Question

The spin only magnetic moment value of $B_2^{+}$ species is $x\times {10}^{-2}B.M.$.
[Given $:\sqrt{3}=1.73$] x is : (Nearest integer)
(JEE (MAIN)-2021)

Answer 1

According to $MOT$, electronic configuration of $B_2^{+}$ is
$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\begin{array}{c}\pi 2p_x^1\\ \pi 2p_y\end{array}\sigma 2p_z$
It has one unpaired electron.
$\left(\mu \right)$ Spin - only magnetic moment $=\sqrt{n(n+2)}B.M.$
$n=$ Number of unpaired electrons

$\mu =\sqrt{1(1+2)}B.M.$
$\mu =1.73B.M.$
$\mu =173\times {10}^{-2}B.M.$
$x=173$