The spin only magnetic moment of Cr^3+ ion in BM is:

The correct answer is Option (2) → 3.87

The spin only magnetic moment depends only on the number of unpaired electrons, n:

    \( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)

Cr (Z = 24) is [Ar]3d⁵4s¹. Removing three electrons (the 4s electron first, then two 3d electrons) gives \( \ce{Cr^{3+}} \): [Ar]3d³, with 3 unpaired electrons:

    \( \displaystyle \mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\ \mathrm{BM} \)