4. **Assess Ligand Strength**:
- \( \text{F}^- \) is a weak field ligand, so it will not cause pairing.
- \( \text{NH}_3 \) is a moderate field ligand, which may cause some pairing.
- \( \text{CN}^- \) is a strong field ligand, which will cause pairing.

5. **Count Unpaired Electrons**:
- For \( \text{CoF}_6^{3-} \): \( 3d^6 \) → 4 unpaired electrons (due to weak field).
- For \( \text{Cr(NH}_3)_6^{3+} \): \( 3d^3 \) → 3 unpaired electrons.
- For \( \text{FeF}_6^{3-} \): \( 3d^5 \) → 5 unpaired electrons (due to weak field).
- For \( \text{Mn(CN)}_6^{4-} \): \( 3d^5 \) → 1 unpaired electron (due to strong field).

6. **Calculate Magnetic Moment**:
- For \( \text{CoF}_6^{3-} \): \( n = 4 \) → \( \mu = \sqrt{4(4 + 2)} = \sqrt{24} \)
- For \( \text{Cr(NH}_3)_6^{3+} \): \( n = 3 \) → \( \mu = \sqrt{3(3 + 2)} = \sqrt{15} \)
- For \( \text{FeF}_6^{3-} \): \( n = 5 \) → \( \mu = \sqrt{5(5 + 2)} = \sqrt{35} \)
- For \( \text{Mn(CN)}_6^{4-} \): \( n = 1 \) → \( \mu = \sqrt{1(1 + 2)} = \sqrt{3} \)

7. **Conclusion**: The complex with the maximum magnetic moment is \( \text{FeF}_6^{3-} \) with 5 unpaired electrons, resulting in the highest magnetic moment of \( \sqrt{35} \).