S2.2 The covalent model

1. The need of each atom to gain a full outer shell by sharing a pair of electrons — Sharing does give each atom a full outer shell, but a 'need' is not a force and cannot hold atoms together. The octet rule describes a tendency; the bond itself is the attraction between the shared pair and both nuclei.
2. The shared pair of electrons itself, acting as a physical link between the atoms — The shared pair is not a rod or hook. It holds the atoms together only because both nuclei are electrostatically attracted to it; the bond is that attraction.
3. The electrostatic attraction of both nuclei to a shared pair of electrons — The shared pair lies between the H and Cl nuclei and is attracted to both of them. This electrostatic attraction holds the two atoms together, and it is what a covalent bond is.
4. The electrostatic attraction between the oppositely charged ions H⁺ and Cl⁻ — HCl molecules contain no ions. The H–Cl bond is polar, with partial charges δ+ and δ−, but the electrons are shared. H⁺ and Cl⁻ form only when HCl reacts with water.

Syllabus statement S2.2.1 · Read this in Learn

1. It is longer and stronger — The triple bond is stronger, but it is shorter, not longer. The extra shared electrons do not take up extra space between the atoms; they attract both nuclei and pull them closer together (120 pm against 154 pm).
2. It is shorter and weaker — The triple bond is shorter, but it is also stronger overall (839 against 346 kJ mol⁻¹). Alkynes react readily because addition breaks only part of the multiple bond: converting C≡C into C=C needs about 839 − 614 = 225 kJ mol⁻¹, less than a whole C–C bond. The C≡C bond as a whole is not weak.
3. It is longer and weaker — The shared pairs do repel one another, but their attraction to both nuclei is the larger effect. The triple bond is shorter (120 pm) and stronger (839 kJ mol⁻¹) than the single bond (154 pm, 346 kJ mol⁻¹).
4. It is shorter and stronger — A triple bond has three shared pairs between the nuclei, against one in the single bond. More shared electrons attract both nuclei more strongly, pulling them closer (C≡C 120 pm, C–C 154 pm) and making the bond harder to break (839 against 346 kJ mol⁻¹).

Syllabus statement S2.2.2 · Read this in Learn

1. Tetrahedral, as there are four electron domains around the nitrogen atom — Tetrahedral is the electron-domain geometry, which includes the lone pair. The molecular geometry describes only the positions of the atoms, and with three H atoms around N it is trigonal pyramidal.
2. Trigonal planar, as only the three bonding pairs repel one another — The lone pair is an electron domain and repels the bonding pairs strongly. Four domains around nitrogen give a tetrahedral arrangement, so the three H atoms are pushed below the N atom into a pyramid.
3. Flat, with 90° bond angles, as the atoms are drawn in its Lewis formula on paper — A Lewis formula is drawn flat and does not show angles. In three dimensions the four domains point to the corners of a tetrahedron, so NH₃ is trigonal pyramidal with angles of about 107°.
4. Trigonal pyramidal, as the lone pair occupies one of four tetrahedral domains — The four electron domains take up a tetrahedral arrangement (electron-domain geometry). Only three are occupied by atoms, so the atoms form a pyramid with N at the apex: trigonal pyramidal, with H–N–H angles of about 107°.

Syllabus statement S2.2.4 · Read this in Learn

1. O–H — O–H is very polar (difference 3.4 − 2.2 = 1.2) and gives hydrogen bonding, but its difference is smaller than that of C–F (1.4). Polarity is set by the electronegativity difference, not by the ability to hydrogen bond.
2. O–F — O and F are the two most electronegative atoms here, but polarity depends on the difference: 4.0 − 3.4 = 0.6. Two strongly attracting atoms share the pair fairly evenly.
3. C–F — The electronegativity difference is 4.0 − 2.6 = 1.4, the largest of the four, so the shared pair is drawn most unequally, towards F.
4. P–H — P has many more protons than H, but electronegativity does not follow atomic number: both values are 2.2, so the difference is 0 and the bond is essentially non-polar.

Syllabus statement S2.2.5 · Read this in Learn

1. It contains carbon ions that move through the layers when a voltage is applied — Graphite contains only covalently bonded carbon atoms, not ions. The mobile charged particles are delocalized electrons within each layer.
2. Its atoms are held by metallic bonding, as its shiny grey appearance suggests — Graphite looks metallic but is a covalent network: each carbon atom is covalently bonded to three others. Its conduction comes from electrons delocalized within the covalent layers, not from metallic bonding.
3. Each carbon atom bonds to three others, leaving one electron per atom delocalized — In each layer every carbon atom forms three covalent bonds. Its fourth valence electron is delocalized across the layer, and these mobile electrons carry the current along the layers.
4. Its layers slide over one another easily, carrying electric charge as they move — Sliding layers explain why graphite is soft and slippery; they are held only by weak London forces. Conduction comes from the delocalized electrons within each layer.

Syllabus statement S2.2.7 · Read this in Learn

1. Ethanol splits into ions that water surrounds, but hexane does not form ions — Ethanol dissolves as intact molecules; its solution does not conduct electricity. It dissolves because it forms hydrogen bonds with water, not because it ionizes.
2. Ethanol can form hydrogen bonds with water molecules, but hexane cannot — The O–H group of ethanol forms hydrogen bonds with water, replacing the hydrogen bonds broken between water molecules. Hexane forms only London forces with water, far weaker than the hydrogen bonds it would have to break, so it does not dissolve.
3. Hexane is less dense than water, so it floats on the water rather than dissolving — Density decides which layer floats once two liquids fail to mix, not whether they mix. Ethanol is also less dense than water, yet it mixes completely.
4. Hexane molecules are too large to fit into the gaps between water molecules — Solubility depends on the attractions formed between solute and solvent, not on fitting into gaps. Sucrose molecules are larger than hexane molecules but dissolve readily because they hydrogen bond with water.

Syllabus statement S2.2.9 · Read this in Learn

1. 2.94 — This is 10.0 ÷ 3.4, the solvent distance divided by the spot distance. RF is the spot distance divided by the solvent distance, so it can never exceed 1.
2. 0.40 — This is 4.4 ÷ 11.0, with both distances measured from the bottom edge of the paper. The spot and solvent both started from the baseline, so measure from there: 3.4 ÷ 10.0 = 0.34.
3. 0.66 — This is (11.0 − 4.4) ÷ 10.0, the gap between the spot and the solvent front. Despite the name 'retardation factor', RF uses the distance the spot moved from the baseline: 3.4 ÷ 10.0 = 0.34.
4. 0.34 — Distance moved by the spot = 4.4 − 1.0 = 3.4 cm; distance moved by the solvent = 11.0 − 1.0 = 10.0 cm. RF = 3.4 ÷ 10.0 = 0.34.

Working Distance moved by spot = 4.4 cm − 1.0 cm = 3.4 cm. Distance moved by solvent = 11.0 cm − 1.0 cm = 10.0 cm. RF = 3.4 cm ÷ 10.0 cm = 0.34 (no units).

Syllabus statement S2.2.10 · Read this in Learn

1. 6 electrons, as boron forms three bonding pairs and has no lone pair — Each of boron's three valence electrons pairs with one from a chlorine atom, giving three B–Cl bonding pairs and no lone pair: 3 × 2 = 6 electrons, fewer than an octet. Each Cl has one bonding pair and three lone pairs, eight electrons.
2. 8 electrons, as one B–Cl bond must be a double bond to give boron an octet — Boron has only three valence electrons, all used in three single B–Cl bonds. The octet rule describes a tendency, not a law: in BCl₃ boron has six electrons, an incomplete octet.
3. 3 electrons, as boron has three valence electrons of its own — Three is the number of valence electrons of a free boron atom. In the molecule each one is paired with an electron from chlorine, and both electrons of every shared pair count for boron: 3 × 2 = 6.
4. 2 electrons, as boron transfers three electrons to chlorine, leaving 1s² — BCl₃ is a covalent molecule with no ions. Treating it as B³⁺ and Cl⁻ leaves boron with only its 1s² electrons, but boron shares three bonding pairs and has six electrons in its valence shell.

Working Boron has 3 valence electrons and forms one single bond to each Cl atom: 3 bonding pairs × 2 = 6 electrons, with no lone pair left. (Check: total valence electrons 3 + 3 × 7 = 24 = 3 bonding pairs (6) + 9 lone pairs on Cl (18).) Boron has 6 electrons, an incomplete octet.

Syllabus statement S2.2.1 · Read this in Learn

1. A lone pair on the aluminium atom of each unit is donated to a chlorine atom of the other unit — The aluminium atom in AlCl₃ has no lone pair: its three valence electrons are all in Al–Cl bonds. It is the acceptor. The donor is chlorine, which has three lone pairs.
2. A lone pair on a chlorine atom of each unit is donated to the aluminium atom of the other — In AlCl₃, aluminium uses all three valence electrons in bonds and has only six electrons, so it can accept a pair. A chlorine atom of the other unit has lone pairs and donates one, forming a coordination bond. Two such bonds join the units.
3. Al³⁺ ions in one unit attract Cl⁻ ions in the other unit, forming two ionic bonds — Aluminium chloride vapour consists of molecules, not ions: the Al–Cl bonds are covalent. The metal/non-metal rule for ionic bonding has exceptions, and aluminium chloride is one.
4. Each new bond forms from one electron of an aluminium atom and one of a chlorine atom — An aluminium atom in AlCl₃ has no unshared electron left to contribute, because all three of its valence electrons are already in Al–Cl bonds. Both electrons of each new bond must therefore come from chlorine.

Syllabus statement S2.2.3 · Read this in Learn

1. About 104.5°, as lone pairs repel more strongly than bonding pairs do — Oxygen has four electron domains (two bonding pairs, two lone pairs) in a tetrahedral arrangement. The two lone pairs repel more strongly than bonding pairs, squeezing the H–O–H angle below 109.5° to about 104.5°.
2. About 109.5°, as all four electron domains around O repel one another equally — The four domains are arranged tetrahedrally, but they do not repel equally. Lone pairs repel more strongly than bonding pairs, so the angle between the bonding pairs is reduced to about 104.5°.
3. About 180°, as the two bonding pairs get as far apart as possible — Oxygen also has two lone pairs, which are electron domains. With four domains the arrangement is tetrahedral, and the molecule is bent, not linear.
4. About 90°, as the H atoms are drawn at right angles in its Lewis formula — The right angle is a feature of how the Lewis formula is drawn on paper. In three dimensions the four domains around oxygen point towards the corners of a tetrahedron, giving about 104.5°.

Syllabus statement S2.2.4 · Read this in Learn

1. About 109.5°, as the two bonds of C=O count as two separate electron domains — A double bond is ONE electron domain, because both shared pairs occupy the same region between C and O. Carbon therefore has three domains, not four, and the arrangement is trigonal planar.
2. Slightly less than 120°, as the C=O double bond repels more than a single bond — Carbon has three electron domains (two C–H bonds and the C=O bond), so the arrangement is trigonal planar. The double bond holds more electron density and repels the C–H bonds more strongly, so the H–C–H angle is squeezed slightly below 120° (to about 116°).
3. Exactly 120°, as all three electron domains around carbon repel equally — The arrangement is trigonal planar, but the domains do not repel equally. The C=O double bond contains more electron density than a C–H bond, so it repels more and the H–C–H angle falls slightly below 120°.
4. About 104.5°, as oxygen's two lone pairs push the two C–H bonds together — The lone pairs on oxygen belong to oxygen, not to carbon, so they are not electron domains of the carbon atom. Only the three domains around carbon set the angle, which is close to 120°.

Syllabus statement S2.2.4 · Read this in Learn

1. Its two C=O bond dipoles are equal and point in opposite directions, so they cancel — Each C=O bond is polar (difference 0.8, δ− on O). Carbon has two electron domains, so the molecule is linear and the two equal bond dipoles point in exactly opposite directions. Their vector sum is zero.
2. Its C=O bonds are non-polar, as the electronegativity difference is below 1.8 — Any electronegativity difference makes a bond polar. A difference of 0.8 gives a clearly polar C=O bond; values near 1.8 are only a rough guide to where bonding becomes mainly ionic.
3. It carries no overall charge, so no separation of charge can exist in the molecule — Many neutral molecules have a net dipole, for example H₂O. A dipole is a separation of partial charges within a neutral molecule. CO₂ has none because its bond dipoles cancel.
4. Its carbon atom has no lone pair, and only lone pairs can produce a net dipole — Lone pairs matter only through their effect on shape. A molecule without a lone pair on its central atom can still be polar (CH₃Cl). CO₂ is non-polar because it is linear and its equal bond dipoles cancel.

Syllabus statement S2.2.6 · Read this in Learn

1. Its O–H bonds are polar, and every molecule with polar bonds has a net dipole — The bonds are polar, but polar bonds do not always give a polar molecule: in CO₂ and CCl₄ they cancel. Water is polar because its bent geometry prevents the dipoles from cancelling.
2. It is bent, so its two O–H bond dipoles do not point in opposite directions — Each O–H bond is polar (δ− on O). Oxygen's four electron domains (two of them lone pairs) make the molecule bent, about 104.5°, so the two bond dipoles add to give a resultant along the bisector of the angle.
3. Its H atoms lie at 90° to each other, as in its Lewis formula, so the dipoles add — The angle is about 104.5°, not 90°. The Lewis formula is drawn flat and does not show angles; VSEPR predicts the bent shape in three dimensions.
4. Its oxygen atom has two lone pairs, which are the only source of its net dipole — The net dipole comes from the polar O–H bonds, which do not cancel. The lone pairs matter because they make the molecule bent; they are not on their own the source of the dipole.

Syllabus statement S2.2.6 · Read this in Learn

1. Diamond has a very high melting point because of strong intermolecular forces — Diamond has no separate molecules, so there are no intermolecular forces to overcome. Each carbon atom is covalently bonded to four others, and melting requires these strong covalent bonds to be broken.
2. C₆₀ fullerene is a giant covalent network, so like diamond it dissolves in no solvent — C₆₀ is a molecular allotrope: separate cage-shaped molecules held together by London forces. It dissolves in non-polar solvents such as methylbenzene and sublimes at a much lower temperature than diamond.
3. Silicon conducts electricity because its atoms are held together by metallic bonding — Silicon has a diamond-like covalent network in which each atom is bonded to four others. It is a semiconductor, not a metal, and has no metallic bonding.
4. Graphene conducts electricity as each atom bonds to three others, leaving delocalized electrons — Graphene is a single layer of graphite: each carbon atom forms three covalent bonds in hexagonal rings, and its fourth valence electron is delocalized across the layer, so the layer conducts electricity.

Syllabus statement S2.2.7 · Read this in Learn

1. Subliming CO₂ breaks its C=O bonds, which are weaker than the Si–O bonds in SiO₂ — Subliming CO₂ does not break any covalent bonds: the vapour consists of intact CO₂ molecules. Only the weak intermolecular forces between molecules are overcome. In fact a C=O bond (804 kJ mol⁻¹) is stronger than an Si–O bond (466 kJ mol⁻¹); SiO₂ melts far higher because melting it breaks many Si–O bonds, while subliming CO₂ breaks none.
2. SiO₂ molecules are heavier than CO₂ molecules, so their London forces are stronger — There are no SiO₂ molecules. Silicon dioxide is a covalent network, and the difference of about 1800 °C is far too large to come from a small difference in London forces.
3. SiO₂ is a giant covalent network, but CO₂ is made of small separate molecules — In SiO₂ each Si atom is covalently bonded to four O atoms and each O atom to two Si atoms, so melting breaks many strong covalent bonds. CO₂ is made of separate molecules, and subliming it overcomes only weak London forces between them.
4. SiO₂ is an ionic solid of Si⁴⁺ and O²⁻ ions, like other oxides with high melting points — A high melting point does not show ionic bonding. SiO₂ is a covalent network in which each Si atom shares electrons with four O atoms; it contains no ions.

Syllabus statement S2.2.7 · Read this in Learn

1. London forces only, as I₂ is a non-polar molecule with no dipole — I₂ has no permanent dipole, but the permanent dipole of HCl can induce one in it. So dipole–induced dipole forces act as well as London forces.
2. London and dipole–induced dipole forces, as HCl polarizes the I₂ molecule — London forces act between all molecules. HCl also has a permanent dipole, which induces a temporary dipole in the non-polar I₂ molecule, giving a dipole–induced dipole attraction.
3. London and dipole–dipole forces, as the HCl molecules are polar — Dipole–dipole forces need a permanent dipole on both molecules. I₂ is non-polar, so the force between HCl and I₂ is dipole–induced dipole, not dipole–dipole.
4. London forces and hydrogen bonds, as HCl has H bonded to electronegative Cl — Cl is electronegative, but hydrogen bonds need H bonded to N, O or F and an N, O or F atom with a lone pair on the other molecule; HCl and I₂ provide neither. The permanent dipole of HCl induces a dipole in I₂, so dipole–induced dipole forces act alongside London forces.

Syllabus statement S2.2.8 · Read this in Learn

1. An attraction between the H of an O–H group and a lone pair on O in another molecule — A hydrogen bond forms when hydrogen, covalently bonded to an electronegative atom (here O), is attracted to a lone pair on an electronegative atom (O) of a neighbouring molecule.
2. The covalent bond between the O atom and the H atom within one methanol molecule — That is a covalent bond inside the molecule. A hydrogen bond is an intermolecular attraction between the δ+ H of one molecule and an O atom of another.
3. An attraction between the H atoms of one molecule and the H atoms of another — Hydrogen atoms bonded to O are δ+, so two of them repel. A hydrogen bond links an H atom to an electronegative atom with a lone pair.
4. An attraction between an H atom of the CH₃ group and an O atom in a neighbouring molecule — Only an H atom bonded to N, O or F takes part. The H atoms of the CH₃ group are bonded to carbon and are not δ+ enough to form hydrogen bonds.

Syllabus statement S2.2.8 · Read this in Learn

1. They include hydrogen bonds, as each propanone molecule contains H and O atoms — All the H atoms in propanone are bonded to carbon, so none can form hydrogen bonds with neighbouring molecules. Hydrogen bonding needs an H atom bonded to N, O or F.
2. They include no London forces, as propanone molecules are polar — London forces act between all molecules, including polar ones. Propanone has London forces as well as dipole–dipole forces.
3. They include London forces, which are the only kind of van der Waals force — London forces are present, but they are not the only kind of van der Waals force. The term includes London, dipole–induced dipole and dipole–dipole forces.
4. They include dipole–dipole forces, which count as van der Waals forces — Propanone is polar: the C=O bond is polar and the bond dipoles do not cancel. So dipole–dipole forces act between its molecules, alongside London forces, and both are included in the term van der Waals forces.

Syllabus statement S2.2.8 · Read this in Learn

1. propane < ethanal < ethanol — With comparable molar masses the type of intermolecular force decides. Propane has only London forces (−42 °C); ethanal is polar and adds dipole–dipole forces (20 °C); ethanol forms hydrogen bonds (78 °C).
2. propane < ethanol < ethanal — Boiling does not break the covalent bonds (such as C=O) inside the molecules; it overcomes the forces between molecules. Ethanol's hydrogen bonding is stronger than ethanal's dipole–dipole forces, so ethanol boils higher.
3. ethanal < propane < ethanol — Ranking by M_r puts ethanal (44.06) below propane (44.11), but the difference is tiny. Ethanal's dipole–dipole forces are stronger than propane's London forces, so it boils at 20 °C against −42 °C.
4. ethanal < ethanol < propane — Boiling point rises with chain length within one homologous series, but these compounds belong to different series. Propane, with only London forces, boils lowest (−42 °C) despite its three carbon atoms.

Syllabus statement S2.2.9 · Read this in Learn

1. Liquid sulfur — Liquid sulfur consists of neutral S₈ molecules with their electrons held in localized bonds and lone pairs. Melting lets the molecules move, but they carry no charge, so there is no current. Molten ionic compounds conduct because their ions become mobile; molecular substances do not.
2. Diamond crystal — Diamond and graphite are both carbon, but their structures differ. In diamond every valence electron is used in four localized C–C bonds, so there are no mobile electrons and it does not conduct.
3. Solid graphite — In graphite each carbon atom is covalently bonded to three others, and its fourth valence electron is delocalized across the layer. These mobile electrons carry a current.
4. Ethanol in water — Ethanol dissolves as intact molecules, held among water molecules by hydrogen bonds. It does not ionize, so the solution contains no mobile ions and does not conduct.

Syllabus statement S2.2.9 · Read this in Learn

1. Relative to the mobile phase, X is attracted more strongly to the stationary phase than Y is — A component moves more slowly when it is attracted more strongly to the stationary phase relative to the mobile phase. X travels only 0.20 of the solvent's distance, so this balance favours the stationary phase more for X than for Y.
2. X has larger molecules than Y, so the fibres of the paper hold X back more than Y — The paper is not a sieve. Separation depends on the relative attractions of each component to the stationary and mobile phases, not on molecular size.
3. X would also reach an RF of 0.75 if the solvent were allowed to run for a longer time — Running longer increases both the spot's distance and the solvent's distance in proportion, so X's RF stays 0.20. That is why RF values can be used to identify components.
4. X is less soluble in the solvent than Y, and the paper plays no part in the separation — The stationary phase is half of the separation. RF depends on the balance of attractions to both phases, so the paper's attraction for each component matters as much as solubility in the solvent.

Syllabus statement S2.2.10 · Read this in Learn

1. A lattice of positive Si ions in a sea of delocalized electrons, as in metals — Silicon looks shiny and conducts a little, but it contains no positive ions in a sea of electrons. It is a covalent network, and its slight conductivity is that of a semiconductor, far below that of a metal.
2. A network in which each Si atom is covalently bonded to four others, as in diamond — Each Si atom forms four covalent bonds to other Si atoms in a tetrahedral three-dimensional network, as C does in diamond. Melting breaks many strong Si–Si bonds, hence 1414 °C, and no solvent can separate the atoms.
3. Separate silicon molecules held together by strong intermolecular forces — A substance of separate molecules held by intermolecular forces melts far below 1414 °C. Silicon has no molecules: melting it breaks covalent Si–Si bonds, which is why its melting point is so high.
4. Flat layers in which each Si atom is bonded to three others, as in graphite — Silicon has no delocalized electron per atom: each Si atom bonds to four others, as in diamond, not three as in graphite. Its slight conductivity comes from its being a semiconductor.

Syllabus statement S2.2.7 · Read this in Learn

1. C₆₀ consists of separate molecules held by London forces; diamond and graphite are covalent networks — C₆₀ is a molecular allotrope: methylbenzene molecules can surround separate C₆₀ molecules, overcoming only the London forces between them. Diamond and graphite are covalent networks; dissolving them would mean breaking strong C–C covalent bonds.
2. C₆₀ splits into ions in methylbenzene, but diamond and graphite cannot form ions — Neither C₆₀ nor methylbenzene forms ions. C₆₀ dissolves as intact molecules, held among the solvent molecules by London forces, and the solution does not conduct.
3. Diamond and graphite are held together by strong intermolecular forces between their atoms — The atoms of diamond, and the atoms within each graphite layer, are joined by covalent bonds, not intermolecular forces. Neither dissolves because every atom is covalently bonded into a network that a solvent cannot break up.
4. Methylbenzene breaks the C–C bonds in C₆₀, which are weaker than those in diamond — Dissolving does not break covalent bonds: C₆₀ dissolves as intact C₆₀ molecules. The C–C bonds in all three allotropes are strong; the difference is between separate molecules and covalent networks.

Syllabus statement S2.2.7 · Read this in Learn

1. 2 lone pairs, as 4 of the 12 valence electrons remain after the C–H and C=O bonds — Methanal has 4 + 2 × 1 + 6 = 12 valence electrons. Two C–H bonds and a C=O double bond use 8, leaving 4, which form two lone pairs on O. O then has 4 + 4 = 8 electrons and C has 8.
2. 4 lone pairs, as O gains two electrons from C and is an O²⁻ ion with an octet — Methanal is a covalent molecule with no ions. O shares two pairs with C in the C=O double bond and holds two lone pairs: 4 + 4 = 8 electrons. Four lone pairs as well as the C=O bond would give O twelve electrons.
3. 3 lone pairs, as the C=O double bond is one shared pair, so O needs three more — A double bond is two shared pairs, four electrons. O has four electrons from C=O, so two lone pairs complete its octet; three would give it ten and use more electrons than the molecule has.
4. 0 lone pairs, as a Lewis formula shows only the bonding pairs between the atoms — A Lewis formula shows all the valence electrons, lone pairs included. After the bonds, 4 of methanal's 12 valence electrons remain, as two lone pairs on O.

Working Valence electrons: C 4 + 2 × H 1 + O 6 = 12. Carbon forms two C–H bonds (4 electrons) and a C=O double bond (4 electrons), 8 electrons in bonds. The remaining 12 − 8 = 4 electrons form 2 lone pairs on O. Check: O has 4 (C=O) + 4 (lone pairs) = 8; C has 2 × 2 + 4 = 8; each H has 2.

Syllabus statement S2.2.1 · Read this in Learn

1. O=O is weaker, as the two shared pairs between the nuclei repel each other strongly — The shared pairs do repel one another, but the shorter bond length shows that the attraction of both nuclei for the extra pair is the larger effect. A shorter bond between the same two atoms is a stronger bond: O=O is about 498 kJ mol⁻¹, O–O about 144 kJ mol⁻¹.
2. O=O is weaker, as oxygen reacts so readily, for example in combustion reactions — Reactivity does not measure bond strength. Combustion releases energy because the C=O and O–H bonds formed are stronger than the bonds broken, not because O=O is weak. The O=O bond (about 498 kJ mol⁻¹) is much stronger than O–O (about 144 kJ mol⁻¹), as its shorter length suggests.
3. O=O is stronger, as two shared pairs attract both nuclei more than one — Two shared pairs between the nuclei mean more negative charge attracting both nuclei, so they are pulled closer (121 pm against 148 pm) and more energy is needed to separate them: O=O about 498 kJ mol⁻¹ against O–O about 144 kJ mol⁻¹. Shorter and stronger go together.
4. The two bonds have the same strength, as both are between two oxygen atoms — The atoms are the same but the number of shared pairs is not. The O=O bond is shorter because two pairs attract the nuclei more strongly, and it is correspondingly stronger: about 498 kJ mol⁻¹ against about 144 kJ mol⁻¹ for O–O.

Syllabus statement S2.2.2 · Read this in Learn

1. The bond formed from the lone pair on nitrogen is longer and weaker than the other three N–H bonds — Where the electrons came from leaves no trace: electrons are indistinguishable. The arrow sometimes drawn for a coordination bond only records its origin. All four N–H bonds in NH₄⁺ are equivalent.
2. The new bond formed from a pair of electrons supplied by the H⁺ ion to nitrogen — A hydrogen ion is a bare proton with no electrons, so it cannot supply a pair. The donor is the atom with the lone pair, nitrogen; H⁺ is the acceptor.
3. The new bond formed from one electron supplied by nitrogen and one electron supplied by H⁺ — H⁺ has no electron to contribute, and the nitrogen atom in NH₃ has no unpaired electron left, only a lone pair. Both electrons of the new bond come from nitrogen's lone pair: it is a coordination bond.
4. All four N–H bonds are identical, although one formed from the lone pair on nitrogen — H⁺ has no electrons, so the new bond is a coordination bond: both electrons come from the lone pair on nitrogen. Once formed, a coordination bond is an ordinary covalent bond, and all four N–H bonds in NH₄⁺ have the same length and strength.

Syllabus statement S2.2.3 · Read this in Learn

1. Each hydrogen atom has given its electron to nitrogen, so N³⁻ and H⁺ ions attract — No electron is transferred: the difference of 0.8 gives unequal sharing, not ions. Ammonia is a molecule with three polar covalent bonds, each with δ− on nitrogen and δ+ on hydrogen.
2. The pair is shared equally, because a difference of 0.8 is too small to make a bond polar — Any difference in electronegativity makes the sharing unequal. A difference of 0.8 gives a clearly polar bond; the N–H bond dipoles are what make ammonia a polar molecule.
3. Both atoms gain a share of the pair, so nitrogen and hydrogen are both δ− — Sharing does not make both atoms negative. The pair is pulled towards nitrogen, so nitrogen has more than its equal share (δ−) and hydrogen less (δ+); the two partial charges are equal and opposite.
4. The shared pair lies closer to nitrogen, so nitrogen is δ− and hydrogen is δ+ — Nitrogen (3.0) attracts the shared pair more strongly than hydrogen (2.2). Electron density shifts towards nitrogen, which carries a partial negative charge, δ−, leaving hydrogen with an equal partial positive charge, δ+.

Syllabus statement S2.2.5 · Read this in Learn

1. The lone pair on nitrogen is missing, so nitrogen is shown with only six electrons — Nitrogen has five valence electrons. Three are in the shared pairs, so two remain as a lone pair, which a Lewis formula must show. With the lone pair, nitrogen has 3 × 2 + 2 = 8 electrons; the diagram shows only six.
2. Nothing: nitrogen forms three bonds, and its three shared pairs complete its valence shell — Three shared pairs give nitrogen only six electrons. A Lewis formula shows all valence electrons, bonding and non-bonding: nitrogen's fifth valence electron and one more form a lone pair, missing here.
3. Each hydrogen atom should also have three lone pairs, so that every atom has eight electrons — Hydrogen has one valence electron and a shell that holds only two, so one shared pair completes it. The octet applies to period 2 atoms such as nitrogen; the missing electrons are nitrogen's lone pair.
4. Nitrogen is shown with six electrons, but it should have five, its number of valence electrons — A free nitrogen atom has five valence electrons, but in a molecule each shared pair counts towards both bonded atoms. Nitrogen's valence shell in NH₃ holds three shared pairs and one lone pair: eight electrons. The fault is the missing lone pair, not too many electrons.

Syllabus statement S2.2.1 · Read this in Learn

1. Cl₂O is linear, because only the two bonding pairs on oxygen affect its shape — Lone pairs are electron domains and repel the bonding pairs. The two lone pairs on oxygen push the two Cl atoms out of a straight line, so Cl₂O is bent; the drawing is flat only because Lewis formulas do not show shape.
2. Cl₂O is tetrahedral, because there are four electron domains around the oxygen atom — Four domains give a tetrahedral electron-domain geometry, but the molecular geometry is named from the positions of the atoms only. With two of the four domains being lone pairs, the three atoms form a bent shape.
3. Cl₂O is linear, because its Lewis formula shows the Cl–O–Cl bond angle as 180° — A Lewis formula shows which atoms are bonded and where the electron pairs are, not the angles between the bonds. Oxygen has four electron domains, two of them lone pairs, so the Cl–O–Cl angle is far below 180° (roughly 110°) and the molecule is bent.
4. Cl₂O is bent, because the two lone pairs on oxygen repel its two bonds out of a straight line — Oxygen has two bonding pairs and two lone pairs: four electron domains, tetrahedral electron-domain geometry. The two Cl atoms occupy two corners of that tetrahedron, so the three atoms do not lie in a straight line: the molecular geometry is bent. Beryllium has only two domains and no lone pairs, so BeCl₂ is linear.

Syllabus statement S2.2.4 · Read this in Learn

1. The molecule is polar, but only because of the C–H bond, since the three C–Cl dipoles cancel each other — Three equal dipoles cancel only when they lie in one plane at 120°, as in BCl₃. Here the drawing shows all three C–Cl bonds pointing below the carbon atom, so their components along the H–C axis add rather than cancel. The net dipole comes mainly from the C–Cl bonds.
2. The molecule is non-polar, because it carries no overall charge, so its charges must balance out — A dipole is a separation of partial charges within a neutral molecule, not a net charge. CHCl₃ is neutral overall but its δ− is concentrated at the chlorine end and its δ+ at the hydrogen end, so it is polar.
3. The molecule is polar, because its three C–Cl dipoles add to give a net dipole towards Cl — The three C–Cl dipoles all point downwards and to one side of the carbon atom, so their vector sum does not vanish: it points along the H–C axis towards the chlorine atoms. The small C–H dipole (H δ+) points the same way, so CHCl₃ has a net dipole with its δ− end at the chlorine side.
4. The molecule is non-polar, because its central carbon atom does not carry a lone pair of electrons — Lone pairs matter only through the geometry they produce. Polarity is decided by whether the bond dipoles cancel: with three C–Cl dipoles on one side of the carbon atom and a C–H bond on the other, they do not, so CHCl₃ is polar.

Syllabus statement S2.2.6 · Read this in Learn

1. London forces between HI molecules are stronger than those between HCl molecules, because HI has more electrons — From HCl to HI the boiling point rises steadily with molar mass. The molecules become less polar down the group, so the rise must come from London (dispersion) forces, which strengthen as the number of electrons in the molecule increases.
2. The rise in boiling point from HCl to HI is due to dipole–dipole forces, which strengthen as the molecules get larger — Every molecule, polar or not, has London forces. From HCl to HI the electronegativity difference falls, so the dipole–dipole forces get weaker, not stronger; the rising boiling point comes from stronger London forces in the larger molecules.
3. Hydrogen bonds between HCl molecules are weaker than those between HF molecules, so HCl boils lower — HCl does not hydrogen bond at all: hydrogen bonding needs H bonded to N, O or F. HCl molecules attract one another by dipole–dipole and London forces only, which is why HCl lies on the same steady trend as HBr and HI while HF lies far above it.
4. HF has the highest boiling point because the H–F bond is a hydrogen bond, the strongest of the four bonds — The H–F bond is a covalent bond and is not broken on boiling. The hydrogen bonds are between HF molecules, from the H of one to a lone pair on the F of another; these strong intermolecular forces lift HF far above the trend of the other three.

Syllabus statement S2.2.9 · Read this in Learn

1. 1.33 — This is 8.0 ÷ 6.0, the solvent distance divided by the spot distance. RF is the distance moved by the spot divided by the distance moved by the solvent, so it can never exceed 1: 6.0 ÷ 8.0 = 0.75.
2. 0.75 — Both distances are measured from the baseline (1.0 cm): spot X moved 7.0 − 1.0 = 6.0 cm and the solvent moved 9.0 − 1.0 = 8.0 cm. RF = 6.0 ÷ 8.0 = 0.75.
3. 0.78 — This is 7.0 ÷ 9.0, with both readings taken from the bottom edge of the paper. The spot and the solvent both started at the baseline, 1.0 cm up, so subtract it from each reading: 6.0 ÷ 8.0 = 0.75.
4. 0.25 — This is (9.0 − 7.0) ÷ 8.0, the gap between the spot and the solvent front divided by the solvent distance. RF uses the distance the spot travelled from the baseline: 6.0 ÷ 8.0 = 0.75.

Working From the scale: baseline at 1.0 cm, solvent front at 9.0 cm, centre of spot X at 7.0 cm. Distance moved by X = 7.0 − 1.0 = 6.0 cm. Distance moved by solvent = 9.0 − 1.0 = 8.0 cm. RF = 6.0 cm ÷ 8.0 cm = 0.75 (no units).

Syllabus statement S2.2.10 · Read this in Learn