Inorganic chemistry I: A-level Chemistry AQA revision notes

3.2.1

Periodicity

Blocks of the periodic table 3.2.1.1

Definitions

s, p, d and f blocks: elements are classified by the subshell that receives their highest-energy electron.
Periodicity: the repeating pattern of physical and chemical properties across periods, caused by the repeating pattern of outer electron configurations.

Key results

Period 3 runs from Na ( $[Ne] 3 s^{1}$ ) to Ar ( $[Ne] 3 s^{2} 3 p^{6}$ ): Na and Mg are s-block; Al to Ar are p-block.

Notes

Every Period 3 atom has the same $[Ne]$ core; only the number of 3s and 3p electrons changes, which is why trends across the period can be explained by nuclear charge and the subshell involved.
The modern table is arranged by atomic number, not atomic mass. Mendeleev ordered by mass but left gaps and occasionally swapped elements (tellurium and iodine) so that elements with similar properties lined up; later work on proton number justified his choices.
Elements in the same group have the same outer electron configuration (e.g. all Group 2 elements end $n s^{2}$ ), which is why they react in similar ways and form ions of the same charge.
Periodicity is visible in many properties at once: atomic radius, ionisation energy, electronegativity, melting point and the acid–base character of the oxides all follow repeating patterns, and each can be explained from the electron configuration.

Atomic radius across Period 3 3.2.1.2

Notes

Atomic radius decreases from Na to Cl. Electrons are added to the same shell ( $n = 3$ ) while the nuclear charge increases, and shielding by inner electrons is similar, so the outer electrons are pulled closer.
The explanation needs all three parts: more protons, same shell, similar shielding.
Approximate covalent radii fall from about 190 pm (Na) to about 100 pm (Cl). Argon's value is not comparable, because it forms no covalent bonds and only a van der Waals radius can be measured.
Ions follow a different pattern. $Na^{+}$ , $Mg^{2 +}$ and $Al^{3 +}$ have lost the whole third shell, so they are much smaller than their atoms and shrink further as nuclear charge rises; $P^{3 -}$ , $S^{2 -}$ and $Cl^{-}$ keep three shells and are much larger.
An isoelectronic series shows the effect of nuclear charge alone: $S^{2 -} > Cl^{-} > K^{+} > Ca^{2 +}$ , since all have 18 electrons but the number of protons increases from 16 to 20.
The decrease in radius explains other trends across the period: electronegativity and first ionisation energy rise because the outer electrons are closer to a more highly charged nucleus.

First ionisation energy across Period 3 3.2.1.2

Key results

General increase from Na (496 kJ mol⁻¹) to Ar (1521 kJ mol⁻¹), with dips at Al (578) and S (1000).

Notes

The general rise has the same cause as the falling radius: greater nuclear charge, same shell, similar shielding.
Al is lower than Mg because aluminium's outer electron is in a 3p orbital, which is higher in energy (and slightly better shielded by the 3s electrons) than magnesium's 3s.
S is lower than P because sulfur's fourth 3p electron is paired; repulsion between the paired electrons makes one of them easier to remove.
Na has the lowest value in the period: its single 3s electron is the furthest from the nucleus and is shielded by ten inner electrons, with an effective attraction of only about +1.
Ar has the highest value: it has the largest nuclear charge in the period, with its outer electrons still in the third shell.
The same pattern of a general rise with two dips repeats across Period 2 (dips at B and O), which is strong evidence that the explanation in terms of s and p subshells and orbital pairing is general.
In exam answers about the dips, name the subshell or orbital the electron is removed from; vague references to 'stability' gain no credit.

Melting points across Period 3 3.2.1.2

Key results

Approximate melting points: Na 371 K, Mg 923 K, Al 933 K, Si 1687 K, P₄ 317 K, S₈ 388 K, Cl₂ 172 K, Ar 84 K.

Notes

Na → Al (giant metallic): the ions become smaller and more highly charged and there are more delocalised electrons per atom, so metallic bonding strengthens.
Si (giant covalent): melting breaks many strong covalent bonds throughout the network, giving the highest value.
P₄, S₈ and Cl₂ (simple molecular): melting overcomes only London forces. S₈ melts highest because it is the largest molecule with the most electrons, so its London forces are strongest; Cl₂ is the smallest.
Ar is monatomic, with very weak London forces between single atoms.
Electrical conductivity follows the same structures: Na, Mg and Al conduct (delocalised electrons); Si is a semiconductor; P, S, Cl and Ar do not conduct.
Al has the most delocalised electrons per atom (three) and the smallest, most highly charged ion of the three metals, but its melting point is only slightly above magnesium's because its crystal structure is less favourable, so do not over-claim a large difference.
For the molecular elements, compare molecules, not atoms: $S_{8}$ (128 electrons) > $P_{4}$ (60) > $Cl_{2}$ (34) > Ar (18), which matches the order of their melting points.
In explanations, state the structure, the type of bonding or force overcome on melting, and its relative strength; this is the three-part answer examiners look for.

Worked example

Explain why silicon (melting point 1414 °C) cannot be classified as a typical metal or a typical simple molecular substance, using both melting point and electrical conductivity as evidence.

Show worked solution

Silicon's melting point is far higher than any of the simple molecular elements in Period 3 (P, S, Cl₂, Ar all melt below 120 °C), which rules out a simple molecular structure, since melting a simple molecular solid only requires overcoming weak van der Waals forces between molecules, not strong covalent bonds.

However, silicon is a poor electrical conductor, not a good one, which rules out a simple metallic classification, since a metal's good conductivity comes from freely delocalised electrons that silicon does not have.

Both observations together are consistent with a giant covalent (macromolecular) structure: every silicon atom is joined to four neighbours by strong covalent bonds throughout a continuous lattice, so a very high temperature is needed to melt it (accounting for the melting point), but there are no mobile charge carriers to conduct electricity (accounting for the poor conductivity).

3.2.2

Group 2, the alkaline earth metals

Trends down Group 2 3.2.2

Key results

Atomic radius increases down the group; first ionisation energy decreases (Mg 738 → Ba 503 kJ mol⁻¹).
Melting point generally decreases from Ca to Ba; magnesium is lower than calcium, an irregularity caused by its different crystal structure.

Notes

Each element has an extra occupied shell, so the outer electrons are further from the nucleus and more shielded; less energy is needed to remove them, despite the greater nuclear charge.
Each atom contributes two delocalised electrons, but larger ions mean the delocalised electrons are further from the ion centres, weakening metallic bonding down the group.
All Group 2 elements form $M^{2 +}$ ions by losing their two outer s electrons.
Reactivity increases down the group because the outer electrons are lost more easily, the same reason ionisation energy falls. This is the opposite trend to Group 7, where reactivity decreases down the group.
Flame tests distinguish the ions: calcium brick red, strontium crimson, barium apple green; magnesium gives no flame colour. The colours come from electrons falling back to lower energy levels after excitation.
Beryllium is atypical: its very small ion polarises anions so strongly that many of its compounds are covalent, and its oxide is amphoteric. It is not examined in detail at A-level but explains why trends often start at magnesium.

Reactions with water 3.2.2

Key results

Ca, Sr, Ba: $M + 2 H_{2} O \to M(OH)_{2} + H_{2}$ , increasingly vigorously down the group.
Magnesium reacts very slowly with cold water, $Mg + 2 H_{2} O (l) \to Mg(OH)_{2} + H_{2}$ , but readily with steam: $Mg + H_{2} O (g) \to MgO + H_{2}$ (white solid).

Notes

Reactivity increases down the group because the outer electrons are lost more easily (lower ionisation energies).
Calcium hydroxide is only slightly soluble, so a cloudy white suspension may form.
Observations for calcium: fizzing, the metal sinks and then rises as bubbles form, a white cloudy suspension, and the solution warms. Barium reacts faster with the same products.
The solutions formed are alkaline: the hydroxides release $OH^{-}$ ions, so universal indicator turns blue or purple, more strongly for barium because its hydroxide is more soluble.
Each reaction is a redox reaction: the metal is oxidised from 0 to +2 and hydrogen is reduced from +1 to 0.
Magnesium with steam burns with a bright white flame; the steam is generated by heating mineral wool soaked in water and passed over heated magnesium ribbon.

Solubility of hydroxides and sulfates 3.2.2

Key results

Hydroxides become more soluble down the group: $Mg(OH)_{2}$ is sparingly soluble; $Ba(OH)_{2}$ is the most soluble.
Sulfates become less soluble down the group: $MgSO_{4}$ is soluble; $BaSO_{4}$ is essentially insoluble.
Approximate solubilities at 20 °C: $Mg(OH)_{2}$ 0.0009 g, $Ca(OH)_{2}$ 0.17 g and $Ba(OH)_{2}$ about 4 g per 100 g water; $MgSO_{4}$ about 35 g, $CaSO_{4}$ 0.2 g and $BaSO_{4}$ 0.0002 g per 100 g water.

Notes

The two trends run in opposite directions; learn them as a pair.
These trends explain the uses of $Mg(OH)_{2}$ , $Ca(OH)_{2}$ and $BaSO_{4}$ and the sulfate test.
Saturated calcium hydroxide solution is limewater, used to test for carbon dioxide: $Ca(OH)_{2} + CO_{2} \to CaCO_{3} + H_{2} O$ gives a white precipitate (the 'milky' appearance).
Adding NaOH solution to solutions of Group 2 ions distinguishes them: $Mg^{2 +}$ gives a white precipitate even with dilute NaOH, $Ca^{2 +}$ only a slight precipitate, and $Ba^{2 +}$ none, matching the hydroxide solubility trend.
Adding dilute sulfuric acid to the same solutions gives the reverse pattern: a thick white precipitate with $Ba^{2 +}$ , little with $Ca^{2 +}$ and none with $Mg^{2 +}$ .

Testing for sulfate ions 3.2.2

Key results

$Ba^{2 +} (a q) + SO_{4}^{2 -} (a q) \to BaSO_{4} (s)$ .

Method

Acidify the sample with dilute hydrochloric acid (or nitric acid).
Add barium chloride solution: a white precipitate of barium sulfate shows sulfate ions.

Notes

The acid removes carbonate ions ( $CO_{3}^{2 -} + 2 H^{+} \to CO_{2} + H_{2} O$ ), which would otherwise give a white barium carbonate precipitate.
Sulfuric acid must not be used to acidify, since it adds sulfate ions.
Barium nitrate solution can be used in place of barium chloride, which is useful when the same sample will afterwards be tested for halide ions with silver nitrate.
Sulfite ions also give a white precipitate with barium ions, but barium sulfite dissolves in the acid, so acidifying first removes this possible interference too.
Write the ionic equation for a positive result rather than the full equation, and give the observation: 'white precipitate' (not 'cloudy' or 'milky').

Uses linked to the chemistry 3.2.2

Key results

$Mg(OH)_{2}$ in indigestion remedies and $Ca(OH)_{2}$ in agriculture neutralise acid: $M(OH)_{2} + 2 H^{+} \to M^{2 +} + 2 H_{2} O$ .
$BaSO_{4}$ is given as a 'barium meal' for X-ray imaging of the gut.

Notes

Barium ions are toxic, but $BaSO_{4}$ is so insoluble that very little barium enters the body; barium atoms absorb X-rays strongly, so the gut shows clearly.
Sparingly soluble antacids are safe because only a little dissolves at a time; more dissolves as hydroxide is used up.
Calcium hydroxide (slaked lime) and calcium carbonate (limestone) raise the pH of acidic soils, e.g. $Ca(OH)_{2} + 2 H^{+} \to Ca^{2 +} + 2 H_{2} O$ . Lime is cheap and only slightly soluble, so it works gradually without making the soil strongly alkaline.
Magnesium hydroxide is preferred to sodium hydroxide in antacids: being sparingly soluble, it never produces a high concentration of hydroxide ions that would damage the stomach lining.
Each use follows from a property in this topic: solubility (barium meal, antacid), basicity (soil treatment, flue gases), and reducing power (titanium extraction). Linking use to property is a common exam demand.

Group 2 in industry 3.2.2

Key results

Titanium extraction: $TiCl_{4} + 2 Mg \to Ti + 2 MgCl_{2}$ , at high temperature in an argon atmosphere.
Flue-gas desulfurisation: $CaO + SO_{2} \to CaSO_{3}$ or $CaCO_{3} + SO_{2} \to CaSO_{3} + CO_{2}$ .

Notes

Magnesium is the reducing agent: titanium goes from +4 to 0 and magnesium from 0 to +2. The argon prevents the hot metals reacting with oxygen and nitrogen.
Basic calcium compounds remove the acidic gas sulfur dioxide, reducing acid rain; the sulfite can be oxidised to calcium sulfate (gypsum).
Titanium dioxide ore is first converted to the chloride: $TiO_{2} + 2 C + 2 Cl_{2} \to TiCl_{4} + 2 CO$ , at about 1100 K. $TiCl_{4}$ is a covalent liquid that can be purified by fractional distillation.
Titanium cannot be made by heating its oxide with carbon, because titanium carbide forms and makes the metal brittle. The multistage magnesium process (the Kroll process) is expensive, which is why titanium is costly despite being abundant.
Desulfurisation uses a slurry of calcium oxide or carbonate sprayed through the flue gases. Removing sulfur dioxide matters because it forms sulfurous and sulfuric acids in the atmosphere, causing acid rain that damages buildings, forests and lakes.

Worked example

A dose of antacid tablet contains 0.500 g of magnesium hydroxide, $Mg(OH)_{2}$ ( $M_{r} = 58.3$ ).

Calculate the volume of stomach acid, of concentration 0.100 mol dm⁻³ HCl, that this dose can neutralise:

Mg(OH)_{2} (s) + 2 HCl (a q) \to MgCl_{2} (a q) + 2 H_{2} O (l)

Show worked solution

Moles of Mg(OH)₂:

= \frac{0.500}{58.3} = 8.58 \times 1 0^{- 3} mol

(to 3 s.f.).

The equation shows a 1:2 ratio of Mg(OH)₂ to HCl, so moles of HCl neutralised:

= 2 \times 8.58 \times 1 0^{- 3} = 1.72 \times 1 0^{- 2} mol

Volume of HCl:

= \frac{n}{c} = \frac{1.72 \times 1 0 ^{- 2}}{0.100} = 0.172 dm^{3} = 172 cm^{3}

(to 3 s.f.).

This shows why even a small mass of Mg(OH)₂ is an effective antacid dose: its low solubility means it is not immediately washed away, so it remains available to neutralise acid released over some time, and its diprotic hydroxide gives two moles of neutralising capacity per mole of compound.

3.2.3

Group 7 (17), the halogens

Trends in the halogens 3.2.3.1

Key results

F₂ pale yellow gas; Cl₂ yellow-green gas; Br₂ red-brown liquid; I₂ grey-black solid.
Boiling point increases down the group (85, 239, 332, 458 K); electronegativity decreases (3.98, 3.16, 2.96, 2.66).

Notes

Boiling point rises because the molecules get larger, with more electrons, so London forces between them are stronger.
Electronegativity falls because the bonding pair is further from the nucleus and more shielded, outweighing the greater nuclear charge.
Fluorine is anomalous: its F–F bond (158 kJ mol⁻¹) is weaker than Cl–Cl (243 kJ mol⁻¹) because the small atoms bring their lone pairs close enough to repel. This weak bond helps make fluorine the most reactive element.
All halogens exist as diatomic molecules with a single covalent bond; as elements they are oxidising agents, gaining one electron per atom to form halide ions with a noble-gas configuration.
Halogens dissolve better in non-polar solvents such as cyclohexane than in water; shaking an aqueous halogen with cyclohexane extracts it into the upper organic layer, where its colour (orange for $Br_{2}$ , violet for $I_{2}$ ) identifies it.

Oxidising ability and displacement 3.2.3.1

Key results

Oxidising ability: $F_{2} > Cl_{2} > Br_{2} > I_{2}$ . Reducing ability of halides: $I^{-} > Br^{-} > Cl^{-} > F^{-}$ .
$Cl_{2} + 2 Br^{-} \to 2 Cl^{-} + Br_{2}$ ; $Cl_{2} + 2 I^{-} \to 2 Cl^{-} + I_{2}$ ; $Br_{2} + 2 I^{-} \to 2 Br^{-} + I_{2}$ .

Notes

A halogen displaces any halide below it. Observations: bromine in water is orange; iodine in aqueous iodide is brown; iodine dissolved in cyclohexane is violet.
Smaller halogen atoms attract an incoming electron more strongly, so they are stronger oxidising agents.
Down the group the atom gets larger and the incoming electron is further from the nucleus and more shielded, so it is attracted less strongly: oxidising power falls.
Conversely, larger halide ions lose an electron more easily, so reducing power rises from fluoride to iodide. This idea links displacement, the silver nitrate test and the reactions with concentrated sulfuric acid.
Writing the ionic equation for each displacement and identifying the oxidising agent (the halogen) and the reducing agent (the halide ion) is a standard exam task.
No reaction occurs when a halogen is added to its own halide or to a halide above it; the solution simply shows the colour of the added halogen.

Identifying halide ions with silver nitrate 3.2.3.1, RP4

Key results

$Ag^{+} (a q) + X^{-} (a q) \to AgX (s)$ .
AgCl: white, dissolves in dilute $NH_{3}$ . AgBr: cream, dissolves only in concentrated $NH_{3}$ . AgI: yellow, insoluble even in concentrated $NH_{3}$ .

Method

Acidify the sample with dilute nitric acid, then add silver nitrate solution.
Note the precipitate colour, then test its solubility in dilute and concentrated ammonia.

Notes

Nitric acid removes carbonate and hydroxide ions that would also precipitate with silver ions; hydrochloric acid cannot be used because it adds chloride.
Silver fluoride is soluble, so fluoride gives no precipitate.
The ammonia test is needed because white and cream are hard to tell apart.
In ammonia, silver ions form the complex $[Ag(NH_{3})_{2}]^{+}$ , which lowers the concentration of free $Ag^{+}$ . AgCl, the most soluble silver halide, dissolves when this happens even in dilute ammonia; AgI, the least soluble, does not dissolve even in concentrated ammonia.
Silver halides darken in sunlight as they decompose to silver: $2 AgX \to 2 Ag + X_{2}$ . This was the basis of photographic film.
Required Practical 4 combines this test with others (Group 2 ions with NaOH and $H_{2} SO_{4}$ , sulfate with barium chloride, ammonium ions with warm NaOH and damp red litmus, carbonate with acid and limewater) to identify unknown ions.
Test for ammonium ions: warm with sodium hydroxide; ammonia gas is released and turns damp red litmus paper blue: $NH_{4}^{+} + OH^{-} \to NH_{3} + H_{2} O$ .

Halides with concentrated sulfuric acid 3.2.3.1

Key results

All halides first: $NaX + H_{2} SO_{4} \to NaHSO_{4} + HX$ (misty fumes).
Bromide also: $2 HBr + H_{2} SO_{4} \to Br_{2} + SO_{2} + 2 H_{2} O$ (S from +6 to +4).
Iodide also reduces sulfur further: to S (0), $6 HI + H_{2} SO_{4} \to 3 I_{2} + S + 4 H_{2} O$ , and to $H_{2} S$ (−2), $8 HI + H_{2} SO_{4} \to 4 I_{2} + H_{2} S + 4 H_{2} O$ .

Notes

Chloride is not a strong enough reducing agent to reduce sulfuric acid; only the acid–base reaction occurs.
Observations with iodide: purple vapour or black solid (iodine), yellow solid (sulfur), smell of bad eggs ( $H_{2} S$ ).
The trend shows reducing power increasing down the group: larger halide ions lose their outer electrons more easily.
Observations to quote: chloride, steamy white fumes of HCl only. Bromide, steamy fumes of HBr, orange-brown fumes of bromine and a colourless choking gas ( $SO_{2}$ ). Iodide, steamy fumes of HI, a black solid or purple vapour of iodine, a yellow solid (sulfur) and a smell of bad eggs ( $H_{2} S$ ).
Write the redox steps as half-equations to check the electrons: for bromide, $2 Br^{-} \to Br_{2} + 2 e^{-}$ and $H_{2} SO_{4} + 2 H^{+} + 2 e^{-} \to SO_{2} + 2 H_{2} O$ .
The first reaction of every halide is acid–base, not redox: the sulfuric acid donates a proton to the halide ion. Only the later reactions involve changes in the oxidation state of sulfur.
Because of these redox reactions, pure HBr and HI cannot be made by this method; phosphoric acid, which is not an oxidising agent, is used instead.

Chlorine, water treatment and bleach 3.2.3.2

Definitions

Disproportionation: a reaction in which the same element is simultaneously oxidised and reduced.

Key results

$Cl_{2} + H_{2} O ⇌ HCl + HClO$ ; in sunlight, $2 Cl_{2} + 2 H_{2} O \to 4 HCl + O_{2}$ .
With cold dilute sodium hydroxide: $Cl_{2} + 2 NaOH \to NaCl + NaClO + H_{2} O$ , giving bleach.

Notes

In these reactions chlorine (0) goes to −1 in chloride and +1 in chlorate(I), $ClO^{-}$ .
Chlorine kills pathogens in drinking water. It is toxic and can form chlorinated organic compounds, but at controlled concentrations the benefit to public health outweighs these risks.
Chloric(I) acid, HClO, is a weak acid; the balance between HClO and $ClO^{-}$ depends on pH.
In the reaction with water, chlorine goes from 0 to −1 (in HCl) and +1 (in HClO), so this is also disproportionation. Chloric(I) acid and the chlorate(I) ion are the species that kill bacteria.
Chlorine water turns universal indicator red (acids formed) and then colourless, because chloric(I) acid bleaches the dye.
Alternatives to chlorine include ozone and ultraviolet light, which kill microorganisms but leave no residual disinfectant in the pipes, which is why chlorine is still commonly added at the end of treatment.
Household bleach is sodium chlorate(I) solution. It must not be mixed with acidic cleaners, because acid reverses the equilibrium and releases toxic chlorine gas.