Hydrogen bonding

Origin of the bond. In a bond such as $\text{O}-\text{H}$, oxygen pulls the shared electron pair strongly toward itself (high electronegativity), leaving the hydrogen nucleus almost bare and carrying a significant partial positive charge, $\delta+$. Because hydrogen has no inner shielding shell, this exposed proton is small enough to approach very close to a lone pair sitting on an electronegative atom of a neighbouring molecule (or another part of the same molecule). That electrostatic pull between the $\delta+$ hydrogen and the lone pair is the hydrogen bond, conventionally drawn as a dotted line:

where: X is the electronegative atom the hydrogen is covalently bonded to, H is the bridging hydrogen carrying the $\delta+$ charge, the solid line is the ordinary covalent bond, the dots are the (weaker) hydrogen bond, and Y is the electronegative atom supplying the lone pair. Two conditions must both hold: a small, highly electronegative atom (F, O, or N only — chlorine is too large and too weakly electronegative despite being in the same electronegativity range) bonded to hydrogen, and an available lone pair on a nearby F, O, or N atom.

Two flavours of hydrogen bonding. Intermolecular hydrogen bonding occurs between separate molecules — water molecules linking to water molecules, $\text{HF}$ to $\text{HF}$, or water to ammonia. Because each molecule can link to several neighbours, this builds extended chains, rings, or three-dimensional networks, and it is this network that must be broken before the substance can boil. Intramolecular hydrogen bonding occurs within a single molecule, when the geometry lets an X–H group reach a lone-pair-bearing atom elsewhere on the same molecule — most commonly forming a stable five- or six-membered ring, as in ortho-nitrophenol. Because the hydrogen's attraction is satisfied internally, it is no longer free to bond with neighbouring molecules, so intramolecular hydrogen bonding does not raise boiling point the way the intermolecular kind does — it can even lower it.

Consequences for physical properties. Extensive intermolecular hydrogen bonding means extra energy must be supplied to break the network before molecules can separate into vapour, so substances capable of it — $\text{H}_2\text{O}$, $\text{HF}$, $\text{NH}_3$ — show anomalously high boiling points compared with the heavier, non-hydrogen-bonded hydrides below them in their groups. The same network resistance shows up in viscosity: liquids with extensive hydrogen bonding, such as glycerol or concentrated $\text{H}_2\text{SO}_4$, flow sluggishly because molecules cannot easily slide past a web of hydrogen bonds.

Explaining why a handful of "FON" hydrides boil so much higher than their heavier group-mates, while chemically similar hydrides made with larger atoms do not, requires tracing the argument from bond polarity through to the energy needed for vaporisation, and this is exactly the reasoning a Class 11 answer is expected to reproduce in full rather than merely assert.

Step 1. Compare electronegativities. Fluorine, oxygen and nitrogen sit at the top-right of the periodic table and are among the most electronegative elements that still form simple covalent hydrides; their atoms are also small, so the electron pair they pull toward themselves in an X–H bond leaves very little electron density shielding the hydrogen nucleus.

Step 2. Locate the resulting charge. Because the atom is both small and strongly electron-withdrawing, the H atom in $\text{O}-\text{H}$, $\text{N}-\text{H}$, or $\text{F}-\text{H}$ carries a large $\delta+$ that is unusually exposed — closer to a bare proton than an ordinary polar-bond hydrogen. A hydrogen bonded to a larger, less electronegative atom (say sulfur, in $\text{H}_2\text{S}$) keeps far more of its electron density and never develops this exposed charge.

Step 3. Identify what the exposed $\delta+$ can reach. That small, sharply positive hydrogen can approach a lone pair on an F, O, or N atom of a neighbouring molecule far more closely than an ordinary dipole–dipole interaction allows, because there is essentially no electron cloud in the way. This closeness is what makes the hydrogen bond markedly stronger than typical van der Waals or dipole–dipole attractions, even though it remains far weaker than a covalent bond.

Step 4. Scale up to the bulk liquid. In liquid water, each molecule can simultaneously donate two hydrogen bonds (through its two H atoms) and accept two (through oxygen's two lone pairs), building an extensive three-dimensional network that links essentially every molecule to its neighbours. Breaking enough of this network for molecules to escape into the vapour phase demands substantially more energy than merely overcoming weak dispersion forces.

Step 5. Conclude the anomaly. Because $\text{H}_2\text{S}$, lacking hydrogen bonding, is held together only by weak dispersion and dipole–dipole forces, it vaporises at a far lower temperature ($\approx 213\ \text{K}$) despite sulfur being heavier than oxygen — mass alone would predict the opposite order. Water's extra network energy, not its molar mass, is what pushes its boiling point to $373\ \text{K}$, and the same logic (electronegative + small donor atom, available lone-pair acceptor, network formation) explains the parallel anomalies in $\text{HF}$ and $\text{NH}_3$.

Explain why $\text{H}_2\text{O}$ (boiling point $373\ \text{K}$) boils at a far higher temperature than $\text{H}_2\text{S}$ (boiling point $\approx 213\ \text{K}$), even though sulfur is the heavier atom, and state what would happen to the anomaly if oxygen's lone pairs were somehow unavailable.

Step 1. Set the expectation from mass alone. If only van der Waals dispersion forces mattered, boiling point should rise smoothly down Group 16 as molar mass increases — $\text{H}_2\text{S} \to \text{H}_2\text{Se} \to \text{H}_2\text{Te}$ does follow that smooth rise, but $\text{H}_2\text{O}$ sits far above where the trend line, extrapolated backward, would place it.

Step 2. Identify the bonding difference. Oxygen is small and highly electronegative, so each $\text{O}-\text{H}$ bond in water is strongly polarised, giving the H atoms a large $\delta+$ and oxygen's two lone pairs a strong pull on neighbouring $\delta+$ hydrogens — the two conditions for hydrogen bonding are both met. Sulfur, though in the same group, is larger and less electronegative, so $\text{S}-\text{H}$ bonds are far less polarised and $\text{H}_2\text{S}$ cannot form significant hydrogen bonds.

Step 3. Connect bonding to energy. Liquid water is held by an extensive network of intermolecular $\text{O}-\text{H}\cdots\text{O}$ hydrogen bonds that must be substantially broken before vaporisation, while liquid $\text{H}_2\text{S}$ relies only on weak dispersion and dipole–dipole attractions that require far less energy to overcome.

Step 4. Answer the "what if" extension. If oxygen's lone pairs were somehow unavailable to accept hydrogen bonds (they are not, in reality, but the thought experiment isolates the cause), water would lose its hydrogen-bond network entirely and would be expected to boil close to where the smooth Group 16 mass trend places it — well below room temperature, similar in spirit to $\text{H}_2\text{S}$. The anomaly is caused by the hydrogen-bond network alone, not by any other property of oxygen, so removing the lone-pair acceptor removes the anomaly.

Competitive stretch

Distinguish the hydrogen bonding in ortho-nitrophenol from that in para-nitrophenol, and state which isomer has the higher melting point. Both isomers carry an $-\text{OH}$ donor and a $-\text{NO}_2$ acceptor on a benzene ring. In ortho-nitrophenol the two groups sit on adjacent carbons, close enough for the $-\text{OH}$ hydrogen to reach the nearby nitro-oxygen's lone pair within the same molecule, forming a stable six-membered ring — intramolecular hydrogen bonding, which "uses up" the hydrogen internally and lowers the melting point. In para-nitrophenol the groups are too far apart for that internal loop, so each molecule instead hydrogen-bonds intermolecularly to its neighbours' $-\text{NO}_2$ groups, building an extended network. Because the intermolecular network needs more energy to break apart, para-nitrophenol has the distinctly higher melting point of the two isomers.

• Assuming any O–H, N–H, or F–H bond automatically means hydrogen bonding is present. The bond alone is not enough — a nearby lone pair on an F, O, or N atom must also be available to accept it.
• Treating the hydrogen bond as a full covalent or ionic bond. It is a weaker intermolecular (or intramolecular) attraction — stronger than typical van der Waals forces, but far weaker than a covalent bond, and it is drawn with a dotted line, never a solid one.
• Assuming intramolecular hydrogen bonding always raises boiling point the way intermolecular bonding does. It can do the opposite, as in ortho- versus para-nitrophenol, because it competes with rather than adds to intermolecular association.
• Extending "FON" hydrogen bonding to any electronegative atom, especially chlorine. Only fluorine, oxygen, and nitrogen are simultaneously small and electronegative enough; chlorine is electronegative but too large for the close approach hydrogen bonding needs.

Ask the tutor: Not sure whether a particular pair of molecules can hydrogen-bond, or want to check your reasoning on an isomer-comparison question — ask the AI tutor to walk through it with you.