Magnetic moment 2.83 BM is given by which of the following ions?
`"At. nos. Ti=22, Cr=24, Mn=25,Ni=28"`

Options

A. `Mn^(2+)`
B. `Cr^(3+)`
C. `Ni^(2+)`
D. `Mn^(4+) and O_(2)`

Explanation

To determine which ion corresponds to a magnetic moment of 2.83 Bohr magnetons (BM), we can use the formula for magnetic moment:

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

Where \( n \) is the number of unpaired electrons.

### Step-by-Step Solution:

1. **Set up the equation:**
Given that the magnetic moment is 2.83 BM, we can square both sides to eliminate the square root:
\[ (2.83)^2 = n(n + 2) \]
This simplifies to:
\[ 8.0089 \approx n(n + 2) \]

2. **Rearrange the equation:**
Rearranging gives us a quadratic equation:
\[ n^2 + 2n - 8.0089 = 0 \]

3. **Estimate \( n \):**
We can approximate \( n \) by testing integer values. We need \( n \) such that \( n(n + 2) \) is close to 8.0089.

- For \( n = 2 \):
\[ 2(2 + 2) = 2 \times 4 = 8 \]
- For \( n = 3 \):
\[ 3(3 + 2) = 3 \times 5 = 15 \quad \text{(too high)} \]

Therefore, \( n = 2 \) is a good candidate.

4. **Identify the ions:**
We need to check which of the given ions has 2 unpaired electrons.

- **Manganese (Mn²⁺):**
- Atomic number = 25
- Electron configuration: \( [Ar] 4s^2 3d^5 \)
- For Mn²⁺: \( 3d^5 \) (5 unpaired electrons)

- **Chromium (Cr³⁺):**
- Atomic number = 24
- Electron configuration: \( [Ar] 4s^1 3d^5 \)
- For Cr³⁺: \( 3d^3 \) (3 unpaired electrons)

- **Nickel (Ni²⁺):**
- Atomic number = 28
- Electron configuration: \( [Ar] 4s^2 3d^8 \)
- For Ni²⁺: \( 3d^8 \) (2 unpaired electrons)

5. **Conclusion:**
The ion that has 2 unpaired electrons is Nickel (Ni²⁺). Therefore, the answer is:

**Nickel ion (Ni²⁺) corresponds to a magnetic moment of 2.83 BM.**