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 of approximately 5 Bohr Magnetons (BM), we need to calculate the magnetic moment using the formula:

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

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

Let's analyze each ion step by step:

### Step 1: Analyze Vanadium (V²⁺)
- Atomic number of Vanadium (V) = 23
- Electron configuration: \([Ar] 3d^3 4s^2\)
- For V²⁺, we lose 2 electrons (from 4s), so the configuration becomes: \[3d^3\]
- Number of unpaired electrons (\(n\)) = 3
- Calculate magnetic moment:
\[ \mu = \sqrt{n(n + 2)} = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \, \text{BM} \]
- Conclusion: V²⁺ does not exhibit a magnetic moment of 5 BM.

### Step 2: Analyze Titanium (Ti²⁺)
- Atomic number of Titanium (Ti) = 22
- Electron configuration: \([Ar] 3d^2 4s^2\)
- For Ti²⁺, we lose 2 electrons (from 4s), so the configuration becomes: \[3d^2\]
- Number of unpaired electrons (\(n\)) = 2
- Calculate magnetic moment:
\[ \mu = \sqrt{n(n + 2)} = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \, \text{BM} \]
- Conclusion: Ti²⁺ does not exhibit a magnetic moment of 5 BM.

### Step 3: Analyze Manganese (Mn²⁺)
- Atomic number of Manganese (Mn) = 25
- Electron configuration: \([Ar] 3d^5 4s^2\)
- For Mn²⁺, we lose 2 electrons (from 4s), so the configuration becomes: \[3d^5\]
- Number of unpaired electrons (\(n\)) = 5
- Calculate magnetic moment:
\[ \mu = \sqrt{n(n + 2)} = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \, \text{BM} \]
- Conclusion: Mn²⁺ does not exhibit a magnetic moment of 5 BM.

### Step 4: Analyze Chromium (Cr²⁺)
- Atomic number of Chromium (Cr) = 24
- Electron configuration: \([Ar] 3d^5 4s^1\)
- For Cr²⁺, we lose 2 electrons (1 from 4s and 1 from 3d), so the configuration becomes: \[3d^4\]
- Number of unpaired electrons (\(n\)) = 4
- Calculate magnetic moment:
\[ \mu = \sqrt{n(n + 2)} = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \, \text{BM} \]
- Conclusion: Cr²⁺ exhibits a magnetic moment approximately equal to 5 BM.

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

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