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I was wondering if the Carbon atom in HCOOH (methanoic/formic acid) forms a positive partial charge by donating its electrons to both the Oxygen atoms, since they both possess a higher electronegativity value, or just one of them.

I was also curious about how the single bond and double bond between Carbon and the Oxygen atoms would affect the electron donation.

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  • $\begingroup$ I'm guessing the OP was trying to render $\ce{HC(=O)OH}$. $\endgroup$ – Oscar Lanzi Jan 21 at 11:49
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Electronic effects are cumulative. For example, the central carbon atom in an orthoester like HC(OEt)3 is surrounded by 3 oxygen atoms which add up.

In the case of formic acid (which is what you described in your question), yes the carbon atom is donating its electron density to BOTH oxygen atoms. However, the oxygen atom involved in the C=O bond will attract more electrons, because the double-bond is shorter.

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There is no straightforward answer to such a question, as it is not always obvious to indicate how one should count electrons being donated to other groups, or how to correctly partition electrons between nuclei.

In 1978, using ab initio calculations F.A. Momany has computed the Mulliken charges of the atoms of formic acid at the STO-3G level, and compared this to the use of an electrostatic potential. If one only looks at the Mulliken partial charges, this would tentatively answer the question.

The partial atomic charges of the C atom, the C=O oxygen and the OH oxygen are respectively is 0.262, -0.294 and -0.256. So it seems that the C might indeed be donating to both of the oxygen atoms. But one should also consider that the H atom bound to the OH oxygen is also donating some of its electron density to the O: the partial charge of the OH H atom is computed at 0.214. Thus, overall, the C atom is only contributing by a net positive charge of 0.042 to the O atom of the OH group.

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