In the `Cr^(+2), Mn^(+3), Fe^(+2)` and `CO^(+3)` ions, number of unpaired electrons and magnetic moment wiII be

Options

A. 3, 3.87
B. 4, 4.90
C. 3, 2.83
D. 1, 1.73

Explanation

To determine the number of unpaired electrons and the magnetic moment for the ions Cr^(+2), Mn^(+3), Fe^(+2), and Co^(+3), we will follow these steps:

### Step 1: Determine the electronic configuration of each ion.

1. **Chromium (Cr)**
- Atomic number = 24
- Ground state electronic configuration: \( \text{Ar} \, 3d^5 \, 4s^1 \)
- For \( \text{Cr}^{2+} \): Remove 2 electrons (1 from 4s and 1 from 3d)
- Configuration: \( \text{Ar} \, 3d^4 \, 4s^0 \)

2. **Manganese (Mn)**
- Atomic number = 25
- Ground state electronic configuration: \( \text{Ar} \, 3d^5 \, 4s^2 \)
- For \( \text{Mn}^{3+} \): Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: \( \text{Ar} \, 3d^4 \, 4s^0 \)

3. **Iron (Fe)**
- Atomic number = 26
- Ground state electronic configuration: \( \text{Ar} \, 3d^6 \, 4s^2 \)
- For \( \text{Fe}^{2+} \): Remove 2 electrons (2 from 4s)
- Configuration: \( \text{Ar} \, 3d^6 \, 4s^0 \)

4. **Cobalt (Co)**
- Atomic number = 27
- Ground state electronic configuration: \( \text{Ar} \, 3d^7 \, 4s^2 \)
- For \( \text{Co}^{3+} \): Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: \( \text{Ar} \, 3d^6 \, 4s^0 \)

### Step 2: Count the number of unpaired electrons for each ion.

1. **Cr^(+2)**:
- Configuration: \( 3d^4 \)
- Unpaired electrons: 4 (1 in each of the first four 3d orbitals)

2. **Mn^(+3)**:
- Configuration: \( 3d^4 \)
- Unpaired electrons: 4 (same as Cr^(+2))

3. **Fe^(+2)**:
- Configuration: \( 3d^6 \)
- Unpaired electrons: 4 (3 in separate orbitals and 1 paired)

4. **Co^(+3)**:
- Configuration: \( 3d^6 \)
- Unpaired electrons: 4 (same as Fe^(+2))

### Step 3: Calculate the magnetic moment using the formula.

The formula for magnetic moment (\( \mu \)) is:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

1. **For all ions (Cr^(+2), Mn^(+3), Fe^(+2), Co^(+3))**:
- \( n = 4 \)
- Magnetic moment:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \, \text{Bohr magneton} \]

### Final Results:
- **Cr^(+2)**: 4 unpaired electrons, magnetic moment = 4.90 Bohr magneton
- **Mn^(+3)**: 4 unpaired electrons, magnetic moment = 4.90 Bohr magneton
- **Fe^(+2)**: 4 unpaired electrons, magnetic moment = 4.90 Bohr magneton
- **Co^(+3)**: 4 unpaired electrons, magnetic moment = 4.90 Bohr magneton