Question

Find the sum of bond order and the number of $\pi$ - bonds in $C_2$ molecule on the basis of molecular orbital theory.

Answer 1

$C_2$ has $12$ electrons.
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4$
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}(8-4)$$=2$
Number of pi bonds in $C_2$ is 2.
Therefore, sum is $2+2=4$