Problem 51 - Chemistry: The Central Science

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? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about Si in terms of hydrid orbitals? Does it make a difference that Si lies in the row below \(\mathrm{C}\) in the periodic table? Explain.

Step-by-Step Solution

Answer

In summary, the Lewis structures, hybridization, planarity, and number of σ and π bonds for ethane, ethylene, and acetylene are as follows: 1. Ethane: H-C-C-H (sp3 hybridized, not planar, 7 σ bonds, 0 π bonds) | | H H 2. Ethylene: H-C=C-H (sp2 hybridized, planar, 5 σ bonds, 1 π bond) | | H H 3. Acetylene: H-C≡C-H (sp hybridized, planar, 3 σ bonds, 2 π bonds) Assuming silicon can form analogs of these molecules, it would show similar hybridizations (sp3, sp2, and sp), despite being in the row below carbon in the periodic table.

Step 1: (a) Drawing Lewis structures

Draw the Lewis structures for ethane, ethylene, and acetylene as follows: 1. Ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\): Carbon atoms are single-bonded to each other and surrounded by three hydrogen atoms each. H-C-C-H | | H H 2. Ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\): Carbon atoms are double-bonded to each other and surrounded by two hydrogen atoms each. H-C=C-H | | H H 3. Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\): Carbon atoms are triple-bonded to each other and surrounded by one hydrogen atom each. H-C≡C-H

Step 2: (b) Hybridization of carbon atoms

The hybridization of carbon atoms in each molecule can be determined as follows: 1. Ethane: Each carbon atom forms 4 single bonds (3 with hydrogen and 1 with another carbon). Therefore, the carbon atoms in ethane are sp3 hybridized. 2. Ethylene: Each carbon atom forms 1 double bond with another carbon and 2 single bonds with hydrogen. Therefore, the carbon atoms in ethylene are sp2 hybridized. 3. Acetylene: Each carbon atom forms 1 triple bond with another carbon and 1 single bond with hydrogen. Therefore, the carbon atoms in acetylene are sp hybridized.

Step 3: (c) Predicting planarity of molecules

Based on the hybridization, we can predict if the molecules are planar or not: 1. Ethane (sp3 hybridization): Tetrahedral geometry; not planar. 2. Ethylene (sp2 hybridization): Trigonal planar geometry; planar. 3. Acetylene (sp hybridization): Linear geometry; planar.

Step 4: (d) Counting σ and π bonds

The number of sigma (σ) and pi (π) bonds in each molecule can be counted as follows: 1. Ethane: 7 σ bonds (6 C-H and 1 C-C) and 0 π bonds. 2. Ethylene: 5 σ bonds (4 C-H and 1 C-C) and 1 π bond (1 C=C). 3. Acetylene: 3 σ bonds (2 C-H and 1 C-C) and 2 π bonds (1 C≡C).

Step 5: (e) Bonding about Si in terms of hybrid orbitals

Assuming silicon can form molecules that are analogs with ethane, ethylene, and acetylene: 1. In ethane-like molecules, Si would be sp3 hybridized (forming 4 single bonds). 2. In ethylene-like molecules, Si would be sp2 hybridized (forming 1 double bond and 2 single bonds). 3. In acetylene-like molecules, Si would be sp hybridized (forming 1 triple bond and 1 single bond). Silicon, lying in the row below carbon in the periodic table, has a larger atomic size and an additional electron shell. But it does not make a considerable difference in hybridization as it can form similar types of hybrid orbitals (sp3, sp2, and sp) with its valence electrons.

Key Concepts

Hybridization of Carbon Atoms

Understanding the hybridization of carbon atoms in different compounds is fundamental in visualizing the shapes and potential reactivity of molecules. Hybridization explains how carbon atoms can form different numbers of bonds with various geometries. In ethane \(\mathrm{C}_{2}\mathrm{H}_{6}\), each carbon atom is sp3 hybridized because it forms four single bonds: three with hydrogen atoms and one with another carbon atom, resulting in a tetrahedral shape.

Ethylene \(\mathrm{C}_{2}\mathrm{H}_{4}\), on the other hand, shows sp2 hybridization in each carbon atom. This is due to the presence of one double bond with another carbon and two single bonds with hydrogen atoms. The sp2 hybridization leads to a trigonal planar arrangement that brings about the planarity of the molecule.

Finally, in acetylene \(\mathrm{C}_{2}\mathrm{H}_{2}\), the carbon atoms are sp hybridized. This is because each carbon forms a triple bond with the other carbon and a single bond with a hydrogen atom, giving rise to a linear geometry. Through hybridization, we can better understand the 3D arrangement of atoms in a molecule and the resulting molecular geometry.

Sigma and Pi Bonds

Sigma \(\sigma\) and pi \(\pi\) bonds are types of covalent bonds that differ in their electron orbital overlap. Sigma bonds involve head-on overlapping and are generally stronger and more stable than pi bonds. Pi bonds, which involve side-by-side overlap, are more reactive. In ethane, which is solely composed of sigma bonds, we find a stable molecular structure with six C-H sigma bonds and one C-C sigma bond.

Ethylene introduces the concept of pi bonds to the mix, with four C-H sigma bonds, one C-C sigma bond, and one additional pi bond formed from p-orbital overlap. This pi bond is responsible for the rigidity and planarity of the molecule.

Acetylene escalates the pi bond count to two, with two C-H sigma bonds and one C-C sigma bond. The presence of two parallel pi bonds in acetylene creates a triple bond with the adjoining carbon, resulting in a molecule with higher potential energy and reactivity than ethane and ethylene.

Molecular Geometry

Molecular geometry is determined by the spatial arrangement of atoms in a molecule. The geometry impacts physical and chemical properties, such as boiling points, polarity, and potential chemical reactions. Ethane exhibits a tetrahedral molecular geometry due to its sp3 hybridization, meaning it is not planar. Ethylene, with sp2 hybridization, adopts a trigonal planar molecular geometry, making it indeed planar, which can be significant in chemical reactions that involve the pi bonds.

Acetylene is distinguished by its linear molecular geometry owing to sp hybridization, also classified as planar. These molecular geometries are crucial for understanding the molecule's bond angles and potential interactions, embedding a deeper comprehension of how structure affects the function and reactivity of different molecules.