# Question #93d0a

Jul 10, 2016

No, there is no need for hybridization in $\text{^(-):"C"-="O} {:}^{+}$.

It's a diatomic linear molecule, and both atoms have all the necessary orbitals to make their bonds. So, no hybridization is required.

• There are one sigma ($\sigma$) and two pi ($\pi$) bonds made between $\text{C}$ and $\text{O}$, since a triple bond contains one $\sigma$ and two $\pi$ bonds.
• The $\sigma$ bond is made via a ${\text{C"_(2p_z)-"O}}_{2 {p}_{z}}$ head-on overlap. That is enough to make this bond, because each atom has one $2 {p}_{z}$ atomic orbital.
• The first $\pi$ bond is made via a ${\text{C"_(2p_x)-"O}}_{2 {p}_{x}}$ sidelong overlap. That is enough to make this bond, because each atom has one $2 {p}_{x}$ atomic orbital.
• The second $\pi$ bond is made via a ${\text{C"_(2p_y)-"O}}_{2 {p}_{y}}$ sidelong overlap. That is enough to make this bond, because each atom has one $2 {p}_{y}$ atomic orbital.

Finally, the nonbonding electron pair on $\text{C}$ and that on $\text{O}$ (otherwise known as "lone pairs") are stored in nonbonding molecular orbitals, which were nonbonding because they:

• could not overlap to form bonds because they weren't compatible with any other orbitals (incorrect symmetry).
• did not hybridize (because they didn't need to)

Lastly, if it were ${\text{CO}}_{2}$, then yes, there would be hybridization, but on $\text{^(-)"C"-="O} {:}^{+}$, no. The molecule has to have more than two atoms to require hybridization.

($: \stackrel{. .}{\text{O"="C"=stackrel(..)"O}} :$ has $s p$ hybridization on carbon and $s {p}^{2}$ hybridization on oxygen.)