Chapter 9, Problem 41E

In each of the following groups of substances, pick the one that has the given property. Justify your answer.

a. highest boiling point: HBr, Kr, or Cl₂

b. highest freezing point: H₂O, NaCl, or HF

c. lowest vapor pressure at 25°C: Cl₂, Br₂, or I₂

d. lowest freezing point: N₂, CO, or CO₂

e. lowest boiling point: CH₄, CH₃CH₃, or CH₃CH₂CH₃

f. highest boiling point: HF, HCl, or HBr

g.

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, highest freezing point, lowest vapor pressure, lowest freezing point and lowest boiling point have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories - 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.

If the strength of intermolecular force is high in a substance then its melting point, freezing point, boiling point will be high and the vapor pressure will be low.

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 41E

The compound with highest boiling point is $\text{HBr}$.  Because this is the only compound in the set that has strong intermolecular force – dipole force.

Explanation of Solution

$\text{Kr,}\text{HBr,}{\text{Cl}}_{\text{2}}$

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

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

$\text{HBr}$ is the only polar covalent compound among the given compounds. Being a polar covalent compound $\text{HBr}$ has dipole forces. Hence among the given compounds only in $\text{HBr}$ the intermolecular force is 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 $\text{HBr}$ 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 $\text{HBr}$ have intermolecular force of low strength.

$\text{Kr}$ and ${\text{Cl}}_{\text{2}}$ are non-polar covalent compounds unlike $\text{HBr}$ . The intermolecular force in them is London dispersion force which is not as strong as dipole forces. London dispersion forces 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 $\text{HBr}$.

Conclusion

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

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, highest freezing point, lowest vapor pressure, lowest freezing point and lowest boiling point have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories - 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.

If the strength of intermolecular force is high in a substance then its melting point, freezing point, boiling point will be high and the vapor pressure will be low.

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 41E

The compound with highest freezing point or melting point is $\text{NaCl}$ due to the strongest electrostatic forces present in the molecule.

Explanation of Solution

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

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

$\text{NaCl}$ has the highest melting point. The intermolecular force in $\text{NaCl}$ is highest in strength.

$\text{NaCl}$ is the only ionic compound among the given compounds. Ionic compounds have the electrostatic forces which are known to be strong. Hence among the given compounds only $\text{NaCl}$ has strong intermolecular force. At normal temperature, the intermolecular forces are not broken. So high temperature is required to break the strong intermolecular forces and allow the substance to melt. So $\text{NaCl}$ has the highest melting point.

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

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

${\text{H}}_{\text{2}}\text{O}$ and $\text{HF}$ are polar covalent compounds. The intermolecular forces in them are dipole forces and hydrogen bonding which are not as strong as electrostatic forces.

Hence ${\text{H}}_{\text{2}}\text{O}$ and $\text{HF}$ have low freezing/melting point than $\text{NaCl}$.

Conclusion

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

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, highest freezing point, lowest vapor pressure, lowest freezing point and lowest boiling point have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories - 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.

If the strength of intermolecular force is high in a substance then its melting point, freezing point, boiling point will be high and the vapor pressure will be low.

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 41E

The compound with lowest vapor pressure is ${\text{I}}_{\text{2}}$ due to the presence of strong hydrogen London dispersion force in it.

Explanation of Solution

${\text{Cl}}_{\text{2}},{\text{Br}}_{\text{2}},{\text{I}}_{\text{2}}$

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

${\text{I}}_{\text{2}}$ has the lowest vapor pressure. Compare to the other compounds, the intermolecular force in ${\text{I}}_{\text{2}}$ is strong.

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. If the intermolecular forces are weak the molecules are not held together strongly that they are able to move freely. The pressure exerted by the molecules is directly proportional to the free movement of molecules. 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.

${\text{I}}_{\text{2}}$ is a large non-polar covalent compound. The London dispersion forces in them are strong due to its large molecular size. Hence, ${\text{I}}_{\text{2}}$ has the lowest 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.

Both ${\text{Cl}}_{\text{2}}\text{and}{\text{Br}}_{\text{2}}$ are non-polar covalent compounds like ${\text{I}}_{\text{2}}.$ The intermolecular force exist in them is London dispersion force. Their molecular size is not as large as ${\text{I}}_{\text{2}}$ that the London dispersion force in them is not strong. Thus they readily exhibit free movement and exert vapor pressure.

Conclusion

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

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, highest freezing point, lowest vapor pressure, lowest freezing point and lowest boiling point have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories - 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.

If the strength of intermolecular force is high in a substance then its melting point, freezing point, boiling point will be high and the vapor pressure will be low.

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 41E

The compound with lowest freezing point is ${\text{N}}_{\text{2}}$ due to the presence of weakest London dispersion force in it.

Explanation of Solution

$\text{(d)}{\text{N}}_{\text{2}},\text{CO,}{\text{CO}}_{\text{2}}$

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

${\text{N}}_{\text{2}}$ has the lowest freezing point. Among the given compounds the intermolecular force in ${\text{N}}_{\text{2}}$ is the weakest.

${\text{N}}_{\text{2}}$ is a non-polar covalent molecule. It has only London dispersion forces. ${\text{N}}_{\text{2}}$ is a small molecule that the strength of the intermolecular force that is London dispersion force is not high. So that the ${\text{N}}_{\text{2}}$ 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 ${\text{N}}_{\text{2}}$ have intermolecular force of high strength.

The other compounds $\text{CO}$ and ${\text{CO}}_{\text{2}}$ are polar covalent compound and non-polar covalent compounds respectively. Both $\text{CO}$ and ${\text{CO}}_{\text{2}}$ have London dispersion forces and they are larger molecular size than ${\text{N}}_{\text{2}}$. Thus they have strong London dispersion forces than that of ${\text{N}}_{\text{2}}$. Further, $\text{CO}$ has dipole-dipole force. So overall, the strength of intermolecular force in $\text{CO}$ and ${\text{CO}}_{\text{2}}$ is higher than that of ${\text{N}}_{\text{2}}$. So their freezing point is not lower than that of ${\text{N}}_{\text{2}}$.

Conclusion

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

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, highest freezing point, lowest vapor pressure, lowest freezing point and lowest boiling point have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories - 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.

If the strength of intermolecular force is high in a substance then its melting point, freezing point, boiling point will be high and the vapor pressure will be low.

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 41E

The compound with lowest boiling point is ${\text{CH}}_{\text{4}}$ due to the presence of weakest London dispersion force in it.

Explanation of Solution

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

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

${\text{CH}}_{\text{4}}$ has the lowest boiling point. The intermolecular force in ${\text{CH}}_{\text{4}}$ is of low strength.

${\text{CH}}_{\text{4}}$ is a non-polar covalent compound. Among the given compounds ${\text{CH}}_{\text{4}}$ has the smallest size. Being a non-polar covalent compound ${\text{CH}}_{\text{4}}$ has London dispersion forces. When the molecule is smaller in size London dispersion forces is not well pronounced. Hence in ${\text{CH}}_{\text{4}}$ the intermolecular force is very weak. The weak intermolecular force leads to its lowest boiling point.

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

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

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{3}}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ are also non-polar covalent compounds like ${\text{CH}}_{\text{4}}.$ But they have larger sizes that the intermolecular force in them are strong and cannot be collapsed easily. So a higher temperature is required to boil them.  Due to this their boiling point is higher than that of ${\text{CH}}_{\text{4}}.$

Conclusion

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

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, highest freezing point, lowest vapor pressure, lowest freezing point and lowest boiling point have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories - 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.

If the strength of intermolecular force is high in a substance then its melting point, freezing point, boiling point will be high and the vapor pressure will be low.

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 41E

The compound with highest boiling point is $\text{HF}$ because it forms the strongest hydrogen bonding unlike other compounds in the given set.

Explanation of Solution

$\text{HF},\text{HCl,}\text{HBr}$

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

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

$\text{HF}$ is the only compound among the given set of compounds that exhibits hydrogen bonding. Hydrogen bonded liquids have high boiling point.

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

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

All the compounds in the given set are polar covalent compounds. Among the given compounds $\text{HF}$ has high polarity due to the high electronegativity of Fluorine. All these compounds have London dispersion forces and dipole forces. But only $\text{HF}$ has hydrogen bonding as additional intermolecular force. As a result, among the given compounds $\text{HF}$ has the strongest intermolecular force and higher boiling point.

Conclusion

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

Interpretation Introduction

Interpretation:

From the given set of compounds, the compounds with highest boiling point, highest freezing point, lowest vapor pressure, lowest freezing point and lowest boiling point have to be identified and the same has to be justified.

Concept Introduction:

Matter is generally classified into three distinct categories - 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.

If the strength of intermolecular force is high in a substance then its melting point, freezing point, boiling point will be high and the vapor pressure will be low.

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 41E

The compound with the lowest vapor pressure is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ because it forms the strongest hydrogen bonding unlike other compounds in the given set.

Explanation of Solution

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

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 hydrogen bonding, the strongest intermolecular force.

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. If the intermolecular forces are weak the molecules are held together loosely. Then they exhibit faster movement. 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 that the molecules of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$ are held together tightly and does not move freely. 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 ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}$ due to the absence of direct $\text{O-H}$ bonding, the hydrogen bonding does not exist and has dipole force and London dispersion forces. ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ is a non-polar covalent compound and has only London dispersion forces. Both the dipole forces and London dispersion 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.