AP Chemistry: 3.1 Intermolecular Forces – Exam Style questions with Answer- MCQ

Questions 

The structural formulas for two isomers of 1,2-dichloroethene are shown above. Which of the two liquids has the higher equilibrium vapor pressure at 20°C, and why?

(A) The cis-isomer, because it has dipole-dipole interactions, whereas the trans-isomer has only London dispersion forces

(B) The cis-isomer, because it has only London dispersion forces, whereas the trans-isomer also has dipole-dipole interactions

(C) The trans-isomer, because it has dipole-dipole interactions, whereas the cis-isomer has only London dispersion forces

(D) The trans-isomer, because it has only London dispersion forces, whereas the cis-isomer also has dipole-dipole interactions

▶️Answer/Explanation

Ans: C

Based on the structural formulas provided for the cis and trans isomers of 1,2-dichloroethene, the correct answer is (C) The trans-isomer, because it has dipole-dipole interactions, whereas the cis-isomer has only London dispersion forces.

In the trans isomer, the two chlorine atoms are on opposite sides of the C=C double bond. This results in a net dipole moment for the molecule due to the electronegativity difference between carbon and chlorine atoms. The trans isomer molecules can experience dipole-dipole interactions in addition to London dispersion forces.

In the cis isomer, however, the two chlorine atoms are on the same side of the C=C double bond. This results in a net zero dipole moment as the dipole moments from the two C-Cl bonds cancel out. The cis isomer molecules only experience London dispersion forces between the instantaneous dipoles.

Dipole-dipole interactions are generally weaker than covalent or ionic bonds but stronger than London dispersion forces. The presence of these dipole-dipole interactions in the trans isomer requires more energy to overcome compared to just London forces in the cis isomer.

Therefore, the trans isomer with dipole-dipole interactions will have a lower vapor pressure at equilibrium compared to the cis isomer which exhibits only London dispersion forces. Substances with weaker intermolecular forces have higher vapor pressures.

Questions 

Solid carbon tetrachloride, \(CCl_4\)(s), is represented by the diagram above. The attractions between the\( CCl_4\) molecules that hold the molecules
together in the solid state are best identified as
(A) polar covalent bonds
(B) nonpolar covalent bonds
(C) intermolecular attractions resulting from temporary dipoles
(D) intermolecular attractions resulting from permanent dipoles

▶️Answer/Explanation

Ans: C

Carbon tetrachloride (\(CCl_4\)) is a nonpolar molecule due to the symmetrical arrangement of its four chlorine atoms around the central carbon atom, which results in a net zero dipole moment.

In the solid state, the interactions between \(CCl_4\) molecules are primarily due to intermolecular forces rather than intramolecular bonds because \(CCl_4\) molecules do not form covalent bonds with each
other.

The correct answer is:

(C) Intermolecular attractions resulting from temporary dipoles

These temporary dipoles arise from the movement of electrons within molecules, leading to temporary imbalances in electron distribution, which induce temporary dipoles in neighboring molecules. This type
of intermolecular attraction is known as London dispersion forces or van der Waals forces.

Questions 

The structural formulas for two isomers of 1,2-dichloroethene are shown above. Which of the two liquids has the higher equilibrium vapor pressure at 20°C, and why?

(A) The cis-isomer, because it has dipole-dipole interactions, whereas the trans-isomer has only London dispersion forces

(B) The cis-isomer, because it has only London dispersion forces, whereas the trans-isomer also has dipole-dipole interactions

(C) The trans-isomer, because it has dipole-dipole interactions, whereas the cis-isomer has only London dispersion forces

(D) The trans-isomer, because it has only London dispersion forces, whereas the cis-isomer also has dipole-dipole interactions

▶️Answer/Explanation

Ans: C

Based on the structural formulas provided for the cis and trans isomers of 1,2-dichloroethene, the correct answer is (C) The trans-isomer, because it has dipole-dipole interactions, whereas the cis-isomer has only London dispersion forces.

In the trans isomer, the two chlorine atoms are on opposite sides of the C=C double bond. This results in a net dipole moment for the molecule due to the electronegativity difference between carbon and chlorine atoms. The trans isomer molecules can experience dipole-dipole interactions in addition to London dispersion forces.

In the cis isomer, however, the two chlorine atoms are on the same side of the C=C double bond. This results in a net zero dipole moment as the dipole moments from the two C-Cl bonds cancel out. The cis isomer molecules only experience London dispersion forces between the instantaneous dipoles.

Dipole-dipole interactions are generally weaker than covalent or ionic bonds but stronger than London dispersion forces. The presence of these dipole-dipole interactions in the trans isomer requires more energy to overcome compared to just London forces in the cis isomer.

Therefore, the trans isomer with dipole-dipole interactions will have a lower vapor pressure at equilibrium compared to the cis isomer which exhibits only London dispersion forces. Substances with weaker intermolecular forces have higher vapor pressures.

Questions 

The structure of haloacetic acids, \(XCH_{2}COOH\) (where X is either F, Cl, Br, or I ), is shown above. The dissociation constants and
molar masses of four haloacetic acids are listed in the table below.

An aqueous solution contains small but equal concentrations of both chloroacetic and fluoroacetic acids. Which statement comparing the percent ionizations of the two acids in the solution is true?
(A) The percent ionization of chloroacetic acid is greater than that of fluoroacetic acid.
(B) The percent ionization of chloroacetic acid is less than that of fluoroacetic acid.
(C) The percent ionizations cannot be compared without knowing the concentrations of the two acids.
(D) The percent ionizations cannot be compared without knowing the pH of the solution.

▶️Answer/Explanation

Ans: A

The compound most likely to have the lower boiling point is chloroacetic acid, and the reason is related to the strength of intermolecular forces.

. London Dispersion Forces: These are the weakest intermolecular forces and are present in all molecules. They are stronger in molecules with more electrons, which generally means larger or heavier
atoms. Iodoacetic acid, with iodine, has more electrons and a larger electron cloud than chloroacetic acid, leading to stronger London dispersion forces.
. Dipole-Dipole Forces: These forces are stronger than London dispersion forces and occur between molecules with permanent dipoles. Both chloroacetic acid and iodoacetic acid have similar
structures, so the difference in dipole-dipole forces would not be significant enough to account for differences in boiling points.

Considering these points, the correct answer is (A) Chloroacetic acid, because the London dispersion forces among its molecules are weaker. The presence of chlorine instead of iodine means
chloroacetic acid has fewer electrons and a smaller electron cloud, resulting in weaker London dispersion forces compared to iodoacetic acid, and thus a lower boiling point.

Questions 

The structure of haloacetic acids, \(XCH_{2}COOH\) (where X is either F, Cl, Br, or I ), is shown above. The dissociation constants and
molar masses of four haloacetic acids are listed in the table below.

Which compound, chloroacetic acid or iodoacetic acid, most likely has the lower boiling point, and why?
(A) Chloroacetic acid, because the London dispersion forces among its molecules are weaker.
(B) Chloroacetic acid, because the dipole-dipole forces among its molecules are weaker.
(C) Iodoacetic acid, because the London dispersion forces among its molecules are stronger.
(D) Iodoacetic acid, because the dipole-dipole forces among its molecules are stronger.

▶️Answer/Explanation

Ans: B

 Let’s explain the answer based on option B, which states that chloroacetic acid has the lower boiling point because the dipole-dipole forces among its molecules are weaker.

In chloroacetic acid (\(CH_2CICOOH\)), the chlorine atom (\(Cl\)) is more electronegative than the carbon atom, creating a polar covalent bond between them. This results in a dipole moment within the
molecule, with the chlorine end being slightly negative (\(8^-\)) and the carbon end being slightly positive (\(8^+\)).

Similarly, in iodoacetic acid (\(CH_2ICOOH\)), the iodine atom (\(I\)) is more electronegative than the carbon atom, creating a polar covalent bond between them, resulting in a dipole moment within the
molecule, with the iodine end being slightly negative (\(8^-\)) and the carbon end being slightly positive (\(8^+\)).

However, chlorine is more electronegative than iodine, resulting in a larger electronegativity difference between the carbon and halogen atoms in chloroacetic acid compared to iodoacetic acid. This larger
electronegativity difference leads to stronger dipole-dipole interactions in chloroacetic acid compared to iodoacetic acid.

Stronger dipole-dipole interactions in chloroacetic acid mean that more energy is required to break these interactions, resulting in a higher boiling point for chloroacetic acid compared to iodoacetic acid.

Therefore, option B is correct because chloroacetic acid has weaker dipole-dipole forces among its molecules compared to iodoacetic acid, leading to a lower boiling point for chloroacetic acid.

Question

Based on the information in the table above, which of the compounds has the highest boiling point, and why?
(A) Butanal, because it can form intermolecular hydrogen bonds
(B) Pentane, because it has the longest carbon chain
(C) Pentane, because it has the most C−H bonds
(D) Propanoic acid, because it can form intermolecular hydrogen bonds

▶️Answer/Explanation

Ans:D

To determine which compound has the highest boiling point, we need to consider the intermolecular forces present in each compound.

  1. Butanal: Butanal is an aldehyde. It can form hydrogen bonds between the carbonyl oxygen and hydrogen atoms of neighboring molecules.

  2. Pentane: Pentane is an alkane. It exhibits only dispersion forces (London forces) between its molecules.

  3. Propanoic acid: Propanoic acid is a carboxylic acid. It can form hydrogen bonds between the carboxyl group and hydrogen atoms of neighboring molecules.

Comparing the intermolecular forces:

  • Butanal and propanoic acid both have the potential to form hydrogen bonds, which are stronger than dispersion forces.
  • Pentane has only dispersion forces.

Therefore, compounds that can form hydrogen bonds typically have higher boiling points because these bonds are stronger than dispersion forces.

So, the compound with the highest boiling point is propanoic acid (option D) because it can form intermolecular hydrogen bonds.

Question

Equimolar samples of \(CH _3OH(l)\) and \(C_2H_5OH(l)\) are placed in separate, previously evacuated, rigid 2.0 L vessels. Each vessel is attached to a pressure gauge, and the temperatures are kept at 300 K. In both vessels, liquid is observed to remain present at the bottom of the container at all times. The change in pressure inside the vessel containing \(CH_3OH(l) \)is shown below.

Which of the following best describes the change that takes place immediately after the\( CH_{3}OH\)(l) is introduced into the previously evacuated vessel?
(A) A chemical change takes place because covalent bonds are broken.
(B) A chemical change takes place because intermolecular attractions are overcome.
(C) A physical change takes place because covalent bonds are broken.
(D) A physical change takes place because intermolecular attractions are overcome.

Answer/Explanation

Ans:D

The correct answer is (D) A physical change takes place because intermolecular attractions are overcome. When \(CH_{3}OH(l)\) is introduced into the evacuated vessel, the molecules of methanol start to evaporate, which is a physical change. This process involves the overcoming of intermolecular forces, specifically hydrogen bonds and van der Waals forces, that hold the liquid molecules together. No covalent bonds within the methanol molecules are broken during this change; instead, the molecules transition from the liquid phase to the gas phase, which results in an increase in pressure as observed in the pressure gauge. This is a typical behavior of a volatile liquid in a closed container at constant temperature.

Question

Equimolar samples of \(CH _3OH(l)\) and \(C_2H_5OH(l)\) are placed in separate, previously evacuated, rigid 2.0 L vessels. Each vessel is attached to a pressure gauge, and the temperatures are kept at 300 K. In both vessels, liquid is observed to remain present at the bottom of the container at all times. The change in pressure inside the vessel containing \(CH_3OH(l) \)is shown below.

Compared to the equilibrium vapor pressure of \(CH_{3}\)OH(l) at 300 K, the equilibrium vapor pressure of \(C_{2}H_{5}\)OH(l) at 300 K is
(A) the same, because both compounds have hydrogen bonding among their molecules
(B) higher, because London dispersion forces among\( C_{2}H_{5}\)OH molecules are greater than those among\( CH_{3}\)OH molecules
(C) lower, because London dispersion forces  among \(C_{2}H_{5}\)OH molecules are greater than those among \(CH_{3}\)OH molecules
(D) lower, because of the larger number of hydrogen bonds among \(C_{2}H_{5}\)OH molecules

▶️Answer/Explanation

Ans:D

Boiling point of \( C_{2}H_{5}OH\) is higher than \( CH_{3}OH\). Although both have hydrogen bonding but the magnitude of hydrogen dispersion forces is higher in \( C_{2}H_{5}OH\) and hence having high boiling boiling point and low vapor pressure. So option D is correct

1. Hydrogen Bonding: Both \(CH_3OH\) and \(C_2H_5OH\) exhibit hydrogen bonding due to the presence of -OH groups in their molecules. Hydrogen bonding occurs when a hydrogen atom covalently bonded to an electronegative atom (in this case, oxygen) is attracted to another electronegative atom in a neighboring molecule. This results in stronger intermolecular forces compared to compounds without hydrogen bonding.

2. Molecular Size and Mass: While both compounds have hydrogen bonding, \(C_2H_5OH\) (ethanol) has a larger molecular size and mass compared to \(CH_3OH\) (methanol). The larger size and mass of ethanol molecules lead to stronger London dispersion forces. London dispersion forces are temporary dipole-dipole attractions that arise due to temporary fluctuations in electron distribution within molecules. Larger and more massive molecules experience greater London dispersion forces, resulting in stronger intermolecular attractions.

3. Boiling Point and Vapor Pressure: The strength of intermolecular forces directly affects the boiling point and vapor pressure of a substance. A higher boiling point indicates stronger intermolecular attractions, while a lower vapor pressure suggests fewer molecules escaping from the liquid phase to the vapor phase at a given temperature.

  • Due to its stronger London dispersion forces resulting from its larger size and mass, \(C_2H_5OH\) has a higher boiling point compared to \(CH_3OH\).
  • Additionally, the stronger intermolecular attractions in \(C_2H_5OH\) lead to lower vapor pressure at the same temperature compared to \(CH_3OH\). This means that fewer ethanol molecules escape into the vapor phase, resulting in a lower equilibrium vapor pressure.

Therefore, even though both compounds exhibit hydrogen bonding, the stronger London dispersion forces in \(C_2H_5OH\) contribute to its higher boiling point and lower vapor pressure compared to \(CH_3OH\).

Question

 Based on Coulomb’s law and the information in the table above, which of the following anions is most likely to have the strongest interactions with nearby water molecules in an aqueous solution?
(A)\( Cl ^−\)
(B)\( I^ −\)
(C) \(S^{2−}\)
(D) \(Te^{2−}\)

▶️Answer/Explanation

Ans:C

To determine which anion is most likely to have the strongest interactions with nearby water molecules in an aqueous solution, we can consider Coulomb’s law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

However, in an aqueous solution, the interactions between ions and water molecules involve ion-dipole interactions, where the partial charges on the water molecule interact with the full charge of the ion. Therefore, we should consider the charge of the ion and its size.

From the options provided:
 \( Cl^- \) has a charge of -1.
 \( I^- \) has a charge of -1.
 \( S^{2-} \) has a charge of -2.
 \( Te^{2-} \) has a charge of -2.

Considering Coulomb’s law, ions with larger charges will generally have stronger interactions with water molecules. Additionally, larger ions are expected to have weaker interactions due to their larger size and lower charge density.

Among the options, \( S^{2-} \) and \( Te^{2-} \) have larger charges compared to \( Cl^- \) and \( I^- \). However, \( S^{2-} \) has a smaller size compared to \( Te^{2-} \), making its charge density higher.

Therefore, \( S^{2-} \) is most likely to have the strongest interactions with nearby water molecules in an aqueous solution. Thus, the correct answer is option:(C) \( S^{2−} \)

Questions

Based on the structures shown above, which of the following statements identifies the compound with the higher boiling point and provides the best explanation for the higher boiling point?
(A) Compound 1, because it has stronger dipole-dipole forces than compound 2
(B) Compound 1, because it forms hydrogen bonds, whereas compound 2 does not
(C) Compound 2, because it is less polarizable and has weaker London dispersion forces than compound 1
(D) Compound 2, because it forms hydrogen bonds, whereas compound 1 does not

▶️Answer/Explanation

Ans: D

To determine which compound has the higher boiling point and the best explanation, we need to consider the intermolecular forces present in each compound.

(A) Compound 1, because it has stronger dipole-dipole forces than compound 2 This statement is incorrect. While trimethylamine (Compound 1) has a permanent dipole due to the electronegativity difference between nitrogen and hydrogen, n-butylamine (Compound 2) also has a permanent dipole due to the N-H bond. The dipole-dipole forces are likely stronger in Compound 2 due to the presence of the polar N-H bond.

(B) Compound 1, because it forms hydrogen bonds, whereas compound 2 does not This statement is incorrect. Trimethylamine (Compound 1) does not have any hydrogen atoms attached to the highly electronegative nitrogen atom, so it cannot form hydrogen bonds. However, n-butylamine (Compound 2) has an N-H bond, allowing it to form hydrogen bonds.

(C) Compound 2, because it is less polarizable and has weaker London dispersion forces than compound 1 This statement is incorrect. London dispersion forces (van der Waals forces) are present in both compounds due to their non-polar hydrocarbon portions. However, n-butylamine (Compound 2) is likely more polarizable due to the presence of the polar N-H bond, leading to stronger dispersion forces compared to trimethylamine (Compound 1).

(D) Compound 2, because it forms hydrogen bonds, whereas compound 1 does not This statement is correct. n-Butylamine (Compound 2) has an N-H bond, which can form hydrogen bonds with other molecules or with itself. Hydrogen bonding is a strong intermolecular force that requires more energy to overcome, resulting in a higher boiling point. On the other hand, trimethylamine (Compound 1) does not have any hydrogen atoms attached to the nitrogen atom, so it cannot form hydrogen bonds.

Therefore, the correct answer is (D) Compound 2, because it forms hydrogen bonds, whereas compound 1 does not.

Questions

At 298 K and 1 atm, Br2 is a liquid with a high vapor pressure, and \(Cl_{2}\) is a gas. Those observations provide evidence that under the given conditions, the
(A) forces among \(Br_{2}\) molecules are stronger than those among \(Cl_{2}\) molecules
(B) forces among \(Cl_{2}\) molecules are stronger than the Cl-Cl bond
(C) Br-Br bond is stronger than the Cl-Cl bond
(D) Cl-Cl bond is stronger than the Br-Br bond

▶️Answer/Explanation

Ans: A

The given conditions indicate that \(Br_{2}\) is a liquid with a high vapor pressure, while \(Cl_{2}\) is a gas at 298 K and 1 atm pressure.

The fact that \(Br_{2}\) is a liquid with a high vapor pressure suggests that the intermolecular forces between \(Br_{2}\) molecules are relatively weak. In contrast, \(Cl_{2}\) exists as a gas under the same conditions, indicating weaker intermolecular forces compared to \(Br_{2}\).

Given that both \(Br_{2}\) and \(Cl_{2}\) consist of diatomic molecules with similar molecular weights, the difference in their physical states at 298 K and 1 atm pressure can be attributed to the strength of intermolecular forces.

Therefore, the correct answer is (A) forces among \(Br_{2}\) molecules are stronger than those among \(Cl_{2}\) molecules.

Question

Which particle diagram shown above best represents the strongest intermolecular force between two ethanol, \(C_2H_6O\), molecules?

A Diagram 1, because it shows hydrogen bonds forming between hydrogen atoms from different ethanol molecules.

B Diagram 1, because it shows strong, directional dipole-dipole forces between two polar ethanol molecules.

C Diagram 2, because it shows the formation of a hydrogen bond between an H atom bonded to an O atom with an O atom from another molecule.

D Diagram 2, because it shows the dipole from an ethanol molecule inducing a dipole in another ethanol molecule.
▶️Answer/Explanation

Ans:C

 Hydrogen bonding is the strongest attractive force between ethanol molecules. It requires an H atom to be bonded to a small atom with a high electronegativity, like N, O, or F. The hydrogen bond forms between atoms with partial positive and partial negative charges. In the OH bond, the H atom has a partial positive charge, and the O atom has a partial negative charge. The hydrogen bond is formed between OH—OH bonds from different molecules, as shown.

Question

 

The diagram above represents four cations, all shown to the same scale. Which cation would be predicted by Coulomb’s law to have the strongest ion-dipole attraction to water, and why?

A \(Li^+\) , because it is the smallest group 1 metal ion.

B \(Mg^{2+}\) , because it has the largest charge-to-size ratio.

C \(Na^+\) , because it has the smallest charge-to-size ratio.

D  \(Ca^{2+}\) , because it is the largest group 2 metal ion.

▶️Answer/Explanation

Ans:B

 Because water is polar, the O atom has a partial negative charge that can interact with cations through coulombic forces. Coulomb’s law predicts that the attractive force between the partial negative charge in the O atom and the cation will be directly proportional to the charge of the ion and inversely proportional to the size or ionic radius. Mg2+ is a relatively small ion and has the most positive charge among the cations shown.

Question

 

A solid compound of a group 1 (alkali) metal and a group 17 (halogen) element dissolves in water. The diagram above represents one type of solute particle present in the solution. Which of the following identifies the solute particle and best helps explain how the solute particle interacts with water molecules?

A The particle is a negative ion, and the interactions are hydrogen bonds.
B  The particle is a negative ion, and the interactions are ion-dipole attractions.
C  The particle is a positive ion, and the interactions are ion-dipole attractions.
D  The particle is a positive ion, and the interactions are dipole-dipole attractions.
▶️Answer/Explanation

Ans: C

 The water molecules are all oriented the same way with respect to the solute particle, with the negative ends of the water molecule dipoles directed toward the solute particle. This can only be the case if the solute particle has a positive charge. The major attractive forces between the polar water molecules and the positive ion are characterized as ion-dipole attractions.

Question

 In which of the following liquids do the intermolecular forces include dipole-dipole
forces?
(A)  \(F_{2}\)(l)
(B) \(CH_{4}\)(l)
(C) \( CF_{4}\)(l)
(D) \( CH_{2}\) \(F_{2}\)

▶️Answer/Explanation

Ans:D

Question

Based on the information in the table above, which liquid, \(CS_{2}\)(l) or \(CCl_{4}\)(l), has the higher equilibrium vapor pressure at 25° C, and why?
(A) \( CS_{2}\)(l), because it has stronger London dispersion forces
(B) \(CS_{2}\)(l), because it has weaker London dispersion forces
(C) \( CCl_{4}\)(l), because it has stronger London dispersion forces
(D)  \(CCl_{4}\)(l), because it has weaker London dispersion forces

▶️Answer/Explanation

Ans:B

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