Electronegativity and Molecules

Question

How many resonance structures exist for the formate ion, HCO₂^-?

Answer 1

Short answer: there are 2 resonance structures for formate.

Start by counting valence electrons, because that is what fixes the skeleton before you draw anything:

H   1 x 1 =  1
C   1 x 4 =  4
O   2 x 6 = 12
charge    = +1
              --
total     = 18 valence electrons = 9 pairs

Nine pairs. Carbon is the central atom - hydrogen can only ever make one bond, so it cannot be in the middle. H-C plus two C-O bonds uses three pairs. Spread the remaining six pairs onto the oxygens and one oxygen comes up short of an octet, so one of its lone pairs becomes a second C-O bond. That leaves one C=O and one single-bonded O carrying the negative charge - and the charge can sit on either oxygen:

   structure I              structure II

        O                       O(-)
       //                      /
  H - C                   H - C
       \                      \\
        O(-)                   O

Those two are the complete set, and they are equivalent: identical in energy, so they contribute equally to the real ion.

Three ways students land on a different number, and how to tell each one is wrong.

Getting 3, by moving the hydrogen. Drawing H on the other oxygen looks like a third form. It is not one. Resonance moves electrons only; the instant an atom changes position you have drawn a different species (a tautomer), not a resonance structure. This is the single most common error on this question type, and it is worth fixing now because it will otherwise follow you into enolates and amides.

Getting 4, by shuffling lone pairs on the page. Redrawing the same structure with an oxygen lone pair pointing a different direction is the same structure. Two Lewis structures are different only if the bond connectivity or the placement of formal charge differs.

Getting 1, by treating the double bond as parked. If you stop as soon as you have a valid octet structure you report one. The check that catches this: ask whether any equivalent atom could have taken the double bond instead. Here a second oxygen could, so you are not finished.

What the two drawings are approximating. The real ion is neither structure - it is a single species in which both C-O bonds are the same. The experimental numbers say so plainly:

C-O single bond      1.43 A
C=O double bond      1.20 A
formate, measured    1.27 A   (BOTH bonds, identical)

bond order 1.5 on each C-O;  formal charge -1/2 on each O

A measured bond length sitting neatly between the single and double values is the physical evidence that resonance is not a bookkeeping trick. Carbon has three groups and no lone pair, so formate is trigonal planar with an O-C-O angle near 126 degrees, and the whole ion including the delocalized pair is flat.

The counting rule generalizes. Count the equivalent positions the double bond can occupy: acetate CH3CO2- gives 2 (the methyl group changes nothing), carbonate CO3(2-) gives 3, nitrate NO3- gives 3. Same skeleton logic every time.

Why this matters past the drawing. Formic acid itself, HCOOH, has two unequal C-O bonds (about 1.20 A and 1.34 A) because its oxygens are not equivalent - one carries the proton. Remove that proton and the two oxygens become identical, the charge spreads over both, and the anion is stabilized. That stabilization is the reason formic acid has a pKa near 3.75 while ethanol, whose conjugate base has nowhere to delocalize, sits around 16. The resonance count you just did is the explanation for a 12-order-of-magnitude difference in acidity.