Chemistry: Atoms First
2nd Edition
ISBN: 9780073511184
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 7, Problem 7.145QP
Interpretation Introduction
Interpretation:
To find which of the molecular geometry has greater stability for
Concept Introduction:
- Molecular geometry is the spatial arrangement of atoms in a molecule. It is the three dimensional arrangement of bonded atoms. Many chemical, physical and even biological properties depend on molecular geometry.
- Using VSEPR theory and Lewis structure, the exact geometry of a molecule can be obtained.
- In VSEPR, the geometry of the molecule is explained based on minimizing electrostatic repulsion between the molecules’ valence electrons around a central atom
- Lewis structures is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Give the hybridization of Rn (atomic number 86) in the following ion: RnI3+
Use the molecular orbital model to fully describe the bonding in O2+, O2, O2-, and O22-.Determine which of the following statements are true and which are false.
True or false? :
Bond length increases with increasing bond order while bond energy decreases.The number of unpaired electrons in O2- and O2 is, respectively, 1 and 2. The electron configuration of O2 is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1. The electron configuration of O2- is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3. The bond order in O2- and O22- is, respectively, 2.5 and 1. The bond lengths increase in the order: O22- < O2- < O2 < O2+.
A student claims that if element D is in group 18, then the molecular geometry of DF4(g) is square planar. Do you agree or disagree with the student? Justify your answer in terms of the VSEPR theory.
Chapter 7 Solutions
Chemistry: Atoms First
Ch. 7.1 - Determine the shapes of (a) SO3 and (b) ICl4.Ch. 7.1 - Determine the shapes of (a) CO2 and (b) SCl2.Ch. 7.1 - (a) From what group must the terminal atoms come...Ch. 7.1 - These four models may represent molecules or...Ch. 7.1 - Acetic acid, the substance that gives vinegar its...Ch. 7.1 - Ethanolamine (HOCH2CH2NH2) has a smell similar to...Ch. 7.1 - The bond angle in NH3 is significantly smaller...Ch. 7.1 - Which of these models represents a species in...Ch. 7.1 - Prob. 7.1.1SRCh. 7.1 - Prob. 7.1.2SR
Ch. 7.1 - Prob. 7.1.3SRCh. 7.1 - Prob. 7.1.4SRCh. 7.2 - Prob. 7.3WECh. 7.2 - Prob. 3PPACh. 7.2 - For each of the following hypothetical molecules,...Ch. 7.2 - Which of these models could represent a polar...Ch. 7.2 - Prob. 7.2.1SRCh. 7.2 - Prob. 7.2.2SRCh. 7.3 - Prob. 7.4WECh. 7.3 - Prob. 4PPACh. 7.3 - Prob. 4PPBCh. 7.3 - Prob. 4PPCCh. 7.3 - Prob. 7.3.1SRCh. 7.3 - Prob. 7.3.2SRCh. 7.4 - Hydrogen selenide (H2Se) is a foul-smelling gas...Ch. 7.4 - Prob. 5PPACh. 7.4 - For which molecule(s) can we not use valence bond...Ch. 7.4 - Which of these models could represent a species...Ch. 7.4 - Prob. 7.4.1SRCh. 7.4 - Prob. 7.4.2SRCh. 7.5 - Prob. 7.6WECh. 7.5 - Use hybrid orbital theory to describe the bonding...Ch. 7.5 - Prob. 6PPBCh. 7.5 - Prob. 6PPCCh. 7.5 - Prob. 7.5.1SRCh. 7.5 - Prob. 7.5.2SRCh. 7.6 - Thalidomide (C13H10N2O4) is a sedative and...Ch. 7.6 - The active ingredient in Tylenol and a host of...Ch. 7.6 - Determine the total number of sigma and pi bonds...Ch. 7.6 - In terms of valence bond theory and hybrid...Ch. 7.6 - In addition to its rise in aqueous solution as a...Ch. 7.6 - Use valence bond theory and hybrid orbitals to...Ch. 7.6 - Use valence bond theory and hybrid orbitals to...Ch. 7.6 - Explain why hybrid orbitals are necessary to...Ch. 7.6 - Prob. 7.6.1SRCh. 7.6 - Prob. 7.6.2SRCh. 7.6 - Prob. 7.6.3SRCh. 7.6 - Prob. 7.6.4SRCh. 7.7 - Prob. 7.9WECh. 7.7 - Use molecular orbital theory to determine whether...Ch. 7.7 - Use molecular orbital theory to determine whether...Ch. 7.7 - For most of the homonuclear diatomic species shown...Ch. 7.7 - Prob. 7.7.1SRCh. 7.7 - Prob. 7.7.2SRCh. 7.7 - Prob. 7.7.3SRCh. 7.7 - Prob. 7.7.4SRCh. 7.8 - It takes three resonance structures to represent...Ch. 7.8 - Use a combination of valence bond theory and...Ch. 7.8 - Use a combination of valence bond theory and...Ch. 7.8 - Prob. 10PPCCh. 7.8 - Prob. 7.8.1SRCh. 7.8 - Prob. 7.8.2SRCh. 7.8 - Prob. 7.8.3SRCh. 7.8 - Prob. 7.8.4SRCh. 7 - Prob. 7.1QPCh. 7 - Sketch the shape of a linear triatomic molecule, a...Ch. 7 - Prob. 7.3QPCh. 7 - Prob. 7.4QPCh. 7 - In the trigonal bipyramidal arrangement, why does...Ch. 7 - Prob. 7.6QPCh. 7 - Predict the geometry of the following molecules...Ch. 7 - Prob. 7.8QPCh. 7 - Predict the geometries of the following species...Ch. 7 - Predict the geometries of the following ions: (a)...Ch. 7 - Prob. 7.11QPCh. 7 - Prob. 7.12QPCh. 7 - Prob. 7.13QPCh. 7 - Describe the geometry about each of the central...Ch. 7 - Prob. 7.15QPCh. 7 - Prob. 7.16QPCh. 7 - Prob. 7.17QPCh. 7 - Prob. 7.18QPCh. 7 - Prob. 7.19QPCh. 7 - Prob. 7.20QPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Explain the term polarizability. What kind of...Ch. 7 - Prob. 7.24QPCh. 7 - What physical properties are determined by the...Ch. 7 - Prob. 7.26QPCh. 7 - Describe the types of intermolecular forces that...Ch. 7 - The compounds Br2 and ICl are isoelectronic (have...Ch. 7 - If you lived in Alaska, which of the following...Ch. 7 - The binary hydrogen compounds of the Group 4A...Ch. 7 - List the types of intermolecular forces that exist...Ch. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Diethyl ether has a boiling point of 34.5C, and...Ch. 7 - Prob. 7.36QPCh. 7 - Which substance in each of the following pairs...Ch. 7 - Prob. 7.38QPCh. 7 - What kind of attractive forces must be overcome to...Ch. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - The following compounds have the same molecular...Ch. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Use valence bond theory to explain the bonding in...Ch. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - Prob. 7.49QPCh. 7 - What is the hybridization of atomic orbitals? Why...Ch. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Describe the bonding scheme of the AsH3 molecule...Ch. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Describe the hybridization of phosphorus in PF5.Ch. 7 - Prob. 7.58QPCh. 7 - Prob. 7.59QPCh. 7 - Prob. 7.1VCCh. 7 - Prob. 7.2VCCh. 7 - Prob. 7.3VCCh. 7 - Prob. 7.4VCCh. 7 - Prob. 7.60QPCh. 7 - Which of the following pairs of atomic orbitals of...Ch. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - Prob. 7.65QPCh. 7 - Prob. 7.66QPCh. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Benzo[a]pyrene is a potent carcinogen found in...Ch. 7 - What is molecular orbital theory? How does it...Ch. 7 - Define the following terms: bonding molecular...Ch. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Draw a molecular orbital energy level diagram for...Ch. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 7 - Prob. 7.79QPCh. 7 - Acetylene (C2H2) has a tendency to lose two...Ch. 7 - Compare the Lewis and molecular orbital treatments...Ch. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Draw the molecular orbital diagram for the cyanide...Ch. 7 - Given that BeO is diamagnetic, use a molecular...Ch. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Both ethylene (C2H4) and benzene (C6H6) contain...Ch. 7 - Chemists often represent benzene with the...Ch. 7 - Determine which of these molecules has a more...Ch. 7 - Nitryl fluoride (FNO2) is used in rocket...Ch. 7 - Describe the bonding in the nitrate ion NO3 in...Ch. 7 - Prob. 7.95QPCh. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - Antimony pentafluoride (SbF5) combines with XeF4...Ch. 7 - Prob. 7.101QPCh. 7 - The molecular model of nicotine (a stimulant) is...Ch. 7 - Predict the bond angles for the following...Ch. 7 - The germanium pentafluoride anion (GeF5) has been...Ch. 7 - Draw Lewis structures and give the other...Ch. 7 - Which figure best illustrates the hybridization of...Ch. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - Cyclopropane (C3H6) has the shape of a triangle in...Ch. 7 - The compound 1,2-dichloroethane (C2H4Cl2) is...Ch. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117QPCh. 7 - Prob. 7.118QPCh. 7 - The amino acid selenocysteine is one of the...Ch. 7 - Prob. 7.120QPCh. 7 - Prob. 7.121QPCh. 7 - Prob. 7.122QPCh. 7 - Gaseous or highly volatile liquid anesthetics are...Ch. 7 - Prob. 7.124QPCh. 7 - Prob. 7.125QPCh. 7 - Two of the drugs that are prescribed for the...Ch. 7 - Prob. 7.127QPCh. 7 - Prob. 7.128QPCh. 7 - The BO+ ion is paramagnetic. Determine (a) whether...Ch. 7 - Use molecular orbital theory to explain the...Ch. 7 - Which best illustrates the change in geometry...Ch. 7 - Prob. 7.132QPCh. 7 - Prob. 7.133QPCh. 7 - Aluminum trichloride (AlCl3) is an...Ch. 7 - Prob. 7.135QPCh. 7 - Prob. 7.136QPCh. 7 - Prob. 7.137QPCh. 7 - Consider an N2 molecule in its first excited...Ch. 7 - The Lewis structure for O2 is Use molecular...Ch. 7 - Draw the Lewis structure of ketene (C2H2O) and...Ch. 7 - The compound TCDD, or...Ch. 7 - Name the kinds of attractive forces that must be...Ch. 7 - Carbon monoxide (CO) is a poisonous compound due...Ch. 7 - Prob. 7.144QPCh. 7 - Prob. 7.145QPCh. 7 - Prob. 7.146QPCh. 7 - Prob. 7.147QPCh. 7 - Prob. 7.148QPCh. 7 - Prob. 7.1KSPCh. 7 - Which of the following species does not have...Ch. 7 - Prob. 7.3KSPCh. 7 - Prob. 7.4KSP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- It is possible to write a simple Lewis structure for the SO42- ion, involving only single bonds, which follows the octet rule. However, Linus Pauling and others have suggested an alternative structure, involving double bonds, in which the sulfur atom is surrounded by six electron pairs. (a) Draw the two Lewis structures. (b) What geometries are predicted for the two structures? (c) What is the hybridization of sulfur in each case? (d) What are the formal charges of the atoms in the two structures?arrow_forward2. Which of the following species has the largest C—N bond order? CN− OCN− CH3NH2 N(CH3)3arrow_forwardSolid sulfur normally consists of crystals of S8 molecules, but when heated strongly, the solid vaporizes to give S2 molecules (among other molecular species). Describe the bonding in S2 in molecular orbital terms, assuming the orbitals are analogous to those of the preceding period. What would you expect to happen to the sulfur sulfur bond length if two electrons were added to give the S22 ion? What would you expect to happen to the bond length if, instead, two electrons were taken away to give S22+?arrow_forward
- Identify the hybrid orbitals used by boron in BCl3 and in BCl4, the ion formed from the reaction of BCl3 and Cl. Explain your choices.arrow_forwardThe sulfamate ion, H2NSO3, can be thought of as having been formed from the amide ion, NH2, and sulphur trioxide, SO3. (a) What are the electron-pair and molecular geometries or the amide ion and or SO3? What are the hybridizations of the N and S atoms, respectively? (b) Sketch a structure for the sulfamate ion, and estimate the bond angles. (c) What changes in hybridization do you expect for N and S in the course of the reaction NH2 + SO3 H2NSO3? (d) Is SO3 the donor of an electron pair or the acceptor of an electron pair in the reaction with amide ion? Does the electrostatic potential map shown below confirm your prediction?arrow_forwardDescribe the hybridization around the central atom and the bonding in SCl2 and OCS.arrow_forward
- Base on molecular orbital diagrams and the bond order, rank CO, NO, and O2 molecules in order of increasing bond strength.arrow_forwardUse the molecular orbital model to fully describe the bonding in O₂+, O₂, O₂™, and 0₂²- Determine which of the following statements are true and which are false. The electron configuration of O₂ is (025)²(0₂s*)²(02p)²(¹2)4(¹²p*)². 2- The bond order in O₂¯ and O₂²- is, respectively, 2.5 and 1. The number of unpaired electrons in O₂ and O₂ is, respectively, 1 and 0. The electron configuration of O₂¯ is (02)²(02s*)²(0₂p)²(12p)4(¹²µ*)³. The bond energies increase in the order: 0₂²- < 0₂¯ < 0₂ < 0₂¹. Bond length increases with increasing bond order while bond energy decreases.arrow_forwardUse molecular orbital theory, write a molecular orbital diagram, calculate bond order, and predict whether or not each ion exists in a relatively stable form, for the following two molecular ions for He2^2+arrow_forward
- What atomic or hybrid orbitals make up the sigma bond between B and F in tetrafluoroborate ion, BF4-arrow_forwardMolecular Orbital Theory -- Homodiatomics Use the molecular orbital model to fully describe the bonding in O₂+, 0₂, O₂¯, and O₂²-. Determine which of the following statements are true and which are false. true The bond order in O₂+ and O₂ is, respectively, 1.5 and 2. false ✓ The electron configuration of O₂¯ is (02s)²(02s*)²(02p)²(12p)4(12p*)³. true ✓ The electron configuration of O₂+ is (0 2s)²(02s*)²(02µ)²(¹2p)4(¹₂µ*)². The bond lengths increase in the order: 0₂²- < 0₂¯ < O₂ < 0₂+. true true Bond length increases with increasing bond order while bond energy decreases. true ✓ The number of unpaired electrons in O₂¯ and O₂ is, respectively, 1 and 2.arrow_forwardWhich of the following compounds is likely to be nonplanar (not planar)? +) IZarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
- Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY