Question

Find the sum of bond order and number of $\pi$ bonds in $B_2$ molecule.

Answer 1

$B_2$ has 10 electrons.
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^2$
So, Bond Order$=\frac{1}{2}(a-b)$
Here, a is the number of electrons in bonding molecular orbitals and b is the number of electrons in antibondng molecular orbitals.
Bond Order$=\frac{1}{2}(6-4)$$=1$ (single bond)
Number of pi bonds in $B_2$ is 0
Therefore, sum is $1+0=1$