Pre AP Chemistry -2.2G Intermolecular Forces and Molecular Polarity- MCQ Exam Style Questions -New Syllabus 2025-2026
Pre AP Chemistry -2.2G Intermolecular Forces and Molecular Polarity- MCQ Exam Style Questions – New Syllabus 2025-2026
Pre AP Chemistry -2.2G Intermolecular Forces and Molecular Polarity- MCQ Exam Style Questions – Pre AP Chemistry – per latest Pre AP Chemistry Syllabus.
Which type of intermolecular force acts between a polar molecule and a nonpolar molecule?
(A) Ion–dipole
(B) Hydrogen bonding
(C) Dipole–induced dipole
(D) Ionic bond
▶️ Answer / Explanation
When a polar molecule (permanent dipole) is near a nonpolar molecule, the electric field of the polar molecule distorts the electron cloud of the nonpolar molecule, inducing a temporary dipole in it. The resulting attraction is called a dipole–induced dipole interaction.
• Ion–dipole requires a full ion.
• Hydrogen bonding requires N–H, O–H, or F–H groups.
• Ionic bonds are intramolecular forces in ionic crystals, not between separate molecules.
Answer: C
A substance has a very low boiling point and is nonpolar. Which intermolecular force is primarily responsible for holding its molecules together in the liquid phase?
(A) Hydrogen bonding
(B) Dipole–dipole forces
(C) London dispersion forces
(D) Ion–dipole forces
▶️ Answer / Explanation
Nonpolar molecules have no permanent dipole, so they cannot experience dipole–dipole interactions or hydrogen bonding. The only IMF present is London dispersion forces (LDFs) — arising from instantaneous fluctuations in electron distribution creating temporary dipoles.
Low boiling point confirms the IMFs are weak, consistent with small nonpolar molecules where LDFs are minimal (few electrons, small size).
Answer: C
\(HCl\) is a polar molecule. Which intermolecular forces are present between two \(HCl\) molecules?
(A) London dispersion forces only
(B) Dipole–dipole forces and London dispersion forces
(C) Hydrogen bonding, dipole–dipole, and London dispersion forces
(D) Hydrogen bonding only
▶️ Answer / Explanation
\(HCl\) is polar, so it experiences dipole–dipole interactions between molecules. All molecules also experience London dispersion forces.
Hydrogen bonding requires H bonded to F, O, or N. Cl is not electronegative enough to support true H-bonding.
Answer: B
Three statements are made about hydrogen bonding and its effects on physical properties.
1. Hydrogen bonding can only occur when a hydrogen atom is covalently bonded to nitrogen, oxygen, or fluorine, and that hydrogen interacts with a lone pair on another N, O, or F atom.
2. \(HCl\) exhibits hydrogen bonding because it contains a hydrogen atom bonded to a very electronegative atom (chlorine), making its boiling point higher than expected from LDFs alone.
3. Water has an anomalously high boiling point of 100°C — far above what the Group 16 hydride trend would predict — because each water molecule can form up to four hydrogen bonds (two as donor, two as acceptor).
Which statements are correct?
(A) 2 and 3 only
(B) 1 only
(C) 1 and 3 only
(D) 1, 2, and 3
▶️ Answer / Explanation
Statement 1 — TRUE
This is the definition of hydrogen bonding. Hydrogen must be covalently bonded to N, O, or F and interact with a lone pair on another N, O, or F atom.
Statement 2 — FALSE
\(HCl\) does not exhibit hydrogen bonding. Chlorine is not electronegative and small enough to produce true hydrogen bonding. \(HCl\) only experiences dipole–dipole and London dispersion forces.
Statement 3 — TRUE
Water’s unusually high boiling point occurs because each molecule can form four hydrogen bonds (two donated and two accepted), creating an extensive hydrogen-bonding network.
Answer: C — Statements 1 and 3 only
Which property of a nonpolar molecule most directly determines the strength of its London dispersion forces?
(A) Its bond angle
(B) Its number of electrons (size / molar mass)
(C) Its dipole moment
(D) Its bond length
▶️ Answer / Explanation
London dispersion forces arise from instantaneous dipoles caused by electron motion. The more electrons a molecule has, the more polarizable its electron cloud becomes, producing stronger LDFs.
Answer: B
Which of the following molecules can form hydrogen bonds with water?
(A) \(CH_4\)
(B) \(CCl_4\)
(C) \(NH_3\)
(D) \(N_2\)
▶️ Answer / Explanation
\(NH_3\) has an N–H bond and a lone pair on nitrogen, allowing it to both donate and accept hydrogen bonds with water.
Answer: C
The boiling points of four substances are listed below. Which ordering of dominant IMF correctly matches each substance?
| Substance | BP (°C) |
|---|---|
| \(CH_4\) | −161 |
| \(CH_3Cl\) | −24 |
| \(CH_3OH\) | +65 |
| \(NaCl\) | +1413 |
(A) LDF < dipole–dipole < H-bonding < ionic bonds
(B) ionic bonds < H-bonding < dipole–dipole < LDF
(C) LDF < H-bonding < dipole–dipole < ionic bonds
(D) dipole–dipole < LDF < H-bonding < ionic bonds
▶️ Answer / Explanation
Matching each substance to its dominant attractive force:
• \(CH_4\) — nonpolar → LDF only → lowest BP
• \(CH_3Cl\) — polar → dipole–dipole
• \(CH_3OH\) — O–H present → hydrogen bonding
• \(NaCl\) — ionic lattice → ionic bonding
Force strength order:
LDF < dipole–dipole < hydrogen bonding < ionic bonding
Answer: A
\(I_2\) is a nonpolar solid that dissolves readily in nonpolar \(CCl_4\) but barely dissolves in polar water. Which principle explains this behavior?
(A) “Like dissolves like” — nonpolar solutes dissolve in nonpolar solvents because similar IMFs allow mixing
(B) \(I_2\) is ionic and ionic compounds don’t dissolve in nonpolar solvents
(C) \(I_2\) forms hydrogen bonds with \(CCl_4\) but not with water
(D) Water molecules are too large to surround \(I_2\) molecules
▶️ Answer / Explanation
The principle is “like dissolves like”.
Nonpolar solutes dissolve best in nonpolar solvents because both are held together by similar intermolecular forces (London dispersion forces).
\(I_2\) and \(CCl_4\) are both nonpolar, so they mix easily.
Answer: A
Both \(HF\) (MW = 20 g/mol) and \(F_2\) (MW = 38 g/mol) contain fluorine. \(HF\) boils at +19.5°C while \(F_2\) boils at −188°C. What is the best explanation?
(A) \(HF\) has more electrons
(B) \(F_2\) is ionic
(C) \(HF\) molecules form strong hydrogen bonds
(D) \(HF\) is nonpolar
▶️ Answer / Explanation
\(HF\) forms strong hydrogen bonds between molecules because hydrogen is bonded directly to fluorine.
\(F_2\) is nonpolar and experiences only weak London dispersion forces.
Therefore \(HF\) has a much higher boiling point.
Answer: C
Which intermolecular forces are present in pure liquid \(CH_3COOH\)?
(A) London dispersion only
(B) Dipole–dipole and LDF
(C) Hydrogen bonding, dipole–dipole, and LDF
(D) Ionic forces
▶️ Answer / Explanation
Acetic acid contains an O–H bond allowing hydrogen bonding.
It is also polar → dipole–dipole interactions.
All molecules also have London dispersion forces.
Answer: C
Three statements are made about the relationship between molecular polarity, structure, and intermolecular forces.
1. \(CCl_4\) and \(CHCl_3\) both have tetrahedral geometry. \(CCl_4\) is nonpolar while \(CHCl_3\) is polar because replacing one Cl with H breaks the symmetry, so \(CHCl_3\) has stronger intermolecular forces solely due to its polarity.
2. Two structural isomers with the same molecular formula can have different boiling points if their shapes differ — a more branched isomer is more compact and has weaker London dispersion forces than a straight-chain isomer.
3. When \(NaCl\) dissolves in water, ionic bonds in the lattice are broken and ion–dipole interactions form between ions and water molecules.
Which statements are correct?
(A) 1 only
(B) 3 only
(C) 2 and 3 only
(D) 1, 2, and 3
▶️ Answer / Explanation
Statement 1 — FALSE
Although \(CHCl_3\) is polar and \(CCl_4\) is nonpolar, the claim that polarity alone determines stronger intermolecular forces is incorrect. \(CCl_4\) actually has a higher boiling point because its larger electron cloud produces stronger London dispersion forces.
Statement 2 — TRUE
Branching makes molecules more compact and reduces surface area contact between molecules, weakening London dispersion forces compared to straight-chain isomers.
Statement 3 — TRUE
When \(NaCl\) dissolves, ionic lattice forces are broken and ion–dipole interactions form between the ions and polar water molecules.
Answer: C — Statements 2 and 3 only
A polar molecule is placed near a nonpolar molecule. Which sequence correctly describes the interaction?
(A) Nonpolar molecule attracts polar molecule directly
(B) Polar molecule induces a temporary dipole in the nonpolar molecule
(C) Both molecules develop permanent dipoles
(D) Ionic bond forms
▶️ Answer / Explanation
This interaction is called a dipole–induced dipole force.
The electric field of the polar molecule distorts the electron cloud of the nonpolar molecule, creating a temporary dipole.
Answer: B
The table shows the boiling points of four molecules with similar molar masses (~32–34 g/mol).
| Molecule | MW (g/mol) | BP (°C) | Polarity |
|---|---|---|---|
| \(CH_3OH\) | 32 | +65 | Polar |
| \(H_2S\) | 34 | −60 | Polar |
| \(PH_3\) | 34 | −88 | Nearly nonpolar |
| \(H_2O\) | 18 | +100 | Polar |
Why does methanol (\(CH_3OH\)) have a dramatically higher boiling point than \(H_2S\), despite similar molar masses?
(A) Methanol has more electrons
(B) Methanol has an O–H group capable of hydrogen bonding
(C) \(H_2S\) is nonpolar
(D) Methanol is ionic
▶️ Answer / Explanation
Methanol contains an O–H group which enables strong hydrogen bonding between molecules.
\(H_2S\) cannot form hydrogen bonds and only experiences dipole–dipole and London dispersion forces.
Answer: B
Fluoromethane (\(CH_3F\)) has a boiling point of −78°C, while chloromethane (\(CH_3Cl\)) has a boiling point of −24°C. Which explanation is correct?
(A) \(CH_3Cl\) has a larger dipole moment
(B) \(CH_3Cl\) has more electrons giving stronger London dispersion forces
(C) \(CH_3F\) forms hydrogen bonds
(D) \(CH_3Cl\) is nonpolar
▶️ Answer / Explanation
\(CH_3Cl\) has greater molar mass and more electrons than \(CH_3F\), leading to stronger London dispersion forces.
These stronger dispersion forces result in the higher boiling point.
Answer: B
Four molecules and their dominant intermolecular forces are listed below.
| Molecule | Dominant IMF |
|---|---|
| \(Xe\) | LDF |
| \(HCl\) | Dipole–dipole |
| \(H_2O\) | Hydrogen bonding |
| \(NH_3\) | Hydrogen bonding |
A student predicts the boiling point order as: \(NH_3 < H_2O < HCl < Xe\). Which statement is correct?
(A) Correct — boiling point increases with molar mass
(B) Incorrect — Xe should have the lowest BP
(C) Incorrect — \(HCl\) should have a lower BP than \(NH_3\) and \(H_2O\)
(D) Incorrect — \(NH_3\) and \(H_2O\) have identical BP
▶️ Answer / Explanation
\(HCl\) lacks hydrogen bonding, while \(NH_3\) and \(H_2O\) exhibit hydrogen bonding.
Therefore \(HCl\) should have a lower boiling point than those molecules.
Answer: C
Glycerol (\(HOCH_2CH(OH)CH_2OH\)) has a boiling point of 290°C while propane has −42°C. Why?
(A) Glycerol is ionic
(B) Glycerol forms extensive hydrogen bonding and has larger molar mass
(C) Propane has stronger dispersion forces
(D) Glycerol has smaller surface area
▶️ Answer / Explanation
Glycerol has three –OH groups allowing extensive hydrogen bonding between molecules.
It also has greater molar mass than propane, strengthening London dispersion forces.
Answer: B
A chemist compares \(C_{20}H_{42}\) with water and claims the hydrocarbon must have a higher boiling point because it is larger. Which explanation is correct?
(A) Boiling point always increases with molar mass
(B) LDFs of large hydrocarbons exceed water’s H-bonding
(C) Water always has the highest BP
(D) For very large molecules, London dispersion forces can exceed hydrogen bonding
▶️ Answer / Explanation
Very large nonpolar molecules have extremely strong London dispersion forces due to their large number of electrons.
These forces can exceed hydrogen bonding in smaller molecules such as water.
Answer: D
Consider the three statements below about intermolecular forces and molecular polarity.
1. London dispersion forces are the only intermolecular force present in nonpolar molecules, and their strength increases as the number of electrons in the molecule increases.
2. \(CO_2\) is a nonpolar molecule because its two polar \(C{=}O\) bond dipoles cancel due to its linear, symmetric geometry — therefore it only experiences London dispersion forces between molecules.
3. A polar molecule always has a higher boiling point than a nonpolar molecule, regardless of their sizes or molar masses.
Which of the following correctly identifies the true statements?
(A) 1 only
(B) 1 and 2 only
(C) 2 and 3 only
(D) 1, 2, and 3
▶️ Answer / Explanation
Evaluate each statement:
Statement 1 — TRUE
Nonpolar molecules have no permanent dipole, so the only intermolecular force present is London dispersion forces. These forces become stronger as the number of electrons increases because the electron cloud becomes more polarizable.
Statement 2 — TRUE
\(CO_2\) is linear, so the two C=O bond dipoles cancel each other. This makes the molecule nonpolar and therefore the only intermolecular force between molecules is London dispersion forces.
Statement 3 — FALSE
A polar molecule does not always have a higher boiling point than a nonpolar molecule. Very large nonpolar molecules can have stronger London dispersion forces than small polar molecules.
Answer: B — Statements 1 and 2 only