Question

The spin-only magnetic moment value of $M ^{3+}$ ion (in gaseous state) from the pairs $Cr ^{3+} / Cr ^{2+}, Mn ^{3+} / Mn ^2, \quad Fe ^{3+} / Fe ^{2+}$ and $Co ^{3+} / Co ^{2+}$ that has negative standard electrode potential, is BM [Nearest integer]

Answer 1

Correct Answer: 4

The spin-only magnetic moment (\(\mu\)) for a transition metal ion can be calculated using the formula: \(\mu = \sqrt{n(n+2)}\) where \(n\) is the number of unpaired electrons. 

We need to determine which \(M^{3+}\) ion from the pairs \(Cr^{3+}/Cr^{2+}, Mn^{3+}/Mn^{2+}, Fe^{3+}/Fe^{2+}, Co^{3+}/Co^{2+}\) has a negative standard electrode potential.

Transition metal ions with negative electrode potential and unpaired electrons:

• \(Cr^{3+}\): Electron configuration: \([Ar]\,3d^3\), \(n=3\)
• \(Mn^{3+}\): Electron configuration: \([Ar]\,3d^4\), \(n=4\)
• \(Fe^{3+}\): Electron configuration: \([Ar]\,3d^5\), \(n=5\)
• \(Co^{3+}\): Electron configuration: \([Ar]\,3d^6\), \(n=4\)

Considering standard electrode potential values, generally, \(Cr^{3+}\) and \(Mn^{3+}\) tend to have negative values. Verification focuses on these ions.

Calculate Spin-Only Magnetic Moment:

Ion	Unpaired Electrons (\(n\))	Magnetic Moment (\(\mu\))
\(Cr^{3+}\)	3	\(\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}\)
\(Mn^{3+}\)	4	\(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}\)

Range provided: 4, 4. Confirmed: \(Mn^{3+}\) has a magnetic moment nearest to 4 BM. Therefore, the solution is \(4 \text{ BM}\).