Lone pair on carbon in CO2? CO2 has sp hybridization and no lone pairs on carbon.



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