Which of the following paramagnetic ions would exhibit a magnetic moment (spin only) of the order of 5 BM?
(At. No : Mn = 25, Cr = 24,V = 23, Ti = 22)

Options

A. `V^(2+)`
B. `Ti^(2+)`
C. `Mn^(2+)`
D. `Cr^(2+)`

Answer

To determine which of the given paramagnetic ions exhibits a magnetic moment (spin only) of the order of 5 BM, we will analyze each ion based on its electronic configuration and the number of unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Ions and Their Atomic Numbers:**
- Vanadium (V) - Atomic Number 23
- Titanium (Ti) - Atomic Number 22
- Chromium (Cr) - Atomic Number 24
- Manganese (Mn) - Atomic Number 25

2. **Determine the Oxidation States:**
- We will consider the +2 oxidation state for each ion.

3. **Electronic Configuration:**
- **Vanadium (V):**
- Atomic number 23: Configuration is [Ar] 3d³ 4s²
- In +2 state: Remove 2 electrons (from 4s): 3d³
- Unpaired electrons (n) = 3
- **Titanium (Ti):**
- Atomic number 22: Configuration is [Ar] 3d² 4s²
- In +2 state: Remove 2 electrons (from 4s): 3d²
- Unpaired electrons (n) = 2
- **Chromium (Cr):**
- Atomic number 24: Configuration is [Ar] 3d⁵ 4s¹
- In +2 state: Remove 1 electron from 4s and 1 from 3d: 3d⁴
- Unpaired electrons (n) = 4
- **Manganese (Mn):**
- Atomic number 25: Configuration is [Ar] 3d⁵ 4s²
- In +2 state: Remove 2 electrons (from 4s): 3d⁵
- Unpaired electrons (n) = 5

4. **Calculate the Magnetic Moment:**
- The formula for spin-only magnetic moment (μ) is given by:
\[ μ = \sqrt{n(n + 2)} \text{ BM} \]
- **For Vanadium (V²⁺):**
\[ μ = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \text{ BM} \]
- **For Titanium (Ti²⁺):**
\[ μ = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \text{ BM} \]
- **For Chromium (Cr²⁺):**
\[ μ = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \text{ BM} \]
- **For Manganese (Mn²⁺):**
\[ μ = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \text{ BM} \]

5. **Conclusion:**
- The magnetic moments calculated are:
- Vanadium (V²⁺): ~3.87 BM
- Titanium (Ti²⁺): ~2.83 BM
- Chromium (Cr²⁺): ~4.90 BM
- Manganese (Mn²⁺): ~5.92 BM
- The ion that exhibits a magnetic moment closest to 5 BM is **Chromium (Cr²⁺)** with a value of approximately 4.90 BM.

### Final Answer:
The paramagnetic ion that would exhibit a magnetic moment of the order of 5 BM is **Chromium (Cr²⁺)**.