CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
14th Edition
ISBN: 9781259327933
Author: Burdge
Publisher: MCG
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Chapter 7, Problem 7.10QP

Predict the geometries of the following ions: (a) NH4+, (b) NH2, (c) CO32−, (d) ICl2, (e) IC14, (f) AlH4, (g) SnCl5 (h) H3O+, (i) BeF42−.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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.

To predict: The geometry for the given molecule.

Answer to Problem 7.10QP

Tetrahedral

Explanation of Solution

Draw the Lewis structure for the molecule (a)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , 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 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.

Finally, the structure does not have any electrons to be placed over the atoms since all the atoms have completed its valance 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 central atom N has bonded with four hydrogen atoms.

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

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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.

To predict: The geometry for the given molecule.

Answer to Problem 7.10QP

(b)

Bent

Explanation of Solution

Draw the Lewis structure for the molecule (b)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , 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 8.

The next step is to subtract the electrons present in the total number of bonds 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 such that each atom have completed 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.

Therefore, the molecular geometry for the given molecules is bent due to the presence of two lone pair of electrons with it.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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.

To predict: The geometry for the given molecule.

Answer to Problem 7.10QP

(c)

Trigonal planar

Explanation of Solution

Draw the Lewis structure for the molecule (c)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , 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 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 then considering the valency of carbon one C-O is denoted as C=O with expense of two valence electrons placed over atoms.

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 trigonal planar since there are 3 oxygen atoms bonded with carbon in CO32-.

There exist no lone pair on central atom hence the molecular geometry for this molecule is also trigonal planar.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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 7.10QP

Answer

(d)

Linear

Explanation of Solution

Draw the Lewis structure for the molecule (d)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , 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 22 since it has one extra negative charge with it.

The next step is to subtract the electrons present in the total number of bonds in the molecule with the total valence electrons such that 4 has to be subtracted with 22 as each bond contains two electrons with it and there are two 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 (d) using VSEPR.

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral due to the presence of two lone pairs.

Therefore, the molecular geometry for the given molecule is bent because of the repulsions produced by the two lone pairs present in the central atom.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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.

To predict: The geometry for the given molecule.

Answer to Problem 7.10QP

(e)

Square planar

Explanation of Solution

Draw the Lewis structure for the molecule (e)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , 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 36.

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

Finally, the 28 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 (e) using VSEPR.

Explanation:

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type octahedral since there are four I-Cl bonds and two lone pair.

Therefore, the geometry for the molecule is square planar due to the presence of two lone pair of electrons.

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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.

To predict: The geometry for the given molecule.

Answer to Problem 7.10QP

(f)

Tetrahedral

Explanation of Solution

Draw the Lewis structure for the molecule (f)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , additional homework tip  6

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 8 which include the presence one negative charge with it.

The next step is to subtract the electrons present in the total number of bonds 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.

Finally, there are no electrons to be placed over the atoms.

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

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral since there are four Al-H bonds in the molecule.

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

(g)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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.

The geometry for the given molecule.

Answer to Problem 7.10QP

Solution

(g)

Trigonal bipyramidal

Explanation of Solution

Draw the Lewis structure for the molecule (g)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , additional homework tip  7

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 40 which include the presence one negative charge with it.

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 10 has to be subtracted with 40 as each bond contains two electrons with it and there are five bonds in the skeletal structure.

Finally, the 30 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 (g) using VSEPR.

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type trigonal bipyramidal since there are five chlorine atoms bonded with the central atom.

There exists no lone pair on central atom hence the molecular geometry for this molecule is also trigonal bipyramidal.

(h)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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 7.10QP

Answer

(h)

Trigonal pyramid

Draw the Lewis structure for the molecule (h)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , additional homework tip  8

Explanation of Solution

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 8 which is included with the positive charge present in the given ion.

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

Finally, the 2 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 (h) using VSEPR.

Explanation:

The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral since there are three O-H bonds and one lone pair of electron.

Therefore, the molecular geometry for the given ion is trigonal pyramidal since there exist one lone pair over O atom which is bonded with three hydrogen atoms.

(i)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: For the given set of ions 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 have 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.

The geometry for the given molecule.

Answer to Problem 7.10QP

(i)

Tetrahedral

Explanation of Solution

Draw the Lewis structure for the molecule (i)

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT, Chapter 7, Problem 7.10QP , additional homework tip  9

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 32 which includes with the two negative charges present in the given ion.

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

Determine the molecular geometry for the molecule (i) 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 is no lone pair of electrons present in the given ion.

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Chapter 7 Solutions

CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT

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
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