Question

What is the bond-order of $CN$?

[Note : Write the answer after multiplying it with 4]

Answer 1

$CN$ has $13$ electrons

MO configuration : $(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi 2p{)}^4,(\sigma 2p{)}^1$

Now,

No. of electrons in BO = 9

No. of electron in AO = 4

So, bond order $=\frac{1}{2}(B-A)$
where,

B is the number of electrons in bonding molecular orbitals and
A is the number of electrons in antibondng molecular orbitals.

$\therefore$Bond Order$=\frac{1}{2}(9-4)=2.5$

Now,

4$\times$ Bond order = 4 $\times$ 2.5 = 10.