2026 and 2025 paper questions

The first and second ionization energies are low, so the first two electrons are valence electrons that are easy to remove. A very large jump at the third value means the third electron comes from the inner shell, which is close to the nucleus and strongly held. Two valence electrons means Group 2 (II-A), so B.

Group 1 would show the big jump between the first and second values, and Group 17 and Group 18 have no early small values at all, since every ionization energy of these elements is already large.

Helium is monatomic, so its molar mass is $4\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and $8\mathrm{g}$ is $8/4=2$ mol. Number of atoms $=2\times 6.02\times 10^{23}=1.204\times 10^{24}$, which is B.

Option D ($1.204\times 10^{23}$) is the count for only $0.2$ mol, ten times too few, which comes from forgetting to divide by $4\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ correctly (using $40$). Options A and C are smaller still.

Use the electronegativity difference (Table 3.1 values). The bond is ionic if the difference is higher than 1.8, polar covalent between 0.4 and 1.8, and nonpolar covalent if 0.4 or lower.

Na is a metal of very low electronegativity and O is highly electronegative, so Na loses its electron to O ($\mathrm{Na}{}^{+}$ and $\mathrm{O}{}^{2-}$). The tempting wrong answer is B (B and Cl), because $\mathrm{BCl}{}_3$ looks like a metal chloride, but boron is a non-metal and the difference of 1.12 is below 1.8, so $\mathrm{BCl}{}_3$ is covalent.

The Maxwell-Boltzmann (Boltzmann distribution) curve plots the number of molecules against their kinetic energy. Every molecule of the sample lies somewhere under the curve, so the total area under the curve is the total number of molecules. This area stays the same when the temperature changes; only the shape of the curve changes (Figure 7.2b of the book).

Temperature (A) is fixed for one curve and appears as the label of the curve, not as its area. Pressure (B) is not plotted. Activation energy (D) is a position on the energy axis, and only the small shaded area beyond $E_a$ gives the molecules able to react, not the whole area.

The bond energy of the hydrogen halides falls from HF to HI as the electronegativity difference falls (Table 3.2): H–F $562$, H–Cl $431$, H–Br $366$, H–I $299\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The correct order is $\mathrm{H}-\mathrm{F}>\mathrm{H}-\mathrm{Cl}>\mathrm{H}-\mathrm{Br}>\mathrm{H}-\mathrm{I}$.

Statements B, C and D agree with this order. Statement A, $\mathrm{H}-\mathrm{Cl}>\mathrm{H}-\mathrm{F}$, reverses it ($431$ is not greater than $562$), so A is false. The tempting wrong choice is B: $\mathrm{H}-\mathrm{F}>\mathrm{H}-\mathrm{Cl}$ looks surprising because fluorine is so small, but the large electronegativity difference (1.8 against 0.9) makes H–F the strongest bond. The weak F–F bond of the halogens (lone-pair repulsion) is a different case and does not apply to H–F.

The rate of a reaction depends on the temperature (A: more molecules with energy at least $E_a$), on the concentration of the reactants (B: more collisions per second) and on a catalyst (C: it lowers $E_a$). The colour of the reactants is only a physical property and has nothing to do with collision frequency, energy or orientation, so it does not affect the rate.

The surface area of a solid reactant is another factor (powdered zinc reacts faster than a lump), but colour is not.

At STP one mole of any gas occupies $22.4\mathrm{d}{\mathrm{m}}^3$, so $67.2\mathrm{d}{\mathrm{m}}^3$ is $67.2/22.4=3$ mol. Number of molecules $=3\times 6.02\times 10^{23}=1.806\times 10^{24}$, which is C.

Option A ($6.02\times 10^{24}$) would be 10 mol and B ($1.204\times 10^{24}$) would be 2 mol. Helium is monatomic, so molecules and atoms are the same here.

Bromine ($Z=35$) and potassium ($Z=19$) are in the same period (period 4). Going from K to Br the nuclear charge rises from $+19$ to $+35$, while the extra electrons go into the same shell. The attraction between the nucleus and an incoming electron therefore becomes much stronger, so more energy is released when Br gains an electron. In the book's Figure 10.7 the values are K $-48$ and Br $-325$. The book gives nuclear charge as the main reason why electron affinity increases from left to right.

Shielding (option A) is wrong because the extra electrons are added to the same shell, so the shielding by inner shells stays about the same across the period and cannot explain the large difference. Atomic size (option B) is the tempting choice: Br is smaller than K, but the smaller size is itself a result of the greater nuclear charge, which is the main cause. Density (option D) plays no part in electron affinity.

An exothermic reaction gives out heat to the surroundings, so the enthalpy of the products is lower than that of the reactants. $\Delta H=H_{\text{products}}-H_{\text{reactants}}$ is therefore negative. A positive $\Delta H$ (A) belongs to an endothermic reaction, where heat is absorbed.

The energy cycle for dissolving gives $\Delta H_{\text{solution}}=\Delta H_{\text{reverse lattice}}+\Delta H_{\text{hydration}}$.

Option B ($-3$) comes from getting the sign backwards. Option D ($-1569$) adds the two sizes with negative signs and C ($+1569$) adds them with positive signs; but the lattice must be broken (energy absorbed, $+786$) and the ions hydrated (energy released, $-783$), so the two terms nearly cancel.

One mole of any gas occupies $22.414\mathrm{d}{\mathrm{m}}^3$ at STP, because the volume of a gas depends on the number of molecules and not on their kind. This is the molar volume, so the answer is A.

Liquids and solids do not follow this rule: the volume of one mole of a liquid or solid is different for each substance, since it depends on the density of the substance.

The forward reaction of $\mathrm{N}{}_2+3\mathrm{H}{}_2\rightleftharpoons 2\mathrm{NH}{}_3+\mathrm{\Delta }$ gives out heat, so it is exothermic. Raising the temperature adds heat, and the system opposes this by shifting to the left, the endothermic direction, so the forward direction is not favoured.

The others all push the equilibrium forward: higher pressure favours the side with fewer gas molecules (4 mol to 2 mol), extra $\mathrm{H}{}_2$ is used up, and removing $\mathrm{NH}{}_3$ makes more of it form.

A salt gives an acidic solution when it comes from a strong acid and a weak base, because the cation of the weak base hydrolyses and produces $\mathrm{H}{}^{+}$. $\mathrm{CuSO}{}_4$ is the salt of $\mathrm{H}{}_2\mathrm{SO}{}_4$ (strong) and $\mathrm{Cu}(\mathrm{OH}){}_2$ (weak), so $\mathrm{Cu}{}^{2+}$ reacts with water.

$\mathrm{Na}{}_2\mathrm{SO}{}_4$ and $\mathrm{KNO}{}_3$ are salts of a strong acid and a strong base, so they do not hydrolyse and $\mathrm{pH}=7$. $\mathrm{Na}{}_2\mathrm{CO}{}_3$ is a salt of a weak acid and a strong base, so it is basic ($\mathrm{pH}>7$).

Propane is an alkane. Its C-H and C-C bonds are non-polar and strong sigma bonds, so alkanes do not react with polar reagents such as acids, bases, oxidizing and reducing agents. Concentrated $\mathrm{HBr}$ is a polar reagent, so there is no reaction and propane remains unchanged (option D).

2-bromopropane (A) is the tempting option, but it is the major product when propene, not propane, adds $\mathrm{HBr}$ by Markovnikov's rule. Propyne (B) would need loss of hydrogen, and 1-bromopropane (C) is only the minor product from propene.

The highest shell is $n=3$, so the element is in period 3. The valence shell is $3s^{2}3p^2$, which holds $2+2=4$ electrons. The last electron enters a p sub-shell, so it is a p-block element and its group number is the number of valence electrons plus $10$, that is $4+10=14$ (group IVA). The element is silicon.

Group 13 (option C) is the tempting wrong choice: it would need three valence electrons ($3s^{2}3p^1$). Groups 1 and 2 (options A and B) are s-block groups and would have only $s$ electrons in the valence shell.

Tollens' reagent is a mild oxidising agent. It oxidises an aldehyde to the carboxylate ion and is itself reduced to silver, which coats the tube as a silver mirror. Ethanal ($\mathrm{CH}{}_3\mathrm{CHO}$) is the only aldehyde in the list.

Propanone is a ketone and is not oxidised by Tollens' reagent. Ethanol is an alcohol and propanoic acid is already an acid, so neither gives a mirror.

Peroxyacetyl nitrate (PAN) is a secondary pollutant of photochemical smog. In sunlight $\mathrm{NO}{}_2$ splits into NO and an oxygen atom, the oxygen atom reacts with unburned hydrocarbons (volatile organic compounds) to give peroxyacyl radicals $\mathrm{RCO}{}_3$, and these react with NO to give PAN, $\mathrm{R}-\mathrm{CO}{}_2\mathrm{ONO}{}_2$. PAN contains carbon, which comes from the hydrocarbons.

Oxides of sulphur (option A) are linked with acid rain, not PAN. Hydrogen sulphide and oxides of phosphorus (options B and C) cannot supply the carbon-containing acyl group of PAN.

Free radical substitution is the characteristic reaction of alkanes, for example chlorination of ethane in ultraviolet light, where a halogen free radical replaces a hydrogen atom. Alkanes are non-polar and have only sigma bonds, so they do not react with ionic or polar reagents, but they do react with free radicals.

Alkenes (A) mainly undergo electrophilic addition at the $\pi$ bond. Alcohols (B) and carboxylic acids (D) react through their polar functional groups, by substitution at the $-\mathrm{OH}$ group, oxidation or esterification, not by a radical chain.

In $\mathrm{CH}{}_4$ the carbon atom has 4 valence electrons and each hydrogen atom adds 1, so there are $4+4=8$ electrons, that is 4 electron pairs. All four are bonding pairs (AX${}_4$, no lone pair). To keep the repulsion between the pairs as small as possible they point to the corners of a tetrahedron, so the shape is tetrahedral with H–C–H angles of $109.5^{\circ }$.

Option B (planar) is the tempting wrong answer because $\mathrm{CH}{}_4$ is drawn flat on paper with four bonds at $90^{\circ }$. A flat arrangement would give bond angles of only $90^{\circ }$ and more repulsion; the real angle is $109.5^{\circ }$ in three dimensions. Linear (A) and bent (D) need two bonding pairs, as in $\mathrm{BeCl}{}_2$ and $\mathrm{H}{}_2\mathrm{O}$.

Concentrated sulphuric acid is an oxidising agent, and it is reduced further when it reacts with a more powerful reducing agent. The reducing power of the halide ions increases down the group, $\mathrm{F}{}^{-}<\mathrm{Cl}{}^{-}<\mathrm{Br}{}^{-}<\mathrm{I}{}^{-}$. The iodide ion is the strongest reducing agent, so it reduces the sulphur from $+6$ in $\mathrm{H}{}_2\mathrm{SO}{}_4$ all the way to $-2$ in $\mathrm{H}{}_2\mathrm{S}$.

Bromide (option C) is the tempting choice, but it is a weaker reducing agent and only reduces the acid to $\mathrm{SO}{}_2$ ($2\mathrm{HBr}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Br}{}_2+\mathrm{SO}{}_2+2\mathrm{H}{}_2\mathrm{O}$, similar to the book's reaction of HI and $\mathrm{H}{}_2\mathrm{SO}{}_4$ to give $\mathrm{SO}{}_2$). Fluoride and chloride (options A and B) are not oxidised by the acid at all, so they cannot reduce it.

A halogen can displace another halogen from its halide only if it is the stronger oxidizing agent. The oxidizing power falls down the group: $\mathrm{F}{}_2>\mathrm{Cl}{}_2>\mathrm{Br}{}_2>\mathrm{I}{}_2$.

Fluorine is stronger than bromine, so $\mathrm{F}{}_2$ oxidizes $\mathrm{Br}{}^{-}$ to $\mathrm{Br}{}_2$ (option C). The other reactions go the wrong way, because a weaker halogen cannot oxidize the halide ion of a stronger one: $\mathrm{Br}{}_2$ and $\mathrm{Cl}{}_2$ cannot oxidize $\mathrm{F}{}^{-}$ (A and B), and $\mathrm{I}{}_2$ cannot oxidize $\mathrm{Br}{}^{-}$ (D).

In \ce{CH3-CH3 + Cl. -> CH3-CH2. + HCl} a radical (chlorine) reacts with a molecule (ethane) and a new radical (ethyl) is formed. A step that uses one radical and produces another keeps the chain going, so this is propagation.

Initiation (A) makes radicals from a molecule, \ce{Cl2 -> 2Cl.}. Termination (D) combines two radicals into a molecule. Substitution (C) names the overall reaction, not a step of the mechanism.

\ce{CH3COONa} is the salt of a weak acid (\ce{CH3COOH}) and a strong base (\ce{NaOH}). The acetate ion is a strong conjugate base and reacts with water, producing \ce{OH-} ions, so the solution is basic (\mathrm{pH}>7).

\ce{Na+} is the ion of a strong base and does not react with water. The solution is not neutral, because the acetate ion does hydrolyse.

Ethyl bromide warmed with \ce{NaOH} in ethanol (a strong base in an alcoholic solvent) undergoes \beta-elimination. \ce{HBr} is removed from adjacent carbons and ethene forms: \ce{CH3CH2Br + NaOH ->[ethanol] CH2=CH2 + NaBr + H2O}.

The substitution product, ethanol, forms with aqueous \ce{NaOH}, not with ethanolic \ce{NaOH}. Sodium ethoxide (B) and ethanal (C) are not formed in this reaction, and vinyl bromide (D) would need \ce{Br2} to be lost from ethene, not added.

A carboxylic acid is reduced to a primary alcohol by lithium aluminium hydride: \ce{RCOOH + 4[H] ->[LiAlH4] RCH2OH + H2O}. \ce{NaBH4} is too mild for a carboxylic acid, which is why \ce{LiAlH4} is the reagent (option C).

\ce{KMnO4} (A) and \ce{K2Cr2O7} (B) are oxidizing agents, so they would oxidise, not reduce. \ce{H2/Pt} (D) hydrogenates a \ce{C=C} double bond.

Amines are derivatives of ammonia with a lone pair of electrons on the nitrogen atom. They can donate this pair to \ce{H+} or to another electron-deficient centre, so they are basic (and nucleophilic) in nature. For example \ce{CH3NH2 + HCl -> CH3NH3+ + Cl-}.

Acidic is the tempting wrong choice only because of the \ce{-NH2} group's hydrogen atoms; but an amine does not release \ce{H+} in water, it accepts it. Amines are organic compounds, so 'metallic' makes no sense, and the lone pair makes them far from neutral.

Ethanol is oxidised, first to ethanal and then to ethanoic acid, by acidified \ce{K2Cr2O7} (or \ce{KMnO4}). The carbon gains oxygen and loses hydrogen, and the number of its bonds to oxygen increases, so the change is oxidation: \ce{CH3CH2OH ->[O] CH3CHO ->[O] CH3COOH}.

Reduction (D) is the reverse change, adding hydrogen or losing oxygen. Elimination (A) removes water to give an alkene, and condensation (C) joins two molecules with loss of a small molecule.

Silver mirror (Tollens') needs an aldehyde; the iodoform test needs the \ce{CH3-CO-} group. Ethanal, \ce{CH3CHO}, is the only compound that has both features.

Propanal is an aldehyde but has no \ce{CH3-CO-} (it is \ce{CH3CH2-CHO}), so it gives no iodoform. Propanone has \ce{CH3-CO-} but is a ketone, so no mirror. Ethanol gives iodoform only (it is oxidised to ethanal by the reagent) but no silver mirror.

\ce{NaBH4} supplies hydride ions that add to the \ce{C=O} group. A ketone gives a secondary alcohol: propanone is reduced to 2-propanol.

1-Propanol (option A) would come from an aldehyde, propanal, because an aldehyde gives a primary alcohol. \ce{NaBH4} is a reducing agent, so an acid or aldehyde cannot be the product.

2,4-DNPH gives a yellow or orange precipitate with any aldehyde or ketone (the \ce{C=O} group) and nothing with alcohols or alkyl halides. In ethanal and ethanol, ethanal gives the precipitate and ethanol does not, so DNPH distinguishes them.

1-Propanol and 2-propanol (B), and ethyl chloride and ethyl bromide (D), contain no \ce{C=O}, so both members of each pair give no precipitate.

London (dispersion) forces are the momentary attractions between an instantaneous dipole and an induced dipole. They act between all molecules, polar and non-polar, so they occur in covalent molecules. They are the only intermolecular force between non-polar molecules such as \ce{Cl2}, \ce{Br2} and \ce{I2}.

Option D (only non-polar) is the tempting wrong answer: London forces are the only force in non-polar molecules, but polar molecules have them as well. Ionic compounds (B) are held by ionic bonds, not by London forces between molecules.

For a first order reaction the half-life is independent of the initial concentration and is t_{1/2}=0.693/k (book section 7.2.3, \ln2=0.693).

Option B, t_{1/2}=[A^\circ]/2k, is the zero order half-life. Option C, t_{1/2}=1/k[A^\circ], is the second order half-life. Option D contains the initial concentration, but a first order half-life does not depend on it, and its numerical factor is not 0.693.

Iodine and sodium hydroxide react with a compound containing the \ce{CH3-CO-} group to give a pale yellow precipitate of **tri-iodomethane (iodoform), \ce{CHI3}**.

\ce{CI4}, \ce{CH2I2} and \ce{CH3I} are not the product, because the methyl group is fully iodinated to \ce{CI3-CO-R} and the \ce{C-C} bond then breaks, so \ce{CHI3} (one H left) is formed.

Fehling's solution is reduced to red \ce{Cu2O} only by an aldehyde. \ce{CH3CH2CHO} (propanal) is the aldehyde.

Ethanol, ethanoic acid and chloroethane are not oxidised by Fehling's solution (a mild oxidising agent), so no red precipitate forms.

\ce{^{29}Cu^2+} (1 mark)

\ce{^{33}As} (1 mark)

Arsenic has 33 electrons. After 3d^{10} the remaining 3 electrons go into 4p, one in each of the three orbitals (Hund's rule).

\ce{^{35}Br-} (1 mark)

Neutral bromine has 35 electrons, 4p^5 at the end. The charge -1 adds one electron to the outermost shell, giving 36 electrons and a full 4p sub-shell.

Metal chlorides such as \ce{NaCl} and \ce{MgCl2} are ionic and simply dissolve in water to give neutral solutions (pH 7). The covalent chlorides react with water (hydrolysis) and form hydrogen chloride, which makes the solution acidic. Three such reactions (1 mark each):

Aluminium chloride (1 mark)

Silicon tetrachloride (1 mark)

Phosphorus pentachloride (1 mark)

In each reaction the acidity comes from the \ce{HCl} formed. The solutions have pH about 3 for \ce{AlCl3} and about 0 for \ce{SiCl4} and \ce{PCl5}.

Compare the two orbitals on three points: shape, azimuthal quantum number and the number of orbitals (with the electrons they hold).

Any three of these points earn full marks. Both kinds of orbital can hold a maximum of two electrons each, and the first shell (n=1) has only an s orbital, while the p sub-shell starts from n=2.

Catenation is the ability of carbon atoms to link with one another to form long chains and rings. The word comes from the Greek "catena", which means chain. Silicon forms only small chains and sulphur only small rings, so the property is strongest in carbon.

Importance in organic chemistry. Catenation is the main reason for the very large number of organic compounds, for three reasons.

1. Carbon makes strong C-C bonds (bond energy 346\,\mathrm{kJ\,mol^{-1}}) because of its small size, and the atoms can join by single, double and triple bonds, which gives chains, branched chains and rings of many shapes and lengths.
2. Organic compounds are kinetically stable, so the long chains do not break down easily.
3. Carbon also attaches to hydrogen, oxygen, nitrogen, halogens and other elements, and the same atoms can be arranged in different ways (isomerism), so one molecular formula can stand for several compounds.

VSEPR theory says that the electron pairs around the central atom repel one another and take up positions as far apart as possible. The shape is named from the atoms only.

\ce{PCl3} (1.5 marks)

\ce{NH4+} (1.5 marks)

A carboxylic acid reacts with an alcohol on heating in the presence of **concentrated \ce{H2SO4} (catalyst) to form an ester and water. This is esterification**, a condensation reaction (two molecules join with loss of a small molecule, water). Ethanoic acid and ethanol give ethyl ethanoate.

The reaction is reversible, so the arrows are written \rightleftharpoons; the concentrated \ce{H2SO4} is also a dehydrating agent that takes up the water formed and so pushes the equilibrium towards the ester. The mechanism is protonation of the carbonyl oxygen, attack by the alcohol, proton transfer and loss of water.

A crystalline solid has a definite, regular, three-dimensional arrangement of its particles (atoms, ions or molecules), as in \ce{NaCl}, diamond, ice and glucose. Three properties of crystalline solids are:

1. Definite geometrical shape (1 mark)

Crystals have a definite, distinctive geometric shape because of the orderly arrangement of their particles in three dimensions. The angles between the faces of a crystal are always the same, however the crystal is grown.

2. Sharp melting point (1 mark)

On heating, the particles vibrate more and more. At the melting point their vibrational energy becomes high enough for all the particles to leave their fixed positions together, so the solid melts completely at one sharp temperature.

3. Cleavage planes (1 mark)

Crystalline solids break along certain definite planes, called cleavage planes. When pressure is applied along a cleavage plane, the crystal splits into smaller crystals of the same shape as the original. The angles between the cleavage planes differ from one crystalline solid to another.

The test is the silver nitrate test. Aqueous silver nitrate is added to the aqueous halide solution, and the halide ion gives a coloured precipitate of the silver halide. The nitrate ion is a spectator ion.

Chloride (1 mark)

A white precipitate of \ce{AgCl}.

Bromide (1 mark)

A cream (off-white) precipitate of \ce{AgBr}, which partly dissolves in aqueous ammonia.

Iodide (1 mark)

A yellow precipitate of \ce{AgI}, which does not dissolve in aqueous ammonia.

Bond energy depends on the electronegativity difference between the bonded atoms, the size of the atoms (bond length) and the bond order. Table 3.3 and Table 3.2 give the values used below.

(a) \ce{Cl2 > F2} (1 mark)

Bond energy: \ce{Cl-Cl}=242 and \ce{F-F}=158\ \mathrm{kJ\,mol^{-1}}. Fluorine atoms are very small, so the bond is short (149\ \mathrm{pm}) and the lone pairs on the two fluorine atoms are close together and repel each other. This weakens the F–F bond and less energy is needed to break it. In \ce{Cl2} the atoms are larger, the bond is longer (199\ \mathrm{pm}) and the lone pair repulsion is smaller, so the bond energy is higher.

(b) \ce{H2 > HBr} (1 mark)

Bond energy: \ce{H-H}=436 and \ce{H-Br}=366\ \mathrm{kJ\,mol^{-1}}. A bromine atom is much larger than a hydrogen atom, so the H–Br bond is longer than the H–H bond. The shorter the bond length, the greater the bond energy, so \ce{H2} has the higher bond energy.

(c) \ce{N2 > O2} (1 mark)

Bond energy: \ce{N2}=946 and \ce{O2}=498\ \mathrm{kJ\,mol^{-1}}. The bond order of \ce{N2} is 3 (triple bond) and that of \ce{O2} is 2 (double bond). The higher the bond order, the shorter (1.09\ \mathrm{\AA} against 1.21\ \mathrm{\AA}) and stronger the bond, so \ce{N2} has the higher bond energy.

A water molecule forms hydrogen bonds with its neighbours: the \ce{H} of one molecule is attracted to the \ce{O} of the next. These are strong intermolecular forces, and they explain the following properties of water.

1. High surface tension (1 mark)

The strong attractions pull the molecules at the surface strongly inward, so the surface resists being stretched. Water has a surface tension of 7.275\times10^{-2}\,\mathrm{N\,m^{-1}}, much higher than methanol, ethanol or benzene.

2. Low vapour pressure (1 mark)

The strong hydrogen bonds make evaporation slow, because few molecules have enough energy to escape. Less vapour is present above the liquid, so water has a low vapour pressure.

3. High heat of vaporization and high boiling point (1 mark)

The hydrogen bonds must be broken before molecules can escape, so much heat is needed: \Delta H_v=+40.7\,\mathrm{kJ\,mol^{-1}}. Because the vapour pressure is low, water must be heated to a higher temperature before its vapour pressure equals the external pressure, so its boiling point is high (100\,^\circ\mathrm{C} at 760\,\mathrm{torr}).

The reducing power of a halide ion is its tendency to lose an electron (to change to the halogen, \ce{2X- -> X2 + 2e-}). Going down the group it increases: \ce{I-} > \ce{Br-} > \ce{Cl-} > \ce{F-}. Three factors explain this.

1. Size of the ion. The halide ion becomes larger down the group (\ce{F-} is the smallest, \ce{I-} the largest), so the outer electron is farther from the nucleus and is lost more easily.
2. Attraction between the nucleus and the valence electrons. The number of shells (and so the shielding) increases down the group, so the nucleus attracts the outer electrons less strongly and the ion loses an electron more easily.
3. Electronegativity. It decreases down the group, so the ion holds its electrons less tightly. \ce{F-} holds its electron most strongly and is the weakest reducing agent; \ce{I-} is the strongest.

Evidence: with concentrated \ce{H2SO4}, \ce{F-} and \ce{Cl-} cannot reduce it, \ce{Br-} reduces it to \ce{SO2} and \ce{I-}, the strongest reducing agent, reduces it further to \ce{H2S}.

The limiting reactant is the reactant that is completely consumed and so limits the amount of product. The non-limiting (excess) reactant is the one left unreacted after the reaction is complete.

Example: for \ce{2H2 + O2 -> 2H2O}, with 1 mol of \ce{H2} and 1 mol of \ce{O2}, hydrogen is limiting and \tfrac12 mol of oxygen is left over.

Use Hess's law: the enthalpy change of a reaction is the same whether it takes place in one step or in several steps, so the given equations can be added after reversing or multiplying them, and their \Delta H^\circ values are treated in the same way.

Solids are of two types. Crystalline solids have a regular, repeating three-dimensional arrangement of particles. Amorphous solids have no regular, orderly arrangement of particles (apart from small ordered regions called crystallites).

Any three of the differences above are enough for 3 marks. A crystalline solid can be made amorphous by melting it and cooling the molten mass rapidly, so that the particles do not get time to arrange themselves.

Shell

Bohr's energy level around the nucleus, given by the principal quantum number n=1,2,3,\dots and named K, L, M, N. A bigger n means a larger and higher-energy shell.

Sub-shell

A part of a shell with the same shape of orbital, given by the azimuthal quantum number l=0 to n-1 and named s, p, d, f. The shell n has n sub-shells.

Orbital

The region of space around the nucleus where the probability of finding an electron is maximum. Its orientation is given by the magnetic quantum number m, and it holds at most two electrons of opposite spin.

Example: the L shell contains the 2s and 2p sub-shells; the 2p sub-shell contains three orbitals 2p_x, 2p_y, 2p_z.

The intermolecular force depends on the type of the molecule: a molecule with \ce{H} bonded to \ce{F}, \ce{O} or \ce{N} forms hydrogen bonds (a strong dipole-dipole attraction); a polar molecule has dipole-dipole forces; a non-polar molecule has only London forces (instantaneous dipole-induced dipole forces). London forces are present in all of them, but the strongest force present decides the order.

Descending order of strength of the intermolecular forces (2 marks for the forces, 1 mark for the order):

Hydrogen bonds are the strongest, dipole-dipole forces are next and London forces are the weakest. Between the two hydrogen-bonded molecules, water forms more hydrogen bonds per molecule than ammonia (two hydrogen atoms and two lone pairs), so the attraction in water is the larger one.

Basis (1.5 marks)

When an ionic solid dissolves, the lattice must be broken (energy absorbed, the reverse of the lattice energy) and the gaseous ions are then hydrated by water molecules (energy released). The enthalpy of solution is the net result.

\ce{NH4NO3} and \ce{CaCl2} (1.5 marks)

For \ce{NH4NO3} the energy needed to break up the lattice is greater than the energy released on hydration, so heat is absorbed from the surroundings and the process is endothermic. For \ce{CaCl2} the doubly charged \ce{Ca^2+} ion of high charge density attracts water molecules strongly (ion–dipole attraction), so the hydration energy exceeds the lattice energy and the process is exothermic.

Acid rain (1 mark)

Acid rain is rain that is more acidic than normal rain because acids formed from the pollutant oxides of sulphur and nitrogen dissolve in rain water. Sulphur dioxide is the main cause, and the oxides of nitrogen add to it.

Formation (2 marks)

Sulphur dioxide is formed when fuels containing sulphur are burned. It is oxidised in the air to sulphur trioxide (the oxides of nitrogen act as catalysts) and the trioxide dissolves in water to form sulphuric acid.

Nitrogen monoxide is formed from nitrogen and oxygen at the high temperature of engines (or by lightning). It is oxidised to nitrogen dioxide, which dissolves in water droplets to form nitric acid.

The sulphuric acid and nitric acid fall to the ground in the rain.

The property is catenation, the self-linking of carbon atoms to form long chains and rings. Carbon is small, so its C-C bonds are strong (346\,\mathrm{kJ\,mol^{-1}}), and this gives stable chains and rings.

1. Carbon makes strong C-C bonds (bond energy 346\,\mathrm{kJ\,mol^{-1}}) because of its small size, and the atoms can join by single, double and triple bonds, which gives chains, branched chains and rings of many shapes and lengths.
2. Organic compounds are kinetically stable, so the long chains do not break down easily.
3. Carbon also attaches to hydrogen, oxygen, nitrogen, halogens and other elements, and the same atoms can be arranged in different ways (isomerism), so one molecular formula can stand for several compounds.

Peroxyacetyl nitrate (PAN) is a harmful component of photochemical smog. Smog of this type forms in sunlight from nitrogen oxides (\mathrm{NO_x}, that is \ce{NO} and \ce{NO2}) and volatile organic compounds (VOCs). The steps below show the role of \mathrm{NO_x}.

PAN is a harmful ingredient of photochemical smog together with ozone; it irritates the eyes and the respiratory tract. In short, \ce{NO2} supplies oxygen atoms for ozone and itself combines with the organic fragments to give PAN, while \ce{NO} recycles \ce{NO2}.

Water is a limited resource, and the book lists steps for conserving and managing it. Any three of the following measures ensure the preservation of water resources.

• Smart use of water. Use water efficiently at home, in industry and in agriculture, through drip irrigation, low-flow fixtures and water-efficient appliances.
• Rainwater harvesting. Collect rainwater for uses such as irrigation and toilet flushing, so the demand for fresh water falls.
• Wastewater treatment and recycling. Clean used water in treatment plants and reuse it for irrigation and industrial processes, which saves freshwater and stops pollution of water bodies.
• Watershed protection. Restore and maintain natural watersheds, with erosion control and reforestation of barren land, for a steady supply of water.
• Legislation and public awareness. Enforce laws to prevent pollution and control water withdrawals, and educate people about saving water.

Other measures in the book are desalination for coastal cities, research into water-saving technology, and international cooperation on trans-boundary water.

Free radical substitution of ethane

Chlorine reacts with ethane in ultraviolet light by a free radical chain mechanism, in three stages.

The iodoform test is applied by warming the alcohol with excess iodine and excess sodium hydroxide. It shows the \ce{CH3CH(OH)-} group, and a pale yellow precipitate of tri-iodomethane (iodoform), \ce{CHI3}, forms. Ethanol is the only primary alcohol that gives it; secondary alcohols of the type \ce{CH3CH(OH)R} also give it.

1. Ethanol

2. A secondary alcohol containing CH3CH(OH)-

For example propan-2-ol gives \ce{CHI3} and sodium ethanoate, \ce{CH3COONa}. Methanol, other primary alcohols such as propan-1-ol, and tertiary alcohols give no yellow precipitate.

Chromium and the chloride ion: count the electrons, then fill the sub-shells in the order 1s,2s,2p,3s,3p,4s,3d. The mass number is not needed for the configuration.

(a) Chromium, Z=24 (1.5 marks)

(b) Chloride ion, \ce{Cl-}, Z=17 (1.5 marks)

Both solutions are explained by hydrolysis, the reaction of the ions of a salt with water. The ion that comes from the weak parent (acid or base) reacts with water and decides the pH.

Na2CO3 is basic (1.5 marks)

\ce{Na2CO3} is the salt of a weak acid (\ce{H2CO3}) and a strong base (\ce{NaOH}). The carbonate ion is a strong conjugate base, so it takes protons from water and leaves an excess of \ce{OH-}. \ce{Na+} does not react with water.

NH4Cl is acidic (1.5 marks)

\ce{NH4Cl} is the salt of a strong acid (\ce{HCl}) and a weak base (\ce{NH4OH}). The ammonium ion is a strong conjugate acid, so it gives protons to water and leaves an excess of \ce{H+}. \ce{Cl-} does not react with water.

Chemical equilibrium is reached by a reversible reaction when the following conditions are met.

1. A reversible reaction. The products can react to give back the reactants, so the reaction goes in the forward and the reverse directions at the same time (shown by \rightleftharpoons).
2. A closed system at constant temperature. Nothing enters or leaves the reaction mixture, and the temperature is not changed, because K_c has only one value at a given temperature.
3. Equal rates. The rate of the forward reaction becomes equal to the rate of the reverse reaction.

When these conditions are met, the concentrations of all the reactants and products stay constant (the colour, pressure and composition no longer change), and the ratio of the concentrations gives the constant K_c.

Ethylamine, \ce{CH3CH2NH2}, is a primary amine.

(a) From an alkyl halide (1 mark)

Ethyl halide is heated with excess alcoholic ammonia in a sealed tube at 100^\circ\mathrm{C} (nucleophilic substitution). Excess ammonia keeps over-alkylation low.

(b) From a nitrile (the printed 'nitrate'; 1 mark)

The reduction of a nitrile gives a primary amine with one more carbon than the alkyl group of the halide. Ethylamine needs the nitrile \ce{CH3CN} (ethanenitrile), reduced by \ce{LiAlH4}, by sodium in ethanol, or catalytically by \ce{H2} with Ni, Pt or Rh-\ce{Al2O3}.

(c) From an amide (1 mark)

Acetamide is reduced by \ce{LiAlH4}: the \ce{C=O} group becomes \ce{CH2} and the carbon chain stays intact.

Sulphur has two unpaired electrons, in 3p_y and 3p_z.

Use the iodoform test. Warm each alcohol with iodine and sodium hydroxide.

The reaction is base catalysed. The base (\ce{OH-}) produces the cyanide ion, which is the nucleophile. The carbonyl carbon of \ce{HCHO} is electrophilic (\delta^+) because oxygen is more electronegative.

(a) Nitrobenzene to aniline (1.5 marks)

Nitrobenzene is reduced by tin and concentrated hydrochloric acid (nascent hydrogen [\mathrm{H}]). The \ce{-NO2} group becomes \ce{-NH2} and aniline (phenylamine) is formed.

(b) 4-Nitroaniline to 1,4-diaminobenzene (1.5 marks)

Iron with sulphuric acid also reduces a nitro group to an amino group. Only the \ce{-NO2} group is reduced; the \ce{-NH2} group already on the ring is unchanged, so the product is 1,4-diaminobenzene (p-phenylenediamine).

The answer lies between 24 and 26 and is close to 24 because \ce{^{24}Mg} is by far the most abundant isotope.

Compounds of sulphur are used in many industries. Three of them (1 mark each):

Sulphuric acid is also used in car batteries, in tanning leather and in pickling of metals, and sulphur itself is used for vulcanising rubber and in gunpowder.

Markovnikov's rule

When an unsymmetrical reagent such as \ce{HBr} adds to an unsymmetrical alkene such as propene, the hydrogen atom bonds to the doubly bonded carbon that already has more hydrogen atoms, and the positive part of the reagent goes where the hydrogens are more. The more stable carbocation is the intermediate, so the major product has the halogen on the carbon with fewer hydrogens.

Example and mechanism: propene and HBr

The letter d means l=2. Each quantum number is then fixed by the rules for allowed values.

Definition (1 mark)

The standard enthalpy of atomization, \Delta H^\circ_\mathrm{at}, is the enthalpy change when one mole of gaseous atoms is formed from the element under standard conditions.

Example (2 marks)

The product is one mole of atoms, which is why only half a mole of a diatomic molecule is written on the left. Atomization breaks bonds (here the Cl–Cl or H–H bond), so it always needs energy and \Delta H^\circ_\mathrm{at} is positive.

Definition (1 mark)

Electronegativity is the ability of an atom to attract the electrons towards itself in a chemical bond.

Trend along a period (2 marks)

Electronegativity increases from left to right across a period. Moving along the period the nuclear charge increases, the electrons are added to the same valence shell, so the shielding stays about the same and the atomic size decreases. The nucleus therefore pulls the shared pair of electrons more strongly. Group 1 elements have the lowest values and the halogens the highest.

Definition

A **\beta-elimination** is an elimination in which a halogen on one carbon (the \alpha-carbon) and a hydrogen on the neighbouring \beta-carbon are removed together, and a \ce{C=C} double bond forms between the two carbons. A base is needed, and the \beta-hydrogen must be present.

Example and mechanism: bromoethane

Using 22.414\,\mathrm{dm^3\,mol^{-1}} gives 56.04\,\mathrm{dm^3}, the same to three significant figures.

The vapour pressure of a liquid at a given temperature depends on the strength of the intermolecular forces: the stronger the forces, the lower the vapour pressure.

• Water molecules are held together by hydrogen bonding, a strong intermolecular force. Few molecules have enough energy to break free from the surface, so evaporation is slow and little vapour is present above the liquid. The vapour pressure of water is therefore low.
• \ce{CCl4} is non-polar and has only weak London forces, and ether has only weak dipole-dipole and London forces with no \ce{O-H} group to form hydrogen bonds. Their molecules escape easily, so their vapour pressures are high.

So water, with the strongest forces, has a lower vapour pressure than \ce{CCl4} and ether.

\Delta S^\circ is positive: an ordered crystal lattice breaks up into free, hydrated ions in solution, so the disorder of the system increases.

The Born–Haber cycle is the application of Hess's law to an ionic compound. The formation of \ce{KCl(s)} from its elements is broken into five steps, the last of which is the formation of the lattice.

By Hess's law the sum of the five steps equals the enthalpy of formation of \ce{KCl}, \ce{K(s) + 1/2 Cl2(g) -> KCl(s)}:

The unit of a first order rate constant is \mathrm{s^{-1}}.

Mass spectrometry (1 mark)

Mass spectrometry is a technique used to determine the molecular composition and structure of a sample by measuring the mass-to-charge ratio (m/z) of its ions. It has three steps: ionization, mass analysis (separation of the ions by m/z) and detection, which together give a mass spectrum.

Construction and working (4 marks)

A mass spectrometer is made of an ion source, an accelerating electric field, a mass analyser (the magnetic field in the figure) and a detector, joined to a system that records the spectrum. It works in four stages.

1. Ionization. The sample is vaporized and ionized in the ion source, for example by bombarding it with high-energy electrons (EI), with a high-voltage field (ESI) or with a laser (MALDI).
2. Ion separation. The ions are accelerated and enter the mass analyser. The magnetic field deflects them, and ions of different m/z follow paths of different radius: the lightest ions are deflected most.
3. Detection. The ions that reach the detector are recorded. The detector notes the time each ion takes to arrive, and this is used to calculate its m/z.
4. Data analysis. The record is plotted as a mass spectrum: m/z on the x-axis and the abundance of ions on the y-axis. Each peak is one kind of ion, its position gives the mass and its height gives the relative abundance.

Average atomic mass from the data (2 marks)

The peaks of an element give the mass number and the relative abundance of each isotope. The average atomic mass is the weighted average:

Example: chlorine has two peaks, \ce{^{35}Cl} at 75.76% and \ce{^{37}Cl} at 24.24%.

This explains why the atomic masses in the periodic table are not whole numbers: each is the weighted average of the isotopes of the element.

1. Molecular orbital theory (3 marks)

According to molecular orbital theory (MOT) the atomic orbitals of the two oxygen atoms combine linearly to form molecular orbitals that belong to the whole molecule. Each combination gives a lower-energy bonding orbital (BMO) and a higher-energy anti-bonding orbital (ABMO). The number of molecular orbitals equals the number of atomic orbitals combined.

The bond order is 2, so the two oxygen atoms are joined by a double bond (bond energy 498\ \mathrm{kJ\,mol^{-1}}, bond length 1.21\ \mathrm{\AA}). The two unpaired electrons in the \pi^* orbitals make \ce{O2} paramagnetic.

2. Valence bond theory (3 marks)

According to valence bond theory (VBT) a covalent bond forms by the overlap of half-filled atomic orbitals, and the electrons of the overlapping orbitals have opposite spins. Each oxygen atom has two half-filled 2p orbitals (2p_y and 2p_z), so two bonds are formed.

3. Why MOT is superior to VBT (1 mark)

VBT shows all the electrons of \ce{O2} as paired (a \sigma bond and a \pi bond, each a pair of opposite spins), so it predicts that \ce{O2} is diamagnetic. In fact liquid oxygen is attracted by a magnet: \ce{O2} is paramagnetic. MOT explains this, because two electrons occupy the two degenerate \pi^*2p_y and \pi^*2p_z orbitals singly with parallel spins. MOT also gives the bond order (2) from the electron count, and it explains why \ce{He2} does not exist (bond order 0). VBT cannot explain the paramagnetic behaviour of \ce{O2}, so MOT is the better theory.

Moles of each reactant

(a) Limiting reactant and mass of hydrogen

(b) Excess reactant left

Information from the equation

• The forward reaction is exothermic (heat is a product).
• The number of gas molecules decreases: 1+2=3 mol of gas on the left and 1 mol on the right.

By Le Chatelier's principle, three changes shift the equilibrium towards methanol and so increase its yield.

Factor 1: high pressure (2 marks)

Three molecules of gas (\ce{CO} and 2\,\ce{H2}) give one molecule of \ce{CH3OH}. Increasing the pressure makes the system reduce its volume by moving to the side with fewer gas molecules, so the equilibrium shifts to the right and more methanol is formed. (The book's Concept assessment 8.5 says the same: decreasing the pressure would lower the yield.)

Factor 2: low temperature (2 marks)

The forward reaction is exothermic, so heat can be written as a product.

Removing heat (lowering the temperature) makes the system produce heat, so the equilibrium shifts to the right and the value of K_c increases. Too low a temperature makes the reaction too slow, so a moderate temperature with a catalyst is used in practice; the catalyst only makes equilibrium come sooner and does not change the yield.

Factor 3: concentration (2 marks)

Adding more \ce{CO} or \ce{H2}, or continually removing the methanol as it is formed, shifts the equilibrium to the right. Removing the product keeps the reaction moving towards methanol. K_c does not change by a change of concentration.

The equilibrium shifts to the left, towards \ce{C3H8} and \ce{O2}, so less \ce{CO2} and \ce{H2O} is present. The value of K_c does not change, because only temperature changes K_c.

Removing a product lowers its concentration, so the system tries to replace it. The forward reaction is favoured, and the equilibrium shifts to the right (towards \ce{SO3}).

More \ce{SO2} and \ce{O2} are used up and more \ce{SO3} is formed, until equilibrium is re-established. K_c does not change.

The reaction is endothermic (heat \Delta is on the reactant side), so heat is a reactant.

Decreasing the temperature removes heat, and the system opposes this by favouring the heat-producing (exothermic) direction, the reverse reaction. The equilibrium shifts to the left, towards \ce{CO2}, and the concentrations of \ce{CO} and \ce{O2} fall. The value of K_c decreases.

Ethanol is a primary alcohol. It is oxidised first to ethanal and then to ethanoic acid by **acidified \ce{K2Cr2O7} (or acidified \ce{KMnO4}) heated under reflux**. Reflux keeps the ethanal in the flask so that it is oxidised all the way to the acid (if it were distilled off the product would stop at the aldehyde).

Chloroethane has two carbons, so the acid made from it is propanoic acid (one carbon more). The reaction is done in two stages.

A nitrile is hydrolysed to the carboxylic acid by boiling with dilute mineral acid, such as dilute \ce{HCl} (or with dilute alkali, followed by acidification). The nitrogen leaves as ammonia, which appears as \ce{NH4Cl} in acid.

With alkali: \ce{CH3CN + NaOH + H2O -> CH3COONa + NH3}, then \ce{CH3COONa + HCl -> CH3COOH + NaCl}.

2-Methylpropan-1-ol is a primary alcohol. Heated under reflux with excess acidified \ce{K2Cr2O7} (or acidified \ce{KMnO4}) it is oxidised through the aldehyde to 2-methylpropanoic acid.

An ester is hydrolysed to the acid and the alcohol by heating with dilute acid (or dilute alkali, followed by acidification). Ethyl ethanoate gives ethanoic acid and ethanol.

Acid hydrolysis is reversible (the reverse of esterification). With alkali the reaction goes to completion and gives the sodium salt (saponification); acidification then frees the acid.

Benzaldehyde is an aldehyde, so it is oxidised to benzoic acid by acidified \ce{K2Cr2O7} or acidified \ce{KMnO4} on heating under reflux.

Possible isomers of \ce{C5H10O}

\ce{C5H10O} has one degree of unsaturation (one double bond or one ring), so it can be a carbonyl compound, an unsaturated alcohol or ether, or a ring compound. The main isomers are drawn below.

Aldehydes (functional group -\ce{CHO}):

Ketones (functional group >\ce{C=O}):

Other functional groups (some further examples):

Types of structural isomerism shown

Halogenoalkanes (\ce{R-X}, general formula \ce{C_nH_{2n+1}X} for a mono derivative) are classified in the following ways.

1. By the number of halogen atoms

2. By the type of halogen atom

They are fluoro-, chloro-, bromo- or iodoalkanes according to whether \ce{X} is \ce{F}, \ce{Cl}, \ce{Br} or \ce{I}, for example \ce{CH3CH2Cl} (chloroethane) and \ce{CH3CH2Br} (bromoethane).

3. By the position of the halogen in the carbon chain

What decides the class is the carbon atom that carries the halogen, and how many alkyl groups are attached to that carbon.

Ethene has a \ce{C=C} double bond made of one strong \sigma bond and one weaker \pi bond. The \pi electrons lie above and below the plane of the molecule and are loosely held and exposed, so electrophiles can attack them easily and ethene undergoes addition reactions.

Ethane has only strong, non-polar \sigma bonds. They are held tightly between the nuclei and give no electrons for attack, so ethane reacts only by free radical substitution in ultraviolet light. Ethene is therefore more reactive than ethane.

This is hydration of ethene (acid-catalysed addition of water). It takes place in two steps and gives ethanol.

This is halohydration. Chlorine reacts with water to give hypochlorous acid, which adds to the double bond, with the solvent water acting as a reactant.

Ethanal contains the aldehyde group, \ce{-CHO}. It is identified in three steps.

Ethanol contains the hydroxyl group (-\ce{OH}) of an alcohol. It is identified by these tests.

Propene contains a carbon-carbon double bond (\ce{C=C}), so it is unsaturated and shows addition reactions. It is identified by two tests.

A saturated hydrocarbon such as propane does not decolourise either reagent in the cold, so the tests separate alkenes from alkanes.

Solubility product (1 mark)

When a slightly soluble ionic compound is in contact with its saturated solution, a dynamic equilibrium exists between the undissolved solid and its ions. The solubility product constant K_{sp} is the product of the equilibrium concentrations of the ions, each raised to a power equal to its coefficient in the balanced equation. The solid itself does not appear in the expression.

Prediction of precipitation of CaF2 (4 marks)

On mixing a solution of \ce{Ca^2+} ions with a solution of \ce{F-} ions, a precipitate of \ce{CaF2} may or may not form. To predict this, the ion product Q' is calculated. It has the same form as K_{sp} but uses the initial concentrations (after mixing) instead of the equilibrium concentrations.

Illustration

Take K_{sp}(\ce{CaF2})=3.45\times10^{-11} (from the solubility 2.05\times10^{-4}\,\mathrm{M}). Mix 10\,\mathrm{cm^3} of 0.010\,\mathrm{M} \ce{CaCl2} with 10\,\mathrm{cm^3} of 0.010\,\mathrm{M} \ce{NaF}. Each solution is diluted to half by the mixing, so [\ce{Ca^2+}]=[\ce{F-}]=0.0050\,\mathrm{M}.

Since Q'>K_{sp}, a precipitate of \ce{CaF2} forms.

Molecular collisions

For a chemical reaction to occur, the reacting atoms or molecules must first collide. Not every collision gives products. A collision is effective only if (i) the colliding particles have enough energy to overcome the repulsion between their electron clouds and (ii) they have the proper orientation, so that the atoms which must form the new bonds meet each other. Most collisions are ineffective, and the particles simply bounce apart unchanged.

Activation energy

The activation energy E_a is the minimum amount of energy, in addition to the average kinetic energy, which the particles must possess for an effective collision. It acts as an energy barrier (a hill) between reactants and products. As the molecules approach, they slow down, their kinetic energy falls and their potential energy rises. Only molecules with energy at least equal to E_a can climb the hill; the others fall back unchanged. The higher the activation energy, the smaller the fraction of effective collisions and the slower the reaction.

Formation of the activated complex

In an effective collision of, for example, \ce{A2} and \ce{B2}, the old bonds begin to break while the new bonds begin to form. An unstable, high-energy species called the activated complex (transition state) is formed at the top of the energy hill. It is short-lived and at once breaks down into the products \ce{2AB}.

Enthalpy changes in the reaction

A potential energy diagram shows the energy of the system along the reaction path. The reaction enthalpy is the difference between the energy of the products and that of the reactants, \Delta H=H_\text{products}-H_\text{reactants}.

• Exothermic reaction: the products are at a lower energy level than the reactants, so energy is released and \Delta H is negative.
• Endothermic reaction: the products are at a higher energy level than the reactants, so energy is absorbed and \Delta H is positive.

In both cases the activation energy is a barrier that must be overcome before the products can form. If the reacting particles do not have it, the reaction does not start. The height of the barrier decides how fast the reaction goes, while \Delta H tells how much heat is given out or taken in.

Hybridization is the mixing of atomic orbitals of different energies and shapes to give the same number of new orbitals of equal energy and shape. In \ce{BF3} the central boron atom is sp^2 hybridised.

Step 1: ground state and promotion

Step 2: sp2 hybridization

One 2s and two 2p orbitals of boron mix to give **three equivalent half-filled sp^2 hybrid orbitals**. They point to the corners of an equilateral triangle in one plane, 120^\circ apart (trigonal planar symmetry).

Step 3: overlap with fluorine

Fluorine (1s^2\,2s^2\,2p^5) has one half-filled 2p orbital. Each sp^2 orbital of boron overlaps head on with a half-filled 2p orbital of one fluorine atom to give a \sigma sp^2–p bond. Three \sigma bonds are formed and boron has no lone pair.

In acetylene (ethyne) each carbon atom is sp hybridised. The molecule is linear, with a \ce{C#C} triple bond.

Step 1: promotion

Step 2: sp hybridization

Each carbon mixes its 2s orbital with one 2p orbital to give **two half-filled sp hybrid orbitals**, arranged in a straight line (180^\circ). The other two 2p orbitals (2p_y and 2p_z) stay unhybridised and are perpendicular to the sp orbitals and to each other.

Step 3: bonds formed

Because the two carbon atoms and the two hydrogen atoms lie in a straight line, the H–C–C bond angle is **180^\circ and the molecule is linear**.

The configuration [\ce{Ne}]\,3s^2\,3p^6\,3d^9 is that of the ion, and it has 10+2+6+9=27 electrons. The ion has a charge of +2, so it has lost two electrons and the neutral atom has 27+2=29 electrons. A neutral atom has equal protons and electrons, so p=29, and neutrons n=63-29=34. The answer is C (e=29, p=29, n=34).

Option A counts the electrons of the ion (27) and wrongly takes this as the proton number, and option B mixes the two. The question asks for the atom, not for the ion.

Sodium chloride gives one \ce{Na+} ion per formula unit (\ce{NaCl -> Na+ + Cl-}), so 0.2 mol of \ce{NaCl} gives 0.2 mol of \ce{Na+}. The number of ions is 0.2\times6.02\times10^{23}=1.204\times10^{23}, which is B.

Option D (2.4\times10^{23}) would need 0.4 mol of \ce{Na+}, which is the case for \ce{Na2SO4} (two \ce{Na+} per formula unit) but not for \ce{NaCl}. Options A and C are ten times too small.

In \ce{H2C=O} the carbon atom is bonded to two hydrogen atoms (two \sigma bonds) and to oxygen by a double bond (one \sigma and one \pi bond). That is three \sigma bond pairs, as the question says. In VSEPR the \pi bond lies in the same region as the \sigma bond of the double bond, so a double bond counts as one group of electrons, as in \ce{SO2} and \ce{CO3^2-} in the book.

Carbon therefore has three groups, no lone pair (AX_3), and the shape is trigonal planar with angles of about 120^\circ, like \ce{BCl3} and \ce{C2H4} (sp^2). The tempting wrong answer is D (tetrahedral), from counting the \pi pair as a fourth pair; it does not add a new direction. Triangular pyramid (A) needs a lone pair, and bent (C) needs two bonded atoms.

Temperature is a measure of the average kinetic energy of the molecules. Raising the temperature increases the average kinetic energy, so the molecules gain kinetic energy and move faster. This is why more molecules reach the activation energy and the reaction rate rises (Maxwell-Boltzmann curve, section 7.3).

They do not lose energy (A) or stop moving (D), and they certainly do not stay the same (B), since a hotter sample has molecules with a higher average speed.

Moles of \ce{CO2} at STP: 0.5/22.4=0.0223 mol. Molar mass of \ce{CO2} is 12+32=44\,\mathrm{g\,mol^{-1}}, so the mass is 0.0223\times44=0.98\,\mathrm{g}, which is B.

Option C (22.4\,\mathrm{g}) confuses the molar volume (22.4\,\mathrm{dm^3}) with a mass, and A (0.49\,\mathrm{g}) is half the correct value.

The slow step is the rate-determining step. Here the equation is given as one slow step, \ce{2A + B -> C}, so the powers in the rate law equal the coefficients of this step: [\ce{A}]^2 for \ce{2A} and [\ce{B}]^1 for \ce{B}. Hence \text{rate}=k[\ce{A}]^2[\ce{B}].

Option B uses the wrong power of A. Option D swaps the powers of A and B. Option C uses the product \ce{C}, which is not a reactant and does not appear in the rate law of the slow step.

London (instantaneous dipole induced dipole) forces get stronger as the molecule gets bigger, with more electrons and a higher molecular mass. They also need close contact between molecules.

\ce{C5H12} has more electrons and a higher molecular mass (M_r=72) than \ce{C4H10} (M_r=58), so the pentanes beat the butanes. Of the two pentanes, n-pentane is a long, straight chain that lies close against its neighbours over its whole length, while iso-pentane is branched and more compact, so it has less surface in contact. The straight-chain n-pentane has the stronger forces (boiling point 36\ ^\circ\mathrm{C} against 28\ ^\circ\mathrm{C} for iso-pentane).

The tempting wrong answer is D (iso-pentane): it has the same formula and mass as n-pentane, but its branching reduces the contact between molecules.

Electron affinity generally increases from left to right along a period because the nuclear charge increases and the atomic size decreases. Chlorine, a halogen at the right of period 3, has the largest value, -349 in the book's Figure 10.7, against -53 for Na and -44 for Al.

Magnesium (option B) has a full 3s^2 sub-shell, and its electron affinity is 0 in the book's table, the lowest of the four. Sodium and aluminium have a much smaller nuclear charge than chlorine, so they release far less energy when they gain an electron.

Combustion (burning in oxygen) always gives out heat, so it is an exothermic process and the standard enthalpy of combustion is always negative (for example -393.5\,\mathrm{kJ\,mol^{-1}} for carbon). It is not endothermic (C), and it is not neutral (B): it has a definite, negative \Delta H.

Reversible (D) describes whether a reaction can go back, not its heat change.

Gibbs free energy is defined as the enthalpy of the system minus the product of its temperature and entropy: G=H-TS. At constant temperature this gives the Gibbs–Helmholtz equation \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ.

The signs matter: option A (H+TS) has the wrong sign, and options B and D combine H and TS by multiplication and division, which is not dimensionally possible because H and TS are both energies and can only be added or subtracted.

The molar mass of \ce{Cl2} is 2\times35.5=71\,\mathrm{g\,mol^{-1}}, so 71\,\mathrm{g} is exactly 1 mol. One mole of any gas occupies 22.4\,\mathrm{dm^3} at STP, so the answer is B.

Option A (11.2\,\mathrm{dm^3}) is the volume of half a mole, which would be 35.5\,\mathrm{g} of \ce{Cl2}, and C (44.8\,\mathrm{dm^3}) is for 2 mol (142\,\mathrm{g}).

The book states that the combination of \ce{H2} and \ce{Cl2} in the presence of sunlight is a zero order reaction. Its rate is independent of the concentrations of both reactants, because the rate is controlled by the light absorbed, so \text{rate}=k[\ce{H2}]^0[\ce{Cl2}]^0=k.

Option A, k[\ce{H2}][\ce{Cl2}], is the tempting wrong answer: it copies the coefficients of the balanced equation, but the order can only be found by experiment and is not read from the equation.

Pressure changes the position of an equilibrium only when the number of gas molecules is different on the two sides (\Delta n\neq0). In \ce{N2 + O2 <=> 2NO} there are 2 mol of gas on each side, so a change in pressure favours neither direction.

The other three have unequal numbers of gas molecules: \ce{N2O4 <=> 2NO2} (1 to 2), \ce{CO + 2H2 <=> CH3OH} (3 to 1) and \ce{N2 + 3H2 <=> 2NH3} (4 to 2), so each is affected.

An electrophile is an electron-deficient species, with a positive charge or an empty orbital, so it accepts an electron pair from its reaction partner (the nucleophile). Examples are \ce{H+} and the carbocation \ce{CH3+}.

Option A describes a nucleophile, which donates an electron pair. Options B and C use the Bronsted-Lowry idea of proton acceptor and donor, which is a different definition; an electrophile is defined by electron pairs, not protons.

\ce{MX} gives one cation and one anion, so \ce{MX(s) <=> M+ + X-} and [\ce{M+}]=[\ce{X-}]=s=2\times10^{-2}\,\mathrm{M}.

The tempting options come from squaring wrongly: (2\times10^{-2})^2 is 4\times10^{-4}, not 2\times10^{-4} (squaring the power of ten but not the 2) and not 8\times10^{-4} (doubling the 2 instead of squaring it).

Fehling's solution is a mild oxidising agent. It oxidises aldehydes (ethanal, methanal, propanal) to carboxylate ions and is reduced to red \ce{Cu2O}. Propanone is a ketone and is not oxidised by it, so it gives no red precipitate.

Methanal is also an aldehyde, so it gives the test.

Across a period the nuclear charge increases while the electrons are added to the same valence shell. The nucleus pulls the electrons in more strongly, so the atomic radius decreases from left to right.

The other three properties all increase across the period: ionization energy increases because the outer electron is held more tightly, electron affinity generally increases (the book gives -53 for Na and -349 for Cl) and electronegativity increases. Electron affinity (option C) is the tempting choice because its values are printed with negative signs, but a more negative value means more energy released, so it increases across the period.

The chain is started when the halogen molecule splits into two free radicals by homolytic fission, \ce{Cl2 ->[h\nu] 2Cl.}. The energy for this is supplied by ultraviolet radiation, so UV light is the condition used in the book (option C).

Absence of light (D) is the opposite condition, because the reaction does not start in the dark at room temperature. Low temperature (A) gives no energy to break the Cl-Cl bond. A very high temperature (B) would also give radicals, but the condition named for the free radical substitution of ethane in the book is ultraviolet light.

Methane has four bonding pairs and no lone pair on carbon (AX_4). The four pairs repel each other equally and move as far apart as possible, to the corners of a regular tetrahedron. The angle between any two C–H bonds is therefore **109.5^\circ**.

120^\circ (C) is the tempting wrong answer: it is the angle for three electron pairs, as in \ce{BCl3}. 180^\circ is for two pairs (\ce{BeCl2}) and 90^\circ is for the octahedral arrangement of six pairs. In \ce{NH3} and \ce{H2O} the lone pairs squeeze the angle below 109.5^\circ (107^\circ and 104.5^\circ), but \ce{CH4} has none.

The formulas \ce{M2O}, \ce{MOH} and \ce{MCl} show that M forms the \ce{M+} ion, so it has one valence electron and belongs to group 1. Of the elements listed only lithium is in group 1. Like the book's example of sodium, a group 1 metal burns in oxygen to give an oxide (\ce{M2O}), reacts with water to give a hydroxide (\ce{MOH}) and burns in chlorine to give a chloride (\ce{MCl}).

Mg and Ca (options C and D) are in group 2 and form \ce{MO}, \ce{M(OH)2} and \ce{MCl2} (compare \ce{MgO}, \ce{Mg(OH)2} and \ce{MgCl2} in the book). Al (option B) is in group 13 and forms \ce{Al2O3}, \ce{Al(OH)3} and \ce{AlCl3}.

Ammonium sulphate, \ce{(NH4)2SO4}, is used as a fertiliser. It is made by neutralising ammonia with sulphuric acid, \ce{2NH3 + H2SO4 -> (NH4)2SO4}, and it supplies both nitrogen and sulphur to the plants.

Copper sulphate (option B), sodium sulphate (option C) and barium sulphate (option D) contain no nitrogen and are not used as the fertiliser compound of sulphur in the book.

A halogen can displace another halogen from its halide only if it is the stronger oxidizing agent. The oxidizing power falls down the group: \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}.

Fluorine is stronger than bromine, so \ce{F2} oxidizes \ce{Br-} to \ce{Br2} (option C). The other reactions go the wrong way, because a weaker halogen cannot oxidize the halide ion of a stronger one: \ce{Br2} and \ce{Cl2} cannot oxidize \ce{F-} (A and B), and \ce{I2} cannot oxidize \ce{Br-} (D).

\ce{NH4Cl} is the salt of a strong acid (\ce{HCl}) and a weak base (\ce{NH4OH}). The ammonium ion is a strong conjugate acid and reacts with water to produce \ce{H+}, so the solution is acidic (\mathrm{pH}<7).

The chloride ion is a very weak conjugate base and does not react with water, so nothing balances the extra \ce{H+}.

Ethene is converted into ethanol by hydration, using \ce{H2SO4/H2O} (x). Ethanol is converted back into ethene by dehydration, using heated \ce{Al2O3} (y), so option C is correct.

Option A has no catalyst for the hydration and \ce{HCl} does not dehydrate an alcohol. \ce{LiAlH4} and \ce{NaBH4} (B and D) are reducing agents, which do not add water to an alkene or remove water from an alcohol.

With \ce{AgNO3} in ethanol the halide ion forms a silver halide. The colour and solubility in ammonia identify it: chloride is white and soluble, bromide is cream and partially soluble, iodide is yellow and insoluble. A yellow precipitate insoluble in \ce{NH4OH} is \ce{AgI}, so the alkyl halide is the iodide, option D.

Bromoethane (C) is the tempting option, since the \ce{AgBr} is pale yellow or cream, but it is partially soluble in ammonia. Chloroethane (B) gives a white precipitate that dissolves, and fluoroethane (A) gives no precipitate because the \ce{C-F} bond is too strong.

Basic strength depends on how easily the nitrogen lone pair is donated. Ammonia has three H atoms and no electron-donating group, so it is the weakest of these: \mathrm{p}K_b of \ce{NH3} is 4.76 against 3.38 for \ce{CH3NH2}. A smaller \mathrm{p}K_b means a stronger base, so methylamine is stronger than ammonia.

The other three options are secondary amines (dimethylamine, methylethylamine, diethylamine) with two electron-donating alkyl groups on nitrogen. A secondary amine is a stronger base than a primary amine such as methylamine, so methylamine is weaker than each of them.

A saturated hydrocarbon has only single bonds. Alkanes have the general formula \ce{C_nH_{2n+2}}, and for n=6 this gives \ce{C6H14}, so option A is saturated.

\ce{C6H12} (B) fits \ce{C_nH_{2n}}, so it has one double bond (an alkene) or a ring; \ce{C6H10} (C) has two fewer hydrogens again (two double bonds or one triple bond); \ce{C6H6} (D) is benzene with double bonds. All of these are unsaturated.

Cold, dilute, acidified (or neutral) \ce{KMnO4} adds two \ce{OH} groups, one on each carbon of the double bond, and converts ethene into the diol ethane-1,2-diol, \ce{HO-CH2-CH2-OH}. The purple \ce{MnO4-} is reduced to the nearly colourless \ce{Mn^2+}.

Ethanol (A) is made from ethene and steam, not by \ce{KMnO4}. Ethanal (B) and ethanoic acid (C) would need hot, concentrated oxidation that breaks the double bond, not the cold dilute reagent.

\ce{NaBH4} reduces the \ce{C=O} of an aldehyde to a primary alcohol, so ethanal gives ethanol.

\ce{NaBH4} is a hydride reducing agent, so the product is not oxidised (ethanoic acid), and it does not remove the oxygen completely (ethene, ethane).

Hydrolysis of a nitrile with dilute \ce{HCl} gives the carboxylic acid with the same number of carbons: the \ce{-CN} carbon becomes the \ce{-COOH} carbon.