Question

How many of the following metal ions have similar value of spin only magnetic moment in gaseous state? (Given: Atomic number: $V, 23; Cr, 24; Fe, 26; Ni, 28$) $V ^{3+} \cdot Ci ^{3+} \cdot Fe ^{2+} \cdot Ni ^{3+}$

Answer 1

Correct Answer: 2

The spin-only magnetic moment (\(\mu\)) is given by:
\[ \mu = \sqrt{n(n+2)} \, \text{BM}, \] where \(n\) is the number of unpaired electrons.
1. V$^{3+}$:
- Electronic configuration: [Ar] 3d\(^2\)
- Number of unpaired electrons (\(n\)) = 2.
- \(\mu = \sqrt{2(2+2)} = \sqrt{8} \, \text{BM}.\)
2. Cr$^{3+}$:
- Electronic configuration: [Ar] 3d\(^3\)
- Number of unpaired electrons (\(n\)) = 3.
- \(\mu = \sqrt{3(3+2)} = \sqrt{15} \, \text{BM}.\)
3. Fe$^{2+}$:
- Electronic configuration: [Ar] 3d\(^6\)
- Number of unpaired electrons (\(n\)) = 4.
- \(\mu = \sqrt{4(4+2)} = \sqrt{24} \, \text{BM}.\)
4. Ni$^{3+}$:
- Electronic configuration: [Ar] 3d\(^7\)
- Number of unpaired electrons (\(n\)) = 3.
- \(\mu = \sqrt{3(3+2)} = \sqrt{15} \, \text{BM}.\)
Result:
Cr$^{3+}$ and Ni$^{3+}$ have the same magnetic moment (\(\sqrt{15} \, \text{BM}\)). The spin-only magnetic moment depends on the number of unpaired electrons (\(n\)). Cr$^{3+}$ and Ni$^{3+}$ have identical magnetic moments because they have the same number of unpaired electrons.