# Drawing bicarbonate ion in different way [closed]

I don't know why can't I draw $$\ce{[HCO3]-}$$ like that:

I tried to draw it like that and checked everything. I checked formal charge and all was good. Why they draw it OH I know its also alright but whats wrong with my draw? 24e checked and formal charge checked. The upper left oxygen has 6 non bonding electrons the next one has 4 non bonding electrons. It's coordinative bond.

• I suggest to review grammar, bring formatting in order and append the relevant comments into the question. Is this structure is what you have in mind? – andselisk Nov 4 '19 at 7:19
• there is no picture its black screen . i mean its simple question everyone can understand with the picture i drew why making life hard ? if i knew the answer and someone asked i'll just tell him simple as that – Razi Awad Nov 4 '19 at 7:21
• It's a PNG with alpha-channel enabled. Change the background from black to another color or just enable checkerboard preview for images in your browser. – andselisk Nov 4 '19 at 7:23
• I have not enough points for a comment, but all the downratings are for the insufficient formatting of the question, not enough information on what you want, no effort in your work. What you drew and i believe you want to draw is the conjugate base (deprotonated form of the acid) of performic acid. If not, then you may have carbonic acid in mind, if so then you drew the wrong structure. Your question is not clear to me, but you may want to try ACD ChemSketch (free) software for drawing molecules. – Andrew Kovács Nov 4 '19 at 7:29
• Do you really suppose the strongly oxidizing hydroperoxide anion group directly bound to reducing aldehyde group would survive their marriage?? It is rather a recipe for an unstable explosive. – Poutnik Nov 4 '19 at 7:36

Who said you couldn’t? You absolutely can. What you have shown is performic acid $$\ce{CH2O3}$$ (or at least its conjugate base). Just to clarify, it is obviously different to carbonic acid $$\ce{H2CO3}$$.