O₃ · ozone
Two equivalent resonance forms place the double bond on either side; the real molecule is a hybrid with two identical bonds of order 1.5, not an oscillation between forms.
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.
Start from the -electron total, because it sets how many electrons the whole structure may contain: 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, for 18 valence electrons in all; finishing with more or fewer than 18 would mean a counting error was made.
All 3 are oxygen, so cannot decide the arrangement. The is the one that forms bonds to two neighbors, and the remaining atoms are terminal atoms bonded only to it.
Connect the to every with a . Because each single bond represents one shared electron pair (2 electrons), the 2 bonds use 4 electrons, leaving 14 of the original 18 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 6 (12 electrons) to reach a full , which leaves 2 electrons.
The last 2 electrons are placed on O as 1 . Every is now accounted for, and none remain.
Recounting the electrons around O 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 O to a full octet while the total stays 18.
Confirm that each has the number of electrons its shell requires. O: 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, O = -1. These sum to 0, 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
Most people's first draft stops at single bonds and leaves O short of an . Always re-count the after placing ; that recount is what tells you a multiple bond is needed.
Determine the using : O 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
Ozone's two bonds are exactly the same length in the real , order 1.5 each, because the is . Twenty kilometers up, this molecule absorbs the UV radiation that would otherwise make sunburn the least of our problems.