Connect 1 Semester Access Card for General Chemistry: The Essential Concepts
Connect 1 Semester Access Card for General Chemistry: The Essential Concepts
7th Edition
ISBN: 9781259692543
Author: Raymond Chang Dr.; Kenneth Goldsby Professor
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.7QP

(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.

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.

(a)

Expert Solution
Check Mark

Answer to Problem 10.7QP

(a)

Trigonal pyramidal

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (a)

Connect 1 Semester Access Card for General Chemistry: The Essential Concepts, Chapter 10, Problem 10.7QP , 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 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)

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.

(b)

Expert Solution
Check Mark

Answer to Problem 10.7QP

(b)

Tetrahedral

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (b)

Connect 1 Semester Access Card for General Chemistry: The Essential Concepts, Chapter 10, Problem 10.7QP , 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 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)

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.

(c)

Expert Solution
Check Mark

Answer to Problem 10.7QP

(c)

Tetrahedral

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (c)

Connect 1 Semester Access Card for General Chemistry: The Essential Concepts, Chapter 10, Problem 10.7QP , 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)

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.

(d)

Expert Solution
Check Mark

Answer to Problem 10.7QP

(d)

See-saw shaped

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (d)

Connect 1 Semester Access Card for General Chemistry: The Essential Concepts, Chapter 10, Problem 10.7QP , 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 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.

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Chapter 10 Solutions

Connect 1 Semester Access Card for General Chemistry: The Essential Concepts

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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