Which transition metal ion M^2+ gives magnetic moment 5.92 BM?

The correct answer is Option (1) → $Mn^{2+}$

To determine which transition metal ion \( M^{2+} \) gives a magnetic moment of 5.92 BM (Bohr Magnetons), we can use the formula for calculating the magnetic moment:

\(\mu = \sqrt{n(n + 2)} \, \text{BM}\)

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

\(Mn^{2+}\):

Electronic configuration: \( [Ar] 3d^5 \)

Unpaired electrons: 5

Magnetic moment: \(\mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \, \text{BM}\)

\(Cr^{2+}\):

Electronic configuration: \( [Ar] 3d^4 \)

Unpaired electrons: 4

Magnetic moment: \(\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \, \text{BM}\)

\(Fe^{2+}\): 

Electronic configuration: \( [Ar] 3d^6 \)

Unpaired electrons: 4 (in a high-spin state)

Magnetic moment: \(\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \, \text{BM}\)

\(Ni^{2+}\): 

Electronic configuration: \( [Ar] 3d^8 \)

Unpaired electrons: 2

Magnetic moment: \(\mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \, \text{BM}\)

Conclusion

The only transition metal ion among the options that has a magnetic moment of approximately 5.92 BM is: \(Mn^{2+}\).