Question

The covalency of nitrogen in $HN{O}_3$ is:

• A. 0
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is C 4

The electronic configuration of $N$ is $1s^{2}2s^2{p}^3$. The 1s orbital is completely filled and unavailable for bonding. That leaves the 2nd and 2p orbitals. The formation of 4 bonds (covalency) is explained by hybridation of the 2s and 2p orbital forming $4s{p}^3$ orbitals.

Normally when the $N$ atom bonds it forms 3 covalent bonds with one pair of nonbonding electrons. Hence, $N$ normally has a covalency of 3 with an available pair of electrons. But, when a proton, or other group that is able to bond with a pair of electrons coordinates with the central $N$ atom a 4th covalent bond is formed. Beyond the 4th bond there is no more capability of the $N$ atom to form additional bonds so it can only have a maximum covalency of 4.