Ten teaching hours, outcomes 2.2.1 to 2.2.10, examined at both levels. The additional higher level outcomes, 2.2.11 to 2.2.16, are in Part B.
Guiding question: What determines the covalent nature and properties of a substance?
Structure 2.2 · The covalent bond
1The covalent bond and Lewis formulas 2.2.1 SL + HL Strong
When two non-metal atoms combine, neither can remove electrons completely from the other: both have high ionization energies and similar electronegativities. Instead, each atom contributes an electron to a pair that is shared between them. The shared pair sits in the region between the two nuclei, and both positively charged nuclei are attracted to it (Figure 2.5).
A covalent bond is the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of the two bonded atoms.
When asked to define or describe a covalent bond, name both components of the attraction: the shared pair of electrons and the nuclei (of the bonded atoms). “Sharing of electrons” alone describes what happens, not what holds the atoms together. Examiner reports on this definition note that the most common omissions are “pair” and “nuclei”.
The octet rule
Atoms of period 2 elements have only four valence orbitals (one 2s and three 2p), which can hold eight electrons. By sharing electrons, atoms such as C, N, O and F reach a total of eight valence electrons, the octet rule: atoms tend to gain a valence shell containing a total of eight electrons. Hydrogen reaches two (the configuration of helium). The octet rule is a pattern, not a law; its limitations are discussed below and, at HL, in 2.2.13.
Lewis formulas
A Lewis formula (electron dot structure) shows all valence electrons in a species: bonding pairs between atoms and non-bonding (lone) pairs on atoms. Pairs may be drawn as dots, crosses or lines; using dots for one atom and crosses for another helps to track where each electron came from, but all electrons are identical.
- Count the total valence electrons: add the group valence electrons of each atom; add one for each negative charge and subtract one for each positive charge.
- Choose the skeleton. The central atom is usually the least electronegative atom (never H; F is always terminal).
- Join each outer atom to the central atom with one shared pair.
- Complete the octets of the outer atoms with lone pairs; place any remaining electrons on the central atom.
- If the central atom has fewer than eight electrons, convert lone pairs on outer atoms into extra shared pairs (double or triple bonds).
- Check: the number of electrons drawn equals the total from step 1; for an ion, enclose the structure in square brackets with the charge.
| PCl3 | Valence electrons: 5 + 3 × 7 = 26 (13 pairs). P is central. Three P–Cl bonds use 3 pairs; three lone pairs on each Cl use 9 pairs; the last pair is a lone pair on P. Every atom has an octet: P has three bonding pairs and one lone pair. |
| NO3− | Valence electrons: 5 + 3 × 6 + 1 = 24 (12 pairs). N is central. Three N–O single bonds (3 pairs) and three lone pairs on each O (9 pairs) use all 12 pairs, but N has only six electrons. Convert one lone pair on one O into a second N=O bond: N now has an octet. Structure: one N=O and two N–O, bracketed with charge −. (At HL, 2.2.11 shows that the double bond can be drawn in three equivalent positions.) |
| Check | 26 and 24 electrons drawn respectively; no atom exceeds an octet. |
Molecules with fewer than an octet
Some atoms form stable molecules with fewer than eight valence electrons. Boron has three valence electrons and forms three bonds in BF3, giving six electrons around boron. Beryllium, whose compounds are largely covalent, has only four electrons around it in gaseous BeCl2. Such species are electron deficient: their central atom has an empty valence orbital that can accept a lone pair from another species (next section).
Exam focus · what the published papers show
A published scheme states: Accept any combination of dots or crosses to represent electrons, or lines to represent electron pairs. So dots, crosses and dashes are equally acceptable and you may mix them. What is marked is whether the electrons are all there and in the right places — which is why the reports about this outcome are almost entirely about missing lone pairs, not about drawing style.
Two sessions report the same omission, and in one of them both versions of the paper are described the same way: Most candidates were able to draw the Lewis structure of PCl₃; weaker candidates missed lone pairs, and some omitted a lone pair. A different session records that Some candidates did not include any lone pairs on the Lewis structure and others had incorrect numbers. This is not a presentation issue. The lone pair on phosphorus is the reason PCl₃ is trigonal pyramidal rather than trigonal planar — leave it out and the next part of the question becomes unanswerable.
Question. Draw the Lewis formula of the nitrate ion. [1]
Count first, and mind the charge. Nitrogen contributes 5, each oxygen 6, and the single negative charge adds one more electron: 5 + 18 + 1 = 24 electrons, which is 12 pairs.
Connect and complete. Three N–O bonds use 3 pairs. The remaining 9 pairs complete the three oxygens — but that leaves nitrogen with only 6 electrons, so one oxygen's lone pair becomes a second bond, giving nitrogen its octet.
Remember the brackets and the charge. The whole formula is enclosed and marked with the overall −1. A report records that a quarter of the students were able to draw the correct Lewis formula of NO₃⁻, and another session lists the same task among the areas candidates found difficult. Forgetting the charge changes the electron count from 24 to 23, and an odd number of electrons cannot be paired at all — which is a useful signal that something has gone wrong.
The syllabus states that molecules containing atoms with fewer than an octet of electrons should be covered. In BeCl₂ the beryllium has just two bonding pairs — four electrons — and that is the correct structure, not an error to be fixed. A report captures the difficulty precisely: around 70 % drew the molecule correctly, but only about half the students noted that, unlike most Lewis structures, the one they had just written did not involve a complete octet for the central atom, and a significant number tried to interpret it in terms of ionic bonding. Drawing it is easier than recognising what you have drawn.
2Single, double and triple bonds 2.2.2 SL + HL Repeated
Atoms may share one, two or three pairs of electrons, giving single, double and triple bonds. As the number of shared pairs increases, the electron density between the nuclei increases, the attraction between the nuclei and the shared electrons increases, and the nuclei are pulled closer together. So bond length decreases and bond strength increases from single to double to triple.
Single, double and triple bonds involve one, two and three shared pairs of electrons. More shared pairs pull the nuclei closer together, so between the same two elements: more bonds → shorter bond → stronger bond. All three parts move together, and a question usually asks for two of them.
| Bond | Shared pairs | Length / pm | Average bond enthalpy / kJ mol−1 | Example |
|---|---|---|---|---|
| C–C | 1 | 154 | 346 | ethane |
| C=C | 2 | 133 | 614 | ethene |
| C≡C | 3 | 120 | 839 | ethyne |
A double bond is stronger than a single bond, but not twice as strong (614 compared with 2 × 346 = 692 kJ mol−1). The second pair of electrons is held less effectively than the first (at HL you will see that it forms a π bond by sideways overlap). This is why alkenes react by breaking only one component of the double bond in addition reactions (Reactivity 2.2 and 3.4).
Bond strength is about the bond within a molecule; it does not decide the melting or boiling point of a molecular substance. Ethyne has the strongest carbon–carbon bond in Table 2.6, yet when liquid ethyne boils only the London forces between molecules are overcome.
Exam focus · what the published papers show
Bond length questions are marked as a consequence of the Lewis formula you drew. A report notes that on one item about 56% of the students predicted a suitable bond length and that some students were awarded the mark as error carried forward from the Lewis formula in part (i) — so a wrong structure can still earn the length mark, provided the length matches it. What loses the mark is inconsistency: another report describes candidates who added single bond lengths which was difficult to understand, especially if their Lewis structure showed multiple bonds. Read the length off the structure in front of you, not off a memory of the molecule.
3Coordination bonds 2.2.3 SL + HL Repeated
A coordination bond (dative covalent bond) is a covalent bond in which both electrons of the shared pair come from the same atom.
A coordination bond forms when an atom with a lone pair (a Lewis base) donates it to an atom or ion with a vacant orbital (a Lewis acid). In the ammonium ion, the lone pair on nitrogen in NH3 is donated to H+, which has no electrons at all. In the oxonium ion, H3O+, oxygen donates one of its two lone pairs to H+. In carbon monoxide, C≡O, two of the three shared pairs are formed from one electron from each atom and the third pair comes entirely from oxygen. In the adduct H3N→BF3, the lone pair on nitrogen fills the empty orbital on the electron-deficient boron atom.
Once formed, a coordination bond is indistinguishable from any other covalent bond. All four N–H bonds in NH4+ have the same length and strength, and the ion is a regular tetrahedron: the arrow is bookkeeping, not a different kind of bond.
A transition element ion has vacant orbitals that can accept lone pairs from surrounding molecules or ions called ligands. In [Cu(H2O)6]2+, six water molecules each donate a lone pair from oxygen to the central Cu2+ ion, forming six coordination bonds in an octahedral arrangement. Ammonia (lone pair on N), chloride ions and cyanide ions act as ligands in the same way. Complex formation is a Lewis acid–base reaction (Reactivity 3.4): the metal ion is the electron-pair acceptor and the ligand the donor.
- Drawing the arrow from the acceptor to the donor. The arrow starts at the atom that supplies the lone pair.
- Drawing a bond between the two metal or boron atoms in dimers such as Al2Cl6 or Be2Cl4. The monomers are joined through bridging chlorine atoms: a lone pair on Cl of one unit is donated to the electron-deficient central atom of the other.
- Stating “a dative bond forms” without saying that a lone pair is donated. The mark is for the origin of the electron pair.
Exam focus · what the published papers show
Asked to identify the coordination bond in the Lewis formula of nitric acid, only 28 % of candidates could do it — the lowest single figure recorded anywhere in these notes. How to find one: look for an atom that has contributed a lone pair to a bond and now has one fewer than you would expect, usually leaving it with a formal positive character. In HNO₃ the nitrogen donates a pair to one oxygen, which brings no electrons of its own to that bond. If a question says a compound "contains a coordination bond", it is telling you the structure cannot be drawn with ordinary shared pairs alone.
How many bonding electrons are there in the urea molecule, H2N–C(=O)–NH2?
Which are the correct sequences of increasing bond strengths and bond lengths between two carbon atoms?
| Bond strength | Bond length | |
|---|---|---|
| A. | C≡C < C=C < C–C | C≡C < C=C < C–C |
| B. | C≡C < C=C < C–C | C–C < C=C < C≡C |
| C. | C–C < C=C < C≡C | C≡C < C=C < C–C |
| D. | C–C < C=C < C≡C | C–C < C=C < C≡C |
Ethyne, C2H2, reacts with oxygen in welding torches.
(i) Deduce the Lewis (electron dot) structure of ethyne. [1]
(ii) Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6. [1]
(iii) Identify the type of interaction that must be overcome when liquid ethyne vaporizes. [1]
White phosphorus is an allotrope of phosphorus and exists as P4. Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds. [1]
In which molecule does the central atom have an incomplete octet of electrons?
Beryllium forms a chloride, BeCl2.
(c)(i) Draw the Lewis (electron dot) structure of the BeCl2 molecule. [1]
(c)(ii) Outline how the Lewis (electron dot) structure of the BeCl2 molecule differs from most Lewis (electron dot) structures. [1]
(d) Beryllium chloride, BeCl2, partially dimerizes in the gas phase to produce this molecule:
(d)(i) Identify the hybridization of the beryllium atom in the dimer, Be2Cl4. [1] AHL
(d)(ii) Describe the interactions between the BeCl2 monomers to form the dimer in Lewis’ acid–base terms. [1]
Which type of bond is formed when a Lewis acid reacts with a Lewis base?
Chlorofluorocarbons (CFCs) contain bonds of the following lengths: C–C = 1.54 × 10−10 m; C–F = 1.38 × 10−10 m; C–Cl = 1.77 × 10−10 m. What is the order of increasing bond strength in the CFC molecule?
Solutions and mark-scheme guidance · Set B
B1 B
Urea has four N–H single bonds, two C–N single bonds and one C=O double bond (two shared pairs): 4 + 2 + 2 = 8 bonding pairs = 16 bonding electrons. Lone pairs (two on O, one on each N) are not bonding electrons.
B2 C
More shared electron pairs between the same two nuclei give a stronger attraction and pull the nuclei closer together: strength C–C < C=C < C≡C; length C≡C < C=C < C–C.
B3 [3]
(i) H–C≡C–H drawn with lines, dots or crosses; each C has four bonding pairs and no lone pair ✓.
(ii) The C≡C bond in ethyne is shorter than C–C in ethane and the reason: more shared electrons (a triple bond) / a stronger bond ✓.
(iii) London (dispersion) forces / instantaneous dipole–induced dipole forces ✓. “Intermolecular forces” alone or “van der Waals forces” alone was not accepted: the specific force must be named.
B4 [1]
Four P atoms in a tetrahedron: each P joined by single bonds to the other three P atoms, with one lone pair on each P ✓ (5 valence electrons = 3 bonding + 2 in the lone pair). Any combination of lines, dots and crosses is acceptable.
B5 D
B has three valence electrons and forms three bonds in BF3, so it is surrounded by only six electrons. In H2Se, PH3 and OF2 the central atom has an octet (bonding pairs + lone pairs).
B6 [4]
(c)(i) Cl–Be–Cl with three lone pairs on each Cl and no lone pair on Be ✓.
(c)(ii) Be does not have a complete valence shell / has an incomplete octet (four electrons) / is electron deficient ✓.
(d)(i) sp2 ✓ (in the dimer each Be has three electron domains).
(d)(ii) A Cl atom on one BeCl2 unit acts as a Lewis base (donates a lone pair) and the Be on the other unit acts as a Lewis acid (accepts it), forming a coordination bond ✓.
B7 A
A Lewis base donates an electron pair to a Lewis acid; the shared pair forms a covalent (coordination, dative) bond.
B8 C
For single bonds to the same carbon atom, a shorter bond is generally a stronger bond: C–F (1.38) is shortest and strongest, C–Cl (1.77) longest and weakest. Order: C–Cl < C–C < C–F.
Structure 2.2 · Shapes and polarity
4The VSEPR model 2.2.4 SL + HL Strong
The valence shell electron pair repulsion (VSEPR) model predicts the shape of a molecule or ion from its Lewis formula. Regions of high electron density around a central atom, called electron domains, repel one another and adopt the arrangement that keeps them as far apart as possible.
An electron domain is a lone pair, a single bond, a double bond or a triple bond around the central atom: a multiple bond counts as one domain. The electron-domain geometry is the arrangement of all domains; the molecular geometry is the arrangement of the atoms only (lone pairs are not part of the shape name).
Applying the model
- Draw the Lewis formula and count electron domains around the central atom.
- Two domains → linear (180°); three → trigonal planar (120°); four → tetrahedral (109.5°).
- Name the molecular geometry from the positions of the atoms.
- Adjust bond angles: lone pairs are held by one nucleus only, spread out closer to the central atom and repel more strongly than bonding pairs. The order of repulsion is lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair. Each lone pair on a tetrahedral centre closes the remaining bond angles by roughly 2–3°.
| Domains | Lone pairs | Electron-domain geometry | Molecular geometry | Bond angle | Examples |
|---|---|---|---|---|---|
| 2 | 0 | linear | linear | 180° | BeCl2, CO2, HCN, C2H2 |
| 3 | 0 | trigonal planar | trigonal planar | 120° | BF3, SO3, CO32−, NO3−, C2H4 |
| 3 | 1 | trigonal planar | bent (V-shaped) | < 120° | SO2, O3 |
| 4 | 0 | tetrahedral | tetrahedral | 109.5° | CH4, NH4+, CCl4 |
| 4 | 1 | tetrahedral | trigonal pyramidal | ≈ 107° | NH3, PCl3, H3O+ |
| 4 | 2 | tetrahedral | bent (V-shaped) | ≈ 104.5° | H2O, H2S, SCl2, NH2− |
Effect of multiple bonds on bond angles
A double or triple bond contains more electron density than a single bond and repels neighbouring domains more strongly. In ethene the two C–H bonds on each carbon are pushed together slightly, so the H–C–H angle is about 117° rather than the ideal 120°. VSEPR predicts the direction of such deviations, not their exact size: bond angles in molecules with lone pairs or multiple bonds must ultimately be measured.
| H3O+ | Valence electrons 6 + 3 − 1 = 8 (four pairs): three O–H bonds and one lone pair on O. Four domains → tetrahedral electron-domain geometry; three atoms bonded → trigonal pyramidal; angle ≈ 107° (one lone pair, like NH3). |
| SO2 | Lewis formula (octet version) O=S–O with one lone pair on S: three domains (the double bond counts as one) → trigonal planar domain geometry; one lone pair → bent, angle slightly less than 120°. |
| NO3− | One N=O and two N–O (Worked example 2.3), no lone pair on N: three domains, all bonding → trigonal planar, 120°. |
| Trend | Bond angles with four domains: CH4 109.5° (no lone pair) > NH3 ≈ 107° (one lone pair) > H2O ≈ 104.5° (two lone pairs). Each extra lone pair squeezes the bonding pairs closer together. |
Exam focus · what the published papers show
How many domains, and how many of them are lone pairs. The first number gives the electron domain geometry — how the domains arrange themselves. The second removes the lone pairs from view, because a molecular shape describes where the atoms are. That is the entire model, and Figure B is all six cases it can produce.
A published scheme for one molecule sets out both, on separate lines and for separate marks:
| Electron domain geometry | trigonal planar |
| Molecular geometry | bent / V-shaped / angular |
with the note Apply ECF from Lewis formula — so the shape is marked against the structure you drew. Another scheme, comparing two molecules, is stricter about the form of the answer: Molecular geometry CS₂: linear AND Molecular geometry H₂S: bent/V-shaped, with Do not accept diagrams for M1 or M2 and the reason «central atom in» H₂S has «two» lone/non-bonding «electron» pairs.
Two cautions. A drawing is not always accepted in place of the word — one scheme refuses diagrams outright while another accepts a bent diagram for M1. Write the name. And a report lists differentiation between electron domain geometry and molecular geometry among the things teachers were asked to address.
Question. Draw the Lewis formula of SO₂, then deduce the electron domain geometry and the molecular geometry. [1 + 2]
Count. 6 + 6 + 6 = 18 electrons, or 9 pairs. Two S–O bonds use 2 pairs; completing both oxygens uses 6 more; one pair is left over and sits on the sulfur as a lone pair.
Two bonding domains plus one lone pair is three domains. Three domains arrange themselves in a trigonal planar electron domain geometry — that is the first answer.
Now hide the lone pair. Only the two oxygens are atoms, so the molecular geometry is bent — the second answer. The bond angle is compressed slightly below 120° by the lone pair, to about 117°. The single leftover pair is what separates SO₂ from CO₂, which has no lone pair on carbon and is therefore linear.
A report records that most candidates recognised that the molecule was bent but many chose the incorrect bond angle of 105 instead of 117. Both numbers are bent-molecule angles — they belong to different domain counts. Three domains with one lone pair gives about 117° (SO₂, O₃). Four domains with two lone pairs gives about 105° (H₂O, H₂S). Answering "bent" identifies the shape but not the number, so count the domains before you quote an angle. The same logic sets the rest: 109.5° with no lone pair, about 107° with one, about 105° with two — each lone pair repels a little harder than a bonding pair and squeezes the angle further.
Two reports describe candidates who knew the shape and could not justify it. On one item, most students wrote the molecular geometry as linear, but a surprising number of students were not able to explain the presence of two electron domains and no lone pair on the carbon atom. On another, the main issue was the failure to argue or explain the shape with number of electron domains. The domain count is the argument. Where a question says "explain" or carries more marks than there are shapes to name, it is asking for the count and what it does.
- Counting only bonding pairs. SCl2 and Cl2O are not linear: the two lone pairs on the central atom make four domains, so the molecules are bent. Examiner reports record many “linear, 180°” answers for exactly these molecules, even from candidates who drew the lone pairs.
- Giving the electron-domain geometry when the molecular geometry is asked (for example “tetrahedral” for NH3). Read which geometry the question asks for.
- Contradictions: stating “trigonal pyramidal” and an angle of 120°, or drawing a flat diagram for a tetrahedral molecule. Your shape name, angle and diagram must agree.
The covalent model — VSEPR
Count the electron domains around the central atom, then take the lone pairs away again. The domains set the geometry; what you can see is only where the atoms are.
Lone pairs are drawn as lobes on the central atom. They occupy more space than a bonding pair, which is why every lone pair you add pushes the bond angle down.
5Bond polarity 2.2.5 SL + HL
Electronegativity is the ability of an atom to attract a bonding pair of electrons towards itself in a covalent bond.
When two atoms of different electronegativity share a pair of electrons, the pair is pulled towards the more electronegative atom. That atom gains a partial negative charge (δ−) and the other a partial positive charge (δ+): the bond is polar and has a bond dipole. A bond between identical atoms (Cl–Cl) or between atoms of equal electronegativity is non-polar. The larger the difference in electronegativity, Δχ, the more polar the bond. A bond dipole can be shown with partial charges or with a vector (an arrow with a crossed tail, pointing towards the δ− atom).
| H | Li | Be | B | C | N | O | F | Na | Mg | Al | Si | P | S | Cl | K | Br | I | Cs |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 2.2 | 1.0 | 1.6 | 2.0 | 2.6 | 3.0 | 3.4 | 4.0 | 0.9 | 1.3 | 1.6 | 1.9 | 2.2 | 2.6 | 3.2 | 0.8 | 3.0 | 2.7 | 0.8 |
In examinations, use the electronegativity values printed in the data booklet.
| Question | Rank the bonds C–H, C–Cl, C–F and C–S in order of increasing polarity and mark partial charges on C–Cl. |
| Data | Δχ: C–H |2.6 − 2.2| = 0.4; C–S |2.6 − 2.6| = 0.0; C–Cl |3.2 − 2.6| = 0.6; C–F |4.0 − 2.6| = 1.4. |
| Answer | C–S < C–H < C–Cl < C–F. In C–Cl, chlorine is more electronegative: Cδ+–Clδ−. |
| Note | Down group 17 electronegativity falls, so C–X bond polarity decreases from C–F to C–I. |
6Molecular polarity 2.2.6 SL + HL Repeated
A molecule is polar if it has a net dipole moment: the centre of positive charge does not coincide with the centre of negative charge. The net dipole is the vector sum of the bond dipoles, so it depends on both bond polarity and molecular geometry.
- If the bond dipoles are arranged symmetrically they cancel and the molecule is non-polar even though its bonds are polar: CO2 (linear), BF3 (trigonal planar), CCl4 (tetrahedral).
- If the geometry is unsymmetrical, or the outer atoms differ, the dipoles do not cancel: H2O (bent), NH3 and PCl3 (pyramidal), CH3Cl and CHF3 (tetrahedral but with different outer atoms).
| SO3 | Three domains, no lone pair on S → trigonal planar. The three S–O bond dipoles are equal and at 120°, so they cancel → non-polar, although every bond is polar. |
| NF3 | Four domains, one lone pair → trigonal pyramidal. The three N–F dipoles point away from N and do not cancel → polar. |
| CH2Cl2 | Tetrahedral, but two C–Cl bonds are much more polar than the two C–H bonds; the dipoles cannot cancel whatever the arrangement → polar. |
| BeCl2 | Two domains → linear; two equal Be–Cl dipoles at 180° cancel → non-polar. |
To explain why a molecule is polar, give two linked points: (1) the bonds are polar because of the electronegativity difference, and (2) the shape is unsymmetrical (for example bent because of lone pairs), so the bond dipoles do not cancel / the charge distribution is uneven. Stating “it has a net dipole” without explaining why does not earn the second mark.
Exam focus · what the published papers show
Bond polarity results from the difference in electronegativities of the bonded atoms. Electronegativity values are in the data booklet, and bond dipoles may be shown either as partial charges or as vectors.
Molecular polarity depends on both bond polarity and molecular geometry. A molecule with polar bonds is polar only if those bond dipoles fail to cancel. Every polar molecule has polar bonds; not every molecule with polar bonds is polar.
For a bond, a published scheme wants the difference and then the direction:
significant/large/0.8 difference in electronegativity / oxygen more electronegative … oxygen «dipole partially» negative / sulfur «dipole partially» positive
with Accept suitable diagram showing the O–S dipole. For a molecule, the two marks are the shape and the cancellation:
| M1 | linear AND two domains «repel» |
| M2 | non-polar AND polar bonds cancel / symmetrical distribution of charge |
and the scheme states the cap explicitly: Award [1] mark if correctly states "linear and non-polar" without correct reasoning. A report on that very item confirms it: Many candidates had 1 mark for saying linear and non-polar, but few had 2 marks by explaining that there were only 2 electron domains and there was a symmetrical distribution of charge in the molecule. Name the shape, then say what the dipoles do.
Published examiner feedback records the confusion directly: A significant number of candidates were confused about molecular polarity, saying that water was non-polar, or that more hydroxyl groups on a molecule made it less polar. An –OH group is strongly polar and adding more of them makes a molecule more polar, not less. The intuition being misapplied is the cancellation rule — but cancellation requires the dipoles to be symmetrically arranged and equal, which scattered hydroxyl groups on an irregular skeleton are not.
A report offers the remedy directly: Encourage students to draw Lewis formulas when determining the polarity of a molecule. Polarity cannot be judged from a molecular formula, because the formula does not show the shape. CO₂ and SO₂ differ by one lone pair and are opposite answers.
(a) Which molecule is most polar? A. CHF3 B. CF4 C. CClF3 D. CCl4
(b) Which combination correctly describes the geometry of the carbonate ion, CO32−?
| Electron domain geometry around C | Molecular geometry around C | |
|---|---|---|
| A. | Trigonal planar | Trigonal pyramidal |
| B. | Tetrahedral | Trigonal planar |
| C. | Trigonal planar | Trigonal planar |
| D. | Tetrahedral | Trigonal pyramidal |
(a) Which molecule is polar? A. BeH2 B. AlH3 C. PH3 D. SiH4
Lewis (electron dot) structures are useful models.
(a) Draw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species. [4]
(b) Predict with a reason, whether the molecule PF3 is polar or non-polar. [1]
(i) Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl3. [3]
(ii) Explain the polarity of PCl3. [1]
(a)(i) Explain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules. [2]
(b) Lewis structures show electron domains and are used to predict molecular geometry. Deduce the electron domain geometry and the molecular geometry for the NH2− ion. [2]
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory. (The molecular geometry at carbon, trigonal planar, is given.) [3]
Deduce the molecular geometries of CS2 and H2S, and the reason why they are different. [2]
What is the correct comparison of H–N–H bond angles in NH2−, NH3 and NH4+?
Solutions and mark-scheme guidance · Set C
C1 (a) A (b) C
(a) CF4 and CCl4 are symmetrical, so their bond dipoles cancel. In CClF3 the polar C–Cl bond points away from the three C–F bonds and partly opposes their resultant dipole. In CHF3 the C–H bond is only weakly polar and its small dipole (C slightly δ−, H δ+) does not oppose the C–F resultant, so the unsymmetrical charge distribution is greatest. CHF3 is therefore the most polar.
(b) CO32−: three electron domains on C, no lone pair, so both geometries are trigonal planar.
C2 (a) C (b) D
(a) PH3 is trigonal pyramidal (three bonding pairs, one lone pair), so its dipoles do not cancel. BeH2 (linear), AlH3 (trigonal planar) and SiH4 (tetrahedral) are symmetrical.
(b) In D both partial charges and dipole arrows are placed correctly: C is δ+ and the halogens δ−; the bond dipoles point from C towards F and Cl, and the resultant molecular dipole points towards the more electronegative side.
C3 [5]
(a) PF3 Lewis structure: three P–F bonds, one lone pair on P, three lone pairs on each F ✓. PF4+: four P–F bonds, no lone pair on P, three lone pairs on each F ✓ (missing brackets or charge ignored; missing lone pairs penalised once). PF3 trigonal pyramidal ✓; PF4+ tetrahedral ✓. The geometry marks are not awarded as error carried forward from an incorrect Lewis structure.
(b) PF3 is polar and the bond dipoles do not cancel / the charge distribution is unsymmetrical ✓.
C4 [4]
(i) Electron domain geometry tetrahedral ✓; molecular geometry trigonal pyramidal ✓; Cl–P–Cl angle about 100° (any value from 91° to 108° accepted) ✓ — less than 109.5° because the lone pair repels bonding pairs more strongly than they repel each other.
(ii) Polar and the charge distribution is unsymmetrical / the dipoles do not cancel ✓.
C5 [4]
(a)(i) The bonds are polar because of the electronegativity difference between H and the group 16 atom ✓; the molecules are bent because of the lone pairs, so the charge distribution is uneven / the dipoles do not cancel ✓. Stating “there is a net dipole moment” without explaining why was not accepted.
(b) NH2− has two bonding pairs and two lone pairs: electron domain geometry tetrahedral ✓, molecular geometry bent (V-shaped, angular) ✓.
C6 [3]
Each N has three bonding pairs and one lone pair: electron domain geometry tetrahedral ✓, molecular geometry trigonal pyramidal ✓. The C has three electron domains (C=O counts as one domain): electron domain geometry trigonal planar ✓. The mark scheme notes that the real geometry of urea is more complex than VSEPR predicts (the N lone pairs are partly delocalized), but the VSEPR answer is what is assessed.
C7 [2]
CS2 is linear and H2S is bent ✓. Reason: S in H2S has two lone pairs (four electron domains) whereas C in CS2 has two electron domains and no lone pair ✓ (sp versus sp3 also accepted).
C8 A
All three species have four electron domains on N. The number of lone pairs is 2 (NH2−), 1 (NH3) and 0 (NH4+). Each lone pair compresses the bond angle further (about 105°, 107°, 109.5°), so NH2− < NH3 < NH4+.
Structure 2.2 · Networks, forces and properties
7Covalent network structures 2.2.7 SL + HL Repeated
Most covalent substances consist of small molecules. Carbon and silicon, however, can each form four covalent bonds and can build covalent network (giant covalent) structures in which every atom is bonded to its neighbours throughout the whole solid. There are no discrete molecules and no intermolecular forces to overcome within the network: to melt or sublime the solid, strong covalent bonds must be broken.
Allotropes are different structural forms of the same element in the same physical state. Because their bonding and structural patterns differ, allotropes have different physical and chemical properties.
| Substance | Structure and bonding | Properties and explanation |
|---|---|---|
| Diamond | Each C bonded to four others by single covalent bonds in a tetrahedral arrangement (109.5°); 3-D network. All four valence electrons used in bonding. | Hardest natural substance and sublimes at very high temperature (about 3500 °C): many strong covalent bonds must be broken. Does not conduct electricity: no delocalized electrons or ions. Conducts heat very well because the rigid network transmits vibrations efficiently. |
| Graphite | Each C bonded to three others in flat hexagonal layers (120°); the fourth valence electron is delocalized across the layer. Layers held together by weak London forces. | Very high sublimation point (strong bonds within layers). Soft and slippery, used as a lubricant and in pencils: layers slide over each other. Conducts electricity along the layers (delocalized electrons), used for electrodes. |
| Graphene | A single layer of graphite, one atom thick: each C bonded to three others, with delocalized electrons. | Extremely strong and tough (no weak inter-layer forces to fail), yet only one atom thick; excellent electrical and thermal conductor. Proposed uses include electronic circuits, electrodes and reinforcing composites. |
| Fullerene C60 | Discrete spherical molecules of 60 C atoms in 12 pentagons and 20 hexagons; each C bonded to three others, with delocalized electrons; London forces between molecules. | Much lower sublimation temperature than diamond or graphite because only intermolecular forces are overcome. Electrons are delocalized within each molecule, but in the bulk solid they cannot move easily from molecule to molecule, so C60 is a far poorer conductor than graphite. |
| Silicon | Same network structure as diamond, each Si bonded to four others. | High melting point (1414 °C); hard. Si–Si bonds are longer and weaker than C–C bonds, so silicon melts at a lower temperature than diamond sublimes. A semiconductor. |
| Silicon dioxide (quartz) | Each Si bonded to four O atoms (tetrahedral); each O bonded to two Si atoms; 3-D network with ratio Si : O = 1 : 2. | High melting point (about 1700 °C), hard, insoluble in water and organic solvents, does not conduct: no mobile charged particles. Used in glass and ceramics. |
Carbon and silicon are both in group 14 and both oxides have the empirical formula XO2. Carbon forms strong C=O double bonds, so each carbon atom is satisfied by two oxygen atoms and discrete, linear, non-polar CO2 molecules result, held together only by weak London forces. Silicon does not form stable Si=O double bonds; instead each Si forms four single Si–O bonds to four different O atoms, building a giant network. Melting CO2 overcomes weak forces between molecules; melting SiO2 breaks strong covalent bonds.
Exam focus · what the published papers show
| Structure | Bonding pattern | Geometry at each carbon | Conducts? |
|---|---|---|---|
| Diamond | Every carbon bonded to four others, one giant three-dimensional network | tetrahedral, 109.5° | No — all four valence electrons are held in bonds |
| Graphite | Every carbon bonded to three others in flat layers, held together only by London forces | trigonal planar, 120° | Yes — the fourth electron is delocalized along the layers |
| Graphene | A single layer of graphite | trigonal planar, 120° | Yes, for the same reason |
| Fullerenes | Closed cages of rings; discrete molecules, not a network | close to trigonal planar | Poorly, and unlike the others they melt as molecules |
| Silicon, SiO₂ | Giant networks like diamond; in SiO₂ each Si is bonded to four O | tetrahedral | No for SiO₂; silicon is a semiconductor |
Graphite's conductivity is a two-mark answer and the reports say candidates give one half of it. One notes that most were able to describe the bonding in graphite though that by itself was not enough to answer why delocalised electrons would allow this material to have electrical conductivity. Another records candidates who only scored one of the two marks, either describing the layered structure of graphite or the delocalized electrons. The published scheme wants both: the layered two-dimensional network, and delocalized electrons «flow along layers». Structure, then consequence — every time. The comfort is that the same paper's easiest question was the mirror image: over 80 % knew that diamond does not conduct.
- “Diamond has van der Waals forces between its molecules.” Diamond has no molecules: there are only covalent bonds throughout.
- SiO2 drawn as discrete O=Si=O molecules, or with silicon bonded to only two oxygen atoms. Each Si bonds to four O atoms.
- Comparing Si–O and S–O bond strengths to explain why SiO2 melts at a much higher temperature than SO3. The bond energies are similar; the difference is that melting SiO2 breaks many covalent bonds, whereas melting SO3 overcomes only weak intermolecular forces.
- “Strong intermolecular forces hold a giant covalent structure together.” In a network solid the atoms are held by covalent bonds, not intermolecular forces.
Which form of carbon is the poorest electrical conductor?
What are the approximate bond angles and structure of crystalline SiO2?
| O–Si–O | Structure | |
|---|---|---|
| A. | 90° | giant molecule |
| B. | 109° | giant molecule |
| C. | 180° | small molecule |
| D. | 180° | giant molecule |
C60 and diamond are allotropes of carbon.
(i) Outline one difference between the bonding of carbon atoms in C60 and diamond. [1]
(ii) Explain why C60 and diamond sublime at different temperatures and pressures. [2]
The HL paper of the same session asked for two differences in part (i) [2].
Graphite rods are sometimes used as inert electrodes. Describe the structure of graphite and explain why graphite conducts electricity. [2]
Carbon and silicon are elements in group 14. Explain why CO2 is a gas but SiO2 is a solid at room temperature. [2]
What is the type of bonding in a compound that has high boiling and melting points, poor electrical conductivity, and low solubility in water?
Solutions and mark-scheme guidance · Set D
D1 C
In diamond all four valence electrons of each carbon are used in localized C–C σ bonds, so there are no delocalized electrons. Graphite, graphene and nanotubes each have one delocalized electron per carbon.
D2 B
Each Si is bonded tetrahedrally to four O atoms (O–Si–O ≈ 109°) and each O bridges two Si atoms, forming a giant covalent network.
D3 [3]
(i) Any one: each C in C60 is bonded to three C atoms whereas each C in diamond is bonded to four; C60 has delocalized electrons (single and double bonds) whereas diamond does not; sp2 versus sp3; trigonal planar (108–120°) versus tetrahedral (109.5°) ✓.
(ii) Diamond is a giant (network) covalent structure and sublimes at a higher temperature because strong covalent bonds must be broken ✓; C60 is molecular, and only the weak London (dispersion) forces between molecules are overcome ✓.
D4 [2]
Graphite has layers of carbon atoms (each bonded to three others in hexagonal rings) in a giant structure ✓; one electron per carbon is delocalized and these electrons are mobile, flowing along the layers ✓.
D5 [2]
CO2 is a non-polar molecule; only weak London forces act between the molecules ✓ (“between” is essential). SiO2 is a giant covalent network (3-D lattice) in which strong Si–O covalent bonds must be broken to melt it ✓.
D6 D
High melting and boiling points rule out molecular covalent substances; poor conductivity rules out metals and (for the liquid) ionic substances; insolubility in water fits a giant covalent structure such as SiO2 or diamond.
8Intermolecular forces 2.2.8 SL + HL Repeated
Molecular covalent substances such as water, iodine and methane consist of discrete molecules. The covalent bonds within each molecule are strong, but the attractions between molecules, called intermolecular forces, are much weaker. It is these intermolecular forces, not the covalent bonds, that are overcome when a molecular substance melts or boils. Which intermolecular forces act depends on the size (number of electrons) and polarity of the molecules.
London (dispersion) forces
Electrons are constantly moving. At any instant the electron cloud of a molecule, even a non-polar one such as F2 or CH4, may be unevenly distributed, creating an instantaneous dipole. This dipole distorts the electron cloud of a neighbouring molecule, inducing a dipole in it, and the two dipoles attract. London forces act between all molecules. They are stronger when the molecules have more electrons (larger, more easily distorted electron clouds) and when the molecules have a larger surface area of contact: straight-chain butane (b.p. −0.5 °C) boils at a higher temperature than its compact, branched isomer 2-methylpropane.
Dipole–induced dipole forces
A polar molecule has a permanent dipole that can induce a temporary dipole in a neighbouring non-polar molecule, and the two attract. This force acts, for example, between water and dissolved oxygen molecules.
Dipole–dipole forces
Polar molecules have permanent dipoles. The δ+ end of one molecule attracts the δ− end of a neighbour, so polar molecules align and attract each other more strongly than non-polar molecules of similar size. Methanal (polar, b.p. −19 °C) boils about 70 °C higher than ethane (non-polar, b.p. −88.5 °C), though the molecules have almost the same number of electrons.
Van der Waals forces is the inclusive term for dipole–dipole, dipole–induced dipole and London (dispersion) forces. When a question asks you to identify a specific force, name it precisely: “van der Waals forces” is not accepted in place of “London forces” or “dipole–dipole forces”.
Hydrogen bonding
A hydrogen bond forms when a hydrogen atom that is covalently bonded to a highly electronegative atom (in practice N, O or F) is attracted to a lone pair on an electronegative atom of a neighbouring molecule. N, O and F are both very electronegative and small, so the hydrogen atom bonded to them carries a large δ+ charge and is left with almost no electron shielding; it can approach closely to a lone pair on another small electronegative atom. The arrangement X–H···Y is close to linear. Hydrogen bonding is the strongest type of intermolecular force for molecules of comparable size, but it is still roughly a tenth of the strength of a typical covalent bond.
The current IUPAC definition describes a hydrogen bond as an attractive interaction between a hydrogen atom covalently bonded to an electronegative atom and a neighbouring electronegative atom; improved spectroscopic and computational evidence led to this broader definition. For examination purposes, look for H bonded to N, O or F and a lone pair on N, O or F of another molecule.
Water molecules each have two O–H hydrogen atoms and two lone pairs, so each can take part in up to four hydrogen bonds. In ice this produces an open, regular lattice in which molecules are held further apart than in liquid water, which is why ice is less dense than liquid water and floats.
Exam focus · what the published papers show
The force between molecules is determined by their size and polarity. The syllabus names London (dispersion), dipole-induced dipole, dipole–dipole and hydrogen bonding, and adds that van der Waals forces should be used as an inclusive term covering the first three — that is, everything except hydrogen bonding.
Hydrogen bonds occur when hydrogen, being covalently bonded to an electronegative atom, has an attractive interaction on a neighbouring electronegative atom. In practice that means the hydrogen must be bonded to N, O or F.
This is the most-repeated misconception in the topic, named in three separate sessions. One report: hydrogen bonds commonly were identified as the covalent bond between any atom and hydrogen, rather than an intermolecular force. Another: some weaker candidates thought bonds between N and H in the molecule were hydrogen bonds. A third simply records that candidates confused intermolecular with intramolecular forces. The N–H bond inside ammonia is a covalent bond. The hydrogen bond is the attraction from that hydrogen to the lone pair on a different ammonia molecule — and it is what a question about boiling point is asking about, because boiling separates molecules without breaking them.
A scheme asking which forces act in an alkane gives London (dispersion) forces «only» and accepts dispersion forces / instantaneous / transient / induced dipole attractions — but adds Do not accept van der Waals' forces for M1. The umbrella term is correct terminology and still loses the mark, because the question wanted the particular force. Use the specific name.
- “The hydrogen bond is the O–H bond in water.” The O–H bond is a covalent bond within the molecule; the hydrogen bond is the attraction between the H of one molecule and a lone pair on O of another.
- “Boiling breaks covalent bonds.” When water boils, the molecules separate but stay intact: only intermolecular forces are overcome. Examiner reports flag this confusion repeatedly.
- Drawing hydrogen bonds carelessly. Show a lone pair on the acceptor atom, the partial charges (δ+ on H, δ− on N/O/F), and use a dashed line roughly in line with the covalent X–H bond. Draw only the number of hydrogen bonds the question asks for.
- H2 or CH4 “form hydrogen bonds because they contain hydrogen”. H must be bonded to N, O or F.
9Relative strengths of intermolecular forces and physical properties 2.2.9 SL + HL Strong
For molecules of comparable molar mass, the relative strengths of intermolecular forces are generally:
London (dispersion) forces < dipole–dipole forces < hydrogen bonding (<< covalent bonds)
The qualifier “of comparable molar mass” is essential. London forces grow with the number of electrons, so for large molecules they can exceed the dipole–dipole or even hydrogen-bonding contributions of smaller molecules. Iodine, I2, has only London forces but is a solid at room temperature, whereas water, with hydrogen bonding, is a liquid.
| Property | Typical behaviour | Explanation |
|---|---|---|
| Volatility | Low melting and boiling points; many are gases or liquids at room temperature; high volatility. | Only weak intermolecular forces must be overcome; covalent bonds are not broken. Stronger intermolecular forces (more electrons, polarity, hydrogen bonding) → lower volatility. |
| Electrical conductivity | Do not conduct as solids or liquids. | Neutral molecules; no mobile ions or delocalized electrons. (Some, such as HCl, react with water to form ions, so their aqueous solutions conduct.) |
| Solubility | “Like dissolves like”: polar and hydrogen-bonding molecules dissolve in water; non-polar molecules dissolve in non-polar solvents such as hexane. | A solute dissolves when the attractions it forms with the solvent are comparable to those it breaks. Small alcohols, carboxylic acids and ammonia form hydrogen bonds with water. As the non-polar hydrocarbon chain grows, solubility in water decreases (butan-1-ol is less soluble than methanol). |
| Question | Ethane, CH3CH3 (b.p. −88.5 °C), methanal, HCHO (−19 °C), and methanol, CH3OH (64.7 °C), each have 18 electrons per molecule. Explain the order of boiling points. |
| Step 1 | Equal numbers of electrons → London forces of similar strength in all three. The differences must come from other forces. |
| Step 2 | Ethane is non-polar (symmetrical, C–H bonds of low polarity): London forces only. Methanal has a polar C=O bond and a trigonal planar but unsymmetrical shape: London + dipole–dipole forces; it has no H on O, so no hydrogen bonding. Methanol has an O–H group: London + dipole–dipole + hydrogen bonding. |
| Answer | The strongest intermolecular force increases ethane (London) < methanal (dipole–dipole) < methanol (hydrogen bonding), so more energy is needed to separate the molecules and the boiling point rises in the same order. |
| Check | Consistent with the measured values in Figure 2.14(b). |
Mark schemes for these questions reward three linked statements: (1) the strongest intermolecular force in substance A, (2) the strongest force in substance B, and (3) a comparison: which force is stronger and therefore which substance is less volatile (has the higher boiling point). Answers based only on “A is more polar” or “A has a larger molar mass” are not credited when the key difference is hydrogen bonding. When the two substances differ greatly in size, discuss London forces explicitly (a larger number of electrons gives stronger London forces).
Exam focus · what the published papers show
Question. Compare and contrast the intermolecular forces in ethenone and carbon dioxide. [2] Mean mark: 0.55 out of 2.
What both have. London dispersion forces — and, because the two molecules have similar molar masses, of similar strength. The report notes that it was rare to find a candidate who recognized the similar strength of the London dispersion forces in the two compounds: that was the "compare" half.
What only one has. Ethenone is a polar molecule, so it also has dipole–dipole forces; carbon dioxide, though its bonds are polar, is linear and symmetrical, so its bond dipoles cancel and it has none. The published answer is exactly ethenone has dipole-dipole forces «and carbon dioxide does not».
Two ways candidates lost this. A significant proportion of the candidates discussed the bonding inside the molecule instead of the intermolecular forces — the intramolecular confusion again — and another group of candidates stated that ethenone had hydrogen bonding, which it cannot: its hydrogens are on carbon.
Published examiner feedback puts a number on it: only about 20% of students realised the critical nature of hydrogen bonding when molecular species dissolve in water. It is apparent that many consider molecular polarity to be the criterion of solubility, even though most polar molecules that do not hydrogen bond to water (e.g. CHCl₃) show low solubility.
Another report says the same in different words: candidates only discussed polarity where the scheme focussed on the formation of hydrogen bonds with water. The published mark is «both can» form hydrogen bonds with water «molecules». Say hydrogen bonding, and say it is with water.
A scheme marking the solubility of a substance whose hydrogens are not on N, O or F accepts forms ion-dipole with water but explicitly not forms hydrogen bond with water. You cannot claim hydrogen bonding for a molecule that has no hydrogen on an electronegative atom, however soluble it is — check the structure before you name the force.
For a homologous series, a scheme gives stronger forces of attraction with increasing chain length / larger electron cloud / molar mass, and adds Accept "more electrons" for "larger electron cloud". The mechanism is that a bigger electron cloud is more easily distorted, producing stronger instantaneous dipoles. Name the force that is actually changing — attributing the trend to dipole–dipole forces, in molecules that have none, does not earn it.
| Volatility | High for covalent molecular substances, because boiling separates molecules and only the weak forces between them must be overcome — never the covalent bonds inside. |
| Conductivity | None, in general: there are no ions and no delocalized electrons. Graphite and graphene are the exceptions, and a few molecules ionise in water. |
| Solubility | In water, if the molecule can hydrogen bond to it. In non-polar solvents, if the molecule is non-polar. Like dissolves like — but state the force, not the label. |
Which is the correct order based on increasing strength?
Which bonds cause the boiling point of water to be significantly greater than that of hydrogen sulfide?
(a) The following compounds have similar relative molecular masses. What is the order of increasing boiling point?
(b) Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. CH3CH2CH2CH2OH
Explain the increase in the boiling point from H2S to H2Te. [2]
(i) Suggest one reason why urea is a solid and ammonia a gas at room temperature. [1]
(ii) Sketch two different hydrogen bonding interactions between ammonia and water. [2]
Nickel catalyses the conversion of propanone to propan-2-ol. Discuss, referring to intermolecular forces present, the relative volatility of propanone and propan-2-ol. [3]
(c) Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2. [2]
(d) Suggest, giving reasons, the relative volatilities of SCl2 and H2O. [3]
Methanoic acid and ethanal (CH3CHO) both contain a carbonyl group and have similar molar masses.
(i) Explain why, in terms of the strongest intermolecular forces between the molecules, ethanal has a much lower boiling point than methanoic acid. [2]
(ii) Outline why ethanal and methanoic acid are both fully miscible with water. [1]
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
| Mr | Boiling point | |
|---|---|---|
| HCN | 27.03 | 26.00 °C |
| HCl | 36.51 | −85.05 °C |
Which compound is both volatile and soluble in water? A. NaCl B. CH3CH2CH3 C. CH3OH D. C12H22O11
Solutions and mark-scheme guidance · Set E
E1 C
For molecules of similar size, dispersion < dipole–dipole < hydrogen bonding, and all intermolecular forces are much weaker than covalent bonds.
E2 D
Water molecules form hydrogen bonds (H bonded to highly electronegative O with lone pairs); H2S cannot, because S is not electronegative enough. The O–H covalent bonds are not broken when water boils.
E3 (a) B (b) D
(a) Propanal has dipole–dipole forces only; propan-1-ol forms hydrogen bonds; ethanoic acid forms more extensive hydrogen bonding (both C=O and O–H take part, and molecules can pair up). Order: CH3CH2CHO < CH3CH2CH2OH < CH3COOH.
(b) Each alcohol has one –OH that hydrogen bonds with water; the longer the non-polar hydrocarbon chain, the more it disrupts the hydrogen-bonded water structure, so butan-1-ol is least soluble.
E4 [2]
The number of electrons increases from H2S to H2Te ✓ (Mr / molecular size increase accepted); therefore the London (dispersion) forces between the molecules increase and more energy is needed to separate them ✓.
E5 [3]
(i) Any one: urea has a greater molar mass / stronger London forces / more hydrogen bonding / is more polar ✓. “Greater intermolecular forces” without naming the reason was not accepted.
(ii) Two correct hydrogen bonds, one mark each ✓✓: an H of water attracted to the lone pair on N of NH3 (O–H···N), and an H of NH3 attracted to a lone pair on O of water (N–H···O). If lone pairs are drawn, the lone pair involved must be on N or O. A hydrogen bond drawn as a solid line is penalised once.
E6 [3]
Propan-2-ol: hydrogen bonding (plus dipole–dipole and London forces) ✓. Propanone: dipole–dipole forces (plus London) — it has no H bonded to O ✓. Propan-2-ol is less volatile and hydrogen bonding is stronger / the sum of its intermolecular forces is stronger ✓.
E7 [5]
(c) Lewis structure with two S–Cl bonds, two lone pairs on S and three on each Cl ✓; bent (V-shaped, angular) ✓.
(d) H2O forms hydrogen bonds, SCl2 does not ✓; SCl2 has much stronger London forces because it has many more electrons (larger molar mass) ✓. Conclusion — either answer is credited if argued: H2O is less volatile and hydrogen bonding is stronger, or SCl2 is less volatile and its dispersion forces could exceed the hydrogen bonding in water ✓. The mark scheme rewards a reasoned comparison, not a memorised rule that hydrogen bonding always wins.
E8 [3]
(i) Methanoic acid forms hydrogen bonds and ethanal has dipole–dipole forces as its strongest intermolecular force ✓; hydrogen bonds are stronger, so more energy is needed to separate methanoic acid molecules and its boiling point is higher ✓. Answers based on polarity or molar mass alone, or calling dipole–dipole forces “van der Waals”, were not credited.
(ii) Both form hydrogen bonds with water molecules ✓ (ethanal through the lone pairs on its carbonyl O).
E9 [1]
HCN has stronger dipole–dipole forces ✓ (a more polar molecule with a larger dipole). “Hydrogen bonds” was not accepted: the H in HCN is bonded to C, not to N, O or F. Note that HCl has the larger Mr, so London forces cannot explain the difference.
E10 C
Methanol is a small molecule, so its intermolecular forces are relatively weak (volatile), and its –OH group hydrogen bonds with water (soluble). NaCl and sucrose are not volatile; propane is not soluble.
10Chromatography 2.2.10 SL + HL Strong
Chromatography separates the components of a mixture using two phases: a stationary phase, which does not move, and a mobile phase, which moves through or over it. Each component is attracted, through intermolecular forces, to both phases. A component that is more strongly attracted to the stationary phase spends more time there and moves more slowly; a component that is more strongly attracted to (more soluble in) the mobile phase is carried further.
- Paper chromatography: the stationary phase is the water held in the cellulose fibres of the paper; the mobile phase is a solvent that rises through the paper by capillary action.
- Thin-layer chromatography (TLC): the stationary phase is a thin layer of silica (SiO2) or alumina (Al2O3) on a glass, plastic or aluminium plate; the mobile phase is a solvent. Polar components are attracted strongly to the polar surface of silica.
RF = distance moved by the component (from the baseline) ÷ distance moved by the solvent front (from the baseline)
Both distances are measured from the baseline to the centre of the spot and to the solvent front. RF has no units and lies between 0 and 1. Under identical conditions (same stationary phase, solvent and temperature) a substance always has the same RF, so comparing RF values with those of known standards helps identify components. A change of solvent changes all the RF values.
| Given | On a TLC plate (silica), the solvent front moved 8.0 cm from the baseline. Spot P moved 2.0 cm and spot Q moved 6.0 cm. |
| Relationship | RF = distance moved by component ÷ distance moved by solvent front. |
| Calculation | RF(P) = 2.0 ÷ 8.0 = 0.25; RF(Q) = 6.0 ÷ 8.0 = 0.75. |
| Interpretation | P is held more strongly by the polar silica stationary phase (for example it can hydrogen bond to the Si–O–H groups at the surface), so it is likely to be the more polar component. Q is more attracted to the mobile phase. |
| Check | Both values lie between 0 and 1 and have no units; the less mobile spot has the smaller value. |
Locating agents (used to make colourless spots visible) and the operation of gas and high-performance liquid chromatographs are not assessed.
Exam focus · what the published papers show
Chromatography sits at the end of this topic because it is the same chain applied to a separation. Which phase does a component prefer? Whichever it can form the stronger intermolecular forces with. A polar component on a polar stationary phase is held back; a non-polar component in a non-polar mobile phase runs with the solvent. Work out the polarity of each substance first, and the chromatogram follows.
Question. A paper chromatogram is marked with four distances: 12 mm above the solvent front, 7 mm from the solvent front down to compound X, 21 mm from compound X down to the origin, and 10 mm from the origin down to the bottom of the paper. What is the retardation factor of compound X?
Both distances are measured from the origin. The compound moved 21 mm. The solvent moved past it and on to the front, so the solvent moved 21 + 7 = 28 mm.
RF = 21 / 28 = 0.75.
Every wrong option is a specific mistake, and the report names the popular one: 72 % answered correctly, and weaker students also included the length below the baseline for the distance travelled by the solvent — that gives 21/38 = 0.55. Adding it to both distances gives 31/38 = 0.82; inverting the fraction gives 28/21 = 1.33. A value above 1 is always wrong, because a spot cannot outrun the solvent that carries it.
Question. A thin-layer chromatogram of naphthalene, C₁₀H₈, and naphthol, C₁₀H₈O, uses a polar silica as the stationary phase, and a non-polar hexane as the mobile phase. Which spot is naphthalene?
Compare the two molecules. Naphthalene is a hydrocarbon: non-polar, London forces only. Naphthol carries an –OH group: polar, and able to hydrogen bond.
Match each to a phase. Naphthol is attracted to the polar silica and is held back. Naphthalene, being non-polar, has little attraction to the silica and travels with the non-polar hexane, so it is the higher spot with the larger RF. A report calls this a very challenging question involving understanding how to calculate RF values and which compound would move further with a non-polar solvent based on structure, adding that many students selected all of the other incorrect answers. The functional group decides it.
A report records that there was some concern that the calculation of retardation factor, RF, values is not in the guide, and answers it plainly: This concept is under Structure 2.2.10 and was brought into the new curriculum from option A (materials) in the legacy curriculum. In practice the calculation appears both as a multiple-choice item and as a structured question. Published answers of 0.566 and 0.451 carry accepted ranges of 0.530–0.580 and 0.430–0.480 — wide, because you are reading a ruler, so a small measurement difference will not cost you the mark. What will is measuring from the wrong place.
Published examiner feedback notes candidates who failed to mention the stationary or mobile phase for the second mark, instead referring to RF values or comparing structures. If the question says "explain", the answer names the two phases and the attraction to each — the number is not an explanation of itself.
The supplied IB papers contain no chromatography questions. The three items below are authentic Cambridge International AS & A Level Chemistry (9701) Paper 4 questions with their published mark schemes; they test exactly the RF skills required by 2.2.10.
The purity of lidocaine can be checked using thin-layer chromatography. Ethyl ethanoate is used as the solvent. The RF values of X and lidocaine are: X 0.49; lidocaine 0.71.
(i) Identify the substances used as the mobile and stationary phases in this thin-layer chromatography experiment. [1]
(ii) Describe how an RF value can be calculated. [1]
(iii) Suggest why the RF value for X is less than that for lidocaine. [1]
Amino acids can be separated by thin-layer chromatography. A mixture of amino acids is analysed using this technique. The chromatogram obtained is shown, drawn to scale. The table shows some RF values for different amino acids in the solvent used.
(i) Use the chromatogram and the RF values to deduce the amino acid responsible for spot A and spot B. [1]
(ii) A second chromatogram of the same mixture is taken using a more polar solvent. Predict the effect on the RF values of the amino acids. Explain your reasoning. [1]
A mixture of tyrosine and lysine can be separated by thin-layer chromatography. Under certain conditions the RF value of lysine is 0.14 and the RF value of tyrosine is 0.45.
(i) Explain what is meant by RF value. [1]
(ii) Suggest an explanation for the difference in RF values. [1]
Solutions and mark-scheme guidance · Set F
F1 [3]
(i) Mobile phase: ethyl ethanoate; stationary phase: silica, SiO2 (or alumina, Al2O3) ✓.
(ii) RF = distance moved by the solute (spot) ÷ distance moved by the solvent front, both measured from the baseline ✓.
(iii) X is more strongly attracted to (adsorbed on) the stationary phase / lidocaine dissolves better in the solvent ✓.
F2 [2]
(i) Solvent front 5.0 cm from the baseline. Spot A ≈ 3.55 cm: RF ≈ 3.55/5.0 = 0.71, leucine. Spot B ≈ 1.45 cm: RF ≈ 0.29, glutamic acid. Both needed ✓.
(ii) The RF values increase and the amino acids are more soluble in the more polar solvent / form more hydrogen bonds with it ✓.
F3 [2]
(i) Distance moved by the amino acid ÷ distance moved by the solvent front ✓ — it must be clear which distance is divided by which.
(ii) Tyrosine is more soluble in the solvent (mobile phase) / lysine is more strongly attracted to the stationary phase ✓. “They have different solubilities” is too vague.
Review · Structure 2.2, SL core
11The examiner’s view, and the topic as one chain
Evidence base: principal examiner reports for Cambridge International AS & A Level Chemistry (2016–2024) on the same chemistry.
Answered well: dot-and-cross diagrams of simple molecules, recognising that HF forms hydrogen bonds, naming the tetrahedral shape and 109.5°, and recalling the definition of RF.
Found difficult: (1) the single most persistent weakness is confusing covalent bonds with intermolecular forces, suggesting that covalent bonds break when a molecular substance melts or boils (reported for SiCl4, halogens, sulfur and phosphorus, CS2 and CO2); (2) incomplete definitions of the covalent bond, omitting the pair of electrons or the attraction to the nuclei; (3) VSEPR answers that ignore lone pairs (SCl2 and Cl2O given as linear), shape names that contradict the stated bond angle, and flat drawings of 3-D molecules; (4) hydrogen bonds drawn without the lone pair, drawn as the H–F covalent bond itself, or drawn as a dative arrow; (5) linking volatility of the halogens to electronegativity or to the nucleus–outer electron attraction instead of to the number of electrons and the strength of instantaneous dipole–induced dipole forces; (6) describing giant covalent structures as held together by “strong intermolecular forces”, or giving diamond van der Waals forces between molecules; (7) in chromatography, not associating slower movement (smaller RF, longer retention) with stronger attraction to the stationary phase, imprecise definitions of RF that do not make clear which distance is divided by which, and predicting that a more polar solvent would decrease RF values, or predicting an increase without explaining it through greater solubility in the mobile phase.
What successful answers did: named the specific force between the particles concerned, compared the forces in both substances explicitly, linked strength of force to energy needed, and kept the level of explanation (between molecules or within molecules) consistent throughout.
| 1 · Count the electrons | All valence electrons, adjusted for charge. Connect, complete the outer atoms, then fix the centre with multiple bonds. Draw every lone pair. |
| 2 · Count the domains | Bonding domains plus lone pairs. A double or triple bond is still one domain. Two numbers, six possible answers. |
| 3 · Name both geometries | Electron domain geometry uses all the domains; molecular geometry ignores the lone pairs. Quote the angle that matches the domain count, not the shape name. |
| 4 · Check the bonds | Different electronegativities make a polar bond. Mark δ+ on the less electronegative atom. |
| 5 · Check the molecule | Do the bond dipoles cancel? They cancel only if the ligands are identical and symmetrically arranged with no lone pair on the centre. |
| 6 · Read off the forces | London always; add dipole–dipole if the molecule is polar; add hydrogen bonding if an H sits on N, O or F. Then compare them. |
| 7 · Explain the property | Volatility, conductivity and solubility all follow from the forces between molecules — never from the bonds inside them. |
If you take one habit from this topic, make it this: whenever a question names a molecule, sketch its Lewis formula in the margin before answering anything — with the lone pairs. That one sketch supplies the domain count, the shape, the angle, whether the dipoles cancel, which intermolecular forces act, and therefore the boiling point, the solubility and the chromatogram. Nearly every failure the reports describe is an answer written without that sketch in front of the candidate, and nearly every two-mark question is the sketch plus one sentence about what it implies.
12Quick check
- How many valence electrons are in the nitrate ion, and how does the charge change the count?
- Draw the Lewis formula of PCl₃. How many lone pairs are on the phosphorus, and what shape does that give?
- Why is the Lewis structure of BeCl₂ not an error, even though beryllium has no octet?
- Give the electron domain geometry, the molecular geometry and the approximate bond angle of SO₂.
- Two molecules are both described as "bent". One has a bond angle near 117°, the other near 105°. What distinguishes them?
- CO₂ and SO₂ both have polar bonds. Why is only one of them a polar molecule?
- Explain, in the two steps a scheme requires, why graphite conducts electricity and diamond does not.
- Which intermolecular forces act in N₂, in HCN, in NH₃ and in PH₃?
- Why is ethanol far more soluble in water than chloroform, although both are polar?
- A spot sits 3.6 cm above the origin; the solvent front is 4.8 cm above the origin; the paper extends 1.5 cm below the origin. What is the RF?
| 1 | 24. 5 from nitrogen, 6 from each of three oxygens, and the 1− charge adds one electron. Ignoring the charge gives 23 — an odd number, which cannot be paired, and is the signal that something is wrong. |
| 2 | 26 electrons, 13 pairs: three P–Cl bonds, nine pairs completing the chlorines, and one lone pair on phosphorus. Four domains, one lone: trigonal pyramidal, about 107°. |
| 3 | The octet rule is a tendency, and the syllabus explicitly requires molecules with fewer than an octet. Beryllium has only two valence electrons and no way to reach eight: four electrons is the correct answer. |
| 4 | Three domains (two bonding, one lone pair) → electron domain geometry trigonal planar; molecular geometry bent; angle about 117°. Both geometry names are needed. |
| 5 | The number of domains. Three domains with one lone pair gives ≈117° (SO₂); four domains with two lone pairs gives ≈105° (H₂O). "Bent" names the shape but not the angle. |
| 6 | CO₂ is linear with no lone pair on carbon, so the two bond dipoles are equal and exactly opposed and cancel — net dipole zero. SO₂ has a lone pair on sulfur, so it is bent, the dipoles are not opposed and they add to a net dipole. |
| 7 | Structure: graphite is a layered two-dimensional network in which each carbon bonds to only three others. Consequence: the fourth valence electron is delocalized and free to flow along the layers. Diamond bonds each carbon to four others, so every valence electron is held in a bond and none is free to move. |
| 8 | N₂: London only — identical atoms, no dipole. HCN: London and dipole–dipole. NH₃: London, dipole–dipole and hydrogen bonding — the H is on nitrogen. PH₃: London and dipole–dipole only; phosphorus is not N, O or F, so no hydrogen bonding. |
| 9 | Ethanol's –OH group lets it hydrogen bond to water. Chloroform is polar but has no hydrogen on an electronegative atom, so it cannot — and the reports name this exact pair as the reason polarity is the wrong criterion. |
| 10 | 0.75. 3.6 / 4.8. The 1.5 cm below the origin belongs to neither distance — including it is the single most-reported error on this outcome. |
13Summary and knowledge organiser
Essential knowledge
- Covalent bonds are directional. More shared pairs between the same two atoms give shorter, stronger bonds. Bond polarity results from an electronegativity difference; molecular polarity depends on bond polarity and shape.
- VSEPR: electron domains around a central atom repel and take up positions as far apart as possible; lone pairs repel more strongly than bonding pairs and compress bond angles by about 2–3° each.
- Covalent network structures (diamond, graphite, graphene, C60, SiO2) are held by covalent bonds throughout; molecular substances are held by weak intermolecular forces between molecules.
- Intermolecular forces in increasing strength for molecules of similar size: London < dipole–induced dipole < dipole–dipole < hydrogen bonding. London forces increase with the number of electrons and can outweigh hydrogen bonding in large molecules.
- Chromatography separates components according to their relative attraction to the stationary and mobile phases.
Essential definitions and relationships
| Term or relationship | Meaning and use |
|---|---|
| Covalent bond | Electrostatic attraction between a shared pair of electrons and the positively charged nuclei. |
| Coordination (dative) bond | A covalent bond in which both shared electrons come from the same atom. |
| Electronegativity | The relative ability of an atom to attract a shared pair of electrons in a covalent bond. |
| Electron domain | A region of electron density around a central atom: a lone pair, a single bond or a multiple bond. |
| Hydrogen bond | Attraction between an H atom bonded to N, O or F and a lone pair on another N, O or F atom. |
| London (dispersion) force | Attraction between an instantaneous dipole and the dipole it induces in a neighbouring molecule. |
| Allotropes | Different structural forms of the same element in the same physical state. |
| RF = dsolute ÷ dsolvent front | Both distances measured from the baseline; no units; always between 0 and 1. A larger RF means greater relative attraction to the mobile phase. |
Essential observations and properties
| Structure type | Melting point | Conductivity | Solubility in water | Mechanical |
|---|---|---|---|---|
| Simple molecular | Low | None (no mobile charged particles) | Depends on polarity and H-bonding | Soft |
| Giant covalent | Very high | None, except graphite and graphene (delocalized electrons) | Insoluble | Diamond very hard; graphite soft, layered |
Essential shapes
| Domains | Lone pairs | Molecular geometry | Angle | Example |
|---|---|---|---|---|
| 2 | 0 | Linear | 180° | CO2, BeCl2, HCN |
| 3 | 0 | Trigonal planar | 120° | BF3, CO32− |
| 3 | 1 | Bent | < 120° (≈ 117°) | SO2, O3 |
| 4 | 0 | Tetrahedral | 109.5° | CH4, NH4+ |
| 4 | 1 | Trigonal pyramidal | ≈ 107° | NH3, PCl3 |
| 4 | 2 | Bent | ≈ 105° | H2O, NH2− |
Common misconceptions
- Melting or boiling a molecular substance breaks covalent bonds. Only the intermolecular forces between molecules are overcome.
- Giant covalent substances such as diamond or SiO2 have “strong intermolecular forces”. There are no molecules; covalent bonds extend throughout the structure.
- The hydrogen bond is the O–H bond itself. It is the attraction between an H on one molecule and a lone pair on N, O or F of another.
- Shapes are decided by bonding pairs only. Lone pairs count as electron domains and determine the electron domain geometry.
- A molecule with polar bonds must be polar. Symmetrical molecules (CO2, BF3, CCl4, PCl5) are non-polar because the bond dipoles cancel.
Examination checklist
- Draw Lewis structures with all lone pairs, including on terminal halogen and oxygen atoms; bracket and charge ions.
- State both the electron domain geometry and the molecular geometry, and keep the bond angle consistent with the shape.
- Explain polarity with two ideas: polar bonds (electronegativity difference) and an unsymmetrical shape (dipoles do not cancel).
- Name the specific intermolecular force (London, dipole–dipole, hydrogen bonding), not just “intermolecular forces” or “van der Waals”, and say it acts between molecules.
- Link London force strength to the number of electrons.
- Draw hydrogen bonds as dashed lines from H to a lone pair on N, O or F, roughly in line with the covalent bond.
- For giant structures, state “giant” or “network” and that strong covalent (or ionic, or metallic) bonds throughout must be broken.
- Define RF making clear which distance is divided by which.
Knowledge organiser · the covalent model
| Model | Particles and attraction | Key facts and trends | Must-remember distinctions and common errors |
|---|---|---|---|
| Covalent 2.2.1–2.2.6 | Shared electron pairs attracted to two nuclei. | Single < double < triple in strength; reverse for length. Octet rule with exceptions (Be, B incomplete). Coordination bond: both electrons from one atom. | Draw all lone pairs. Polarity needs polar bonds and an unsymmetrical shape. |
| VSEPR | Electron domains repel. | 2 linear 180°; 3 trigonal planar 120°; 4 tetrahedral 109.5°; each lone pair reduces the angle by ≈ 2–3°. | Multiple bond = one domain. Electron domain geometry ≠ molecular geometry when lone pairs are present. |
| Covalent networks 2.2.7 | Covalent bonds throughout a giant structure. | Diamond: 4 bonds, sp3, hard, non-conductor. Graphite: 3 bonds, layers, delocalized e−, conducts. Graphene: single layer. C60: molecular. SiO2: each Si to 4 O, each O to 2 Si. | C60 melts by overcoming London forces; diamond by breaking covalent bonds. |
| Intermolecular forces 2.2.8–2.2.10 | Between molecules: London, dipole–induced dipole, dipole–dipole, hydrogen bonding. | Stronger IMF → higher bp, lower volatility. Solubility: “like dissolves like”. RF = dsolute/dsolvent. | H-bond needs H–N/O/F and a lone pair on N/O/F. HCN: dipole–dipole, not H-bonding. |