Propylene, \(\mathrm{C}_{3} \mathrm{H}_{6},\) is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

In order to find the total number of valence electrons in the propylene molecule, we can simply add the number of valence electrons from each individual atom. Carbon has 4 valence electrons, and hydrogen has 1 valence electron. Since there are 3 carbon atoms and 6 hydrogen atoms, the total number of valence electrons will be: \(3\times4(valence\,electrons\,from\,carbon) + 6\times1(valence\,electrons\,from\,hydrogen) = 12 + 6 = 18\). There are 18 valence electrons in the propylene molecule.

A single bond is a σ bond. In the propylene molecule, we have the following bonds: - 2 C-C bonds (C1-C2 and C2-C3) - 6 C-H bonds In total, there are 8 σ bonds in the propylene molecule. Since each bond represents 2 electrons, this implies that 16 valence electrons are used to form σ bonds.

A double bond consists of one σ bond and one π bond. In propylene, there is a double bond between carbon-1 and carbon-2 (C1=C2). So, we have: 1 π bond in the molecule Since each bond represents 2 electrons, this would mean that 2 valence electrons are used to form π bonds in the molecule.

As we've identified that 16 valence electrons are used to form σ bonds and 2 valence electrons are used to form π bonds, any remaining valence electrons must be nonbonding pairs. From (a), we know there are a total of 18 valence electrons in the molecule, and we've accounted for 16 + 2 = 18 electrons in (b) and (c), so no valence electrons remain as nonbonding pairs in the propylene molecule.

There are 3 carbon atoms, each with a different hybridization state: 1. Carbon-1 (C1): It forms a single bond with one hydrogen (C-H), and a double bond with carbon-2 (C=C). This means it has 3 σ bonds (including one from the double bond) and one π bond, which requires an sp2 hybridization. 2. Carbon-2 (C2): It forms a single bond with carbon-3 (C-C) and two single bonds with hydrogen atoms (C-H). This means it has 3 σ bonds and one π bond (C1-C2 double bond), which also requires an sp2 hybridization. 3. Carbon-3 (C3): It forms a single bond with carbon-2 (C-C) and two single bonds with hydrogen atoms (C-H). This means it has 3 σ bonds, which requires an sp3 hybridization. Remember that in this case there is no π bond. So, the hybridization states of the three carbon atoms in the propylene molecule are: C1: sp2, C2: sp2, and C3: sp3.