Question

The calculated spin only magnetic moment of \(C r^{2+}\) ion is :
A \(\quad 3.87 \mathrm{BM}\)
B \(4.90 \mathrm{BM}\)
c \(5.92 \mathrm{BM}\)
D \(2.84 \mathrm{BM}\)

Answer 1

Solution:
Electronic configuration of \(C r=[A r] 3 d^{5} 4 s^{1}\)
Electronic configuration of \(C r^{+2}=[A r] 3 d^{4}\)
Number of unpaired electrons \(=4\)
Spin only magnetic moment \(=\sqrt{n(n+2)} B M\) as \(n=4\)
$$
\begin{aligned}
&\therefore \mu=\sqrt{4(4+2)} \\
&=\sqrt{24} \\
&=4.90 B M
\end{aligned}
$$