Questions: Hydrogen Bonding: Energetics and Thermodynamics

Question 1

Water (H₂O) has a much higher boiling point than hydrogen sulfide (H₂S) despite H₂S being heavier. What is the primary reason?

• A. H₂O molecules have stronger dispersion forces due to a higher electron density
• B. Oxygen's higher electronegativity creates a larger partial positive charge on hydrogen, enabling stronger hydrogen bonds
• C. H₂O is a smaller molecule, so molecules pack more tightly and require more energy to separate
• D. H₂S has a lower dipole moment but stronger London dispersion forces that destabilize the liquid

Answer: (B)

Oxygen is far more electronegative than sulfur (3.44 vs 2.58 on the Pauling scale), creating a large δ+ on the hydrogen. This enables strong O–H···O hydrogen bonds (≈20 kJ/mol) between water molecules. H₂S can only form weak S–H···S contacts because sulfur's lower electronegativity makes the hydrogen much less positive. The high boiling point of water (100°C vs −60°C for H₂S) directly reflects the energy cost of breaking these hydrogen-bond networks.

Question 2

Any two molecules with an O–H bond and a lone pair on a nearby electronegative atom are typically engaged in a significant hydrogen bond.

• T. True
• F. False

Answer: False

Geometry matters enormously. A hydrogen bond requires a near-linear X–H···Y angle (ideally 170–180°) and a short H···Y distance (typically < 2.5 Å). Many O–H···O contacts found in crystal structures are too bent or too long to be energetically meaningful. A highly distorted geometry means poor orbital overlap and weak electrostatic alignment, reducing the interaction to something barely above van der Waals strength.

Question 3

Hydrogen bond formation is enthalpically favorable but entropically unfavorable. Explain why, and what consequence this has for hydrogen bond stability at high temperature.

Model answer: Hydrogen bond formation releases energy (negative ΔH, attractive interaction), but it restricts the translational and rotational freedom of both partners — two molecules that were independent become partially ordered around the interaction geometry (negative ΔS). The Gibbs free energy change is ΔG = ΔH − TΔS. At low temperature, the enthalpic term dominates and the bond is stable; at high temperature, the −TΔS term (which is positive and destabilizing) grows and can overcome ΔH, breaking the bond. This is why hydrogen-bonded structures like DNA double helices and protein folds denature at elevated temperatures.

This thermodynamic balance explains why biological hydrogen bond networks are temperature-sensitive: the structures are not held together by very strong bonds, but by many moderately favorable ones whose collective enthalpic gain barely beats the entropy penalty. Calorimetric measurements (ΔH and ΔS of complex formation) are needed to fully characterize any hydrogen-bonded system.