The carbon atom has ##sp## hybridization; the ##”O”## atoms have ##sp^2## hybridization.

You must first draw the Lewis structure for ##”CO”_2##.

According to theory, we can use the steric number ##(“SN”)## to determine the hybridization about an atom.

##”SN”## = number of lone pairs + number of atoms directly attached to the atom.

We see that the ##”C”## atom has ##”SN = 2″##. It has no lone pairs, but it is attached to two other atoms. It is ##sp## hybridized.

Each ##”O”## atom has ##”SN = 3″##. It has 2 lone pairs and is attached to 1 ##”C”## atom.

Just as the carbon atom hybridized to form the best bonds, so do the oxygen atoms.

The valence electron configuration of ##”O”## is ##[“He”] 2s^2 2p^4##.

To accommodate the two lone pairs and the bonding pair, it will also form three equivalent ##sp^2## hybrid orbitals.

Two of the ##sp^2## orbitals contain lone pairs, while the remaining ##sp^2## orbital and the unhybridized ##p## orbital have one electron each.

We can see this arrangement in the ##”C=O”## bond of formaldehyde, which is equivalent to the right hand side of the ##”O=C=O”## molecule.

(from www.slideshare.net)

There is a similar arrangement on the left side of the ##”O=C=O”## molecule, but the ##pi## bond is horizontal rather than vertical.

Here is a video on the hybridization of carbon dioxide.

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