Write the Lewis dot structure of the nitrite ion `(NO_(2)^(Θ))` .

Explanation

Step 1: The valence shell electron configurations of N and O atoms are `2s^2 2p^3` and `2s^2 2p^4`, respectively, and the ion itself has one negative charge. Thus, the total number of valence electrons is `5+(2xx6)+1=18`.
Step 2: The best skeleton structure is
`O-N-O`
which avoids `O-O` bond and ring structure.
Step 3: Since the skeleton structure involves two single bonds, the number of valence electrons remaining is `18-4=14`.
Step 4: Since each O atom is forming one bond and N atom is forming two bonds, the number of valence electrons needed to complete all octets is 6 for each O atom and 4 for N atom, i.e., 16 electrons.
Step 5: Since the number of remaining electrons is two less than the number of electrons needed to complete all octets, one additional bond is required. It is placed between the N atom and either of the O atoms:
`O=underset(I)N-O` or `O-underset(II)N=O`
Completing the octets by placing the dots around the appropriate symbols in Lewis structure gives
`underset(I) [[underset(..)overset(..)O=N-overset(..)underset(..)O:]]^(-)` or `underset(II)[[:underset(..)overset(..)O-N=overset(..)underset(..)O]]^(-)`
Step 6: The Lewis structure drawn for nitrite ion accounts for the 18 valence electrons and indicates the negative charge.