Final answer:
The student's question involves drawing structural formulas for the isomers of C3H8O and C4H10, which include alcohols, ethers, and alkanes with their corresponding Lewis structures, predicting geometry around carbon atoms, which is tetrahedral, and determining carbon atom hybridization, which is sp3.
Step-by-step explanation:
The student is asking to draw structural formulas for isomers of certain compounds. Starting with C3H8O, there are three possible isomers: 1-propanol, 2-propanol, and methoxyethane (ethyl methyl ether). The isomers differ by the positioning of the -OH group (hydroxyl functional group) and the structure of the carbon skeleton.
For C4H10, which is butane, there are two isomers: n-butane and isobutane (methylpropane). These isomers differ in the arrangement of carbon atoms where n-butane is a straight chain and isobutane is branched.
For each isomer, a Lewis structure would involve placing the correct number of electrons around the carbon and oxygen atoms to reflect bonding and non-bonding pairs. The geometry about each carbon atom in alkanes (like butane) will be tetrahedral. In the case of alcohols and ethers with C3H8O formula, the geometry will also be roughly tetrahedral, but steric factors may cause slight deformations.
Hybridization of each type of carbon atom in these compounds would be sp3, as they are all saturated with single bonds. Hybridization reflects the mixing of atomic orbitals to form new hybrid orbitals suitable for bonding in the given geometry.