Chapter 10, Problem 42E

(a)

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, lowest freezing point, smallest vapor pressure, greatest viscosity, greatest heat of vaporization, smallest enthalpy of fusion have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories viz., solid state, liquid state, gaseous state. In all of these three states of matter, the constituents (molecules or ions) of the matter do possess forces between them which are not the same in each state. These forces are called intermolecular forces.

The intermolecular forces are relatively strong in the solids than liquids and weaker in the gaseous substances. This variation influences many of the properties of all the three distinct states of the matter. Compounds with strong intermolecular forces have high melting point, boiling point, freezing point, viscosity enthalpy of vaporization, enthalpy of fusion and low vapor pressure.

The strength of intermolecular forces is,

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

Answer to Problem 42E

The compound with highest boiling point is ${\text{CBr}}_{\text{4}}$. London dispersion is stronger in these molecules due to the larger size of molecule leading to the highest boiling point.

Explanation of Solution

${\text{CCl}}_{\text{4}},{\text{CBr}}_{\text{4}},{\text{CF}}_{\text{4}}$

Identify the compound which has highest boiling point and justify it.

${\text{CBr}}_{\text{4}}$ has the highest boiling point. Compare to the other compounds, the   intermolecular force in ${\text{CBr}}_{\text{4}}$ is high in strength.

${\text{CBr}}_{\text{4}}$ is a non-polar covalent compound. Among the given compounds ${\text{CBr}}_{\text{4}}$ has the largest size. Being a non-polar covalent compound ${\text{CBr}}_{\text{4}}$ has London dispersion forces. When the molecule is large in size London dispersion forces becomes extensive in it. Hence in ${\text{CBr}}_{\text{4}}$ the intermolecular force is quite strong. At normal temperature, the intermolecular forces are not broken. So at high temperature the intermolecular forces are broken allowing the substance to boil. So ${\text{CBr}}_{\text{4}}$ has the highest boiling point.

Analyze why the other compounds don’t have the highest boiling point and justify the same.

The compounds other than ${\text{CBr}}_{\text{4}}$ have intermolecular force of low strength.

${\text{CCl}}_{\text{4}}$ and ${\text{CF}}_{\text{4}}$ also non-polar covalent compounds like ${\text{CBr}}_{\text{4}}$. But they have small sizes that the intermolecular force in them are not much strong and can be collapsed easily. So a higher temperature is not required to boil them. Due to this their boiling point is lower than that of ${\text{CBr}}_{\text{4}}.$

Conclusion

The compound with the highest boiling point is identified and the same is justified.

(b)

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, lowest freezing point, smallest vapor pressure, greatest viscosity, greatest heat of vaporization, smallest enthalpy of fusion have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories viz., solid state, liquid state, gaseous state. In all of these three states of matter, the constituents (molecules or ions) of the matter do possess forces between them which are not the same in each state. These forces are called intermolecular forces.

The intermolecular forces are relatively strong in the solids than liquids and weaker in the gaseous substances. This variation influences many of the properties of all the three distinct states of the matter. Compounds with strong intermolecular forces have high melting point, boiling point, freezing point, viscosity enthalpy of vaporization, enthalpy of fusion and low vapor pressure.

The strength of intermolecular forces is,

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

Answer to Problem 42E

The compound with lowest freezing point is ${\text{F}}_{\text{2}}$ due to the weakest London dispersion force in small molecule.

Explanation of Solution

$\text{LiF}\text{,}{\text{F}}_{\text{2}}\text{and}\text{HCl}$

Identify the compound which has lowest freezing point and justify it.

The compound with lowest freezing point is ${\text{F}}_{\text{2}}$ due to the weakest London dispersion force in small molecule.

${\text{F}}_{\text{2}}$ is a non-polar covalent molecule. It has only London dispersion forces. ${\text{F}}_{\text{2}}$ is a small molecule that the strength of the intermolecular force that is London dispersion force is not high. So that the ${\text{F}}_{\text{2}}$ molecules readily freeze and it has the lowest freezing point.

Analyze why the other compounds don’t have the lowest freezing point and justify the same.

The compounds other than ${\text{F}}_{\text{2}}$ have intermolecular force of high strength.

The other compounds $\text{LiF}$ and $\text{HCl}$ are ionic compound and polar covalent compound respectively. $\text{LiF}$ has electrostatic forces which is too strong to break. $\text{HCl}$ has dipole forces in addition to London dispersion forces. Thus the intermolecular forces in $\text{LiF}$ and $\text{HCl}$ are strong than that of found in ${\text{F}}_{\text{2}}$. Hence ${\text{F}}_{\text{2}}$ has the lowest freezing point.

Conclusion

The compound with the lowest boiling point is identified and the same is justified.

(c)

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, lowest freezing point, smallest vapor pressure, greatest viscosity, greatest heat of vaporization, smallest enthalpy of fusion have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories viz., solid state, liquid state, gaseous state. In all of these three states of matter, the constituents (molecules or ions) of the matter do possess forces between them which are not the same in each state. These forces are called intermolecular forces.

The intermolecular forces are relatively strong in the solids than liquids and weaker in the gaseous substances. This variation influences many of the properties of all the three distinct states of the matter. Compounds with strong intermolecular forces have high melting point, boiling point, freezing point, viscosity enthalpy of vaporization, enthalpy of fusion and low vapor pressure.

The strength of intermolecular forces is,

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

Answer to Problem 42E

The compound with lowest vapor pressure is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ due to the presence of strong hydrogen bonding in it.

Explanation of Solution

${\text{CH}}_{\text{3}}{\text{OCH}}_{\text{3}},{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH,}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$

Identify the compound which has lowest vapor pressure and justify it.

The compound ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ has the lowest vapor pressure because the molecules of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ exhibit stronger hydrogen bonding.

A boiling liquid partly vaporizes and the vaporized molecules do exist in equilibrium with the liquid molecules. The pressure exerted by the vaporized molecules is termed as vapor pressure. Molecules are able to move freely if the intermolecular forces are weak. The more the free movement of molecules the more will be the pressure exerted by them. If the intermolecular forces are strong, the movement of the molecules is restricted to some extent that the pressure exerted by them will be low. Thus increase in the strength of intermolecular forces in a substance decreases its vapor pressure.

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ is a hydroxyl compound. It exhibits hydrogen bonding to a very large  extent.  So the molecules of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ exhibits low vapor pressure.

Analyze why the other compounds don’t have the lowest vapor pressure and justify the same.

The intermolecular forces exist in them are not of high strength.

In ${\text{CH}}_{\text{3}}{\text{OCH}}_{\text{3}}$ due to the absence of direct $\text{O-H}$ bonding, the hydrogen bonding does not exist. ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ is a non-polar covalent compound and has only London dispersion forces. Both of the intermolecular forces are weaker than hydrogen bonding. Thus these compounds can have vapor pressure higher than that of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}\text{.}$

Conclusion

The compound with the lowest vapor pressure is identified and the same is justified.

(d)

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, lowest freezing point, smallest vapor pressure, greatest viscosity, greatest heat of vaporization, smallest enthalpy of fusion have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories viz., solid state, liquid state, gaseous state. In all of these three states of matter, the constituents (molecules or ions) of the matter do possess forces between them which are not the same in each state. These forces are called intermolecular forces.

The intermolecular forces are relatively strong in the solids than liquids and weaker in the gaseous substances. This variation influences many of the properties of all the three distinct states of the matter. Compounds with strong intermolecular forces have high melting point, boiling point, freezing point, viscosity enthalpy of vaporization, enthalpy of fusion and low vapor pressure.

The strength of intermolecular forces is,

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

Answer to Problem 42E

The compound with greatest viscosity is ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ due to the presence of strong hydrogen bonding in it.

Explanation of Solution

${\text{H}}_{\text{2}}\text{S,}\text{HF,}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

     Identify the compound which has highest viscosity and justify it.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ has the highest viscosity due the presence of strong hydrogen bonding in their molecules.

Viscosity of a liquid is its resistance to flow. A liquid is said to be highly viscous if it hardly flows. When the intermolecular forces are strong, the molecules are unable to move freely. The strong hydrogen bonding in the molecules of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ bound them together tightly that their viscosity becomes high.

Analyze why the other compounds don’t have the highest viscosity and justify the same.

The intermolecular forces exist in them are not of high strength.

${\text{H}}_{\text{2}}\text{S}$ is a polar covalent compound. It has London dispersion force and dipole force. They are weak in strength when compared to hydrogen bonding. $\text{HF}$ also exhibits hydrogen bonding but ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ has more hydrogen bonds due to the presence of two direct $\text{O-H}$ bonds. So $\text{HF}$ and ${\text{H}}_{\text{2}}\text{S}$ have low viscosity compared to ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ because ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ has the intermolecular forces of highest strength among them.

Conclusion

The compound with the highest viscosity is identified and the same is justified.

(e)

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, lowest freezing point, smallest vapor pressure, greatest viscosity, greatest heat of vaporization, smallest enthalpy of fusion have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories viz., solid state, liquid state, gaseous state. In all of these three states of matter, the constituents (molecules or ions) of the matter do possess forces between them which are not the same in each state. These forces are called intermolecular forces.

The intermolecular forces are relatively strong in the solids than liquids and weaker in the gaseous substances. This variation influences many of the properties of all the three distinct states of the matter. Compounds with strong intermolecular forces have high melting point, boiling point, freezing point, viscosity enthalpy of vaporization, enthalpy of fusion and low vapor pressure.

The strength of intermolecular forces is,

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

Answer to Problem 42E

The compound with greatest heat of vaporization is ${\text{H}}_{\text{2}}\text{CO}$ due to the presence of dipole forces in it.

Explanation of Solution

${\text{H}}_{\text{2}}\text{CO,}{\text{CH}}_{\text{4}},{\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}}$

      Identify the compound which has highest heat of vaporization and justify it.

The compound with highest heat of vaporization is ${\text{H}}_{\text{2}}\text{CO}$ due to the presence of dipole forces in it.

${\text{H}}_{\text{2}}\text{CO}$ is a polar covalent compound and so dipole forces exist in the ${\text{H}}_{\text{2}}\text{CO}$ molecules.  Stronger the intermolecular force, more will be the heat required to evaporate the compound.    Thus ${\text{H}}_{\text{2}}\text{CO}$ molecules require more heat energy to vaporize. As a result ${\text{H}}_{\text{2}}\text{CO}$ has high heat  of vaporization.

Analyze why the other compounds don’t have the highest heat of vaporization and justify the same.

The low strength of intermolecular forces in ${\text{CH}}_{\text{4}}\text{ and}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}}$ is responsible for their low heat of vaporization.

Unlike ${\text{H}}_{\text{2}}\text{CO}$, ${\text{CH}}_{\text{4}}\text{and}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}}$ are non-polar covalent compounds. They have weak London dispersion forces. Due to the low strength of their intermolecular forces, their respective heat of vaporization is not as high as that of ${\text{H}}_{\text{2}}\text{CO}$.

Conclusion

The compound with the highest heat of vaporization is identified and the same is justified.

(f)

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, lowest freezing point, smallest vapor pressure, greatest viscosity, greatest heat of vaporization, smallest enthalpy of fusion have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories viz., solid state, liquid state, gaseous state. In all of these three states of matter, the constituents (molecules or ions) of the matter do possess forces between them which are not the same in each state. These forces are called intermolecular forces.

The intermolecular forces are relatively strong in the solids than liquids and weaker in the gaseous substances. This variation influences many of the properties of all the three distinct states of the matter. Compounds with strong intermolecular forces have high melting point, boiling point, freezing point, viscosity enthalpy of vaporization, enthalpy of fusion and low vapor pressure.

The strength of intermolecular forces is,

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

Answer to Problem 42E

The compound with lowest enthalpy of fusion is ${\text{I}}_{\text{2}}$ because it has only London dispersion forces unlike other compounds $\text{CsBr}\text{and}\text{CaO}$ which have stronger ionic forces.

Explanation of Solution

${\text{I}}_{\text{2}},\text{CsBr,}\text{CaO}$

     Identify the compound which has lowest enthalpy of fusion and justify it.

The compound with lowest enthalpy of fusion is ${\text{I}}_{\text{2}}$ because it has only London dispersion forces unlike other compounds $\text{CsBr}\text{and}\text{CaO}$ which have stronger ionic forces.

${\text{I}}_{\text{2}}$ is a non-polar covalent compound. Molecules of ${\text{I}}_{\text{2}}$ have merely London dispersion forces between them. This intermolecular force is usually not very strong that ${\text{I}}_{\text{2}}$ has the lowest enthalpy of fusion.

Analyze why the other compounds don’t have the lowest enthalpy of vaporization and justify the same.

The intermolecular forces in the compounds given except ${\text{I}}_{\text{2}}$, have high strength.

$\text{CsBr}\text{and}\text{CaO}$ both are ionic compounds. Ionic compounds have the strongest electrostatic forces in their molecules. Thus they have high enthalpy of fusion.

Conclusion

The compound with the lowest enthalpy of fusion is identified and the same is justified.