Question

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?

Short answer

The propylene molecule, \(\mathrm{C}_{3} \mathrm{H}_{6}\), has a total of 18 valence electrons. There are 8 sigma bonds in the molecule, using 16 valence electrons, and 1 pi bond using 2 valence electrons. No valence electrons remain in nonbonding pairs. The hybridization at each carbon atom in the molecule is: first carbon - sp², second carbon - sp³, and third carbon - sp³.

Step-by-step solution

(a) Finding the total number of valence electrons in propylene

To find the total number of valence electrons in the propylene molecule, we'll look at the individual atoms, \(\mathrm{C}\) and \(\mathrm{H}\). Carbon has 4 valence electrons, and hydrogen has 1 valence electron. In propylene, there are three carbon atoms and six hydrogen atoms. Therefore, the total number of valence electrons can be calculated as: Total valence electrons = (3 C atoms × 4 e⁻ per C) + (6 H atoms × 1 e⁻ per H) = 12 e⁻ + 6 e⁻ = 18 e⁻

(b) Finding the number of valence electrons used to make sigma bonds

The propylene molecule consists of a carbon-carbon double bond (C=C) and a carbon-carbon single bond (C-C). There are also six carbon-hydrogen single bonds (C-H). Sigma bonds are present in single bonds and one in a double bond. Therefore, in propylene, there are: Total sigma bonds = 1 (C-C) + 1 (C=C) + 6 (C-H) = 8 sigma bonds Each sigma bond consists of 2 valence electrons. The total number of valence electrons used in sigma bonds is: Electrons used in sigma bonds = 8 sigma bonds × 2 e⁻ per bond = 16 e⁻

(c) Finding the number of valence electrons used to make pi bonds

Pi bonds are present in double and triple bonds, and in the case of propylene, we have one carbon-carbon double bond (C=C). Since there is only one pi bond in the double bond, the number of valence electrons used in pi bonds is: Electrons used in pi bonds = 1 pi bond × 2 e⁻ per bond = 2 e⁻

(d) Finding the number of valence electrons remaining in nonbonding pairs

To find the number of valence electrons remaining in nonbonding pairs, we will subtract the valence electrons used in sigma and pi bonds from the total number of valence electrons: Nonbonding pairs of electrons = Total valence electrons - (Electrons used in sigma bonds + Electrons used in pi bonds) = 18 e⁻ - (16 e⁻ + 2 e⁻) = 0 e⁻

(e) Finding the hybridization of carbon atoms in propylene

To determine the hybridization of carbon atoms in propylene, we must first understand their bonding environment. There are three carbons in propylene: one is part of a double bond (C=C), one is part of a single bond (C-C), and one is bonded to both (C-C-C). - The carbon atom in the C=C double bond forms three sigma bonds (1 C=C and 2 C-H bonds) and one pi bond. Three sigma bonds require three hybrid orbitals, which means this carbon atom has sp² hybridization. - The carbon atom in the C-C single bond forms four sigma bonds (1 C-C, 1 C=C and 2 C-H bonds). Four sigma bonds require four hybrid orbitals, which means this carbon atom has sp³ hybridization. - The central carbon atom also forms four sigma bonds (1 C-C, 1 C=C, and 2 C-H bonds) and has the same environment as the second carbon atom. Therefore, it has sp³ hybridization as well. In conclusion, the hybridization at each carbon atom in the propylene molecule is: first carbon - sp², second carbon - sp³, and third carbon - sp³.

Key concepts

Valence Electrons

Valence electrons are crucial in determining how atoms interact within a molecule. They are the electrons found in the outermost shell of an atom and are responsible for forming chemical bonds.
To calculate the total number of valence electrons in a molecule like propylene \((\text{C}_3\text{H}_6)\), we need to consider all the atoms present. Carbon has 4 valence electrons, while hydrogen has 1 valence electron.
In propylene, there are 3 carbon atoms contributing a total of 12 valence electrons (4 electrons per carbon times 3) and 6 hydrogen atoms contributing 6 valence electrons (1 electron per hydrogen times 6). Adding these gives us a total of 18 valence electrons, which will be used in forming bonds or remain as nonbonding electron pairs.

Sigma Bonds

Sigma bonds are formed through the head-to-head overlap of atomic orbitals and are the strongest type of covalent bonds. They are pivotal in maintaining the skeletal structure of a molecule.
In a molecule like propylene, sigma bonds are present not only in single bonds but also as one component of double bonds.

• The propylene molecule comprises 8 sigma bonds: 1 for each single bond (1 C-C and 6 C-H) and 1 for the C=C double bond.

Each sigma bond involves 2 valence electrons. Hence, in propylene, 16 of the 18 valence electrons are used for sigma bond formation. This leaves us analyzing how the remaining electrons are utilized in pi bonds or as nonbonding electrons.

Pi Bonds

Pi bonds arise from the side-by-side overlap of p orbitals. They add to the double or triple bonding character in molecules, contributing to the stability and reactivity of the molecule.
In propylene, there is one carbon-carbon double bond \((\text{C} = \text{C})\), which includes a pi bond in addition to a sigma bond. This means that 2 of the valence electrons are employed in forming the pi bond.
It's important to remember that pi bonds are generally weaker than sigma bonds due to their side-to-side overlap, but they are essential for creating the double and triple bonds that influence the properties of organic molecules like propylene.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for pairing electrons to form chemical bonds.
In propylene, different carbon atoms exhibit different hybridization states based on their bonding environments:

• The carbon atom involved in the double bond \((\text{C} = \text{C})\) combines three sigma bonds \((1 \text{ C} = \text{C} + 2 \text{ C-H})\) requiring sp² hybridization.
• The second and third carbons forming four sigma bonds \((1 \text{ C-C}, 1 \text{ C=C}, 2 \text{ C-H})\) each have sp³ hybridization because they use a combination of their s and p orbitals to engage in bonding.

Understanding hybridization helps explain the geometry and spatial orientation of molecules, which is essential to predicting their physical and chemical properties.