Find the bond order in ${\mathrm{B}}_2$ molecule. $(\mathrm{at}. \mathrm{no}. \mathrm{of} \mathrm{B}=5).$

no. of electrons is

${\mathrm{B}}_2=(2\times$ at. no. of $\mathrm{B})=2\times 5=10$

$\therefore$ $\mathrm{M}$.$\mathrm{O}$. electron configuration of

$\begin{array}{l}{\mathrm{B}}_2={\left({\mathrm{\sigma }}_{1\mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{1\mathrm{s}}^{\ast }\right)}^2{\left({\mathrm{\sigma }}_{2\mathrm{s}}\right)}^2{\left({\mathrm{\sigma }}_{2\mathrm{s}}^{\ast }\right)}^2(\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1 =\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1)\\ \end{array}$

$\therefore$Bond order $=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}=\frac{6-4}{2}=\mathbf{1}$.