Which of the following statements about intermolecular forces is/are correct? (ONE OR MORE)

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Question

Which of the following statements about intermolecular forces is/are correct? (ONE OR MORE)

Accepted Answer

Correct Answer: C and DStep-by-step Solution:Option A: Potential energy of two point charges vs dipole-charge as r → ∞:Ion-ion (charge-charge): U ∝ 1/r. This decays as 1/r.Ion-dipole: U ∝ 1/r². This decays as 1/r² (faster than 1/r).So the ion-ion potential energy (∝ 1/r) approaches zero MORE SLOWLY than ion-dipole (∝ 1/r²). The statement says charge-charge approaches zero MORE RAPIDLY than dipole-charge — this is FALSE (it approaches zero more slowly, not more rapidly). ✗Option B: Average potential energy of two rotating polar molecules ∝ 1/r³ or 1/r⁶?For two rotating polar molecules: the Keesom (orientation) interaction. At high temperatures (free rotation): U_avg ∝ −(μ₁²μ₂²)/(3kT·r⁶) ∝ 1/r⁶. So the average PE has 1/r⁶ dependence. This seems TRUE... but option B says 1/r³. The INSTANTANEOUS dipole-dipole energy ∝ 1/r³, but the AVERAGE (Boltzmann-averaged over all orientations) is ∝ 1/r⁶. So option B states 1/r³ for average, which is INCORRECT. ✗Option C: Dipole-induced dipole interaction energy is independent of temperature:The Debye (dipole-induced dipole) interaction: U = −μ²α/(r⁶) where α is polarizability. This does NOT involve thermal averaging (unlike the Keesom interaction). Therefore U is independent of temperature. ✓ TRUE.Option D: Nonpolar molecules attract each other:Yes — nonpolar molecules experience London dispersion forces (induced dipole-induced dipole interaction). Instantaneous fluctuations in electron distribution create temporary dipoles that induce corresponding dipoles in neighbouring molecules, resulting in net attraction. This explains why noble gases and nonpolar molecules like CH₄ condense at low temperatures. ✓ TRUE.Why Options A and B are incorrect:A: The 1/r potential decays more slowly to zero than 1/r². The charge-charge interaction has longer range than the charge-dipole interaction. Students often confuse "rapidly" with large value — a 1/r dependence is larger than 1/r² at the same r, meaning it approaches zero more slowly (less rapidly). So the premise is reversed.B: The confusion is between instantaneous and average energy. The instantaneous dipole-dipole energy is ∝ 1/r³, but when averaged over all orientations at temperature T (Boltzmann weighting), the AVERAGE energy becomes ∝ 1/r⁶·(1/kT). Option B says the average has 1/r³ dependence, which is wrong — the average has 1/r⁶ dependence, not 1/r³.Key Concepts:Types of van der Waals interactions and their r-dependences:Ion-ion (Coulombic): U ∝ 1/rIon-dipole: U ∝ 1/r²Dipole-dipole (Keesom, average for rotating molecules): U ∝ 1/r⁶ (T-dependent: ∝ 1/(kTr⁶))Dipole-induced dipole (Debye): U ∝ 1/r⁶ (T-independent)Induced dipole-induced dipole (London/dispersion): U ∝ 1/r⁶ (T-independent)London dispersion forces: Present in ALL molecules (polar and nonpolar). Arise from instantaneous electron density fluctuations. Magnitude: U = −(3/4)·(α₁α₂)·(I₁I₂)/(I₁+I₂)·(1/r⁶) where α is polarizability and I is ionisation energy. Stronger for larger, more polarizable molecules.Temperature dependence of intermolecular forces: Only the Keesom interaction (between permanent dipoles of freely rotating molecules) is temperature-dependent (∝ 1/T). Debye and London interactions are temperature-independent.

Which of the following statements about intermolecular forces is/are correct? (ONE OR MORE)

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Which of the following statements about intermolecular forces is/are correct? (ONE OR MORE)

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