Question #93d0a

No, there is no need for hybridization in `""^(-):"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 `"C"` and `"O"`, since a triple bond contains one `sigma` and two `pi` bonds.
• The `sigma` bond is made via a `"C"_(2p_z)-"O"_(2p_z)` head-on overlap. That is enough to make this bond, because each atom has one `2p_z` atomic orbital.
• The first `pi` bond is made via a `"C"_(2p_x)-"O"_(2p_x)` sidelong overlap. That is enough to make this bond, because each atom has one `2p_x` atomic orbital.
• The second `pi` bond is made via a `"C"_(2p_y)-"O"_(2p_y)` sidelong overlap. That is enough to make this bond, because each atom has one `2p_y` atomic orbital.

Finally, the nonbonding electron pair on `"C"` and that on `"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 `"CO"_2`, then yes, there would be hybridization, but on `""^(-)"C"-="O":^(+)`, no. The molecule has to have more than two atoms to require hybridization.

(`:stackrel(..)"O"="C"=stackrel(..)"O":` has `sp` hybridization on carbon and `sp^2` hybridization on oxygen.)