The value of the spin only magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is

Options

A. `Fe^(2+)`
B. `Ni^(2+)`
C. `Mn^(2+)`
D. `Co^(3+)`

Explanation

To determine the ion based on the given spin-only magnetic moment of 2.83 Bohr magneton, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**: The spin-only magnetic moment (μ) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Set Up the Equation**: We know the magnetic moment is 2.83 Bohr magneton, so we can set up the equation:
\[ 2.83 = \sqrt{n(n + 2)} \]

3. **Square Both Sides**: To eliminate the square root, we square both sides:
\[ (2.83)^2 = n(n + 2) \]
Calculating \( (2.83)^2 \):
\[ 2.83^2 = 8.0089 \]
So we have:
\[ 8.0089 = n(n + 2) \]

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

5. **Solve the Quadratic Equation**: We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -8.0089 \):
\[ n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8.0089)}}{2 \cdot 1} \]
\[ n = \frac{-2 \pm \sqrt{4 + 32.0356}}{2} \]
\[ n = \frac{-2 \pm \sqrt{36.0356}}{2} \]
\[ n = \frac{-2 \pm 6}{2} \]

This gives us two possible solutions for \( n \):
\[ n = \frac{4}{2} = 2 \quad \text{(valid, as n must be non-negative)} \]
\[ n = \frac{-8}{2} = -4 \quad \text{(not valid)} \]

6. **Identify the Ion**: Now that we have \( n = 2 \), we need to identify the ion with 2 unpaired electrons. We can look at the electronic configurations of common transition metal ions:
- Fe²⁺: [Ar] 4s⁰ 3d⁶ (4 unpaired electrons)
- Ni²⁺: [Ar] 4s⁰ 3d⁸ (2 unpaired electrons)
- Mn²⁺: [Ar] 4s⁰ 3d⁵ (5 unpaired electrons)
- Co³⁺: [Ar] 4s⁰ 3d⁶ (4 unpaired electrons)

The only ion with 2 unpaired electrons is Ni²⁺.

### Conclusion:
The ion is **Nickel (Ni²⁺)**.