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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.

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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
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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:

$\text{O}-\text{C}-\text{O}$

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:

$:\ddot{\text{O}}=\text{C}=\ddot{\text{O}}:$

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.

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  • $\begingroup$ 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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