Step 1: The valence shell electron configurations for N and O are `2s^2 2p^3` and `2s^2 2p^4`, respectively. Thus, the total number of valence electrons is `5+6=11`.
Step 2: The only possible skeleton for NO, a diatomic molecule, is
`N-O`
Step 3: Since only one bond exists in the structure, the number of valence electrons left to be accounted for is `11-2=9`.
Step 4: The number of valence electrons needed to complete octet is `6+6=12`.
Step 5: Since the number of remaining electrons is three `(2+1)` less than the number of electrons needed to complete all octets, one more bond is required.
`N=O`
Completing the octets with the dots gives the Lewis structure:
`overset(..)underset(..)N=overset(..)underset(..)O`
Step 6: The number of valence electrons in the Lewis structure (4 lone pairs + 2 bond pairs), 12, is one more than the number of valence electrons counted in step 1. Thus, we must take out `1 e^(-)` out of the Lewis structure to give
`underset(I)(overset(..)underset(.)N=overset(..)underset(..)O)` or `underset(II)(overset(..)underset(..)N=overset(..)underset(.)O)`
as the final Lewis structure of nitric oxide. In (I), N atom has one unpaired electron and its octet is incomplete while in (II), O atom has one unpaired electron and its octet is incomplete. Since N is less electronegative than O, we prefer (I) against (II).
