Chemistry
Chemistry
13th Edition
ISBN: 9781259911156
Author: Raymond Chang Dr., Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.7QP

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

(a)

Expert Solution
Check Mark
Interpretation Introduction

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

Concept Introduction:

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

VSEPR Theory:

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

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

If the molecules (Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  1) 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 Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  2 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  3 will have shape like tetrahedral, and geometry of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  4 will have trigonal bipyramidal and Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  5will 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 10.7QP

(a)

Trigonal pyramidal

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (a)

Chemistry, Chapter 10, Problem 10.7QP , 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 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 theChemistry, Chapter 10, Problem 10.7QP , additional homework tip  7is bonded with three chlorine atoms and one lone pair of electron with it.

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

(b)

Expert Solution
Check Mark
Interpretation Introduction

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

Concept Introduction:

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

VSEPR Theory:

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

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

If the molecules (Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  8) 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 Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  9 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  10 will have shape like tetrahedral, and geometry of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  11 will have trigonal bipyramidal and Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  12will 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 10.7QP

(b)

Tetrahedral

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (b)

Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  13

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

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

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

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

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

(c)

Expert Solution
Check Mark
Interpretation Introduction

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

Concept Introduction:

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

VSEPR Theory:

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

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

If the molecules (Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  14) 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 Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  15 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  16 will have shape like tetrahedral, and geometry of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  17 will have trigonal bipyramidal and Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  18will 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 10.7QP

(c)

Tetrahedral

Explanation of Solution

To predict: The geometry for the given molecule.

Draw the Lewis structure for the molecule (c)

Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  19

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

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

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

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

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

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

(d)

Expert Solution
Check Mark
Interpretation Introduction

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

Concept Introduction:

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

VSEPR Theory:

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

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

If the molecules (Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  20) 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 Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  21 will have shape like trigonal planar and if the molecules has lone pairs then the molecular geometry will be tetrahedral.

The molecules of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  22 will have shape like tetrahedral, and geometry of type Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  23 will have trigonal bipyramidal and Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  24will 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 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)

Chemistry, Chapter 10, Problem 10.7QP , additional homework tip  25

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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