General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
Question
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Chapter 12, Problem 12.16QP

(a)

Interpretation Introduction

Interpretation:

The substance with higher boiling point in the given pairs of substances should be determined.

Concept introduction:

  • Polarity of a bond is due to the difference in electro-negativities of atoms presented in it. The polarities of bonds are represented by using vectors.
  • If the result of all bond polarities or vector sum is non-zero in a molecule, then the molecule is called as polar molecule.
  • If the result of all bond polarities or vector sum is zero in a molecule, then the molecule is called as nonpolar molecule.
  • Intermolecular force is the set of repulsive and attractive forces between molecules that result from the polarity between neighboring molecules. There are four types of intermolecular forces.
  • Dipole – Dipole interaction: This force takes place between polar compounds.
  • Hydrogen bonding is a type of dipole-dipole interaction of molecules when the hydrogen is bonded to strong electronegative atom (F, O, N, etc) in the molecules.
  • Dispersion force is a weak force and this force is present in all compounds force.
  • Boiling point is depending upon the strength of inter molecular forces. 

(a)

Expert Solution
Check Mark

Answer to Problem 12.16QP

Xe has higher boiling point than .

Explanation of Solution

Xe and Ne are mono-atomic gases, so they are nonpolar molecules.

Therefore, only dispersion forces are presented in these molecules; dispersion forces is depends upon the molecular weight. Xe has greater molecular weight as compare to Ne

Boiling point is depending upon the strength of inter molecular forces. 

Hence,

Xe has higher boiling point than Ne.

(b)

Interpretation Introduction

Interpretation:

The substance with higher boiling point in the given pairs of substances should be determined.

Concept introduction:

  • Polarity of a bond is due to the difference in electro-negativities of atoms presented in it. The polarities of bonds are represented by using vectors.
  • If the result of all bond polarities or vector sum is non-zero in a molecule, then the molecule is called as polar molecule.
  • If the result of all bond polarities or vector sum is zero in a molecule, then the molecule is called as nonpolar molecule.
  • Intermolecular force is the set of repulsive and attractive forces between molecules that result from the polarity between neighboring molecules. There are four types of intermolecular forces.
  • Dipole – Dipole interaction: This force takes place between polar compounds.
  • Hydrogen bonding is a type of dipole-dipole interaction of molecules when the hydrogen is bonded to strong electronegative atom (F, O, N, etc) in the molecules.
  • Dispersion force is a weak force and this force is present in all compounds force.
  • Boiling point is depending upon the strength of inter molecular forces. 

(b)

Expert Solution
Check Mark

Answer to Problem 12.16QP

(b)

CS2 has higher boiling point than CO2.

Explanation of Solution

In CS2 molecule,

There are two C-S presented in CS2 molecule. Sulfur atom has more electronegativity as compared to carbon atom; so all the bonds in CS2 have polarities.

The result of all the bond polarities are the sum of all the vectors associated with each bonds.

The directions of C-S bond vectors are opposite to each other, so they cancel each other.

Hence,

The vector sum or the result of bond polarities for CS2 molecule is zero, so CS2 is a non-polar molecule.

Therefore,

Only dispersion forces are present in CS2.

In CO2 molecule,

There are two C-O presented in CO2 molecule. Sulfur atom has more electronegativity as compared to carbon atom; so all the bonds in CO2 have polarities.

The result of all the bond polarities are the sum of all the vectors associated with each bonds.

The directions of C-O bond vectors are opposite to each other, so they cancel each other.

Hence,

The vector sum or the result of bond polarities for CO2 molecule is zero, so CO2 is a non-polar molecule.

Therefore,

Only dispersion forces are present in CO2.

Dispersion forces is depends upon the molecular weight. CS2 has greater molecular weight as compare to CO2. So CS2 has greater dispersion forces as compare to CO2

Boiling point depends upon the strength of inter molecular forces.

Hence,

CS2 has higher boiling point than CO2.

(c)

Interpretation Introduction

Interpretation:

The substance with higher boiling point in the given pairs of substances should be determined.

Concept introduction:

  • Polarity of a bond is due to the difference in electro-negativities of atoms presented in it. The polarities of bonds are represented by using vectors.
  • If the result of all bond polarities or vector sum is non-zero in a molecule, then the molecule is called as polar molecule.
  • If the result of all bond polarities or vector sum is zero in a molecule, then the molecule is called as nonpolar molecule.
  • Intermolecular force is the set of repulsive and attractive forces between molecules that result from the polarity between neighboring molecules. There are four types of intermolecular forces.
  • Dipole – Dipole interaction: This force takes place between polar compounds.
  • Hydrogen bonding is a type of dipole-dipole interaction of molecules when the hydrogen is bonded to strong electronegative atom (F, O, N, etc) in the molecules.
  • Dispersion force is a weak force and this force is present in all compounds force.
  • Boiling point is depending upon the strength of inter molecular forces. 

(c)

Expert Solution
Check Mark

Answer to Problem 12.16QP

Cl2 has higher boiling point than CH4.

Explanation of Solution

The Cl-Cl bond in the Cl2 molecule has no bond polarity. So Cl2 molecule is a nonpolar molecule.

Therefore,

Only dispersion forces are present in Cl2.

In CH4 molecule,

There are four C-H presented in CH4 molecule. Carbon atom has more electronegativity as compared to hydrogen; so all the bonds in CH4 have polarities.

The result of all the bond polarities are the sum of all the vectors associated with each bonds.

The directions of C-H bond vectors are opposite to each other, so they cancel each other.

Hence,

The vector sum or the result of bond polarities for CH4 molecule is zero, so CH4 is a non-polar molecule.

Since the CH4 is a nonpolar molecule, it exhibit only one type of intermolecular force, which is dispersion forces.

Dispersion forces is depends upon the molecular weight. Cl2 has greater molecular weight as compare to CH4. So Cl2 have greater dispersion forces as compare to CH4

Boiling point is depending upon the strength of inter molecular forces.

Hence,

Cl2 has higher boiling point than CH4.

(d)

Interpretation Introduction

Interpretation:

The substance with higher boiling point in the given pairs of substances should be determined.

Concept introduction:

  • Polarity of a bond is due to the difference in electro-negativities of atoms presented in it. The polarities of bonds are represented by using vectors.
  • If the result of all bond polarities or vector sum is non-zero in a molecule, then the molecule is called as polar molecule.
  • If the result of all bond polarities or vector sum is zero in a molecule, then the molecule is called as nonpolar molecule.
  • Intermolecular force is the set of repulsive and attractive forces between molecules that result from the polarity between neighboring molecules. There are four types of intermolecular forces.
  • Dipole – Dipole interaction: This force takes place between polar compounds.
  • Hydrogen bonding is a type of dipole-dipole interaction of molecules when the hydrogen is bonded to strong electronegative atom (F, O, N, etc) in the molecules.
  • Dispersion force is a weak force and this force is present in all compounds force.
  • Boiling point is depending upon the strength of inter molecular forces. 

(d)

Expert Solution
Check Mark

Answer to Problem 12.16QP

LiF has higher boiling point than F2.

Explanation of Solution

The F-F bond in the F2 molecule has no bond polarity. So F2 molecule is a nonpolar molecule.

Therefore,

Only dispersion forces are present in F2.

LiF is an ionic compound, in which Li+ and F- ions are electrostatically attracted each other.

Therefore,

Ionic forces are present in LiF prevalently.

Since ionic forces stronger than dispersion forces, then LiF has greater intermolecular forces as compared to F2.

Boiling point depends upon the strength of inter molecular forces.

Hence,

LiF has higher boiling point than F2.

(e)

Interpretation Introduction

Interpretation:

The substance with higher boiling point in the given pairs of substances should be determined.

Concept introduction:

  • Polarity of a bond is due to the difference in electro-negativities of atoms presented in it. The polarities of bonds are represented by using vectors.
  • If the result of all bond polarities or vector sum is non-zero in a molecule, then the molecule is called as polar molecule.
  • If the result of all bond polarities or vector sum is zero in a molecule, then the molecule is called as nonpolar molecule.
  • Intermolecular force is the set of repulsive and attractive forces between molecules that result from the polarity between neighboring molecules. There are four types of intermolecular forces.
  • Dipole – Dipole interaction: This force takes place between polar compounds.
  • Hydrogen bonding is a type of dipole-dipole interaction of molecules when the hydrogen is bonded to strong electronegative atom (F, O, N, etc) in the molecules.
  • Dispersion force is a weak force and this force is present in all compounds force.
  • Boiling point is depending upon the strength of inter molecular forces. 

(e)

Expert Solution
Check Mark

Answer to Problem 12.16QP

NH3 has higher boiling point than PH3.

Explanation of Solution

In ammonia (NH3) molecule,

Three N-H bonds are presented and due to the difference in electronegativities of nitrogen and hydrogen, it has bond polarity. So NH3 molecule is a polar molecule.

Polar molecules exhibit dipole-dipole interactions.

Since the hydrogen atom is bonded to nitrogen, then hydrogen bonding will be presented in between NH3 molecules.

In PH3 molecule,

Three P-H bonds are presented and due to the difference in electronegativities of phosphorus and hydrogen, it has bond polarity. So PH3 molecule is a polar molecule.

Polar molecules exhibit dipole-dipole interactions.

NH3 has dipole-dipole interaction with Hydrogen bonding; but PH3 has only dipole-dipole interactions.

Boiling point depends upon the strength of inter molecular forces.

Hence,

NH3 has higher boiling point than PH3.

Conclusion

The molecules of higher boiling point in the given pairs of molecules are determined according to the polarities or molecular weights of molecules.

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Chapter 12 Solutions

General Chemistry

Ch. 12.7 - Prob. 1RCCh. 12 - Prob. 12.1QPCh. 12 - Prob. 12.2QPCh. 12 - Prob. 12.3QPCh. 12 - Prob. 12.4QPCh. 12 - Prob. 12.5QPCh. 12 - Prob. 12.6QPCh. 12 - Prob. 12.7QPCh. 12 - Prob. 12.8QPCh. 12 - Prob. 12.9QPCh. 12 - Prob. 12.10QPCh. 12 - Prob. 12.11QPCh. 12 - Prob. 12.12QPCh. 12 - Prob. 12.13QPCh. 12 - 12.14 Diethyl ether has a boiling point of 34.5°C,...Ch. 12 - Prob. 12.15QPCh. 12 - Prob. 12.16QPCh. 12 - Prob. 12.17QPCh. 12 - 12.18 What kind of attractive forces must be...Ch. 12 - Prob. 12.19QPCh. 12 - Prob. 12.20QPCh. 12 - Prob. 12.21QPCh. 12 - Prob. 12.22QPCh. 12 - Prob. 12.23QPCh. 12 - Prob. 12.24QPCh. 12 - Prob. 12.25QPCh. 12 - Prob. 12.26QPCh. 12 - Prob. 12.27QPCh. 12 - Prob. 12.28QPCh. 12 - Prob. 12.29QPCh. 12 - Prob. 12.30QPCh. 12 - Prob. 12.31QPCh. 12 - Prob. 12.32QPCh. 12 - Prob. 12.33QPCh. 12 - Prob. 12.34QPCh. 12 - Prob. 12.35QPCh. 12 - 12.36 A solid is hard, brittle, and electrically...Ch. 12 - Prob. 12.37QPCh. 12 - Prob. 12.38QPCh. 12 - Prob. 12.39QPCh. 12 - Prob. 12.40QPCh. 12 - Prob. 12.41QPCh. 12 - Prob. 12.42QPCh. 12 - Prob. 12.43QPCh. 12 - Prob. 12.44QPCh. 12 - Prob. 12.45QPCh. 12 - Prob. 12.46QPCh. 12 - Prob. 12.47QPCh. 12 - Prob. 12.48QPCh. 12 - Prob. 12.49QPCh. 12 - Prob. 12.50QPCh. 12 - Prob. 12.51QPCh. 12 - Prob. 12.52QPCh. 12 - Prob. 12.53QPCh. 12 - Prob. 12.54QPCh. 12 - Prob. 12.55QPCh. 12 - Prob. 12.56QPCh. 12 - Prob. 12.57QPCh. 12 - Prob. 12.58QPCh. 12 - Prob. 12.59QPCh. 12 - Prob. 12.60QPCh. 12 - Prob. 12.61QPCh. 12 - Prob. 12.62QPCh. 12 - 12.63 What is the relationship between...Ch. 12 - Prob. 12.64QPCh. 12 - 12.65 Why is solid carbon dioxide called dry ice? Ch. 12 - Prob. 12.66QPCh. 12 - 12.67 Referring to Figure 12.28, estimate the...Ch. 12 - Prob. 12.68QPCh. 12 - Prob. 12.69QPCh. 12 - Prob. 12.70QPCh. 12 - Prob. 12.71QPCh. 12 - Prob. 12.72QPCh. 12 - Prob. 12.73QPCh. 12 - Prob. 12.74QPCh. 12 - 12.75 These compounds are liquid at −10°C; their...Ch. 12 - 12.76 Freeze-dried coffee is prepared by freezing...Ch. 12 - Prob. 12.77QPCh. 12 - 12.78 Steam at 100°C causes more serious burns...Ch. 12 - 12.79 Vapor pressure measurements at several...Ch. 12 - Prob. 12.80QPCh. 12 - Prob. 12.81QPCh. 12 - Prob. 12.82QPCh. 12 - Prob. 12.83QPCh. 12 - Prob. 12.84QPCh. 12 - Prob. 12.85QPCh. 12 - Prob. 12.86QPCh. 12 - Prob. 12.87QPCh. 12 - Prob. 12.88QPCh. 12 - Prob. 12.89QPCh. 12 - 12.90 Name the kinds of attractive forces that...Ch. 12 - Prob. 12.91QPCh. 12 - Prob. 12.92QPCh. 12 - Prob. 12.93QPCh. 12 - Prob. 12.94QPCh. 12 - Prob. 12.95QPCh. 12 - Prob. 12.96QPCh. 12 - Prob. 12.97QPCh. 12 - Prob. 12.98QPCh. 12 - 12.99 The liquid-vapor boundary line in the phase...Ch. 12 - Prob. 12.100QPCh. 12 - Prob. 12.101QPCh. 12 - Prob. 12.102QPCh. 12 - Prob. 12.103QPCh. 12 - Prob. 12.104QPCh. 12 - Prob. 12.105QPCh. 12 - Prob. 12.106QPCh. 12 - 12.107 The following graph shows approximate plots...Ch. 12 - Prob. 12.108QPCh. 12 - Prob. 12.109QPCh. 12 - Prob. 12.110QPCh. 12 - Prob. 12.111QPCh. 12 - Prob. 12.112QPCh. 12 - Prob. 12.113QPCh. 12 - Prob. 12.114QPCh. 12 - 12.115 Use the concept of intermolecular forces to...Ch. 12 - Prob. 12.116QPCh. 12 - 12.117 What is the origin of dark spots on the...Ch. 12 - Prob. 12.118QPCh. 12 - 12.119 The electrical conductance of copper metal...Ch. 12 - Prob. 12.120SPCh. 12 - Prob. 12.121SPCh. 12 - Prob. 12.122SPCh. 12 - Prob. 12.123SPCh. 12 - Prob. 12.124SPCh. 12 - 12.125 The boiling point of methanol is 65.0°C and...Ch. 12 - Prob. 12.126SPCh. 12 - Prob. 12.127SPCh. 12 - Prob. 12.128SPCh. 12 - Prob. 12.129SP
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