Question

Write the Lewis structure of the nitrite ion, $N{O}_2^{-}.$

Answer 1

Step 1. Valence electrons

Count the total number of valence electrons in nitrogen and oxygen atoms and the additional one negative charge (equal to one electron).

$N:\left(2s^{2}2p^3\right)$

$O:\left(2s^{2}2p^4\right)$

Total valence electrons in $N{O}_2^{–}=$ Valence $e-$ of $N+2\times$ (Valence $e-$ of $O$) + Charge on anion

Total valence electrons in $N{O}_2^{–}=5+(2\times 6)+1=18$
electrons

Step 2. Skeletal structure

The skeletal structure of $N{O}_2^{–}$ is written as:
$ONO$

Step 3. Octet completion

Draw a single bond (one shared electron pair) between the nitrogen and each of the oxygen atoms. This, however, does not complete the octet on nitrogen if the remaining two electrons constitute lone pair on it.

Hence, we have to resort to multiple bonding between nitrogen and one of the oxygen atoms (in this case a double bond).


Final Answer: