The calculated spin only magnetic moment of $\mathrm{Cr}{}^{2+}$ ion is:

Solution & Explanation

\textbf{Answer:} (D) $4.90$ BM \textbf{Solution:} The spin-only magnetic moment is $\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. $\mathrm{Cr}{}^{2+}$ has the configuration $3d^4$, giving $n=4$ unpaired electrons. $\mu =\sqrt{4(4+2)}=\sqrt{24}=4.90$ BM. Hence the answer is option D.