Inorganic chemistry II: A-level Chemistry AQA revision notes

3.2.4

Properties of Period 3 elements and their oxides

Reactions of Period 3 elements 3.2.4

Key results

With oxygen: $4 Na + O_{2} \to 2 Na_{2} O$ ; $2 Mg + O_{2} \to 2 MgO$ ; $4 Al + 3 O_{2} \to 2 Al_{2} O_{3}$ ; $Si + O_{2} \to SiO_{2}$ ; $P_{4} + 5 O_{2} \to P_{4} O_{10}$ ; $S + O_{2} \to SO_{2}$ .
With water: sodium reacts rapidly, $2 Na + 2 H_{2} O \to 2 NaOH + H_{2}$ ; magnesium reacts very slowly with cold water but readily with steam to give MgO.

Notes

The oxidation state of the element in its highest oxide equals its group's number of outer electrons, from +1 (Na₂O) to +6 (SO₃).
Sulfur burns to $SO_{2}$ ; further oxidation to $SO_{3}$ needs a $V_{2} O_{5}$ catalyst (the Contact process).
Flame colours when burning: sodium yellow, magnesium brilliant white, sulfur blue.
Observations on burning in oxygen: sodium burns with a yellow-orange flame to a white solid; magnesium with a brilliant white flame to a white powder; aluminium is protected by a thin oxide layer and burns only when powdered; silicon needs strong heating; white phosphorus ignites spontaneously with a white flame and white smoke; sulfur burns with a blue flame to a colourless choking gas.
With chlorine the elements form NaCl, $MgCl_{2}$ , $AlCl_{3}$ , $SiCl_{4}$ and $PCl_{5}$ ; the bonding changes from ionic to covalent across the period, as it does for the oxides.
Sodium's reaction with water is far more vigorous than magnesium's because sodium loses one electron (lower ionisation energy) rather than two, and its hydroxide is soluble, so no protective layer forms.

Structure and melting points of the oxides 3.2.4

Key results

Approximate melting points: $Na_{2} O$ 1548 K, MgO 3125 K, $Al_{2} O_{3}$ 2345 K, $SiO_{2}$ 1883 K, $P_{4} O_{10}$ 613 K (sublimes), $SO_{3}$ 290 K, $SO_{2}$ 200 K.

Notes

$Na_{2} O$ , MgO and $Al_{2} O_{3}$ are giant ionic and have high melting points. MgO is highest because $Mg^{2 +}$ is smaller and more highly charged than $Na^{+}$ , giving stronger electrostatic attraction.
$Al_{2} O_{3}$ is lower than expected for its charges because its bonding has some covalent character.
$SiO_{2}$ is giant covalent: each Si bonds to four O and each O to two Si, so melting breaks many strong covalent bonds.
$P_{4} O_{10}$ and $SO_{3}$ are molecular: only intermolecular forces are overcome, so they melt at low temperatures.
MgO's very high melting point makes it useful as a refractory lining for furnaces.
Molten ionic oxides conduct electricity because their ions become mobile; this is how aluminium is extracted, by electrolysis of alumina dissolved in molten cryolite. The molecular and giant covalent oxides do not conduct when molten.
$P_{4} O_{10}$ melts higher than $SO_{3}$ because its molecules are larger, with more electrons and therefore stronger London forces.

The oxides with water 3.2.4

Key results

$Na_{2} O + H_{2} O \to 2 NaOH$ (pH about 14). $MgO + H_{2} O \to Mg(OH)_{2}$ (pH about 9–10, since it is only slightly soluble).
$P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} PO_{4}$ ; $SO_{2} + H_{2} O \to H_{2} SO_{3}$ ; $SO_{3} + H_{2} O \to H_{2} SO_{4}$ (all acidic, pH about 0–3).
$Al_{2} O_{3}$ and $SiO_{2}$ are insoluble and do not change the pH of water.

Notes

Whether an oxide dissolves in water is separate from whether it is acidic or basic: alumina and silica are insoluble but still react with bases (and, for alumina, acids).
pH values depend on concentration; the pattern (alkaline on the left, acidic on the right) is what matters.
Sodium oxide is the oxide of a reactive metal: its $O^{2 -}$ ions are strong bases that take protons from water, $O^{2 -} + H_{2} O \to 2 OH^{-}$ .
Sulfur dioxide dissolving in rainwater lowers its pH; this is the origin of acid rain from burning sulfur-containing fuels.
Phosphorus(V) oxide reacts so readily with water that it is used as a drying agent. The phosphoric acid formed is a weak acid, but at the concentrations produced the solution is still strongly acidic.
Write equations with state symbols where asked: e.g. $SO_{3} (g) + H_{2} O (l) \to H_{2} SO_{4} (a q)$ .

Basic, amphoteric and acidic oxides 3.2.4

Key results

Basic: $Na_{2} O + 2 H^{+} \to 2 Na^{+} + H_{2} O$ ; $MgO + 2 H^{+} \to Mg^{2 +} + H_{2} O$ .
Amphoteric: $Al_{2} O_{3} + 6 H^{+} \to 2 Al^{3 +} + 3 H_{2} O$ and $Al_{2} O_{3} + 2 OH^{-} + 3 H_{2} O \to 2 [Al(OH)_{4}]^{-}$ .
Acidic: $SiO_{2} + 2 OH^{-} \to SiO_{3}^{2 -} + H_{2} O$ (hot, concentrated alkali); $P_{4} O_{10} + 12 OH^{-} \to 4 PO_{4}^{3 -} + 6 H_{2} O$ ; $SO_{2} + 2 OH^{-} \to SO_{3}^{2 -} + H_{2} O$ ; $SO_{3} + 2 OH^{-} \to SO_{4}^{2 -} + H_{2} O$ .

Notes

The change from basic to acidic across the period follows the change in bonding: ionic oxides contain $O^{2 -}$ ions, which accept protons; covalent oxides react with water or hydroxide to form oxyanions.
With less alkali, partial neutralisation gives hydrogen salts such as $NaHSO_{3}$ or $NaH_{2} PO_{4}$ .
Aluminium oxide is amphoteric because its bonding is intermediate: the small, highly charged $Al^{3 +}$ polarises the oxide ions, giving partial covalent character. It reacts as a base with acids and as an acid with hydroxide ions.
Silicon dioxide reacts only with hot concentrated alkali (or molten bases) because its giant covalent lattice must be broken; it does not react with acids (other than HF).

Oxyacids and their anions 3.2.4

Key results

$H_{3} PO_{4}$ → $PO_{4}^{3 -}$ (tetrahedral); $H_{2} SO_{3}$ → $SO_{3}^{2 -}$ (trigonal pyramidal, one lone pair on S); $H_{2} SO_{4}$ → $SO_{4}^{2 -}$ (tetrahedral).

Notes

Each acidic H is bonded to an oxygen; losing protons in turn gives intermediate ions such as $H_{2} PO_{4}^{-}$ , $HPO_{4}^{2 -}$ , $HSO_{3}^{-}$ and $HSO_{4}^{-}$ .
In the anions the negative charge is delocalised over the oxygens, so the S–O (or P–O) bonds are equivalent.
' $H_{2} SO_{3}$ ' is conventional notation: aqueous sulfur dioxide is a mixture of dissolved $SO_{2}$ , $HSO_{3}^{-}$ and $SO_{3}^{2 -}$ .
Sulfuric acid's first dissociation is essentially complete, but $HSO_{4}^{-} ⇌ H^{+} + SO_{4}^{2 -}$ is only partial ( $K_{a} \approx 1 \times 1 0^{- 2}$ ). Phosphoric acid is weak even in its first step (p $K_{a}$ about 2.1).
Draw the structures with each acidic hydrogen on an oxygen: $H_{3} PO_{4}$ is $O=P(OH)_{3}$ ; $H_{2} SO_{4}$ is $(HO)_{2} SO_{2}$ . The oxidation state of the central atom (+5 for P, +6 for S) matches the number of electrons it uses in bonding.
The salts formed depend on the amount of alkali: 1, 2 or 3 moles of NaOH per mole of $H_{3} PO_{4}$ give $NaH_{2} PO_{4}$ , $Na_{2} HPO_{4}$ or $Na_{3} PO_{4}$ .

3.2.5

Transition metals

Transition metals and ligands 3.2.5.1–3.2.5.2

Definitions

Transition metal: an element that forms at least one stable ion with a partially filled d subshell (Ti to Cu in Period 4).
Ligand: a molecule or ion that donates a lone pair to a metal ion, forming a co-ordinate bond. Complex: a central metal ion surrounded by ligands.
Co-ordination number: the number of co-ordinate bonds to the central metal ion.

Key results

Monodentate ligands: $H_{2} O$ , $NH_{3}$ , $Cl^{-}$ . Bidentate: ethane-1,2-diamine (en), ethanedioate $C_{2} O_{4}^{2 -}$ . Multidentate: $EDTA^{4 -}$ (six donor atoms).
$[M(H_{2} O)_{6}]^{2 +} + 3 en \to [M(en)_{3}]^{2 +} + 6 H_{2} O$ .

Notes

Sc and Zn are d-block but not transition metals: $Sc^{3 +}$ is $3 d^{0}$ and $Zn^{2 +}$ is $3 d^{10}$ .
The chelate effect: replacing monodentate ligands with multidentate ones increases the number of free particles (4 → 7 above), so $Δ S$ is positive. $Δ H$ is small because similar bonds are broken and made, so $Δ G$ is negative and the chelate complex is more stable.
Characteristic properties of transition metals: complex formation, formation of coloured ions, variable oxidation states and catalytic activity. All arise from the incompletely filled d subshell.
Worked example of the chelate effect: $[Cu(H_{2} O)_{6}]^{2 +} + EDTA^{4 -} \to [Cu(EDTA)]^{2 -} + 6 H_{2} O$ goes from 2 particles to 7, a large increase in entropy. EDTA is used to treat heavy-metal poisoning and to remove $Ca^{2 +}$ in water softening.
In haem, an Fe(II) ion is held by four nitrogen atoms of a porphyrin ring (a multidentate ligand), a fifth N from the globin protein, and a sixth site that binds $O_{2}$ or water.
The charge on a complex is the metal ion's charge plus the ligand charges: $[Fe(CN)_{6}]^{4 -}$ contains Fe(II), and $[CuCl_{4}]^{2 -}$ contains Cu(II).

Shapes and isomerism of complexes 3.2.5.3

Key results

Octahedral (CN 6, 90°), e.g. $[Co(H_{2} O)_{6}]^{2 +}$ ; tetrahedral (CN 4, 109.5°), e.g. $[CoCl_{4}]^{2 -}$ ; square planar (CN 4, 90°), e.g. $[Pt(NH_{3})_{2} Cl_{2}]$ ; linear (CN 2), e.g. $[Ag(NH_{3})_{2}]^{+}$ in Tollens' reagent.
Cis–trans isomerism in square planar and octahedral complexes with two identical ligands; optical isomerism in octahedral complexes with three bidentate ligands, e.g. $[Co(en)_{3}]^{3 +}$ .

Notes

Large ligands such as $Cl^{-}$ fit fewer around the metal, which is why chloro complexes are often tetrahedral.
Cisplatin is the cis isomer of $[Pt(NH_{3})_{2} Cl_{2}]$ ; only the cis form is an anticancer drug, because both chlorides must be replaced by neighbouring sites on DNA.
Octahedral complexes with two ligands of one kind, such as $[Co(NH_{3})_{4} Cl_{2}]^{+}$ , also show cis–trans isomerism: the two chloride ligands are at 90° (cis) or 180° (trans).
Optical isomers of $[Ni(en)_{3}]^{2 +}$ are non-superimposable mirror images: the three bidentate ligands wrap around the metal like a propeller in either a left-handed or right-handed sense.
When drawing octahedral complexes, use wedges and dashes to show 3D arrangement, and draw the co-ordinate bonds from the donor atom (e.g. the N of $NH_{3}$ ), not from a hydrogen.

Ligand substitution 3.2.5.2

Key results

$[Cu(H_{2} O)_{6}]^{2 +} + 4 NH_{3} \to [Cu(NH_{3})_{4} (H_{2} O)_{2}]^{2 +} + 4 H_{2} O$ : pale blue to deep blue; co-ordination number unchanged.
$[Co(H_{2} O)_{6}]^{2 +} + 4 Cl^{-} ⇌ [CoCl_{4}]^{2 -} + 6 H_{2} O$ : pink octahedral to blue tetrahedral; co-ordination number falls from 6 to 4.

Notes

$H_{2} O$ and $NH_{3}$ are similar in size and both uncharged, so they swap without changing the co-ordination number; $Cl^{-}$ is larger.
Haemoglobin contains Fe(II) held by four nitrogen donors in haem. Carbon monoxide binds more strongly than oxygen and replaces it, reducing oxygen transport: this is why CO is toxic.
Adding concentrated hydrochloric acid to copper(II) sulfate solution turns it from blue to yellow-green as $[CuCl_{4}]^{2 -}$ forms, again with co-ordination number falling from 6 to 4 because chloride ions are larger.
The cobalt reaction is an equilibrium: adding water, or cooling, shifts it back towards the pink hexaaqua ion. This is used in humidity indicators.
Partial substitution is common: in excess ammonia only four of copper's six water ligands are replaced, giving $[Cu(NH_{3})_{4} (H_{2} O)_{2}]^{2 +}$ , not $[Cu(NH_{3})_{6}]^{2 +}$ .

Coloured ions and colorimetry 3.2.5.4

Key results

$Δ E = h ν = \frac{h c}{λ}$ , where $Δ E$ is the energy gap between split d orbitals.

Notes

Ligands split the d orbitals into two energy levels. A d electron absorbs visible light of energy $Δ E$ and is promoted; the colour seen is the complementary colour of the light not absorbed.
Changing the metal, its oxidation state, the ligands or the co-ordination number changes $Δ E$ and therefore the colour.
Ions with empty or full d subshells ( $3 d^{0}$ , $3 d^{10}$ ) cannot have d–d transitions and are colourless.
Colorimetry: absorbance is proportional to concentration over a working range. Choose a filter of the colour most absorbed, zero with a blank, and make a calibration curve from standard solutions.
Worked example: a complex that absorbs most strongly at 600 nm has $Δ E = h c / λ = (6.63 \times 1 0^{- 34} \times 3.00 \times 1 0^{8}) / (6.00 \times 1 0^{- 7}) = 3.32 \times 1 0^{- 19}$ J per ion, or 200 kJ mol⁻¹ when multiplied by $N_{A}$ .
An ion that absorbs orange-red light appears blue-green: copper(II) aqua ions absorb in the red, so their solutions look blue.
To measure iron in a sample by colorimetry, iron(III) can be converted to its intensely red thiocyanate complex; the more strongly coloured species gives better sensitivity at low concentrations.
Concentration is read from the calibration graph, not calculated from a single standard, because the absorbance–concentration line may not pass exactly through the origin.

Variable oxidation states 3.2.5.5

Key results

Vanadium reduced by zinc in acid: $VO_{2}^{+}$ (+5, yellow) → $VO^{2 +}$ (+4, blue) → $V^{3 +}$ (+3, green) → $V^{2 +}$ (+2, violet).
Manganate(VII) titrations: $MnO_{4}^{-} : Fe^{2 +} = 1 : 5$ ; $MnO_{4}^{-} : C_{2} O_{4}^{2 -} = 2 : 5$ .

Notes

Variable oxidation states are possible because the 3d and 4s electrons are close in energy, so different numbers can be lost or shared.
Manganate(VII) titrations need no indicator: the endpoint is the first permanent pale pink as excess $MnO_{4}^{-}$ appears. The acid used is dilute sulfuric acid; hydrochloric acid would be oxidised to chlorine.
The redox potential of a transition metal ion depends on pH and on the ligands: for example, Ag(I) is a milder oxidant in Tollens' reagent, $[Ag(NH_{3})_{2}]^{+}$ , than as aqueous $Ag^{+}$ .
Worked example: 25.0 cm³ of iron(II) solution needs 22.40 cm³ of 0.0200 mol dm⁻³ $KMnO_{4}$ . $n (MnO_{4}^{-}) = 4.48 \times 1 0^{- 4}$ mol, so $n (Fe^{2 +}) = 5 \times 4.48 \times 1 0^{- 4} = 2.24 \times 1 0^{- 3}$ mol, and $[Fe^{2 +}] = 0.0896 mol dm^{- 3}$ .
Chromium also changes colour with oxidation state: orange dichromate(VI) is reduced by zinc in acid to green $Cr^{3 +}$ and then blue $Cr^{2 +}$ , which is re-oxidised by air. In alkali, dichromate converts to yellow chromate(VI).
Cobalt(II) in ammonia solution is easily oxidised to cobalt(III), by air or by hydrogen peroxide, because ammonia ligands stabilise the +3 state.
In the ethanedioate titration the mixture must be warmed to about 60 °C, because the reaction is slow at the start until $Mn^{2 +}$ builds up.

Catalysis 3.2.5.6

Definitions

Heterogeneous catalyst: in a different phase from the reactants; reaction happens at active sites on its surface (adsorption, reaction, desorption).
Homogeneous catalyst: in the same phase as the reactants; it works through an intermediate species.
Autocatalysis: a product of the reaction catalyses it.

Key results

Contact process: $V_{2} O_{5} + SO_{2} \to 2 VO_{2} + SO_{3}$ , then $2 VO_{2} + \frac{1}{2} O_{2} \to V_{2} O_{5}$ .
$Fe^{2 +}$ catalyses $S_{2} O_{8}^{2 -} + 2 I^{-} \to 2 SO_{4}^{2 -} + I_{2}$ via $S_{2} O_{8}^{2 -} + 2 Fe^{2 +} \to 2 SO_{4}^{2 -} + 2 Fe^{3 +}$ , then $2 Fe^{3 +} + 2 I^{-} \to 2 Fe^{2 +} + I_{2}$ .
$Mn^{2 +}$ autocatalyses the reaction of manganate(VII) with ethanedioate ions.

Notes

Transition metals are good catalysts because they change oxidation state easily, letting them take part in redox steps and then be regenerated.
The uncatalysed persulfate–iodide reaction is slow because two negative ions repel; each catalysed step involves oppositely charged ions.
A support (such as a ceramic honeycomb) increases surface area and so saves expensive catalyst; poisons such as sulfur or lead block active sites and reduce efficiency.
An autocatalysed reaction starts slowly and then speeds up as the catalyst (here $Mn^{2 +}$ ) accumulates.
Other important heterogeneous catalysts: iron in the Haber process, nickel for hydrogenating vegetable oils, and platinum, palladium and rhodium in catalytic converters. Adsorption must be strong enough to weaken bonds but weak enough for the products to leave: tungsten adsorbs too strongly and silver too weakly to be useful.

Worked example

A 25.0 cm³ sample of acidified iron(II) sulfate solution required 22.40 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) solution to reach the end point.

Calculate the concentration of Fe²⁺ in the original solution.

Show worked solution

Moles of MnO₄⁻ used:

n = \frac{22.40}{1000} \times 0.0200 = 4.48 \times 1 0^{- 4} mol

The balanced equation,

MnO_{4}^{-} + 8 H^{+} + 5 Fe^{2 +} \to Mn^{2 +} + 4 H_{2} O + 5 Fe^{3 +}

shows 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺, so moles of Fe²⁺:

= 5 \times 4.48 \times 1 0^{- 4} = 2.24 \times 1 0^{- 3} mol

Concentration of Fe²⁺:

= \frac{2.24 \times 1 0 ^{- 3}}{25.0/1000} = 0.0896 mol dm^{- 3}

3.2.6

Reactions of ions in aqueous solution

Aqua ions and their acidity 3.2.6

Key results

$[M(H_{2} O)_{6}]^{3 +} + H_{2} O ⇌ [M(H_{2} O)_{5} (OH)]^{2 +} + H_{3} O^{+}$ (M = Al or Fe).
Colours: $[Fe(H_{2} O)_{6}]^{2 +}$ pale green; $[Cu(H_{2} O)_{6}]^{2 +}$ blue; $[Fe(H_{2} O)_{6}]^{3 +}$ pale violet, usually seen as yellow-brown; $[Al(H_{2} O)_{6}]^{3 +}$ colourless.

Notes

A 3+ ion has a higher charge density than a 2+ ion, so it polarises the O–H bonds of its water ligands more strongly and releases $H^{+}$ more readily: 3+ aqua ions are more acidic.
Iron(III) solutions usually look yellow-brown rather than violet because some of the hydrolysed species are present.
Typical pH values for 0.1 mol dm⁻³ solutions: iron(III) about 2, aluminium about 3, iron(II) and copper(II) about 4–6. The 3+ ions are as acidic as some carboxylic acids.
The reaction is a Brønsted–Lowry acid reaction: a co-ordinated water molecule donates a proton to a solvent water molecule. The metal ion stays bonded to the oxygen, now as an $OH^{-}$ ligand.
Adding acid shifts the equilibria to the left, which is why iron(III) solutions are kept acidic to stop brown iron(III) hydroxide forming.

Tests with hydroxide and ammonia 3.2.6

Key results

$Fe^{2 +}$ : green precipitate, unchanged in excess NaOH or $NH_{3}$ ; turns brown on standing in air.
$Cu^{2 +}$ : blue precipitate; unchanged in excess NaOH; dissolves in excess $NH_{3}$ to a deep blue solution.
$Fe^{3 +}$ : brown precipitate, unchanged in excess of either.
$Al^{3 +}$ : white precipitate; dissolves in excess NaOH to a colourless solution; unchanged in excess $NH_{3}$ .

Notes

Add reagent dropwise, then in excess, using a fresh portion of the solution for each reagent.
Iron(II) hydroxide darkens because air oxidises it: $4 Fe(OH)_{2} + O_{2} + 2 H_{2} O \to 4 Fe(OH)_{3}$ .
With dilute ammonia the equations are written with $NH_{3}$ as the base, e.g. $[Fe(H_{2} O)_{6}]^{3 +} + 3 NH_{3} \to [Fe(H_{2} O)_{3} (OH)_{3}] + 3 NH_{4}^{+}$ .
To distinguish aluminium from a Group 2 or other colourless ion, use excess NaOH: only aluminium hydroxide dissolves, because it is amphoteric.
Unknown ions are often identified by combining tests: e.g. a pale green solution that gives a green precipitate with NaOH turning brown at the surface contains $Fe^{2 +}$ ; confirming with $Ba^{2 +}$ and acid shows whether it is iron(II) sulfate.
Record colours and states precisely: 'green precipitate' and 'deep blue solution' are expected; 'goes green' is not enough.

Why the precipitates form and dissolve 3.2.6

Key results

$[M(H_{2} O)_{6}]^{2 +} + 2 OH^{-} \to [M(H_{2} O)_{4} (OH)_{2}] + 2 H_{2} O$ ; with $NH_{3}$ the by-product is $NH_{4}^{+}$ .
$[M(H_{2} O)_{6}]^{3 +} + 3 OH^{-} \to [M(H_{2} O)_{3} (OH)_{3}] + 3 H_{2} O$ .
Amphoteric aluminium hydroxide: $Al(OH)_{3} + 3 H^{+} \to Al^{3 +} + 3 H_{2} O$ and $Al(OH)_{3} + OH^{-} \to [Al(OH)_{4}]^{-}$ .
Excess ammonia with copper: $[Cu(H_{2} O)_{4} (OH)_{2}] + 4 NH_{3} \to [Cu(NH_{3})_{4} (H_{2} O)_{2}]^{2 +} + 2 H_{2} O + 2 OH^{-}$ .

Notes

Hydroxide and ammonia first act as bases, removing protons from coordinated water until a neutral (uncharged) complex precipitates.
In excess, ammonia then acts as a ligand towards copper(II); this is ligand substitution, not redox: copper stays +2.
The neutral complex precipitates because, with no overall charge, it is no longer strongly hydrated by water molecules and the complexes can pack into a solid.
Aluminium hydroxide dissolves in excess hydroxide because a fourth proton is removed: $[Al(H_{2} O)_{3} (OH)_{3}] + OH^{-} \to [Al(H_{2} O)_{2} (OH)_{4}]^{-} + H_{2} O$ , giving a soluble anion.
Iron and copper hydroxides do not dissolve in excess hydroxide because they are not amphoteric; the extra $OH^{-}$ cannot remove further protons to any useful extent.

Tests with carbonate ions 3.2.6

Key results

2+ ions give metal carbonates: $M^{2 +} + CO_{3}^{2 -} \to MCO_{3}$ ( $FeCO_{3}$ green; $CuCO_{3}$ blue-green).
3+ ions give hydroxides and carbon dioxide: $2 [M(H_{2} O)_{6}]^{3 +} + 3 CO_{3}^{2 -} \to 2 [M(H_{2} O)_{3} (OH)_{3}] + 3 CO_{2} + 3 H_{2} O$ (brown Fe, white Al, with effervescence).

Notes

The difference reflects acidity: 3+ aqua ions are acidic enough to protonate carbonate, releasing carbon dioxide, so their carbonates do not form.
Effervescence is the observation that distinguishes a 3+ ion in this test.
The 3+ reaction can be understood in two stages: the acidic aqua ions release $H_{3} O^{+}$ , which reacts with carbonate to give $CO_{2}$ and water; removing $H_{3} O^{+}$ pulls the hydrolysis equilibria to the right until the neutral hydroxide precipitates.
The gas can be confirmed as carbon dioxide with limewater.
Summary for identification: $Cu^{2 +}$ blue-green precipitate; $Fe^{2 +}$ green precipitate; $Fe^{3 +}$ brown precipitate with fizzing; $Al^{3 +}$ white precipitate with fizzing.

Worked example

Two colourless test tubes contain, separately, aqueous Fe²⁺ and aqueous Al³⁺.

Describe and explain what would be observed on adding NaOH(aq) dropwise until in excess to each, and state how the two results distinguish the ions.

Show worked solution

With Fe²⁺: a dirty green precipitate of Fe(OH)₂ forms immediately and does not redissolve as more NaOH is added; left standing in air, the precipitate gradually darkens towards brown as Fe(OH)₂ is oxidised to Fe(OH)₃.

With Al³⁺: a white precipitate of Al(OH)₃ forms initially, but as NaOH is added in excess this precipitate dissolves again, giving a colourless solution containing the soluble aluminate ion $[Al (OH)_{4}]^{-}$ — Al(OH)₃ is amphoteric and reacts with the excess hydroxide as an acid, whereas Fe(OH)₂ is not amphoteric and cannot do this.

The distinguishing observation is therefore behaviour in excess NaOH specifically (redissolves versus stays as a precipitate), not the initial precipitate colour alone, since colour can be affected by concentration and cannot always be judged reliably by eye.