📝 Weak chemical interactions van der Waals (13 MCQs)

📖 From Campbell Biology • 2. The Chemistry of Life • 13 questions available

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What is Weak chemical interactions van der Waals?

Definition:
Van der Waals interactions are weak, short-range attractive forces between atoms or molecules due to transient dipole moments, caused by fluctuations in electron distribution, and they include London dispersion forces (induced dipoles) and dipole-dipole interactions, and while individually weak, these forces are significant in large numbers and play a crucial role in biological processes like protein folding, molecular recognition, and membrane stability.

Working:
Van der Waals forces work when electron distribution in an atom or molecule momentarily becomes asymmetrical, creating a temporary dipole that induces a dipole in a neighboring molecule, leading to attraction; the force is proportional to the polarizability and the distance ( $F \propto \frac{1}{r^7}$ ), and the potential energy is described by the Lennard-Jones potential $V = 4\epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^6 \right]$ ; these forces are additive, and they contribute to the specificity of biological interactions, such as enzyme-substrate binding and the stability of lipid bilayers.

Example:
A simple example is the attraction between gecko's toe hairs and surfaces, which is due to van der Waals forces, allowing geckos to climb walls; another example is the stacking interactions between bases in DNA, where van der Waals forces help stabilize the double helix, illustrating their role in biology.

Reason:
Van der Waals interactions, though weak, are critical for many biological processes, including molecular recognition, cell adhesion, and macromolecular assembly, and understanding them is essential for drug design and molecular biology.

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📝 All Weak chemical interactions van der Waals MCQs

Q1. A molecule contains several polar bonds, but its overall shape causes the bond dipoles to partially cancel. Which interaction is most likely to dominate when this molecule approaches another nonpolar molecule?

A.Ionic attraction

B.Hydrogen bonding

C.London dispersion forces ✅

D.Permanent dipole-dipole attraction

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: London dispersion forces arise from temporary fluctuations in electron distribution and can occur between all atoms and molecules, including nonpolar molecules. If permanent dipoles are absent or effectively canceled, these instantaneous induced attractions become the relevant intermolecular interaction.

Q2. Two molecules can form hydrogen bonds only when particular atoms and bond polarities are appropriately arranged. Which feature is essential for a strong hydrogen-bonding interaction?

A.A hydrogen attached to a highly electronegative atom interacting with an electronegative atom nearby ✅

B.Two carbon atoms sharing electrons equally

C.Two nonpolar groups approaching closely

D.A complete transfer of electrons between molecules

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Hydrogen bonding requires a strongly polarized hydrogen, commonly bonded to oxygen or nitrogen, and an electronegative atom with available electron density on another molecule. The resulting electrostatic attraction is directional and stronger than ordinary dispersion interactions.

Q3. A researcher replaces several polar groups on the surface of a molecule with nonpolar groups. The molecule becomes less soluble in water and more likely to associate with other nonpolar molecules. What best explains both observations?

A.Nonpolar groups form stronger ionic bonds with water

B.Nonpolar groups reduce favorable interactions with water while promoting hydrophobic association ✅

C.Nonpolar groups become permanently charged in water

D.Nonpolar groups increase hydrogen bonding with water

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Replacing polar groups with nonpolar groups removes opportunities for favorable hydrogen bonding and dipole interactions with water. Nonpolar surfaces tend to cluster together in aqueous environments, reducing their exposure to water and producing hydrophobic association.

Q4. A drug molecule fits into a binding pocket through many individually weak attractions. A mutation removes one hydrogen-bond donor but leaves the overall shape unchanged. What is the most reasonable prediction?

A.Binding must become impossible

B.Binding may weaken because one favorable interaction has been lost ✅

C.The drug will automatically form an ionic bond instead

D.The mutation cannot affect binding because weak interactions are irrelevant

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Biological recognition often depends on the combined effect of many weak interactions rather than a single extremely strong bond. Removing one hydrogen-bond donor can decrease binding affinity while still allowing other interactions to maintain partial association.

Q5. Two molecules have similar molecular masses. Molecule X has a highly polarizable electron cloud, whereas molecule Y has a much less polarizable cloud. If both are nonpolar, which molecule is expected to experience stronger London dispersion forces?

A.Molecule X ✅

B.Molecule Y

C.Both must experience exactly equal forces

D.Neither molecule can experience dispersion forces

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: London dispersion forces become stronger when an electron cloud is more easily distorted. Greater polarizability produces larger temporary dipoles and therefore stronger induced attractions. Thus, even among nonpolar molecules of similar mass, polarizability matters.

Q6. A student claims: 'Because hydrogen bonds are called bonds, they must always be stronger than every interaction involving charged particles.' Which correction is most scientifically justified?

A.Hydrogen bonds are stronger than all ionic attractions

B.Hydrogen bonds are not real attractions

C.Hydrogen bonds are directional intermolecular attractions, but their strength depends on the interacting species and environment ✅

D.Hydrogen bonds occur only inside atoms

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The term hydrogen bond does not imply that it universally exceeds electrostatic interactions involving full charges. Hydrogen bonds are relatively strong weak interactions, but their strength depends on geometry, polarity, distance, and surrounding environment.

Q7. A protein contains many nonpolar side chains exposed to water. After folding, most of these side chains become buried inside the protein while polar groups remain more exposed. Why can this rearrangement favor folding?

A.It eliminates all molecular motion

B.It reduces unfavorable exposure of nonpolar surfaces to water while allowing favorable water interactions elsewhere ✅

C.It converts every nonpolar group into an ion

D.It guarantees that covalent bonds will form between all side chains

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Protein folding can be favored when nonpolar groups become less exposed to water and cluster within the interior. This rearrangement reduces unfavorable water organization around hydrophobic surfaces while preserving opportunities for polar interactions at the surface.

Q8. A student compares two situations: in situation 1, two polar molecules are aligned so their opposite partial charges face each other; in situation 2, the same molecules are misaligned. Which conclusion is most appropriate?

A.Alignment cannot influence weak interactions

B.The aligned arrangement can produce stronger electrostatic attraction because favorable charge regions are closer ✅

C.Misalignment always produces a covalent bond

D.Both arrangements must have identical interaction energies

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Weak electrostatic interactions depend strongly on spatial arrangement. When opposite partial charges are positioned favorably, attraction increases. Misalignment can reduce favorable interactions or introduce repulsive contributions, showing why molecular geometry is important in biological recognition.

Q9. A graph records interaction strength between two nonpolar molecules as their separation decreases. The measured attraction becomes increasingly strong until the molecules are extremely close, after which the net interaction becomes strongly repulsive. Which interpretation best fits the graph?

A.Attraction is independent of distance

B.Repulsion at very short distances is consistent with electron-cloud and nuclear overlap effects ✅

C.The graph proves that nonpolar molecules cannot attract each other

D.The molecules must have formed an ionic compound

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Nonpolar molecules can attract through London dispersion forces at moderate distances. At extremely short distances, electron clouds cannot overlap indefinitely, and strong repulsive effects emerge. Therefore, an attraction minimum followed by steep repulsion is physically reasonable.

Q10. A membrane-associated molecule contains a polar region and a large nonpolar region. In water, several molecules spontaneously arrange so their nonpolar regions cluster away from water while polar regions remain exposed. Which combination best explains this behavior?

A.Only covalent bonding and electron transfer

B.Hydrogen bonding, electrostatic interactions, and hydrophobic effects acting together ✅

C.Only ionic bonding between carbon atoms

D.Only nuclear forces

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The arrangement reflects multiple weak effects operating simultaneously. Polar regions can interact favorably with water through hydrogen bonding and electrostatic attractions, while nonpolar regions tend to minimize their exposure to water, producing organized molecular association.

Q11. Three molecular samples are tested. Sample A has permanent dipoles, Sample B is nonpolar but highly polarizable, and Sample C is nonpolar with low polarizability. If all samples are compared under otherwise similar conditions, which ranking of expected weak intermolecular attraction is most defensible?

A.C > B > A

B.B > C > A

C.A and B may both show substantial attraction, while C is expected to have the weakest dispersion contribution ✅

D.C > A > B

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Permanent dipoles can create dipole-dipole attractions, while highly polarizable nonpolar molecules can exhibit relatively strong dispersion forces. A low-polarizability nonpolar molecule has weaker dispersion contributions, so C is generally expected to have the weakest attraction.

Q12. A mutation simultaneously removes one hydrogen-bond donor and increases the exposed nonpolar surface of a protein. Which prediction most logically follows if the protein interacts with an aqueous environment and a specific binding partner?

A.Both changes necessarily strengthen binding

B.Hydrogen bonding may decrease while altered hydrophobic association could either increase or decrease overall binding depending on geometry ✅

C.Only covalent bonding can change

D.The mutation must have no measurable effect

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The mutation changes two different contributions to molecular association. Losing a hydrogen-bond donor can weaken a specific favorable contact, while changing nonpolar surface exposure can alter hydrophobic association. The net effect therefore depends on molecular context and geometry.

Q13. Two very large nonpolar molecules have identical shapes, but molecule P contains a more easily distorted electron cloud than molecule Q. At the same separation, which prediction is most likely, and why?

A.Q experiences stronger dispersion because its electrons are less mobile

B.P experiences stronger dispersion because greater polarizability produces larger temporary dipoles ✅

C.Neither experiences dispersion because both are nonpolar

D.P must form hydrogen bonds with Q

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: London dispersion forces do not require permanent polarity. A highly polarizable electron cloud can undergo larger temporary fluctuations, generating stronger instantaneous dipoles and stronger induced attractions. Therefore, P can interact more strongly despite both molecules being nonpolar.

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