Question

write a lewis structure for the formate ion (hco₂⁻). include all resonance structures. finally, draw a resonance hybrid structure for hco₂⁻ and determine the c - o bond order in the formate ion.

Explanation:

Step1: Draw resonance structures

The formate ion ($HCO_2^-$) has two resonance - contributing structures. In one structure, the carbon is double - bonded to one oxygen and single - bonded to the other oxygen with a negative charge. In the second structure, the double and single bonds are reversed.
The two resonance structures are:
$H - C(\bond{=}O)-O^-$ and $H - C(\bond{-}O^-)=O$

Step2: Draw resonance hybrid

The resonance hybrid has a partial double - bond character between carbon and both oxygen atoms. The negative charge is delocalized over the two oxygen atoms.

Step3: Calculate bond order

The bond order of a bond in a resonance hybrid is calculated as the average of the bond orders in the contributing resonance structures. In the two resonance structures of $HCO_2^-$, for the $C - O$ bonds, one structure has a $C - O$ single bond (bond order = 1) and a $C=O$ double bond (bond order = 2).
The bond order of the $C - O$ bonds in the resonance hybrid of $HCO_2^-$ is $\frac{1 + 2}{2}=1.5$

Answer:

The bond order of the $C - O$ bonds in the formate ion ($HCO_2^-$) is 1.5. The resonance hybrid structure has a carbon atom double - bonded to one oxygen and single - bonded to the other in a delocalized sense, with the negative charge spread over the two oxygen atoms.