The calculated magnetic moment using the 'spin only' formula \( μ = \sqrt{n(n+2)} \) for \( Cr^{2+} \) and \( Fe^{2+} \) respectively is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Both 4.89
The spin-only magnetic moment depends only on the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
The configuration of each ion is obtained from Z, removing the 4s electrons first:
\( \ce{Cr^2+} \): Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \). Losing the 4s electron and one 3d electron gives \( \mathrm{[Ar]\,3d^{4}} \): 4 unpaired electrons.
\( \ce{Fe^2+} \): Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \). Losing the two 4s electrons gives \( \mathrm{[Ar]\,3d^{6}} \): five electrons go singly into the five 3d orbitals and the sixth pairs up, leaving 4 unpaired electrons.
\( \displaystyle \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89\ \mathrm{BM} \)
So both ions have a magnetic moment of about 4.89 BM.
Shortcut: \( \mathrm{d^{4}} \) and \( \mathrm{d^{6}} \) both have four unpaired electrons (four up, or five up and one paired), so the two values must be equal; only Options (2) and (3) say so, and 3.87 BM (\( \sqrt{15} \)) belongs to three unpaired electrons.