CO2 has sp hybridization and no lone pairs on carbon.

enter image description here

For some reason calculation with VSEPR gave sp2 hybridization and lone pair on carbon.


closed as unclear what you're asking by Mithoron, jerepierre, airhuff, Tyberius, aventurin May 15 '18 at 18:38

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    $\begingroup$ Actually, from where did you read that "It has 1 lone pair in reality"? The picture you've attached shows no lone pair either. $\endgroup$ – Gaurang Tandon May 15 '18 at 15:36
  • $\begingroup$ Why do you think the hybridization is $sp^{2}$? $\endgroup$ – Zhe May 15 '18 at 15:49

The carbon atom in $CO_2$ has zero lone pairs of nonbonding electrons.

[By the way, the carbon has $sp$ hybridization, as it requires the bonding orbitals to have the same lobe sign on each side to make two identical $\sigma$ bonds. The $2p_z$ orbital alone is not sufficient to make two identical bonds, because the molecule would then lose the reflection symmetry it ought to have.]

Let's construct the Lewis structure ourselves:

  • Carbon has four valence electrons. We accept that.
  • Each oxygen has six valence electrons. We also accept that.

Therefore, the total number of valence electrons is $4+6+6=16$. Carbon is the larger atom, so this is a more usual structure with the largest atom in the center. The skeletal structure so far is:


We place three lone pairs of electrons on each oxygen next, to account for $2+2$ bonding and $2\times3+2\times3=12$ (currently) nonbonding electrons, for a total of $16$ as we needed.

Lastly, we use one lone pair from each oxygen to make a $\pi$ bond. This only redistributes electrons, without adding any extra or losing any. That results in:


And are the formal charges (FCs) minimized?

  • Carbon owns 4 valence electrons and came in with 4 valence electrons, so $FC = 0$.
  • Oxygen owns 6 valence electrons and came in with 6 valence electrons, so $FC = 0$ as well.

This must be the actual Lewis structure, and it also is in the correct image you have placed in your question.

  • $\begingroup$ SORRY I PUT 6 VALENCE ELECTRONS IN THE VSEPR FORMULA WHEN IT'S 4. AND IT'S HYBRIDISATATION IS sp. And it has no lone pair as L=H-X-D =2-0-2=0. Sorry for the mistake. @mithoron $\endgroup$ – user187604 May 15 '18 at 22:52
  • $\begingroup$ If you can delete the original question it would be better as the question means no sense now @mithoron $\endgroup$ – user187604 May 15 '18 at 22:57
  • $\begingroup$ @user187604 Well, now it's a bit trickier to delete it. I can flag it for mods to consider it, but it would also mean deleting the answer. $\endgroup$ – Mithoron May 16 '18 at 0:21
  • $\begingroup$ @Mithoron does the answerer lose the reputation for deleting the question? If that does happen don't delete it. If not please do. $\endgroup$ – user187604 May 16 '18 at 0:50
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    $\begingroup$ @user187604 I might just have asked, "How do you determine the Lewis structure of CO2 from scratch? I tried this and got one lone pair on carbon for some reason." I can edit my answer to reflect that. No one is trying to embarrass anyone here. $\endgroup$ – timaeus222 May 17 '18 at 0:49

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