NO₃⁻ · nitrate ion
Three equivalent resonance forms rotate one N=O double bond among three identical oxygens, giving each bond an order of 1⅓ and a perfectly symmetric planar ion.
The small red + and blue − numbers beside atoms are , not the ion's overall charge. The small numbered circles just tell identical atoms apart, so “O #2” in the table is the same oxygen as “2” in the drawing.
Begin by counting , the outer-shell electrons that participate in bonding: nitrogen is a Group 15 element, so it has 15 - 10 = 5 valence electrons; each oxygen is a Group 16 element, so it has 16 - 10 = 6 valence electrons, and with 3 of them that is 6 × 3 = 18; the overall -1 charge means 1 additional electron were gained, so add 1, which totals 24 valence electrons, the only electrons available to build the structure, so every bond and placed later must be drawn from this count.
The is the least electronegative , since it must share its electrons among the most neighbors, and here that atom is nitrogen. A more electronegative atom holds its electrons too tightly to serve as the hub, so it stays terminal.
Connect the to every with a . Because each single bond represents one shared electron pair (2 electrons), the 3 bonds use 6 electrons, leaving 18 of the original 24 to place.
Distribute the remaining electrons to the terminal first, because terminal atoms are more electronegative and complete their octets more readily. The O atoms receive 9 (18 electrons) to reach a full , which leaves 0 electrons.
That places all 24 , leaving none for on N at this point.
Recounting the electrons around N gives only 6, short of a full , and no remain to add. The resolution is to convert 1 on the terminal into 1 , forming a double bond. This creates no new electrons: the same pair is simply shared, and a shared pair counts toward the octet of BOTH bonded atoms, which raises N to a full octet while the total stays 24.
Before continuing, check the total electron count surrounding each . N: 8 shared + 0 nonbonding = 8 (a full ); O: 4 shared + 4 nonbonding = 8 (a full octet); each O: 2 shared + 6 nonbonding = 8 (a full octet).
Determine the of each (, minus lone-pair electrons, minus half of the ): N = +1, O = -1, O = -1. Their sum is -1, matching the species' overall charge. Formal charge is a bookkeeping device rather than a real ionic charge, and the negative formal charge is placed on the most electronegative atom, which is where it is most stable, so this arrangement is the preferred structure.
This is one of 3 equivalent , because the multiple bond has no single preferred location. Averaging the 3 structures gives every equivalent bond an identical of 1.33. Resonance means one real, delocalized that no single can fully represent, not a molecule flipping between the drawings.
Watch out
Before anything else, adjust the count for the -1 charge: add 1. Most wrong answers to NO3^- trace back to skipping that single step.
Determine the using : N is surrounded by 3 bonding domains and 0 , for 3 in total. Because electron domains repel one another, 3 domains arrange themselves as far apart as possible, giving trigonal planar and sp² .
No are hiding in that arrangement, so the you see is the shape you get: trigonal planar, with 120° .
: N and O pull almost equally on their , so there's no meaningful dipole anywhere. Nonpolar.
Worth knowing
In the real nitrate all three N-O bonds are identical, each about order 4/3, because the is smeared across all three positions. Nitrate is the workhorse nitrogen source in fertilizer.