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(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?

Short Answer

Expert verified
The Lewis structures, hybridizations, planar nature, and σ/π bonds for the given molecules are as follows: **Chloromethane (CH3Cl):** Lewis structure: ``` H | H - C - Cl | H ``` Hybridization: sp³, Planarity: Not planar, σ bonds: 4, π bonds: 0 **Chloroethene (C2H3Cl):** Lewis structure: ``` H2C = CH - Cl ``` Hybridization: sp², Planarity: Planar, σ bonds: 3, π bonds: 1 **Chloroethyne (C2HCl):** Lewis structure: ``` HC ≡ C - Cl ``` Hybridization: sp, Planarity: Planar, σ bonds: 3, π bonds: 2

Step by step solution

01

Draw Lewis Structures

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 ```
02

Determine Carbon Hybridization

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).
03

Predict Planar Molecules

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.
04

Calculate the Number of σ and π Bonds

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hybridization
Hybridization is a fundamental concept in understanding chemical bonding because it explains the shapes and bond angles in molecules. When atomic orbitals mix to form new, identical hybrid orbitals, this process is known as hybridization. This occurs to maximize the number of bonds an atom can form, increasing the molecule's stability.
To determine the hybridization for a central atom, count the number of atoms it is directly bonded to plus any lone pairs it possesses.
  • For chloromethane \((CH_3Cl)\), the carbon is bonded to three hydrogens and one chlorine, making the hybridization sp³.
  • In chloroethene \(C_2H_3Cl\), each carbon is bonded to two other atoms and has a double bond, resulting in sp² hybridization.
  • Chloroethyne \(C_2HCl\) features a carbon bonded just to hydrogen or chlorine and another carbon, forming sp hybridization due to a triple bond and linear shape.
Understanding these hybridization states helps predict molecule geometry and reactivity.
Planarity
Planarity refers to the flatness of a molecule, which is highly dependent on its geometry and the hybridization state of its central atom. It plays a key role in chemical reactions and interactions, as planar molecules can more readily participate in reactions.
For a molecule to be planar:
  • It must have a linear or trigonal planar geometry.
  • The atoms involved typically exhibit specific hybridizations that favor planarity.
Chloromethane, with its sp³ hybridization and tetrahedral geometry, is not planar. In contrast, chloroethene is planar because it has a trigonal planar shape due to sp² hybridization. Meanwhile, chloroethyne, with a linear arrangement from sp hybridization, is also planar. Recognizing planarity aids in foreseeing physical properties and interaction capabilities.
Sigma and Pi Bonds
Sigma (σ) and pi (π) bonds are essential in understanding different bond types in molecular chemistry. Sigma bonds are formed by the end-to-end overlap of orbitals, while pi bonds are created by the side-to-side overlap of \(p\) orbitals. These bonds greatly influence a molecule's strength and stability.
  • A single bond is always a σ bond.
  • A double bond consists of one σ and one π bond.
  • A triple bond comprises one σ and two π bonds.
For chloromethane, there are four single \(C-H\) and \(C-Cl\) bonds, all σ bonds. Chloroethene features a \(C=C\) double bond with 1 σ and 1 π bond, in addition to \(σ\) bonds from \(C-H\) and \(C-Cl\) connections. With one \(C≡C\) triple bond, chloroethyne includes 1 σ and 2 π bonds, supplemented by σ bonds from \(C-H\) and \(C-Cl\) single bonds. Knowing the type and number of σ and π bonds helps with understanding bond strength and reactivity.
Chemical Bonding
Chemical bonding describes how atoms connect to form molecules, which essentially forms the architecture of molecules. It involves the sharing or transfer of electrons to achieve full valence shells, leading to stable structures. There are various types of chemical bonds, primarily categorized into covalent and ionic bonds.
  • In covalent bonding, atoms share electrons to reach a stable octet, like in all three examples given \(CH_3Cl, C_2H_3Cl, C_2HCl\).
  • Ionic bonds, in contrast, form when electrons are transferred between atoms, creating charged ions.
The understanding of chemical bonds extends to recognizing how hybridization forms stable molecules with optimal shapes. Each bond type influences a molecule’s physical properties, chemical reactivity, and likeability to certain reactions. Grasping these bonding principles lays the foundation for further studies in molecular structure and reactions.

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Most popular questions from this chapter

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

Consider the \(\mathrm{H}_{2}^{+}\) ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\) ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\) ? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher- energy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\) ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}\), or \(\mathrm{Be}_{2}^{+}\) be expected to exist? Explain.

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_4) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(O\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of OSF \(_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

The structure of borazine, \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(B\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(B\) and \(N,\) do the formal charges seem favorable or unfavorable? (d) Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the \(\mathrm{B}\) and \(\mathrm{N}\) atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(\mathrm{B}-\mathrm{N}\) bonds in the borazine molecule are all identical in length at \(144 \mathrm{pm} .\) Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are \(151 \mathrm{pm}\) and \(131 \mathrm{pm},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of botazine?

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