(a) Draw Lewis structures for chloromethane $\left(\mathrm{CH}_{3} \mathrm{Cl}\right),\( chloroethene \)\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\right)\(, and chloroethyne \)\left(\mathrm{C}_{2} \mathrm{HCl}\right) .(\mathbf{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?

First, we'll draw the Lewis structures for each molecule. Recall that Lewis structures represent the arrangement of valence electrons around atoms in a molecule. In each structure, make sure that every atom has a full octet (except hydrogen, which can have a maximum of 2 electrons). **Chloromethane (CH3Cl):** The Lewis structure for chloromethane looks like this: ``` H | H - C - Cl | H ``` **Chloroethene (C2H3Cl):** The Lewis structure for chloroethene looks like this: ``` H2C = CH - Cl ``` **Chloroethyne (C2HCl):** The Lewis structure for chloroethyne looks like this: ``` HC ≡ C - Cl ```

Next, we'll find the hybridization of the carbon atoms in each molecule. Hybridization is the process of combining atomic orbitals from the same atom to create hybrid orbitals for chemical bonding. The hybridization of an atom can be determined by the following formula: Hybridization = Number of atoms bonded to the central atom + lone pairs on the central atom **Chloromethane (CH3Cl):** In chloromethane, the carbon atom forms four single bonds to three hydrogen atoms and one chlorine atom. Hence, the hybridization is sp³ since there are four orbitals interacting (one s orbital, three p orbitals). **Chloroethene (C2H3Cl):** In chloroethene, each carbon atom is double bonded to the other carbon and single bonded to either a chlorine or a hydrogen atom. Thus, it has a trigonal planar geometry and the hybridization is sp² (one s orbital, two p orbitals). **Chloroethyne (C2HCl):** In chloroethyne, one carbon atom is triple bonded to the other carbon and single bonded to a hydrogen or chlorine atom. This leads to a linear molecular geometry, so the hybridization is sp (one s orbital, one p orbital).

Now, we'll predict which molecules are planar. Planarity is determined by the molecule's geometry and the hybridization of its central atom. **Chloromethane (CH3Cl):** Chloromethane has tetrahedral geometry due to its sp³ hybridization. Tetrahedral molecules are not planar; therefore, chloromethane is not a planar molecule. **Chloroethene (C2H3Cl):** Chloroethene has trigonal planar geometry because of its sp² hybridization. Trigonal planar molecules are planar, so chloroethene is a planar molecule. **Chloroethyne (C2HCl):** Chloroethyne is linear due to its sp hybridization. Linear molecules are planar, so chloroethyne is a planar molecule.

Lastly, we'll find the number of σ and π bonds in each molecule. Sigma bonds (σ) are single bonds formed from hybrid orbitals overlapping, while pi bonds (π) are formed from unhybridized p orbitals overlapping. **Chloromethane (CH3Cl):** Chloromethane has four single bonds (C-H and C-Cl). All single bonds are σ bonds, so it has 4 σ bonds and 0 π bonds. **Chloroethene (C2H3Cl):** Chloroethene has one C=C double bond and two single bonds (C-H and C-Cl). The double bond consists of one σ and one π bond, and the single bonds are σ bonds. Therefore, it has 3 σ and 1 π bond. **Chloroethyne (C2HCl):** Chloroethyne has one C≡C triple bond and two single bonds (C-H and C-Cl). The triple bond comprises of one σ and two π bonds, and the single bonds are σ bonds. So it has 3 σ and 2 π bonds.