Question
Water (H₂O) boils at 100°C while hydrogen sulphide (H₂S) boils at −60°C. Both are hydrides of Group 16 elements. Explain why H₂O has such an unusually high boiling point compared to H₂S.
Solution — Step by Step
Oxygen has electronegativity 3.5, hydrogen has 2.1 — a difference of 1.4. Sulphur has electronegativity 2.5, so the S–H difference is only 0.4. The O–H bond is far more polar than S–H.
For intermolecular hydrogen bonding to form, you need: (1) a hydrogen atom bonded to a highly electronegative atom (N, O, or F), and (2) a lone pair on that electronegative atom in a neighbouring molecule. Oxygen satisfies both conditions strongly; sulphur barely qualifies.
Each H₂O molecule has two O–H bonds (two H-bond donors) and two lone pairs on oxygen (two H-bond acceptors). This means every water molecule can form up to four hydrogen bonds simultaneously, creating a three-dimensional network throughout the liquid.
H₂S has much weaker S–H bonds (S is larger, less electronegative), so only weak van der Waals/London dispersion forces act between H₂S molecules. No significant intermolecular H-bonds form. Much less energy is needed to separate H₂S molecules from each other.
Boiling point reflects how much energy we must supply to overcome intermolecular forces. Water’s extensive H-bond network requires significantly more energy to break, hence the boiling point shoots up to 100°C — nearly 160°C higher than H₂S.
Answer: H₂O has a much higher boiling point than H₂S because water molecules form strong intermolecular hydrogen bonds (due to the high electronegativity of O and the small size of the O–H bond), whereas H₂S only has weak van der Waals forces between molecules.
Why This Works
The key principle here is that boiling point is a measure of intermolecular force strength, not molecular weight. If we only looked at molecular mass, H₂S (34 g/mol) should actually boil higher than H₂O (18 g/mol) — but the opposite is true.
H-bond forms when: X–H···Y, where X and Y are N, O, or F
Bond strength order: F–H···F > O–H···O > N–H···N
Oxygen’s small atomic size concentrates its negative charge, making the lone pairs highly accessible for H-bond formation. Sulphur is larger — its electron cloud is more diffuse, and the S–H bond is not polar enough to create a meaningful partial positive charge on hydrogen.
The four-H-bond network in water also explains other anomalous properties: high surface tension, high specific heat capacity, and the fact that ice floats (the H-bond network in ice holds molecules further apart than in liquid water).
Alternative Method — Trend Analysis in Group 16
We can see the anomaly clearly by looking at the boiling point trend across Group 16 hydrides:
| Hydride | Boiling Point |
|---|---|
| H₂O | 100°C |
| H₂S | −60°C |
| H₂Se | −41°C |
| H₂Te | −2°C |
If we ignore water, the trend is: boiling point increases going down the group (H₂S < H₂Se < H₂Te) because molecular mass and London dispersion forces increase. Water completely breaks this trend — it sits far above where a “normal” Group 16 hydride of that mass should be. This anomaly is entirely explained by H-bonding.
This “anomalous position in periodic trend” approach is very NCERT-style and comes up in board exams. Examiners love asking you to identify the odd one out in a group trend and explain why.
Common Mistake
Many students write that “H₂S also forms hydrogen bonds, just weaker ones.” This is incorrect for this context. While there is some theoretical debate, for CBSE and JEE purposes, H₂S does not form intermolecular hydrogen bonds because sulphur is not electronegative or small enough. Writing “weak H-bonds in H₂S” will cost you marks in a board exam answer. Stick to: H₂O forms H-bonds; H₂S does not — only van der Waals forces.
The other trap is confusing intramolecular and intermolecular hydrogen bonds. This question is entirely about intermolecular H-bonds (between different water molecules), not bonds within a single molecule. Boiling point is always about breaking intermolecular forces, not intramolecular ones.