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 given complexes and find the number of unpaired electrons in each. The magnetic moment is maximized when the number of unpaired electrons is maximized. Let's go through the steps systematically.

### Step 1: Analyze the first complex - \( \text{CoF}_6^{3-} \)

1. **Determine the oxidation state of cobalt (Co)**:
- The overall charge is -3.
- Each fluoride ion (F) has a charge of -1, and there are 6 fluoride ions.
- Therefore, \( x + 6(-1) = -3 \) leads to \( x - 6 = -3 \) or \( x = +3 \).
- Thus, Co is in the +3 oxidation state.

2. **Determine the electron configuration**:
- The electron configuration of Co is \( [Ar] 4s^2 3d^7 \).
- For Co in +3 state, it loses 2 electrons from 4s and 1 from 3d: \( [Ar] 3d^6 \).

3. **Count the unpaired electrons**:
- The configuration \( 3d^6 \) can be represented as: ↑↓ ↑↓ ↑ ↑ ↑ (4 unpaired electrons).
- Therefore, the number of unpaired electrons = 4.

### Step 2: Analyze the second complex - \( \text{Cr}^{3+} \)

1. **Determine the oxidation state of chromium (Cr)**:
- The overall charge is +3.
- Since the ligand is neutral, Cr is in the +3 oxidation state.

2. **Determine the electron configuration**:
- The electron configuration of Cr is \( [Ar] 4s^2 3d^5 \).
- For Cr in +3 state, it loses 2 electrons from 4s and 1 from 3d: \( [Ar] 3d^3 \).

3. **Count the unpaired electrons**:
- The configuration \( 3d^3 \) can be represented as: ↑ ↑ ↑ (3 unpaired electrons).
- Therefore, the number of unpaired electrons = 3.

### Step 3: Analyze the third complex - \( \text{FeF}_6^{3-} \)

1. **Determine the oxidation state of iron (Fe)**:
- The overall charge is +3.
- Each fluoride ion has a charge of -1, and there are 6 fluoride ions.
- Therefore, \( x + 6(-1) = -3 \) leads to \( x - 6 = -3 \) or \( x = +3 \).
- Thus, Fe is in the +3 oxidation state.

2. **Determine the electron configuration**:
- The electron configuration of Fe is \( [Ar] 4s^2 3d^6 \).
- For Fe in +3 state, it loses 2 electrons from 4s and 1 from 3d: \( [Ar] 3d^5 \).

3. **Count the unpaired electrons**:
- The configuration \( 3d^5 \) can be represented as: ↑ ↑ ↑ ↑ ↑ (5 unpaired electrons).
- Therefore, the number of unpaired electrons = 5.

### Step 4: Analyze the fourth complex - \( \text{Mn(CN)}_6^{4-} \)

1. **Determine the oxidation state of manganese (Mn)**:
- The overall charge is +2.
- Each cyanide ion (CN) has a charge of -1, and there are 6 cyanide ions.
- Therefore, \( x + 6(-1) = -4 \) leads to \( x - 6 = -4 \) or \( x = +2 \).
- Thus, Mn is in the +2 oxidation state.

2. **Determine the electron configuration**:
- The electron configuration of Mn is \( [Ar] 4s^2 3d^5 \).
- For Mn in +2 state, it loses 2 electrons from 4s: \( [Ar] 3d^5 \).

3. **Count the unpaired electrons**:
- The configuration \( 3d^5 \) can be represented as: ↑ ↑ ↑ ↑ ↑ (5 unpaired electrons).
- However, since CN is a strong field ligand, it causes pairing: ↑↓ ↑↓ ↑ (1 unpaired electron).
- Therefore, the number of unpaired electrons = 1.

### Step 5: Conclusion

Now we summarize the number of unpaired electrons for each complex:
- \( \text{CoF}_6^{3-} \): 4 unpaired electrons
- \( \text{Cr}^{3+} \): 3 unpaired electrons
- \( \text{FeF}_6^{3-} \): 5 unpaired electrons
- \( \text{Mn(CN)}_6^{4-} \): 1 unpaired electron

The complex with the maximum magnetic moment is \( \text{FeF}_6^{3-} \) with 5 unpaired electrons.

### Final Answer

The complex compound having the maximum magnetic moment is \( \text{FeF}_6^{3-} \).

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