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.)