Worked examples — Hydrogen bonding — intermolecular, intramolecular; consequences (boiling points, water density)

Before we start, two measuring sticks we will reuse.

Electronegativity (see Electronegativity) is how hard an atom pulls shared electrons toward itself, on the Pauling scale. The numbers we lean on:

Partial charges ${\delta }^{+}$ / ${\delta }^{-}$. When two atoms of different electronegativity share a bond, the hungrier atom holds the electron cloud closer, so it becomes slightly negative and its partner slightly positive.

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The scenario matrix

Every hydrogen-bonding question is really one of these case classes. The colour-coded map below shows all eight cells and which example fills each; the table repeats it for reference.

Cell	Case class	What makes it tricky	Example
A	Both partners qualify (F/O/N + H on F/O/N)	Confirm the bond, name donor/acceptor	Ex 1
B	Electronegative but WRONG atom (Cl)	High $\chi$ but too big → fails	Ex 2
C	H attached to C only	No polar X–H at all → fails	Ex 3
D	Count H-bonds per molecule	Why H₂O beats HF (the "degenerate winner" case)	Ex 4
E	Intra- vs intermolecular isomers	Same formula, opposite b.p. effect	Ex 5
F	Numeric limiting value	Ice density / volume change	Ex 6
G	Real-world word problem	Lake freezing top-down	Ex 7
H	Exam twist / "which is odd one out"	Mix of traps A–D at once	Ex 8

We cover all eight cells. Let's go.

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Example 1 — Cell A: does it H-bond, and who is who?

Step 1 — Check the donor condition. Is there an H covalently bonded to F, O, or N? In ${\text{NH}}_3$ the H atoms are bonded to N. ✓ Why this step? The donor must be an $\text{X–H}$ with X = F/O/N; otherwise H is never "bare" enough. N qualifies because $\chi (\text{N})=3.0\gg \chi (\text{H})=2.1$, a big pull that leaves H as ${\delta }^{+}$.

Step 2 — Check the acceptor condition. Is there a lone pair on an F/O/N to receive the bare proton? Nitrogen in ${\text{NH}}_3$ has one lone pair. ✓ Why this step? The $\cdots \text{Y}$ end needs a concentrated negative cloud to grab the ${\delta }^{+}$ hydrogen.

Step 3 — Assemble. So a hydrogen bond $\text{N–H}\cdots \text{N}$ forms between molecules: the donor is the N–H of one molecule, the acceptor is the lone pair of the nitrogen on a neighbour.

Recall Verify

${\text{NH}}_3$ boils at $-33{}^{\circ }\text{C}$, far above ${\text{PH}}_3$ at $-88{}^{\circ }\text{C}$ — despite ${\text{PH}}_3$ being heavier. That $55{}^{\circ }\text{C}$ jump is exactly the H-bond energy the parent note promised. Consistent → answer stands. (See Boiling point trends of hydrides.)

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Example 2 — Cell B: high electronegativity, wrong size

Step 1 — Criterion 1: electronegativity. $\chi (\text{Cl})=3.0=\chi (\text{N})$. So the polarity test passes; HCl is genuinely polar (a dipole–dipole attractor). Why this step? We must not dismiss Cl too fast — it really does pull on H.

Step 2 — Criterion 2: size / charge concentration. The atomic radius of Cl ($\approx 99\text{pm}$) is far larger than N ($\approx 65\text{pm}$). A big atom spreads its lone-pair charge over a bigger volume, so the field the bare proton feels is diluted. Why this step? Coulomb attraction scales like $1/r^2$; a diffuse cloud means the effective $r$ never gets small, so the "closeness trick" that makes H-bonds strong never fires.

Step 3 — Verdict. Both conditions are required; Cl fails size. So HCl interacts by ordinary dipole–dipole (van der Waals family), not a strong H-bond.

Recall Verify

HCl boils at $-85{}^{\circ }\text{C}$, below HF ($+20{}^{\circ }\text{C}$) despite HCl being heavier. Heavier-but-lower-boiling is the fingerprint of "no H-bond bonus." Consistent. This is exactly the parent's steel-manned mistake.

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Example 3 — Cell C: H on carbon only

Step 1 — Screen ${\text{CH}}_4$ for a donor. H is bonded to C. Since $\chi (\text{C})=2.5$ is only $0.4$ above $\chi (\text{H})=2.1$, the C–H bond is barely polar. No ${\delta }^{+}$ proton is exposed. Why this step? Without a polar X–H there is simply no hydrogen bond, regardless of any lone pairs elsewhere. And ${\text{CH}}_4$ has no lone pairs anyway.

Step 2 — Screen ${\text{H}}_2\text{O}$. Two O–H donors + two lone-pair acceptors → strong H-bonding (Cell A, revisited).

Step 3 — Compare. Both feel London forces (comparable, similar mass). Only water adds H-bonds → water boils far higher.

Recall Verify

${\text{CH}}_4$ boils at $-161{}^{\circ }\text{C}$; ${\text{H}}_2\text{O}$ at $+100{}^{\circ }\text{C}$. A gap of $261{}^{\circ }\text{C}$ for nearly equal mass — the entire gap is the H-bond network. Answer confirmed.

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Example 4 — Cell D: count the "hands" (why H₂O > HF)

Step 1 — Each H-bond needs one donor H AND one acceptor lone pair. A hydrogen bond is one ${\delta }^{+}$ hydrogen linked to one lone pair: $\text{X–H}\cdots \text{Y}$. So a molecule can only make as many H-bonds as it can supply matched pairs of (a donor H) with (an acceptor lone pair on a neighbour). Why this step? This is the rule that turns "how electronegative?" into "how many bonds?" — the whole resolution hinges on it.

Step 2 — Tally donors and acceptors on each molecule.

• ${\text{H}}_2\text{O}$: 2 O–H hydrogens (donors) and 2 lone pairs on O (acceptors). Donors and acceptors are balanced at 2 each, so a molecule can be a donor twice and an acceptor twice → it participates in 4 H-bonds, forming a 3-D network.
• $\text{HF}$: 1 H (donor) and 3 lone pairs on F (acceptors). Donors and acceptors are unbalanced. Why this step? We need the raw supply of donors and acceptors on each molecule before we can decide how many bonds can actually close — the count of hands is what the whole paradox turns on, and it must be gathered before comparing per-bond strengths.

Step 3 — Apply the limiting factor to HF. Across a large sample, the number of H-bonds that actually form is set by the scarcer resource. HF has 1 donor but 3 acceptors, so donors run out first: every HF can donate its single H to one neighbour and simultaneously accept one H into one of its lone pairs → about 2 H-bonds per molecule (a zig-zag chain). The other 2 lone pairs are wasted — no partner H exists to fill them. Why this step? This is the "degenerate" case: extra acceptors do nothing once donors are exhausted. It is exactly why more electronegative F does not mean more bonds.

Step 4 — Multiply count × per-bond strength (rough estimate). Representative H-bond energies from data (see Boiling point trends of hydrides): an O–H⋯O bond is $\approx 21\text{–}25\text{kJ/mol}$ and an F–H⋯F bond is stronger, $\approx 29\text{–}30\text{kJ/mol}$ (HF's is famously among the strongest neutral H-bonds). Take round mid-values $25$ and $30\text{kJ/mol}$. $E_{\text{water}}\approx \frac{1}{2}\times 4\times 25=50\text{kJ/mol},E_{\text{HF}}\approx \frac{1}{2}\times 2\times 30=30\text{kJ/mol}.$ Why the $\frac{1}{2}$? Each H-bond is shared by two molecules, so per molecule you count only half of each. Even after giving HF the stronger $30\text{kJ/mol}$ per bond, water's larger count (4 vs 2) wins.

Step 5 — Conclude. $E_{\text{water}}>E_{\text{HF}}$ ⇒ water boils higher, matching ${\text{H}}_2\text{O}>\text{HF}>{\text{NH}}_3$.

Recall Verify

$50>30$, so the count argument gives the right order even when HF's per-bond energy is set higher. Real b.p.: water $100{}^{\circ }\text{C}>$ HF $20{}^{\circ }\text{C}$. ✓

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Example 5 — Cell E: ortho vs para (geometry decides)

Step 1 — Look at the ortho geometry (left figure). In the ortho isomer the $-\text{OH}$ and one oxygen of $-{\text{NO}}_2$ are neighbours on the ring, at just the right distance to close a strain-free 6-membered ring: $\text{O–H}\cdots \text{O}$ inside the same molecule. Why this step? A 6-membered ring has near-ideal bond angles, so the intramolecular bond forms readily and "uses up" the OH hydrogen.

Step 2 — Consequence for ortho. With its H internally satisfied, an ortho molecule has little left to offer neighbours → weak intermolecular attraction → lower b.p., steam-volatile, less water-soluble. Why this step? Boiling point is set by attraction between molecules; if the donor H is already spent on an internal partner, there is nothing left to grip neighbours, so less heat is needed to separate them.

Step 3 — Look at the para geometry (right figure). In para the groups sit on opposite ends of the ring — far too far to reach each other. The OH hydrogen must instead bond to other molecules → intermolecular network. Why this step? We must inspect the geometry directly: only when the donor and acceptor cannot meet inside the molecule are they forced to reach outside, which is what builds an intermolecular network.

Step 4 — Consequence for para. Extensive intermolecular H-bonding → higher b.p., non-volatile, more soluble. Why this step? Every OH hydrogen now grips a neighbour, so pulling the molecules apart to boil costs the full H-bond energy — hence the higher boiling point, the mirror image of the ortho case.

Recall Verify

Predicted: ortho (intramolecular) $<$ para (intermolecular). Data: $216{}^{\circ }\text{C}<279{}^{\circ }\text{C}$, a gap of $63{}^{\circ }\text{C}$. ✓ Matches the "ortho closes the door" mnemonic.

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Example 6 — Cell F: the ice-density number

Step 1 — Set up volume from density. Take mass $m=1\text{g}$. Volume $V=m/\rho$, so $V_{\text{water}}=\frac{1}{1.000}=1.000{\text{cm}}^3,V_{\text{ice}}=\frac{1}{0.917}=1.0905{\text{cm}}^3.$ Why this step? Freezing conserves mass; only volume changes. Fixing $m=1\text{g}$ turns density directly into volume.

Step 2 — Percent expansion. $\frac{V_{\text{ice}}-V_{\text{water}}}{V_{\text{water}}}\times 100\%=\frac{1.0905-1.000}{1.000}\times 100\%\approx 9.05\%.$ Why this step? The open tetrahedral lattice (each molecule locked to 4 neighbours, big hexagonal holes) inflates the volume — this number is the holes.

Step 3 — Fraction above water (Archimedes). A floating body displaces its own weight of water, so the submerged fraction equals the density ratio: $f_{\text{submerged}}=\frac{{\rho }_{\text{ice}}}{{\rho }_{\text{water}}}=\frac{0.917}{1.000}=\mathrm{0.917.}$ Therefore above the surface: $f_{\text{above}}=1-0.917=0.083\approx 8.3\%.$ Why this step? Buoyancy balances weight; the ratio of densities is exactly the sunk fraction. Only the small "airy" excess pokes up.

Recall Verify

Expansion $\approx 9\%$ and "tip of the iceberg" $\approx 8\%$ — both are the famous textbook numbers, and ${\rho }_{\text{ice}}<{\rho }_{\text{water}}$ ⇒ ice floats, no contradiction. Units: $\text{g}/({\text{g/cm}}^3)={\text{cm}}^3$. ✓ (See Water — anomalous properties.)

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Example 7 — Cell G: real-world word problem

Step 1 — Recall the density peak. Water is densest at $4{}^{\circ }\text{C}$ (parent note). Above $4{}^{\circ }\text{C}$, cooling raises density (normal contraction). Below $4{}^{\circ }\text{C}$, cooling lowers density (open ice-like clusters begin forming). Why this step? "Densest sinks" only makes sense once we know density is non-monotonic — the tug-of-war peaks at $4{}^{\circ }\text{C}$, so we must pin that peak down before tracking what sinks.

Step 2 — Cool from warm down to 4 °C. As surface water cools, it gets denser and sinks; warmer bottom water rises to be cooled. This convection continues until the whole lake reaches $4{}^{\circ }\text{C}$. Why this step? Above $4{}^{\circ }\text{C}$ colder = denser, so gravity keeps overturning the lake; mixing only stops once cooling no longer increases density.

Step 3 — Cool the surface below 4 °C. Now surface water below $4{}^{\circ }\text{C}$ is less dense than the $4{}^{\circ }\text{C}$ water beneath it, so it stays on top (no more sinking). It cools further, reaches $0{}^{\circ }\text{C}$, and freezes into ice — which, being $\approx 8\%$ less dense (Ex 6), floats. Why this step? Below $4{}^{\circ }\text{C}$ the density trend reverses, so the coldest water can no longer sink — this reversal is the whole reason freezing happens at the surface, not the bottom.

Step 4 — Result. Ice forms as an insulating cap on top; the densest $4{}^{\circ }\text{C}$ water settles at the bottom, where fish survive. Why this step? Tying the density argument back to the biology answers the question asked: the $4{}^{\circ }\text{C}$ water is the densest, so it is exactly the layer that ends up at the bottom.

Recall Verify

Consistent chain: densest ($4{}^{\circ }\text{C}$) sinks → colder water floats → ice caps the top → bottom stays liquid at $4{}^{\circ }\text{C}$. Every step uses only ${\rho }_{4^{\circ }}>{\rho }_{0^{\circ }}>{\rho }_{\text{ice}}$, i.e. $1.000>0.9999>0.917$. ✓

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Example 8 — Cell H: exam twist ("which is the odd one out")

Step 1 — Screen each for a donor (an H bonded to F/O/N).

• Ethanol ${\text{C}}_2{\text{H}}_5\text{OH}$: has an O–H → it can donate a ${\delta }^{+}$ H and accept on the O lone pairs → donor + acceptor ✓ (Cell A).
• Dimethyl ether ${\text{CH}}_3{\text{OCH}}_3$: O is present, but every H is on carbon — there is no O–H. So it has lone pairs (acceptor) but no donor → cannot build its own H-bond network (Cell C trap).
• Propane ${\text{C}}_3{\text{H}}_8$: only C–H bonds and no lone pairs → no H-bond at all, only London forces (Cell C).
• HF: one H on F → donor + acceptor ✓, but donor-limited to chains (Cell D).

Why this step? This is the exam's core trap: an O atom alone is not enough — you need an O–H to donate. Screening every molecule the same way exposes which ones qualify.

Step 2 — Rank the two H-bonders (ethanol vs HF). Both H-bond, so we compare total attraction. Ethanol has an O–H network plus a sizeable carbon skeleton giving strong London forces; HF has stronger individual bonds but forms only 2-bond chains and is a small molecule (weak London). Why this step? Boiling point = sum of ALL intermolecular forces, not just the strongest single bond (the Ex 4 lesson). Ethanol's network + London beats HF's chain. Result so far: ethanol > HF.

Step 3 — Rank the two non-H-bonders (dimethyl ether vs propane). Dimethyl ether has no O–H, but its bent C–O–C still makes it a polar molecule → dipole–dipole attraction. Propane is essentially non-polar → London forces only. Why this step? With H-bonding ruled out for both, the tiebreaker is the next force down the ladder: dipole–dipole beats pure London. Result: dimethyl ether > propane.

Step 4 — Assemble the full order. H-bonders sit above non-H-bonders, so: $\text{ethanol}>\text{HF}>\text{dimethyl ether}>\text{propane}.$

Step 5 — Name the trap in each.

• Ethanol: no trap — genuine strong H-bonder, the winner.
• HF: trap "F is most electronegative so it must top the list" — false, ethanol's network + London outrank it.
• Dimethyl ether: trap "it has O so it H-bonds" — false, no O–H donor; it only manages dipole–dipole.
• Propane: trap "similar mass ⇒ similar b.p." — false, no polar bonds at all, so it sinks to the bottom.

Recall Verify

Real b.p.: ethanol $78{}^{\circ }\text{C}$, HF $20{}^{\circ }\text{C}$, dimethyl ether $-24{}^{\circ }\text{C}$, propane $-42{}^{\circ }\text{C}$. Order $78>20>-24>-42$ matches our prediction. ✓ All three traps (O⇒H-bond, F⇒highest, mass⇒b.p.) are sprung.

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Related deeper reading: Solubility and 'like dissolves like', DNA structure, Van der Waals forces.

Recall One-line self-test

Why does dimethyl ether NOT hydrogen-bond to itself despite having oxygen? ::: It has no O–H (a donor); its H atoms are all on carbon, so no molecule can donate a ${\delta }^{+}$ hydrogen even though the O could accept one.