Match List - I with List - II and select the correct answer using codes given below the lists -
`{:("List - 1 (Metal ions)","List - II Magnetic moment (B.M.)"),((1)Cr^(3+),(A)sqrt(35)),((2)Fe^(2+),(B) sqrt(30)),((3)Ni^(2+),(C)sqrt(24)),((4)Mn^(2+),(D)sqrt(15)),(,(E)sqrt8):}`

Options

A. 1 - (B), 2 - (C), 3 - (E), 4 - (D)
B. 1 - (B), 2 - (C), 3 - (E), 4 - (A)
C. 1 - (D), 2 - (C), 3 - (E), 4 - (A)
D. 1 - (D), 2 - (E), 3 - (C), 4 - (A)

Answer

To solve the problem of matching the metal ions with their respective magnetic moments, we will follow these steps:

### Step 1: Determine the electronic configuration of each metal ion.
1. **Cr³⁺ (Chromium)**: The electronic configuration of Cr is [Ar] 3d⁵ 4s¹. For Cr³⁺, we remove 3 electrons (2 from 4s and 1 from 3d), resulting in 3d³.
2. **Fe²⁺ (Iron)**: The electronic configuration of Fe is [Ar] 3d⁶ 4s². For Fe²⁺, we remove 2 electrons from 4s, resulting in 3d⁶.
3. **Ni²⁺ (Nickel)**: The electronic configuration of Ni is [Ar] 3d⁸ 4s². For Ni²⁺, we remove 2 electrons from 4s, resulting in 3d⁸.
4. **Mn²⁺ (Manganese)**: The electronic configuration of Mn is [Ar] 3d⁵ 4s². For Mn²⁺, we remove 2 electrons from 4s, resulting in 3d⁵.

### Step 2: Count the number of unpaired electrons for each ion.
1. **Cr³⁺ (3d³)**: There are 3 unpaired electrons.
2. **Fe²⁺ (3d⁶)**: There are 4 unpaired electrons.
3. **Ni²⁺ (3d⁸)**: There are 2 unpaired electrons.
4. **Mn²⁺ (3d⁵)**: There are 5 unpaired electrons.

### Step 3: Calculate the magnetic moment using the formula.
The formula for magnetic moment (μ) is given by:
\[ \mu = \sqrt{n(n + 2)} \text{ B.M.} \]
where \( n \) is the number of unpaired electrons.

1. **Cr³⁺**:
\[ n = 3 \Rightarrow \mu = \sqrt{3(3 + 2)} = \sqrt{15} \text{ B.M.} \]
2. **Fe²⁺**:
\[ n = 4 \Rightarrow \mu = \sqrt{4(4 + 2)} = \sqrt{24} \text{ B.M.} \]
3. **Ni²⁺**:
\[ n = 2 \Rightarrow \mu = \sqrt{2(2 + 2)} = \sqrt{8} \text{ B.M.} \]
4. **Mn²⁺**:
\[ n = 5 \Rightarrow \mu = \sqrt{5(5 + 2)} = \sqrt{35} \text{ B.M.} \]

### Step 4: Match the results with List II.
- **Cr³⁺** matches with (D) \(\sqrt{15}\)
- **Fe²⁺** matches with (C) \(\sqrt{24}\)
- **Ni²⁺** matches with (E) \(\sqrt{8}\)
- **Mn²⁺** matches with (A) \(\sqrt{35}\)

### Final Matching:
1. Cr³⁺ → D (√15)
2. Fe²⁺ → C (√24)
3. Ni²⁺ → E (√8)
4. Mn²⁺ → A (√35)

### Answer Code:
The correct answer code is:
1 - D, 2 - C, 3 - E, 4 - A