NO₂⁻ · nitrite ion
Two equivalent resonance forms delocalize the double bond across both N–O bonds; nitrogen's lone pair bends the ion to about 115°.
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.
A must account for exactly the , so total them first: 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 2 of them that is 6 × 2 = 12; the overall -1 charge means 1 additional electron were gained, so add 1, giving 18 valence electrons that the finished structure has to use in full, with none added and none left over.
Nitrogen is the because it is the least electronegative of the present. The least electronegative atom holds its electrons most loosely and therefore shares them with several neighbors most readily, which is exactly what a central atom must do. (Hydrogen and fluorine are never central.)
Begin the structure with a from N to each of the 2 terminal . Each single bond is a shared pair, so these bonds together use 4 electrons and 14 of the 18 are still available.
Place on the terminal before the , since a completes its with lone pairs once its is drawn. The O atoms take 6 lone pairs in total (12 electrons), and 2 electrons remain.
Assign the remaining 2 electrons to the as 1 on N. All 18 are now placed, so the count returns to zero.
N is left with only 6 electrons, below an , and every has already been placed. Rather than adding electrons, 1 already on the terminal is redrawn as 1 , turning 1 single bond into a double bond. Because both atoms count the shared electrons, N now reaches a full octet without changing the total of 18.
Confirm that each has the number of electrons its shell requires. N: 6 shared + 2 nonbonding = 8 (a full ); O: 4 shared + 4 nonbonding = 8 (a full octet); O: 2 shared + 6 nonbonding = 8 (a full octet).
Assign , where each 's formal charge is its minus its minus half its (bonding electrons are halved because a shared pair is split evenly between the two atoms): O = -1. These sum to -1, which equals the overall charge of the species, 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.
The multiple bond shown here could be drawn in 1 other, equally valid position, giving 2 in all. The true species is not any one of these drawings but a single resonance hybrid, the weighted average of them, in which every equivalent bond has the same order of 1.50. The electrons are delocalized over all positions at once; they do not shift back and forth between structures.
Watch out
The classic slip with NO2^-: forgetting to add 1 electron for the -1 charge. Get the count wrong by even one and every later step quietly breaks.
Determine the using : N is surrounded by 2 bonding domains and 1 , for 3 in total. Because electron domains repel one another, 3 domains arrange themselves as far apart as possible, giving trigonal planar and sp² .
A holds its ground but never shows up in the we name. Trace only the and this is bent, angles pinched to about ≈118° by the extra .
Finally, . A shape can't balance its bond pulls, so they stack into a net dipole pointing toward O. This is polar, and this net dipole governs much of its physical behavior.
Worth knowing
Nitrite is the -ite sibling of nitrate: same charge, one less oxygen. It is what keeps cured meats pink and is a real player in the nitrogen cycle.