JEE MAIN - Chemistry (2023 - 24th January Morning Shift - No. 8)
The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is
Ti$$^{2+}$$
V$$^{2+}$$
Cr$$^{2+}$$
Mn$$^{2+}$$
Explanation
$$
\begin{aligned}
& \mathrm{Cr}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^4, 4 \mathrm{~s}^0 \mathrm{n}=4, \mu=\sqrt{4(4+2)}=\sqrt{24} =4.89 \mathrm{BM} \\\\
& \mathrm{Mn}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^5, 4 \mathrm{~s}^0 \mathrm{n}=5, \mu=\sqrt{5(5+2)}=\sqrt{35} =5.91 \mathrm{BM} \\\\
& \mathrm{V}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^3, 4 \mathrm{~s}^0 \mathrm{n}=3, \mu=\sqrt{3(3+2)}=\sqrt{15} =3.87 \mathrm{BM} \\\\
& \mathrm{Ti}^{+2}:[\mathrm{Ar}], 3 \mathrm{~d}^2, 4 \mathrm{~s}^0 \mathrm{n}=2, \mu=\sqrt{2(2+2)}=\sqrt{8} =2.82 \mathrm{BM}
\end{aligned}
$$
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