The calculated 'spin-only' magnetic moment of $\mathrm{Ti}{}^{2+}$ ($3\mathrm{d}{}^2$) is:

Solution & Explanation

\textbf{Answer:} (c) $2.84$ BM \textbf{Solution:} $\mathrm{Ti}{}^{2+}$ has the configuration $3\mathrm{d}{}^2$, so it has $n=2$ unpaired electrons. Using the spin-only formula $\mu =\sqrt{n(n+2)}$: $\mu =\sqrt{2(2+2)}=\sqrt{8}=2.84$ BM. Hence the spin-only magnetic moment of $\mathrm{Ti}{}^{2+}$ is $2.84$ BM.