What is the hybridization of oxygen in formaldehyde?

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In formaldehyde (CH2O), the hybridization of the oxygen atom is sp2. The oxygen atom has six valence electrons, with one used to form a sigma bond with carbon, two existing as lone pairs, and one participating in a pi bond. This configuration necessitates three hybrid orbitals, which corresponds to sp2 hybridization. The presence of a double bond between carbon and oxygen confirms this hybridization type. Therefore, oxygen in formaldehyde is indeed sp2 hybridized.
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This isn't a homework problem, but rather a question that I thought about. Given a Formaldehyde CH_{2}O molecule, what would the hybridization of the oxygen atom be?

I know that the hybridization of the carbon atom would be:

C = \frac{\uparrow}{2sp^{2}} \frac{\uparrow}{2sp^{2}} \frac{\uparrow}{2sp^{2}} \frac{\uparrow}{2p}

I also know that given the double bond between carbon and oxygen, a sigma bond and pi bond will be required. As thus, this is what made me think about the configuration of oxygen's orbitals. If oxygen only makes 2 bonds, then what does that imply about its hybridization? Is it sp hybridization? Does it even hybridize?

O = \frac{\uparrow\downarrow}{2sp} \frac{\uparrow\downarrow}{2sp} \frac{\uparrow}{2p} \frac{\uparrow}{2p}
 
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Hybridization of Oxygen is sp2. You know that the no of valence electrons in Oxygen atom is 6.

One is used to make a sigma bond. (1) Two are lone pairs (1 + 1 = 2) and one is used for pi bond. Total hybrid orbitals required = 3. And sp2 has 3 hybrid orbitals.
 

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