Magnetic moments `2.84 BM` is given by :
(At. nos. ni = 28, Ti = 22, Cr = 24, Co = 27).

Options

A. `Ni^(2+)`
B. `Ti^(3+)`
C. `Cr^(3+)`
D. `Co^(2+)`

Explanation

To solve the question regarding which transition element has a magnetic moment of 2.84 Bohr Magneton (BM), we will follow these steps:

### Step 1: Understanding the Formula for Magnetic Moment
The magnetic moment (μ) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Setting Up the Equation
Given that the magnetic moment is 2.84 BM, we can set up the equation:
\[ 2.84 = \sqrt{n(n + 2)} \]

### Step 3: Squaring Both Sides
To eliminate the square root, we square both sides:
\[ (2.84)^2 = n(n + 2) \]
Calculating \( (2.84)^2 \):
\[ 8.0656 = n(n + 2) \]

### Step 4: Rearranging the Equation
Rearranging gives us a quadratic equation:
\[ n^2 + 2n - 8.0656 = 0 \]

### Step 5: Solving the Quadratic Equation
We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -8.0656 \):
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-8.0656)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ n = \frac{-2 \pm \sqrt{4 + 32.2624}}{2} \]
\[ n = \frac{-2 \pm \sqrt{36.2624}}{2} \]
\[ n = \frac{-2 \pm 6.02}{2} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{4.02}{2} = 2.01 \) (approximately 2)
2. \( n = \frac{-8.02}{2} \) (not valid since \( n \) cannot be negative)

Thus, we conclude that the number of unpaired electrons \( n \) is approximately 2.

### Step 6: Identifying the Element
Now, we need to check which of the given elements has 2 unpaired electrons:
- **Nickel (Ni, atomic number 28)**: Configuration for Ni²⁺ is \( [Ar] 3d^8 \) (2 unpaired electrons).
- **Titanium (Ti, atomic number 22)**: Configuration for Ti³⁺ is \( [Ar] 3d^1 \) (1 unpaired electron).
- **Chromium (Cr, atomic number 24)**: Configuration for Cr³⁺ is \( [Ar] 3d^3 \) (3 unpaired electrons).
- **Cobalt (Co, atomic number 27)**: Configuration for Co²⁺ is \( [Ar] 3d^7 \) (3 unpaired electrons).

### Conclusion
The element that has 2 unpaired electrons is Nickel (Ni²⁺).

### Final Answer
The answer is **Nickel (Ni²⁺)**.

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