(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? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

To draw the Lewis structures for these compounds, we first write down the central atoms (carbons) and the surrounding atoms (hydrogens). Then we connect the atoms with single, double or triple bonds, following the valence rules and distributing the electrons accordingly. - Ethane (\(C_2H_6\)): 1. Place two carbon atoms in the center. 2. Surround each carbon atom with three hydrogen atoms. 3. Connect the carbon atoms with a single bond, and each hydrogen atom with a single bond to the corresponding carbon atom. - Ethylene (\(C_2H_4\)): 1. Place two carbon atoms in the center. 2. Surround each carbon atom with two hydrogen atoms. 3. Connect the carbon atoms with a double bond, and each hydrogen atom with a single bond to the corresponding carbon atom. - Acetylene (\(C_2H_2\)): 1. Place two carbon atoms in the center. 2. Attach one hydrogen atom to each carbon atom. 3. Connect the carbon atoms with a triple bond, and each hydrogen atom with a single bond to the corresponding carbon atom. ##Step 2: Determine Hybridization##

The hybridization of carbon atoms in each compound can be determined based on the number of electron domains (bonding and non-bonding regions) around the carbon atom. - Ethane: Each carbon atom has 4 electron domains (three single bonds and no lone pairs), so the carbon atoms in ethane are \(sp^3\) hybridized. - Ethylene: Each carbon atom has 3 electron domains (two single bonds and one double bond), so the carbon atoms in ethylene are \(sp^2\) hybridized. - Acetylene: Each carbon atom has 2 electron domains (one single bond and one triple bond), so the carbon atoms in acetylene are \(sp\) hybridized. ##Step 3: Predict Planarity##

Molecules can be deemed planar if all their atoms lie in the same plane. - Ethane: Due to its tetrahedral (\(sp^3\)) hybridization, the carbon atoms in ethane have a bond angle of \(109.5^{\circ}\), which does not allow the molecule to be planar. - Ethylene: As the carbon atoms in ethylene have trigonal planar (\(sp^2\)) hybridization and bond angles of \(120^{\circ}\), the ethylene molecule is planar. - Acetylene: The carbon atoms in acetylene have linear (\(sp\)) hybridization, with bond angles of \(180^{\circ}\), so the acetylene molecule is also planar. ##Step 4: Count Sigma and Pi Bonds##

To count the \(\sigma\) and \(\pi\) bonds in each molecule, we need to examine the bonding nature of each bond present in the molecule. - Ethane: There are 7 single bonds (\(6\: C-H\text{ and }1\: C-C\)) so there are 7 \(\sigma\) bonds and 0 \(\pi\) bonds. - Ethylene: There are 5 bonds total (\(4\: C-H\text{ and }1\: C-C\)) and a double bond consists of 1 \(\sigma\) and 1 \(\pi\) bond, so there are 5 \(\sigma\) bonds and 1 \(\pi\) bond. - Acetylene: There are 3 bonds total (\(2\: C-H\text{ and }1\: C-C\)) and a triple bond consists of 1 \(\sigma\) and 2 \(\pi\) bonds, so there are 3 \(\sigma\) bonds and 2 \(\pi\) bonds. ##Step 5: Bonding Around Silicon and \(\pi\) Bond Formation##

If silicon formed molecules analogous to ethane, ethylene, and acetylene, its hybrid orbitals would be similar to carbon’s \(sp^3\), \(sp^2\), and \(sp\) hybrid orbitals, respectively. However, silicon does not readily form molecules with \(\pi\) bonds (like in ethylene and acetylene analogs) because of its larger size compared to carbon. The larger size results in inefficient overlap between orbitals, making the formation of \(\pi\) bonds energetically less favorable.