# Hybridization of carbon in CO2

I know that if there exists a carbon atom with single bonds only (with other atoms),then it is $\mathrm{sp^3}$ hybridized. And if at least one of the bond is double bond then its $\mathrm{sp^2}$ hybridized (not talking about two double bond on same carbon atom). And if there is at least one triple bond then its $\mathrm{sp}$ hybridized. But then carbon in carbon dioxide should be $\mathrm{sp^2}$ hybridized. But if I am right our teacher told it was $\mathrm{sp}$ hybridized. How is that possible if the carbon atom bears no triple bond?

• Carbon uses one of its non-hybridized p-atoms to bond to one oxygen, the other to bond to another oxygen. These bonds are pi-bonds, they lie in the form of a cross, at right angles to each other. Besides this, carbon is bonded to each of the oxygens via sigma-bonds, using its two sp-hybridized atoms. – CowperKettle May 4 '16 at 3:24
• – Martin - マーチン May 4 '16 at 8:19

Hybridization is determined by molecular geometry. Carbons participating in triple bonds like those in acetylene have two regions of electron density. Carbon dioxide is a linear molecule, and also has two regions of electron density. The fact that both have two regions of electron density are what make these carbons $\mathrm{sp}$ hybridized.

To reenforce hybridization is determined by molecular geometry, consider the benzyne intermediate. It has a triple bond, but is $\mathrm{sp^2}$ hybridized because it is too strained to become linear and only have two regions of electron density.

• How is it that we can compare two regions electron density in acetylene with CO2? – Abhishek P G May 5 '16 at 1:43
• The carbon atoms in both $\ce{CO2}$ and $\ce{HCCH}$ are participating in two $\sigma$ bonds (hence the two regions of electron density), and two $\pi$ bonds. The only difference between them is that in $\ce{CO2}$, the two $\pi$ bonds are not to the same atom. – ringo May 5 '16 at 1:48
• Well in that case its correct..............still the picture of double bond and triple bond confuses me..... – Abhishek P G May 5 '16 at 1:57

In ground state, the electron configuration of Carbon is $1s^2\ 2s^2 \ 2p^2$. You can consider one of the $2s$ electrons to be excited to fill the other empty $2p$ orbital to give a configuration of $1s^2\ 2s^1 \ 2p^3$. Each $2p$ orbital, $2p_x\ 2p_y$ and $2p_z$ now holds one electron. The $2s$ and one of the $2p$ orbitals, say the $2p_y$ can hybridize and form 2 sp hybrid orbitals.

Oxygen has the ground state electron configuration $1s^2\ 2s^2 \ 2p^4$. Two of the $2p$ orbitals, say the $2p_x$ and the $2p_z$ only hold one electron. The $2p_x$ can now overlap with one of the sp hybrids from the carbon to form a σ bond. The $2p_z$ can now overlap with the unhybridized $2p_z$ on the carbon to form a $\pi$ bond.

The same thing can happen on the other side of the carbon forming another $\sigma$ bond with Oxygen’s $2p_z$ and a $\pi$ bond with the $2p_y$ orbitals from each atom.

• Can the downvoter please re-check now? And if any error, do comment on the mistake. – ABcDexter May 29 '16 at 6:13

A triple bond takes two $p$ orbitals out of the sigma-bonding network and into pi-bonds, leaving just one $p$ orbital to combine with the $s$ orbital in the sigma bonding. Two double bonds do the same thing and thus also leave the atom, in this case carbon, $sp$ hybridized.