0
$\begingroup$

Which of the given compounds is most acidic? $\ce{CO}$ or $\ce{CO2}$ ?

How do I solve this question, and how do I explain it?

I took the oxidation state of carbon in both. The oxidation state in $\ce{CO2}$ is higher, therefore it is more electronegative, so therefore $\ce{CO2}$ may be more acidic.
For $\ce{CO }$ I don't know about the charge so that I can use it to understand if it would give electron or accept electrons, so that I can apply Lewis acid/base theory.

$\endgroup$
0

1 Answer 1

5
$\begingroup$

Carbon dioxide is a much stronger Lewis acid than carbon monoxide. In carbon dioxide, the carbon has a much higher oxidation state (formally, $+4$ vs. the $+2$ of carbon monoxide), and the two oxygen atoms heavily withdraw electron density from the carbon atom, leaving a very large partial positive charge ($\delta^{+}$). This is rigorously confirmed by electrostatic potential maps and calculated orbital coefficients. As a consequence, the electron-deficient carbon of $\ce{CO2}$ is highly electrophilic, and reacts with a variety of nucleophiles (e.g., Grignard reagents, organolithium reagents, etc.).

Carbon monoxide, on the other hand, actually contains substantial electron density at the carbon atom (in spite of the bonds being generally polarized towards the oxygen atom), which possesses a fairly localized and active lone pair. Consequently, a few notable exceptions aside, carbon monoxide mostly functions as a Lewis base.

$\endgroup$
1
  • $\begingroup$ Formal charge analysis reinforces the fact that carbon monoxide is often a Lewis base; the carbon portion has a lone pair of electrons and a negative formal charge. Also see the case of the cyanide ion, in which the carbon is the more nucleophilic of the two constituent atoms. $\endgroup$
    – Dissenter
    Commented Jun 22, 2014 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.