Question

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule?

Short answer

In summary, (a) the Lewis structures for ethane, ethylene, and acetylene are: Ethane: H H \ / C | H-C | / \ H H Ethylene: H \ C=C | H Acetylene: H \ C≡C \ H (b) The hybridization for the carbon atoms are sp³ (ethane), sp² (ethylene), and sp (acetylene). (c) The planar molecules are ethylene and acetylene. (d) Number of σ and π bonds: Ethane: σ-bonds: 7 π-bonds: 0 Ethylene: σ-bonds: 5 π-bonds: 1 Acetylene: σ-bonds: 4 π-bonds: 2

Step-by-step solution

Draw Lewis structures for each molecule

The Lewis structure of a molecule represents the arrangement of atoms in the molecule and the bonds between them. To draw the structure, follow these steps for each molecule: 1. Count the total valence electrons of the molecule, 2. Arrange the atoms in the molecule, with the least electronegative atom in the center, 3. Add lone pairs and bonds according to the octet rule. Ethane: 1. Total valence electrons: 2(4 from C) + 6(1 from H) = 14 electrons 2. Arrangement: H₃C - CH₃ 3. Lewis structure: H H \ / C | H-C | / \ H H Ethylene: 1. Total valence electrons: 2(4 from C) + 4(1 from H) = 12 electrons 2. Arrangement: H₂C=CH₂ 3. Lewis structure: H \ C=C | H Acetylene: 1. Total valence electrons: 2(4 from C) + 2(1 from H) = 10 electrons 2. Arrangement: HC≡CH 3. Lewis structure: H \ C≡C \ H

Determine the hybridization of the carbon atoms

Hybridization is the mixing of atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons. The type of hybridization can be determined by counting the steric number (SN), which is the number of σ-bonds, and lone pairs around the central atom. Then, identify the hybridization according to the SN value. Ethane: SN (C) = 1 (C-C single bond) + 3 (C-H σ bonds) = 4 Hybridization: sp³ Ethylene: SN (C) = 2 (C-C double bond) + 2 (C-H σ bonds) = 3 Hybridization: sp² Acetylene: SN (C) = 3 (C-C triple bond) + 1 (C-H σ bonds) = 2 Hybridization: sp

Predict which molecules are planar

A molecule is considered planar if all atoms lie in the same plane. We can determine the molecule's planarity by looking at the hybridization of the central atom. Ethane: Carbon is sp³ hybridized, which leads to a tetrahedral geometry. This molecule is not planar. Ethylene: Carbon is sp² hybridized, which leads to a planar trigonal planar geometry. This molecule is planar. Acetylene: Carbon is sp hybridized, which leads to a linear geometry. This molecule is planar.

Determine the number of σ and π bonds in each molecule

In covalent molecules, σ bonds are formed by the overlap of hybridized orbitals, while π bonds are formed by lateral overlap of unhybridized orbitals. A double bond consists of one σ bond and one π bond, and a triple bond consists of one σ bond and two π bonds. Ethane: σ-bonds: 1 (C-C single bond) + 6 (C-H σ bonds) = 7 π-bonds: 0 Ethylene: σ-bonds: 2 (C-C double bond) + 4 (C-H σ bonds) = 5 π-bonds: 1 (C-C double bond) Acetylene: σ-bonds: 3 (C-C triple bond) + 2 (C-H σ bonds) = 4 π-bonds: 2 (C-C triple bond)

Key concepts

Hybridization

Hybridization is a concept in chemistry that describes how atomic orbitals mix to form new, hybrid orbitals. These hybrids are better suited for pairing electrons and typically involve central atoms in a molecule.
To determine the hybridization, count the steric number (SN), which includes both sigma bonds and lone pairs around the atom:

• **Ethane** has an SN of 4 (3 C-H bonds and 1 C-C bond), leading to an sp³ hybridization which corresponds to a tetrahedral arrangement, meaning its carbon atoms are each surrounded by four regions of electron density.


• **Ethylene** presents an SN of 3 (2 C-H bonds and 1 C=C bond). Therefore, each carbon atom is sp² hybridized, aligning with a trigonal planar shape. This indicates three regions of electron density.


• **Acetylene** has an SN of 2 (1 C-H bond and 1 C≡C bond), corresponding to sp hybridization. This lets the carbon atoms form a linear arrangement with two areas of electron density.

Understanding hybridization helps predict the geometry of molecules.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It's closely tied to the hybridization of the central atom and affects the molecule's properties such as polarity and reactivity.

• In **ethane**, with sp³ hybridization, the carbon atoms have a tetrahedral geometry. This geometry reflects a three-dimensional distribution where each carbon atom forms four sigma bonds, but the overall molecule is not planar due to its 3D arrangement.


• **Ethylene**, with sp² hybridization, adopts a trigonal planar geometry around each carbon atom. This planarity results because all atoms bonded to each carbon atom fall into the same plane.


• **Acetylene**, exhibiting sp hybridization, reveals a linear geometry. This results from the straight line that forms between the carbon and hydrogen atoms, and between the two carbon atoms.

By analyzing molecular geometry, you can infer a lot about a molecule's behavior and interactions.

Sigma and Pi Bonds

Sigma (\( \sigma \) ) and pi (\( \pi \) ) bonds are two types of covalent bonds that define how atoms connect within molecules.

• **Sigma bonds** are the strongest type of covalent chemical bond and they form because of the direct overlap of orbitals. They are present in all single bonds and are identified as the first bond in double or triple bonds. Therefore, all three molecules (ethane, ethylene, and acetylene) contain sigma bonds.


• **Pi bonds** arise from the side-to-side overlap of p orbitals. They add additional bonding power and are found in double and triple bonds, but never in single bonds. In the molecules discussed, ethylene has one pi bond (due to the double bond), and acetylene has two pi bonds (because of its triple bond).

By distinguishing between sigma and pi bonds, it's easier to understand a molecule's strength and stability.

Molecular Planarity

Molecular planarity examines whether all atoms in a molecule lie on the same plane.

• In **ethane**, due to its sp³ hybridization forming a tetrahedral structure, the molecule is inherently three-dimensional and thus not planar, due to the spatial orientation involving four bonding directions.


• **Ethylene** exhibits sp² hybridization leading to trigonal planar geometry. Thus, the molecule is planar since the bonds restrict all atoms to align in one flat plane, making it stable along one dimension.


• **Acetylene**, with its linear sp hybridization, is technically planar as well, although it consists merely of two points defining a line. This simplicity limits rotational freedom.

Planarity impacts molecular interactions with other substances and influences molecular reactivity.