Determine the correct hybridization (from left to right) about each central atom in CH3CH2OH.
A) 1st C sp3 ; 2nd C sp2 ; O sp B) 1st C sp3 ; 2nd C sp3 ; O sp3
C) 1st C sp ; 2nd C sp2 ; O sp3 D) 1st C sp3 ; 2nd C sp2 ; O sp2

B) 1st C sp3 ; 2nd C sp3 ; O sp3

Explanation:

Both carbons in ethanol have an sp³ configuration of their orbitals, given that each carbon atom binds to four other atoms, as can be seen on the extended formula in the attachment.

In these carbons (1st and 2nd) one s orbital and three p orbitals are hybridized to form four hybrid orbitals sp³. These orbitals form four bonds and their bond angles are close to the tetrahedral value of 109.5º,as can be seen for methanol which also has an sp³ configuration.

The oxygen atom can also be described as having sp³ hybrid orbitals. Two of the four hybrid sp³ orbitals in oxygen are occupied by lone electron pairs, and two are used to form bonds with other atoms (in this case the oxygen is bonded to a carbon atom and a hydrogen atom).

Therefore, the three central atoms have an sp³ configuration.