Chemistry: Atoms First
Chemistry: Atoms First
2nd Edition
ISBN: 9780073511184
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 7, Problem 7.9QP

Predict the geometries of the following species using the VSEPR method: (a) PCl3, (b) CHCl3, (c) SiH4, (d) TeCl4.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.

Concept Introduction:

Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).

VSEPR Theory:

As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,

  • The first step is to draw the correct Lewis structure for the molecule.
  • Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
  • Finally, the geometry is predicted by using the orientation of atoms.

If the molecules (AB2) have two atoms bonded with the central atom then the molecular geometry by considering the orientation of atoms is linear if it has lone of electron over the central atom then the molecular geometry will be bent.

The molecules of type AB3 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type AB4 will have shape like tetrahedral, and geometry of type AB5 will have trigonal bipyramidal and AB6 will have shape like octahedral respectively.

Lewis structure for any molecule is drawn by using the following steps,

First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.

Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.

Answer to Problem 7.9QP

(a)

Trigonal pyramidal

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (a)

Chemistry: Atoms First, Chapter 7, Problem 7.9QP , additional homework tip  1

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 26.

The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 6 has to be subtracted with 26 as each bond contains two electrons with it and there are three bonds in the skeletal structure.

Finally, the 20 electrons got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.

Determine the molecular geometry for the molecule (a) using VSEPR.

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral since the P is bonded with three chlorine atoms and one lone pair of electron with it.

The molecular geometry for the given molecule is trigonal pyramidal due to the presence of one lone pair around the central atom.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.

Concept Introduction:

Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).

VSEPR Theory:

As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,

  • The first step is to draw the correct Lewis structure for the molecule.
  • Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
  • Finally, the geometry is predicted by using the orientation of atoms.

If the molecules (AB2) have two atoms bonded with the central atom then the molecular geometry by considering the orientation of atoms is linear if it has lone of electron over the central atom then the molecular geometry will be bent.

The molecules of type AB3 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type AB4 will have shape like tetrahedral, and geometry of type AB5 will have trigonal bipyramidal and AB6 will have shape like octahedral respectively.

Lewis structure for any molecule is drawn by using the following steps,

First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.

Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.

Answer to Problem 7.9QP

(b)

Tetrahedral

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (b)

Chemistry: Atoms First, Chapter 7, Problem 7.9QP , additional homework tip  2

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 26.

The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 8 has to be subtracted with 26 as each bond contains two electrons with it and there are four bonds in the skeletal structure.

Finally, the 18 electrons got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.

Determine the molecular geometry for the molecule (b) using VSEPR.

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral as there is no lone pair of electron over the central metal atom and hence the molecular geometry for the given molecule is also Tetrahedral.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.

Concept Introduction:

Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).

VSEPR Theory:

As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,

  • The first step is to draw the correct Lewis structure for the molecule.
  • Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
  • Finally, the geometry is predicted by using the orientation of atoms.

If the molecules (AB2) have two atoms bonded with the central atom then the molecular geometry by considering the orientation of atoms is linear if it has lone of electron over the central atom then the molecular geometry will be bent.

The molecules of type AB3 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type AB4 will have shape like tetrahedral, and geometry of type AB5 will have trigonal bipyramidal and AB6 will have shape like octahedral respectively.

Lewis structure for any molecule is drawn by using the following steps,

First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.

Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.

Answer to Problem 7.9QP

Answer

(c)

Tetrahedral

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (c)

Chemistry: Atoms First, Chapter 7, Problem 7.9QP , additional homework tip  3

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 8.

The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 8 has to be subtracted with 8 as each bond contains two electrons with it and there are four bonds in the skeletal structure.

There are no remaining electrons hence all the atoms in the molecules are fulfilled the octet rule that is each atom involves in bonding in order to fill their valence with eight electrons.

Determine the molecular geometry for the molecule (c) using VSEPR.

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral since central atom does not contain any lone pair of electron with it.

The molecular geometry for the molecule is also tetrahedral as there are four atoms bonded with the central metal atom and there is absence of lone pair of electrons.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.

Concept Introduction:

Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).

VSEPR Theory:

As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,

  • The first step is to draw the correct Lewis structure for the molecule.
  • Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
  • Finally, the geometry is predicted by using the orientation of atoms.

If the molecules (AB2) have two atoms bonded with the central atom then the molecular geometry by considering the orientation of atoms is linear if it has lone of electron over the central atom then the molecular geometry will be bent.

The molecules of type AB3 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type AB4 will have shape like tetrahedral, and geometry of type AB5 will have trigonal bipyramidal and AB6 will have shape like octahedral respectively.

Lewis structure for any molecule is drawn by using the following steps,

First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.

Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.

Answer to Problem 7.9QP

(d)

See-saw shaped

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (d)

Chemistry: Atoms First, Chapter 7, Problem 7.9QP , additional homework tip  4

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 34.

The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 8 has to be subtracted with 8 as each bond contains two electrons with it and there are four bonds in the skeletal structure.

Then the 26 electrons got after the subtractions should be placed over the atoms present in the molecule such that each atom contains eight electrons in the valence shell.

Determine the molecular geometry for the molecule (d) using VSEPR.

The electron domain for the given molecule is obtained by viewing the Lewis structure shows that it contains five electron domains since it has 4 chlorine atoms and one lone pair with it.

The molecular geometry for the molecule is see-saw shape due to the present of that one lone pair of electron.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Predict the geometries of the following species using the VSEPR method: (a) PCl3, (b) CHCl3, (c) SiH4, (d)TeCl4
Give the electron-domain and molecular geometries for the following molecules and ions: (a) HCN, (b) SO3, (c) SF4, (d) PF6, (e) NH3CI*, (f) N3. 2-
Use partial orbital diagrams to show how the atomic orbitals of the central atom lead to hybrid orbitals in (a) AsCl₃; (b) SnCl₂; (c) PF₆⁻.

Chapter 7 Solutions

Chemistry: Atoms First

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
Background pattern image
Chemistry
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
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