General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 10, Problem 10.8QP

(a)

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.

(a)

Expert Solution
Check Mark

Answer to Problem 10.8QP

(a)

Trigonal planar

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (a)

General Chemistry, Chapter 10, Problem 10.8QP , 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 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 (a) 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 three chlorine atoms bonded with Al.

There exist no lone pair on carbon central atom therefore, the molecular geometry for this molecule is trigonal planar.

(b)

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.

(b)

Expert Solution
Check Mark

Answer to Problem 10.8QP

 (b)

Tetrahedral

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (b)

General Chemistry, Chapter 10, Problem 10.8QP , additional homework tip  2

Explanation of Solution

First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 31.

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

Finally, the 23 electrons got after subtractions which is added with one electrons due to the presence of one negative charge in the given molecule, which is totally 24 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 that is the Aluminium atom contains four chlorine atoms and no lone pair of electrons over central atom hence the molecular geometry for the molecule is also tetrahedral.

(c)

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.

(c)

Expert Solution
Check Mark

Answer to Problem 10.8QP

 (c)

Linear shape

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (c)

General Chemistry, Chapter 10, Problem 10.8QP , 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 18.

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

Finally, the 14 electrons got after subtractions has to be equally distributed over the chlorine 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 linear since central atom does not contain any lone pair of electron with it.

The molecular geometry for the molecule is also linear due to the absence of lone pair of electron around zinc atom.

(d)

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.

(d)

Expert Solution
Check Mark

Answer to Problem 10.8QP

(d)

Tetrahedral

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (d)

General Chemistry, Chapter 10, Problem 10.8QP , 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 30.

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

Finally, the 22 electrons got after subtractions plus two electrons due to the charge -2 over the given molecule which is totally 24 electrons has to be equally distributed over chlorine atoms 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 since it contains four chlorine atoms gets bonded with the central metal atom.

Therefore, the molecular geometry for the given molecule is tetrahedral since there are no lone pair of electrons over the zinc central metal atom

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Chapter 10 Solutions

General Chemistry

Ch. 10.6 - Prob. 1RCCh. 10.6 - Prob. 2RCCh. 10.6 - Prob. 1PECh. 10 - Prob. 10.1QPCh. 10 - Prob. 10.2QPCh. 10 - 10.3 How many atoms arc directly bonded to the...Ch. 10 - 10.4 Discuss the basic features of the VSEPR...Ch. 10 - 10.5 In the trigonal bipyramidal arrangement, why...Ch. 10 - 10.6 The geometry of CH4 could be square planar,...Ch. 10 - Prob. 10.7QPCh. 10 - Prob. 10.8QPCh. 10 - Prob. 10.9QPCh. 10 - Prob. 10.10QPCh. 10 - 10.11 Describe the geometry around each of the...Ch. 10 - 10.12 Which of these species are tetrahedral?...Ch. 10 - 10.13 Define dipole moment. What are the units and...Ch. 10 - 10.14 What is the relationship between the dipole...Ch. 10 - 10.15 Explain why an atom cannot have a permanent...Ch. 10 - 10.16 The bonds in beryllium hydride (BeH2)...Ch. 10 - 10.17 Referring to Table 10.3. arrange the...Ch. 10 - 10.18 The dipole moments of the hydrogen halides...Ch. 10 - 10.19 List these molecules in order of increasing...Ch. 10 - 10.20 Docs the molecule OCS have a higher or lower...Ch. 10 - 10.21 Which of these molecules has a higher dipole...Ch. 10 - 10.22 Arrange these compounds in order of...Ch. 10 - 10.23 What is valence bond theory? How does it...Ch. 10 - 10.24 Use valence bond theory to explain the...Ch. 10 - 10.25Draw a potential energy curve for the bond...Ch. 10 - 10.26 What is the hybridization of atomic...Ch. 10 - 10.27 How does a hybrid orbital differ from a pure...Ch. 10 - 10.28 What is the angle between these two hybrid...Ch. 10 - 10.29 How would you distinguish between a sigma...Ch. 10 - 10.30 Which of these pairs of atomic orbitals of...Ch. 10 - 10.31 The following potential energy curve...Ch. 10 - 10.32 What is the hybridization state of Si in...Ch. 10 - 10.33 Describe the change in hybridization (if...Ch. 10 - 10.34 Consider the reaction Describe the changes...Ch. 10 - 10.35 What hybrid orbitals are used by nitrogen...Ch. 10 - Prob. 10.36QPCh. 10 - 10.37 Specify which hybrid orbitals are used by...Ch. 10 - 10.38 What is the hybridization state of the...Ch. 10 - 10.39 The allene molecule H2C=C=CH2 is linear (the...Ch. 10 - 10.40 Describe the hybridization of phosphorus in...Ch. 10 - 10.41 How many sigma bonds and pi bonds are there...Ch. 10 - 10.42 How many pi bonds and sigma bonds are there...Ch. 10 - 10.43 Give the formula of a cation comprised of...Ch. 10 - 10.44 Give the formula of an anion comprised of...Ch. 10 - 10.45 What is molecular orbital theory? How does...Ch. 10 - 10.46 Define these terms: bonding molecular...Ch. 10 - 10.47 Sketch the shapes of these molecular...Ch. 10 - 10.48 Explain the significance of bond order. Can...Ch. 10 - 10.49 Explain in molecular orbital terms the...Ch. 10 - Prob. 10.50QPCh. 10 - Prob. 10.51QPCh. 10 - Prob. 10.52QPCh. 10 - Prob. 10.53QPCh. 10 - Prob. 10.54QPCh. 10 - Prob. 10.55QPCh. 10 - 10.56 Compare the Lewis and molecular orbital...Ch. 10 - Prob. 10.57QPCh. 10 - 10.58 Compare the relative stability of these...Ch. 10 - Prob. 10.59QPCh. 10 - Prob. 10.60QPCh. 10 - Prob. 10.61QPCh. 10 - Prob. 10.62QPCh. 10 - Prob. 10.63QPCh. 10 - Prob. 10.64QPCh. 10 - Prob. 10.65QPCh. 10 - Prob. 10.66QPCh. 10 - Prob. 10.67QPCh. 10 - Prob. 10.68QPCh. 10 - 10.69 Draw Lewis structures and give the other...Ch. 10 - Prob. 10.70QPCh. 10 - Prob. 10.71QPCh. 10 - Prob. 10.72QPCh. 10 - Prob. 10.73QPCh. 10 - Prob. 10.74QPCh. 10 - Prob. 10.75QPCh. 10 - Prob. 10.76QPCh. 10 - Prob. 10.77QPCh. 10 - Prob. 10.78QPCh. 10 - Prob. 10.79QPCh. 10 - Prob. 10.80QPCh. 10 - Prob. 10.81QPCh. 10 - Prob. 10.82QPCh. 10 - Prob. 10.83QPCh. 10 - 10.84 The ionic character of the bond in a...Ch. 10 - Prob. 10.85QPCh. 10 - 10.86 Aluminum trichloride (AlCl3) is an...Ch. 10 - Prob. 10.87QPCh. 10 - Prob. 10.88QPCh. 10 - 10.90 Progesterone is a hormone responsible for...Ch. 10 - Prob. 10.91SPCh. 10 - Prob. 10.92SPCh. 10 - Prob. 10.93SPCh. 10 - 10.94 The molecule benzyne (C6H4) is a very...Ch. 10 - Prob. 10.95SPCh. 10 - 10.96 As mentioned in the chapter, the Lewis...Ch. 10 - Prob. 10.97SPCh. 10 - Prob. 10.98SPCh. 10 - Prob. 10.99SPCh. 10 - Prob. 10.100SPCh. 10 - Prob. 10.101SPCh. 10 - Prob. 10.102SP
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