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Ch.8 - Basic Concepts of Chemical Bonding
Brown - Chemistry: The Central Science 14th Edition
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Tutti i libri di testoBrown 14th EditionCh.8 - Basic Concepts of Chemical BondingProblema 94a
Capitolo 8, Problema 94a

Ortho-Dichlorobenzene, C6H4Cl2, is obtained when two of the adjacent hydrogen atoms in benzene are replaced with Cl atoms. A skeleton of the molecule is shown here. (a) Complete a Lewis structure for the molecule using bonds and electron pairs as needed.

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1
Identify the basic structure of benzene, which is a hexagonal ring consisting of six carbon atoms with alternating double bonds and each carbon atom bonded to one hydrogen atom.
Replace two adjacent hydrogen atoms in the benzene ring with chlorine (Cl) atoms. This modification will occur at the ortho positions, which are the positions next to each other on the benzene ring.
Draw single bonds from the carbon atoms to the chlorine atoms, as chlorine typically forms single bonds in organic compounds.
Complete the remaining bonds in the benzene ring. Ensure that each carbon atom forms three bonds in total and maintains the alternating double bond structure where possible. Adjust the double bonds if necessary to accommodate the electron pairs shared with the chlorine atoms.
Add lone pairs to the chlorine atoms. Each chlorine atom should have three pairs of lone electrons, which are not involved in bonding but influence the shape and reactivity of the molecule.

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Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They use dots to represent valence electrons and lines to represent covalent bonds. Understanding how to draw Lewis structures is essential for visualizing molecular geometry and predicting reactivity.
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Lewis Dot Structures: Ions

Covalent Bonding

Covalent bonding occurs when two atoms share one or more pairs of electrons, allowing them to achieve a full outer electron shell. In ortho-dichlorobenzene, the carbon atoms in the benzene ring form covalent bonds with chlorine atoms, replacing hydrogen atoms. This concept is fundamental for understanding molecular structure and stability.
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Chemical Bonds

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The geometry is influenced by the number of bonds and lone pairs around the central atoms, which can affect the molecule's physical and chemical properties. In ortho-dichlorobenzene, the arrangement of chlorine and carbon atoms will determine the molecule's shape and reactivity.
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Molecular Geometry with Two Electron Groups
Pratica correlata
Domanda del libro di testo

The following three Lewis structures can be drawn for N2O:

(b) The N—N bond length in N2O is 1.12 Å, slightly longer than a typical N≡N bond; and the N—O bond length is 1.19 Å, slightly shorter than a typical N═O bond (see Table 8.4). Based on these data, which resonance structure best represents N2O?

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Domanda del libro di testo

Ortho-Dichlorobenzene, C6H4Cl2, is obtained when two of the adjacent hydrogen atoms in benzene are replaced with Cl atoms. A skeleton of the molecule is shown here. (b) Are there any resonance structures for the molecule? If so, sketch them.

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Domanda del libro di testo

Consider the hypothetical molecule B-A=B. Are the following statements true or false? (a) This molecule cannot exist. (b) If resonance was important, the molecule would have identical A–B bond lengths.

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Domanda del libro di testo

An important reaction for the conversion of natural gas to other useful hydrocarbons is the conversion of methane to ethane. 2 CH4(g) → C2H6(g) + H2(g) In practice, this reaction is carried out in the presence of oxygen, which converts the hydrogen produced into water. 2 CH4(g) + 12 O2(g) → C2H6(g) + H2O(g) Use Table 8.3 to estimate H for these two reactions. Why is the conversion of methane to ethane more favorable when oxygen is used? Why is the conversion of methane to ethane more favorable when oxygen is used?

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Domanda del libro di testo

The following three Lewis structures can be drawn for N2O:

(a) Using formal charges, which of these three resonance forms is likely to be the most important?

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