Question

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{4}\) 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?

Short answer

(a) The total number of valence electrons in the propylene molecule is 16. (b) 8 valence electrons are used to make σ bonds in the molecule. (c) 2 valence electrons are used to make π bonds in the molecule. (d) 6 valence electrons remain in nonbonding pairs in the molecule. (e) The hybridization at each carbon atom in the molecule is sp², sp², and sp².

Step-by-step solution

Determine the total number of valence electrons for propylene.

Propylene has the molecular formula \(\mathrm{C}_{3}\mathrm{H}_{4}\), which means it contains three carbon atoms and four hydrogen atoms. Each carbon atom has 4 valence electrons and each hydrogen atom has 1 valence electron. So, the total number of valence electrons in propylene is: $$\text{Total valence electrons} = 3 (\text{carbon atoms}) \times 4 (\text{valence electrons/carbon}) + 4 (\text{hydrogen atoms}) \times 1 (\text{valence electrons/hydrogen}) = 12 + 4 = 16$$

Count the number of electrons involved in σ bonds in the molecule.

Each single bond in the molecule involves two electrons. A σ bond is a single bond formed between two atoms. In propylene, we have three single \(\mathrm{C-H}\) bonds and one single \(\mathrm{C-C}\) bond. So, to find the number of electrons involved in σ bonds, we count the total number of single bonds and multiply by 2: $$\text{Electrons in σ bonds} = (3 \times \text{C-H single bonds} + 1 \times \text{C-C single bond}) \times 2 = (3 + 1) \times 2 = 4 \times 2 = 8$$

Count the number of electrons involved in π bonds in the molecule.

A π bond is formed between two atoms when there is a double or triple bond present. In propylene, there is one double \(\mathrm{C-C}\) bond which involves π bond. We have one π bond in the molecule, and each π bond involves 2 electrons: $$\text{Electrons in π bonds} = 1 \times \text{C-C π bond} \times 2 \ \text{electrons} = 1 \times 2 = 2$$

Calculate the number of nonbonding electrons in the molecule.

To find the number of nonbonding electrons, we subtract the total electrons used in both σ bonds and π bonds from the total number of valence electrons: $$\text{Nonbonding electrons} = \text{Total valence electrons} - (\text{Electrons in σ bonds} + \text{Electrons in π bonds}) = 16 - (8 + 2) = 16 - 10 = 6$$

Determine the hybridization at each carbon atom in the molecule.

In propylene, the first carbon atom is connected to two hydrogen atoms and forms a single bond with the second carbon atom. Hence, it forms three σ bonds, and its hybridization is \(\mathrm{sp^2}\). The second carbon atom makes two σ bonds (one single bond with the first carbon and one single bond with the third carbon) and one π bond (double bond with the third carbon atom). Thus, it also has \(\mathrm{sp^2}\) hybridization. The third carbon atom forms one σ bond with the second carbon atom and one σ bond with a hydrogen atom. It is also involved in a π bond (double bond) with the second carbon atom. So, its hybridization is \(\mathrm{sp^2}\) as well. In summary: (a) Total number of valence electrons in propylene: 16 (b) Number of electrons involved in σ bonds: 8 (c) Number of electrons involved in π bonds: 2 (d) Number of nonbonding electrons: 6 (e) Hybridization at each carbon atom: sp², sp², and sp².

Key concepts

Valence Electrons in Propylene

Valence electrons are the outermost electrons of an atom and are crucial for forming chemical bonds. In our example of propylene, an organic molecule with the formula \( \mathrm{C}_{3} \mathrm{H}_{4} \), we analyze its valence electron count to predict bonding patterns and reactivity.

In the case of propylene, each carbon atom contributes four valence electrons, as carbon is in group 14 of the periodic table. Hydrogen, being in group 1, contributes one valence electron per atom. Therefore, the total valence electron count for propylene is the sum of the contributions from all the carbon and hydrogen atoms:

\[\text{Total valence electrons} = 3 \times 4 + 4 \times 1 = 12 + 4 = 16\]
This count is fundamental for understanding the bonding and structure within the molecule, which in turn affects its physical and chemical properties.

Sigma and Pi Bonds in Propylene

Understanding \( \sigma \) and \( \pi \) bonds is crucial for deciphering molecular geometry and stability. \( \sigma \) bonds are single covalent bonds that form between two atoms and constitute the primary bond holding the atoms together.

In propylene, \( \sigma \) bonds include one bond from each carbon to hydrogen (\( \mathrm{C-H} \) bonds) as well as the bond between carbon atoms (\( \mathrm{C-C} \) bonds). Every single bond is a \( \sigma \) bond and involves two electrons. With three \( \mathrm{C-H} \) bonds and one \( \mathrm{C-C} \) bond, the number of electrons in \( \sigma \) bonds totals eight.

\( \pi \) bonds, on the other hand, refer to the type of covalent bond which occurs alongside a \( \sigma \) bond in double bonds (and also in triple bonds, but not in the context of propylene). In propylene, there is one double bond between carbon atoms containing one \( \sigma \) bond and one \( \pi \) bond. The \( \pi \) bond in the double bond involves two electrons, contributing to the unique chemical reactivity and physical properties of the molecule.

Hybridization in Propylene

Hybridization is the process of mixing atomic orbitals to form new hybrid orbitals that can overlap to form covalent bonds. This concept helps explain the geometry of molecular bonding beyond the classical valence-shell electron-pair repulsion (VSEPR) theory.

In propylene, each carbon atom is \( \mathrm{sp^2} \) hybridized, which means that one s orbital has mixed with two p orbitals to form three equivalent sp2 hybrid orbitals. These hybrid orbitals can form \( \sigma \) bonds, while the remaining p orbital that is not hybridized forms the \( \pi \) bond. This hybridization gives rise to a trigonal planar layout around each carbon atom, which impacts the molecule's overall shape and reactivity. A clear grasp of hybridization is essential for prediction of molecular structure and bonding characteristics in organic chemistry.