CO₃²⁻ · carbonate ion
One C=O and two C–O⁻ bonds in each of three equivalent resonance forms; the true ion carries the −2 charge spread evenly across all three oxygens.
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: carbon is a Group 14 element, so it has 14 - 10 = 4 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 -2 charge means 2 additional electrons were gained, so add 2, for 24 valence electrons in all; finishing with more or fewer than 24 would mean a counting error was made.
Carbon is the . With four it can form four bonds, more than any other present, so it is best suited to connect to several neighbors at once. Hydrogen cannot be central, because a single valence electron lets it form only one bond.
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.
With the terminal octets complete, no electrons remain, so C receives no at this stage.
C 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, C now reaches a full octet without changing the total of 24.
Confirm that each has the number of electrons its shell requires. C: 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).
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 -2, 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 2 other, equally valid positions, giving 3 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 1.33. The electrons are delocalized over all positions at once; they do not shift back and forth between structures.
Watch out
The classic slip with CO3^2-: forgetting to add 2 electrons for the -2 charge. Get the count wrong by even one and every later step quietly breaks.
Apply by counting the around C: 3 bonding domains plus 0 gives 3 electron domains. Domains positioned to minimize their mutual repulsion adopt , which corresponds to sp² .
No are hiding in that arrangement, so the you see is the shape you get: trigonal planar, with 120° .
: each C-O bond is genuinely polar on its own, but the arrangement points those pulls symmetrically against each other. They cancel to nothing. Nonpolar overall.
Worth knowing
Carbonate built the White Cliffs of Dover. Limestone, marble, chalk, seashells, and eggshells are all carbonate minerals, and the fizz when acid hits them is carbonate giving up CO2.