Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259638138
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 7, Problem 7.7QP

Predict the geometry of the following molecules and ion using the VSEPR model: (a) CBr4, (b) BCl3, (c) NF3, (d) H2Se, (e) NO2.

(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.

The molecules with considering the domains of type AB2 will tend to have shape like linear or bent if the central atom have lone pair of electrons with it, type AB3 will have shape like trigonal planar, type AB4 will have shape like tetrahedral or square planar, 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 7.7QP

Tetrahedral geometry

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 7.7QP , 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 32.

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 32 as each bond contains two electrons with it and there are four bonds in the skeletal structure.

Finally, the 24 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 that is four atoms gets bonded with the central atom in the given molecule.

There exist no lone pair on carbon central atom then the molecular geometry for this molecule is tetrahedral.

(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.

The molecules with considering the domains of type AB2 will tend to have shape like linear or bent if the central atom have lone pair of electrons with it, type AB3 will have shape like trigonal planar, type AB4 will have shape like tetrahedral or square planar, 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 7.7QP

(b)

Trigonal planar

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 7.7QP , 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 24.

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 24 as each bond contains two electrons with it and there are three 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 trigonal planar that is the boron atom contains three chlorine atoms and no lone pair of electrons over boron atoms hence the molecular geometry for the molecule is also trigonal planar.

(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.

The molecules with considering the domains of type AB2 will tend to have shape like linear or bent if the central atom have lone pair of electrons with it, type AB3 will have shape like trigonal planar, type AB4 will have shape like tetrahedral or square planar, 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 7.7QP

Answer

Trigonal pyramidal

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (c)

Chemistry: Atoms First, Chapter 7, Problem 7.7QP , additional homework tip  3

Explanation of Solution

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 over all the atoms present in the molecule such that each atom contains eight electrons in its valence shell.

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 nitrogen contains three fluorine atoms and one lone pair of electron.

The molecular geometry for the molecule is trigonal pyramidal because of one lone pair of electron it contain.

(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.

The molecules with considering the domains of type AB2 will tend to have shape like linear or bent if the central atom have lone pair of electrons with it, type AB3 will have shape like trigonal planar, type AB4 will have shape like tetrahedral or square planar, 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 7.7QP

(d)

Bent 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 7.7QP , 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 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 4 has to be subtracted with 8 as each bond contains two electrons with it and there are two bonds in the skeletal structure.

Finally, the 4 electrons got after subtractions has to be equally distributed over selenium atom such that each atom contains eight electrons in its 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 which is of type tetrahedral due to presence of two lone pair of electrons with it but the molecular geometry is bent since due to the presence of two lone pair of electrons over the central Se atom.

(e)

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.

The molecules with considering the domains of type AB2 will tend to have shape like linear or bent if the central atom have lone pair of electrons with it, type AB3 will have shape like trigonal planar, type AB4 will have shape like tetrahedral or square planar, 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 7.7QP

Bent shaped

Explanation of Solution

Explanation

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (e)

Chemistry: Atoms First, Chapter 7, Problem 7.7QP , additional homework tip  5

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 17 which is added with one electron due to the presence of charge -1 in the given molecule.

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 18 as each bond contains two electrons with it and there are three bonds in the skeletal structure.

Finally, the 12 electrons got after subtractions has to be equally distributed over the atoms such that each atom contains eight electrons in its valence shell.

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

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type trigonal planar since there are two atoms and one lone pair electron around the central atom but the molecular geometry according to VSEPR theory is bent.

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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 - Which of the following exhibits significant...Ch. 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 - Calculate the bond order of N22+, and determine...Ch. 7.7 - Which of the following species is paramagnetic?...Ch. 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 - Which of the following contain one or more...Ch. 7.8 - Which of the atoms in BCl3 need hybrid orbitals to...Ch. 7.8 - Which of the following can hybrid orbitals be used...Ch. 7.8 - Which of the following enables us to explain the...Ch. 7 - Prob. 7.1KSPCh. 7 - Which of the following species does not have...Ch. 7 - Prob. 7.3KSPCh. 7 - Prob. 7.4KSPCh. 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. 7.73QPCh. 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. 7.145QP
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