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

Options

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

Explanation

To determine which element corresponds to a magnetic moment of 2.84 Bohr magnetons (B.M.), we can use the formula for magnetic moment:

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

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

### Step 1: Set up the equation
Given that the magnetic moment \( \mu = 2.84 \) B.M., we can set up the equation:

\[ \sqrt{n(n + 2)} = 2.84 \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the equation:

\[ n(n + 2) = (2.84)^2 \]

Calculating \( (2.84)^2 \):

\[ (2.84)^2 = 8.0656 \]

So, we have:

\[ n(n + 2) = 8.0656 \]

### Step 3: Rearrange the equation
Rearranging gives us a quadratic equation:

\[ n^2 + 2n - 8.0656 = 0 \]

### Step 4: Solve the quadratic equation
To solve for \( n \), we can use the quadratic formula:

\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = 2 \), and \( c = -8.0656 \).

Calculating the discriminant:

\[ b^2 - 4ac = 2^2 - 4(1)(-8.0656) = 4 + 32.2624 = 36.2624 \]

Now applying the quadratic formula:

\[ n = \frac{-2 \pm \sqrt{36.2624}}{2} \]

Calculating \( \sqrt{36.2624} \):

\[ \sqrt{36.2624} \approx 6.02 \]

So,

\[ n = \frac{-2 \pm 6.02}{2} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{4.02}{2} = 2.01 \) (approximately 2)
2. \( n = \frac{-8.02}{2} = -4.01 \) (not physically meaningful)

Thus, we find \( n \approx 2 \).

### Step 5: Identify the element with 2 unpaired electrons
Now we need to find which of the given elements has 2 unpaired electrons:

- **Ni (Atomic number 28)**: Electronic configuration is \( [Ar] 3d^8 4s^2 \) → 2 unpaired electrons in \( 3d^8 \).
- **Ti (Atomic number 22)**: Electronic configuration is \( [Ar] 3d^2 4s^2 \) → 2 unpaired electrons in \( 3d^2 \).
- **Cr (Atomic number 24)**: Electronic configuration is \( [Ar] 3d^5 4s^1 \) → 6 unpaired electrons in \( 3d^5 \).
- **Co (Atomic number 27)**: Electronic configuration is \( [Ar] 3d^7 4s^2 \) → 3 unpaired electrons in \( 3d^7 \).

From the analysis, **Ni²⁺** has 2 unpaired electrons.

### Conclusion
The element that corresponds to a magnetic moment of 2.84 B.M. is **Nickel (Ni) in its +2 oxidation state**.

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