Question 7.9: The nitrate ion NO3^-, has three equivalent oxygen atoms, an......
The nitrate ion NO_{3}^{-} , has three equivalent oxygen atoms, and its electronic structure is a resonance hybrid of three electron-dot structures. Draw them.
STRATEGY
Begin as you would for drawing any electron-dot structure. There are 24 valence elec-trons in the nitrate ion: 5 from nitrogen, 6 from each of 3 oxygens, and 1 for the nega-tive charge. The three equivalent oxygens are all bonded to nitrogen, the less electronegative central atom:
\begin{matrix} \underset{|}{O}\\ N\\ O \ ^{^{\diagup }} \quad\quad \ ^{^{\diagdown}}O\end{matrix} 6 of 24 valence electrons assigned
Distributing the remaining 18 valence electrons among the three terminal oxygen atoms completes the octet of each oxygen but leaves nitrogen with only 6 electrons.
\begin{bmatrix} :\overset{..}{\underset{|}{O}}:\\ N\\ :\overset{..}{\underset{..}{O}} \ ^{^{\diagup }} \quad\quad \ ^{^{\diagdown}}\overset{..}{\underset{..}{O}}: \end{bmatrix}^{-}To give nitrogen an octet, one of the oxygen atoms must use a lone pair to form an N=O double bond. But which one? There are three possibilities, and thus three electron-dot structures for the nitrate ion, which differ only in the placement of bond-ing and lone-pair electrons. The connections between atoms are the same in all three structures, and the atoms have the same positions in all structures.
