The correct answer is option 1: 3.87 BM.
The spin-only magnetic moment (\(μ_{\text{spin}}\)) can be calculated using the formula:
\[μ_{\text{spin}} = \sqrt{n(n+2)}\]
where \(n\) is the number of unpaired electrons.
For \(Cr^{3+}\), we need to consider the electronic configuration of \(Cr^{3+}\). The atomic number of chromium (\(Cr\)) is 24. The electronic configuration of neutral chromium is \(1s^22s^22p^63s^23p^64s^13d^5\). When \(Cr\) loses three electrons to form \(Cr^{3+}\), the three electrons are lost from the 4s and two from the 3d subshell.
The electronic configuration of \(Cr^{3+}\) is \(1s^22s^22p^63s^23p^64s^03d^3\).
Now, let's find the number of unpaired electrons (\(n\)) in \(Cr^{3+}\) which is 3.
\[μ_{\text{spin}} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{BM}\]
Therefore, the correct option is (2) 3.87 BM. |
