Which of the following outer orbital complex has the highest magnetic moment ?

Options

A. `[Mn(NH_3)_6]Cl_3`
B. `[Cr(NH_3)_6]Cl_3`
C. `[Ni(CO)_4]`
D. `[Co(CN)_6]^(4-)`

Answer

To determine which outer orbital complex has the highest magnetic moment, we need to analyze the oxidation states and the number of unpaired electrons for each complex. The magnetic moment can be calculated using the formula:

\[ \text{Magnetic Moment} = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons. Let's analyze the complexes one by one.

### Step 1: Analyze the first complex (MnCl3)
1. **Determine the oxidation state of Mn**:
- Chlorine (Cl) has an oxidation state of -1.
- For three Cl atoms, the total contribution is -3.
- To make the compound neutral, Mn must be +3.
- Therefore, the oxidation state of Mn in MnCl3 is +3.

2. **Electronic configuration of Mn**:
- Atomic number of Mn = 25.
- Electronic configuration: \( [Ar] 4s^2 3d^5 \).
- In the +3 oxidation state, Mn loses 2 electrons from 4s and 1 electron from 3d, resulting in \( 3d^4 \).

3. **Count unpaired electrons**:
- The \( 3d^4 \) configuration has 4 unpaired electrons.

### Step 2: Analyze the second complex (Cr(NH3)6Cl3)
1. **Determine the oxidation state of Cr**:
- NH3 is a neutral ligand, and Cl is -1 (3 Cl = -3).
- To make the compound neutral, Cr must be +3.
- Therefore, the oxidation state of Cr is +3.

2. **Electronic configuration of Cr**:
- Atomic number of Cr = 24.
- Electronic configuration: \( [Ar] 4s^2 3d^4 \).
- In the +3 oxidation state, Cr loses 2 electrons from 4s and 1 from 3d, resulting in \( 3d^4 \).

3. **Count unpaired electrons**:
- The \( 3d^4 \) configuration has 4 unpaired electrons.

### Step 3: Analyze the third complex (Ni(CO)4)
1. **Determine the oxidation state of Ni**:
- CO is a neutral ligand.
- Therefore, the oxidation state of Ni is 0.

2. **Electronic configuration of Ni**:
- Atomic number of Ni = 28.
- Electronic configuration: \( [Ar] 4s^2 3d^8 \).
- In the 0 oxidation state, all electrons remain, so the configuration is \( 4s^2 3d^8 \).

3. **Count unpaired electrons**:
- In \( 3d^8 \), all electrons are paired due to the strong field ligand CO.
- Therefore, there are 0 unpaired electrons.

### Step 4: Analyze the fourth complex (Co(CN)6)
1. **Determine the oxidation state of Co**:
- CN is -1 (6 CN = -6).
- To make the compound neutral, Co must be +2.
- Therefore, the oxidation state of Co is +2.

2. **Electronic configuration of Co**:
- Atomic number of Co = 27.
- Electronic configuration: \( [Ar] 4s^2 3d^7 \).
- In the +2 oxidation state, Co loses 2 electrons from 4s, resulting in \( 3d^7 \).

3. **Count unpaired electrons**:
- In \( 3d^7 \), there are 3 unpaired electrons.

### Summary of Unpaired Electrons
- **MnCl3**: 4 unpaired electrons
- **Cr(NH3)6Cl3**: 4 unpaired electrons
- **Ni(CO)4**: 0 unpaired electrons
- **Co(CN)6**: 3 unpaired electrons

### Conclusion
Both MnCl3 and Cr(NH3)6Cl3 have the highest number of unpaired electrons (4 each). Therefore, they have the highest magnetic moment among the given complexes.

### Final Answer
**The outer orbital complex with the highest magnetic moment is MnCl3 or Cr(NH3)6Cl3.**