What is the current order of spin-only magnetic moment (in B.M.) of \(Mn^{2+}, Cr^{2+}\) and \(V^{2+}\)?

\(Mn^{2+} > Cr^{2+} > V^{2+}\)

The correct answer is option 3. \(Mn^{2+} > Cr^{2+} > V^{2+}\)

The spin-only magnetic moment of a transition metal ion can be calculated using the formula:

\(\mu_{\text{spin-only}} = \sqrt{n(n+2)} \, \text{B.M.} \)

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

Let us determine the number of unpaired electrons for each ion:

\(Mn^{2+}\) (Manganese ion)

Atomic number of Mn: 25

Electronic configuration of Mn: \([Ar] 3d^5 4s^2\)

For \(Mn^{2+}\): Remove \(2\) electrons from \(4s\), giving \([Ar] 3d^5\)

Number of unpaired electrons: 5

Spin-only magnetic moment: \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.}\)

\(Cr^{2+}\) (Chromium ion)

Atomic number of Cr: 24

Electronic configuration of Cr: \([Ar] 3d^5 4s^1\)

For \(Cr^{2+}\): Remove \(2\) electrons, one from \(4s\) and one from \(3d\), giving \([Ar] 3d^4\)

Number of unpaired electrons: 4

Spin-only magnetic moment: \(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{B.M.}\)

\(V^{2+}\) (Vanadium ion)

Atomic number of V: 23

Electronic configuration of V: \([Ar] 3d^3 4s^2\)

For \(V^{2+}\): Remove \(2\) electrons from \(4s\), giving \([Ar] 3d^3\)

Number of unpaired electrons: 3

Spin-only magnetic moment: \(\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.}\)

Therefore, the order is: \(Mn^{2+} > Cr^{2+} > V^{2+}\)