Domanda del libro di testo
Write a hybridization and bonding scheme for each molecule. Sketch the molecule, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6 and 10.7.a. CH2Br2 b. SO2 c. NF3 d. BF3
Ch.10 - Chemical Bonding II: Molecular Shapes & Valence Bond Theory
Tro4th EditionChemistry: A Molecular ApproachISBN: 9780134112831
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TOC.tableOfContents
- Ch.1 - Matter, Measurement & Problem Solving136
- Ch.2 - Atoms & Elements112
- Ch.3 - Molecules, Compounds & Chemical Equations159
- Ch.4 - Chemical Quantities & Aqueous Reactions140
- Ch.5 - Gases118
- Ch.6 - Thermochemistry106
- Ch.7 - Quantum-Mechanical Model of the Atom62
- Ch.8 - Periodic Properties of the Elements103
- Ch.9 - Chemical Bonding I: The Lewis Model104
- Ch.10 - Chemical Bonding II: Molecular Shapes & Valence Bond Theory89
- Ch.11 - Liquids, Solids & Intermolecular Forces73
- Ch.12 - Solids and Modern Material59
- Ch.13 - Solutions110
- Ch.14 - Chemical Kinetics140
- Ch.15 - Chemical Equilibrium78
- Ch.16 - Acids and Bases155
- Ch.17 - Aqueous Ionic Equilibrium176
- Ch.18 - Free Energy and Thermodynamics108
- Ch.19 - Electrochemistry118
- Ch.20 - Radioactivity and Nuclear Chemistry82
- Ch.21 - Organic Chemistry150
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Tro 4th EditionCh.10 - Chemical Bonding II: Molecular Shapes & Valence Bond TheoryProblema 63d
Tro 4th EditionCh.10 - Chemical Bonding II: Molecular Shapes & Valence Bond TheoryProblema 63dCapitolo 10, Problema 63d
Write a hybridization and bonding scheme for each molecule or ion. Sketch the structure, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6 and 10.7. d. I3-
Guida verificata passo dopo passo1
Identify the central atom in the I_3^- ion, which is iodine (I).
Determine the total number of valence electrons in the I_3^- ion. Iodine has 7 valence electrons, and there are three iodine atoms plus one extra electron due to the negative charge, giving a total of 22 valence electrons.
Arrange the atoms in a linear structure: I-I-I, with the central iodine atom bonded to the two outer iodine atoms.
Assign lone pairs to each iodine atom to satisfy the octet rule. The central iodine will have three lone pairs, and each terminal iodine will have three lone pairs.
Determine the hybridization of the central iodine atom. It forms two sigma bonds and has three lone pairs, resulting in sp^3d hybridization. The bonding scheme includes two sigma bonds (I-I) and three lone pairs on the central iodine.
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Hybridization
Hybridization is the process of combining atomic orbitals to form new hybrid orbitals that are suitable for the pairing of electrons to form chemical bonds. In the case of I<sub>3</sub><sup>-</sup>, the central iodine atom undergoes sp<sup>3</sup>d hybridization to accommodate the three iodine atoms and the lone pairs, resulting in a trigonal bipyramidal arrangement.
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00:51
Hybridization
Bonding and Molecular Geometry
The bonding and molecular geometry of a molecule are determined by the arrangement of atoms and the types of bonds formed between them. For I<sub>3</sub><sup>-</sup>, the molecular geometry is linear due to the three iodine atoms surrounding the central iodine, with bond angles of approximately 180 degrees, influenced by the presence of lone pairs.
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01:33Molecular Geometry with Two Electron Groups
Orbital Overlap
Orbital overlap refers to the interaction of atomic orbitals from different atoms as they approach each other, leading to the formation of covalent bonds. In I<sub>3</sub><sup>-</sup>, the overlap of the hybrid orbitals from the central iodine with the p orbitals of the terminal iodines results in sigma bonds, while any lone pairs are represented as non-bonding interactions.
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03:06Molecular Orbital Theory
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Write a hybridization and bonding scheme for each molecule or ion. Sketch the structure, including overlapping orbitals, and label all bonds using the notation shown in Examples 10.6 and 10.7.
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