Question

consider the bicarbonate ion (also called the hydrogen carbonate ion). after drawing the correct lewis dot structure(s), you would see:
a two equivalent resonance forms.
b three equivalent resonance forms.
c three single bonds around the central carbon atom.
d four single bonds around the central carbon atom.
e two double bonds around the central carbon atom.

Explanation:

The bicarbonate ion (\(HCO_3^-\)) has a central carbon atom. Carbon has 4 valence electrons, hydrogen has 1, each oxygen has 6, and there is an extra electron due to the - 1 charge. The total number of valence electrons is \(1 + 4+3\times6 + 1=24\).
Carbon forms one single bond with \(H\), one single bond with one \(O\) (with three lone pairs), one single bond with another \(O\) (with three lone pairs and a negative charge), and one double bond with the third \(O\) (with two lone pairs).
Resonance occurs when the double - bond can be delocalized among the three oxygen atoms. But in \(HCO_3^-\), because of the presence of the \(H\) atom, there are two equivalent resonance forms (the double bond can be between \(C\) and two of the three \(O\) atoms, while the third \(O\) is bonded singly and has a negative charge).
Option C is wrong because there is a double bond. Option D is wrong as there is a double bond. Option E is wrong as there is only one double bond in each resonance form. Option B is wrong because of the \(H\) atom (not three equivalent resonance forms).

Answer:

A. two equivalent resonance forms.