What is the Lewis structure for a nitrate ion?

May 5, 2016

Answer:

${N}^{+} \left(= O\right) {\left({O}^{-}\right)}_{2}$

Explanation:

Electrons to distribute: $\left(3 \times 6 + 5 + 1\right) = 24 \cdot {e}^{-}$ from the oxygens, the nitrogen atom, and from the negative charge.

If you count up the valence electrons, the number is indeed 24 on the above representation. For the nitrate ion, the nitrogen is formally QUATERNIZED, i.e. it bears a formal positive charge.

The two singly bound oxygens bear a formal negative charge, and this representation is consistent with the negative charge of the ion (i.e. each of these 2 oxygen atoms is associated with 7 electrons rather than the 6 associated with the formally neutral, formally doubly bound O atom). Of course, all the oxygen atoms are formally equivalent.