Write the Lewis structure of the nitrite ion,$NO_{2}^{-}$

(N/A) Step $1$. Count the total number of valence electrons of the nitrogen atom,the oxygen atoms,and the additional one negative charge (equal to one electron).
$N (2s^{2} 2p^{3}), O (2s^{2} 2p^{4})$
$5 + (2 \times 6) + 1 = 18$ electrons.
Step $2$. The skeletal structure of $NO_{2}^{-}$ is written as: $O-N-O$.
Step $3$. Draw a single bond (one shared electron pair) between the nitrogen and each of the oxygen atoms. Completing the octets on oxygen atoms uses $16$ electrons. The remaining $2$ electrons constitute a lone pair on the nitrogen atom.
Step $4$. To complete the octet on the nitrogen atom,one of the oxygen atoms forms a double bond with the nitrogen atom.
The resulting resonance structures are:
$\left[ \ddot{O} = \ddot{N} - \ddot{O}: \right]^{-} \leftrightarrow \left[ :\ddot{O} - \ddot{N} = \ddot{O} \right]^{-}$