hybridization

The carbon atom in CO2 molecule is sp hybridized.

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Basic knowledge:

(1)
․ A single covalent bond consists of 1 σ bond.
․ A double bond consists of 1 σ bond and 1 π bond.
․ A triple bond consists of 1 σ bond and 2 π bonds.

(2)
Each C atom has 4 valence orbitals (2s 2px 2py 2pz).
․ In sp hybridization, 2 sp orbitals are formed and 2 p orbitals are left.
․ In sp2 hybridization, 3 sp2 orbitals are formed and 1 p orbital is left.
․ In sp3 hybridization, 4 sp3 orbitals are formed.

(3)
In a molecule, each C atom usually forms 4 covalent bonds.
․ Each hybridized sp/sp2/sp3 orbitals would form 1 σ bond.
․ Each unhybridized p orbital left would form 1 π bond.

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Reason for the answer:

In CO2 (O=C=O) molecules, the C atom forms 2 C=O double bonds. In the 2 C=O double bonds, there are 2 σ bond and 2 π bonds. Therefore, the C atom should have 2 hybridized (sp/sp2/sp3) orbitals to form the 2 σ bonds, and 2 unhybridized p orbitals to form the 2 π orbitals. Since the C atom have 2 hybridized orbitals, it is sp hybridized.

2008-06-23 23:30:38 補充：
The theory of hybridization is used to decribe the electronic state of the valence orbitals (to form covalent bonds) of central atoms (中央原子). There is no hybrization in terminal atoms (端邊原子).

In CO2, the two O atoms are terminal atoms. The two O atoms are NOT hybridized.

2008-06-23 23:30:47 補充：
The electronic configuration of each O atom is 1s2 2p4, and without hybridization. In each O atom, one p orbital (px) is paired up, and two 2p orbitals (py and pz) have unpaired electrons. The py orbital forms a σ bond and pz orbital form a π bond.

The carbon atom is sp hybridized, but the oxygen atom in CO2 is sp2 hybridized.

There are one sigma-bond, one pi-bond and 2 lone pairs on O-atom, the sigma electrons and the lone pair electrons come from hybrid orbitals and hence it is sp2 hybridized.

However, we uauslly talk about the state of hybridization of the central atom (C in CO2) only since the hybridization of other atoms do not affect the shape or chemistry of the molecule.