3.2 Inorganic chemistry — printable coverage pack

3.2.1.1 · Classification

An element is assigned to the s, p, d or f block from its position in the Periodic Table; that position is fixed by proton number.
Use the subshell receiving the differentiating electron: an outer $s$ subshell indicates the s block, an outer $p$ subshell the p block, and filling a $(n-1)d$ subshell the d block.
For example, $\mathrm{[Ne]3s^2}$ is an s-block arrangement, whereas $\mathrm{[Ne]3s^2 3p^4}$ places the element in the p block.
A common error is to classify every element with occupied d orbitals as d block; a filled inner $d^{10}$ subshell does not override the subshell associated with the element's position.

Tier 1 · Easy

1. An element has the electron configuration $\mathrm{[Ne]3s^2}$ . State its block in the Periodic Table.[1 mark]
Answer
- s block
Method: The differentiating electrons occupy an $s$ subshell, so the element is in the s block.

Tier 2 · Standard

1. Element X has proton number $34$ and electron configuration $\mathrm{[Ar]3d^{10}4s^2 4p^4}$ . Identify the block to which X belongs and give a reason.[2 marks]
Answer
- X is a p-block element because its differentiating electrons occupy the $4p$ subshell.
Method: Read the occupied subshell at the end of the configuration. Although X contains a filled $3d$ subshell, the differentiating electrons are in $4p$ , so its position is in the p block.

Tier 3 · Hard

1. Elements Y and Z have consecutive proton numbers. Y has configuration $\mathrm{[Ar]4s^2}$ and Z has configuration $\mathrm{[Ar]3d^1 4s^2}$ . State the block of each element and explain why the classifications differ even though both contain $4s$ electrons.[3 marks]
Answer
- Y is s block; Z is d block; the differentiating electron in Z enters the $3d$ subshell.
Method: Y ends the filling of the $4s$ subshell, so it occupies the s block. The next proton and electron in Z begin occupation of $3d$ . Block classification follows the differentiating subshell, making Z d block despite its retained $4s^2$ electrons.

3.2.1.2 · Physical properties of Period 3 elements

Across Na to Ar, atomic radius generally decreases because proton number and nuclear attraction increase while added electrons enter the same principal shell with similar shielding.
First ionisation energy generally increases, with a fall from Mg to Al because a $3p$ electron is higher in energy than a $3s$ electron, and a fall from P to S because paired $3p$ electrons repel.
Melting points reflect structure and bonding: stronger metallic bonding raises the values from Na to Al, giant covalent Si is very high, and the molecular or atomic species $\mathrm{P_4}$ , $\mathrm{S_8}$ , $\mathrm{Cl_2}$ and Ar depend on London forces.
A common error is to explain an across-period trend by extra shells or extra shielding; the relevant electrons are being added to the same main shell.

Tier 1 · Easy

1. Arrange Na, Mg and Cl in order of decreasing atomic radius and explain the order.[3 marks]
Answer
- $\mathrm{Na>Mg>Cl}$ ; nuclear charge increases across the period while shielding is similar, so attraction to the outer shell increases.
Method: All three atoms add their outer electrons to the third shell. Proton number rises from Na to Cl without an additional inner shell, so effective nuclear attraction increases and pulls the outer shell closer. Therefore the decreasing order is $\mathrm{Na>Mg>Cl}$ .

Tier 2 · Standard

1. Explain why the first ionisation energy falls from Mg to Al and also falls from P to S, despite the general increase across Period 3.[4 marks]
Answer
- Al loses a higher-energy, more shielded $3p$ electron rather than Mg's $3s$ electron.
- S has a pair of electrons in one $3p$ orbital, so electron-electron repulsion makes one easier to remove than an unpaired $3p$ electron from P.
Method: Compare the subshells first: Mg ends $3s^2$ but Al ends $3p^1$ , and a $3p$ electron is higher in energy and slightly more shielded. Then compare orbital occupancy: P is $3p^3$ with one electron per orbital, while S is $3p^4$ and contains a repelling pair. Each effect lowers the energy needed for the stated removal.

Tier 3 · Hard

1. Three Period 3 elements have melting points of approximately $1410\,{}^\circ\mathrm{C}$ , $115\,{}^\circ\mathrm{C}$ and $-189\,{}^\circ\mathrm{C}$ . Identify the elements as Si, S or Ar and explain the large differences.[6 marks]
Answer
- $1410\,{}^\circ\mathrm{C}$ : Si; $115\,{}^\circ\mathrm{C}$ : S; $-189\,{}^\circ\mathrm{C}$ : Ar.
- Si has a giant covalent lattice, $\mathrm{S_8}$ molecules have London forces between molecules, and monatomic Ar has still weaker London forces.
Method: Assign the giant covalent structure first: many strong Si-Si covalent bonds must be broken, giving the highest value. Sulfur exists as relatively large $\mathrm{S_8}$ molecules, so its London forces are appreciable but far weaker than covalent bonds. Argon is monatomic with the least polarizable electron cloud of the two simple species, so it has the weakest attractions and the lowest melting point.

3.2.2 · Group 2, the alkaline earth metals

From Mg to Ba, atomic radius increases and first ionisation energy decreases because each step adds a shell; reactions with water become more vigorous, producing $\mathrm{M(OH)_2}$ and $\mathrm{H_2}$ , while Mg reacts readily with steam to form MgO.
Group 2 hydroxides become more soluble down the group, but Group 2 sulfates become less soluble; sparingly soluble $\mathrm{Mg(OH)_2}$ and insoluble $\mathrm{BaSO_4}$ anchor these trends.
Uses follow the chemistry: $\mathrm{Mg(OH)_2}$ neutralises stomach acid, $\mathrm{Ca(OH)_2}$ neutralises acidic soil, CaO or $\mathrm{CaCO_3}$ removes $\mathrm{SO_2}$ from flue gas, and insoluble $\mathrm{BaSO_4}$ is used as an X-ray contrast medium.
In the sulfate test, acidify before adding $\mathrm{BaCl_2(aq)}$ so carbonate is removed; otherwise a white carbonate precipitate can be mistaken for $\mathrm{BaSO_4}$ .

Tier 1 · Easy

1. Steam is passed over heated magnesium ribbon. Give the chemical equation and the visible change.[2 marks]
Answer
- $\mathrm{Mg(s)+H_2O(g)\rightarrow MgO(s)+H_2(g)}$ ; magnesium burns or glows brightly and a white solid forms.
Method: Steam oxidises Mg to MgO and is reduced to hydrogen. One Mg atom and one water molecule balance the equation; the white product and bright reaction are the expected observations.

Tier 2 · Standard

1. A solution may contain both sulfate and carbonate ions. Describe how to test specifically for sulfate ions using barium chloride solution, including the reason for the first reagent and the final observation.[4 marks]
Answer
- Add dilute hydrochloric acid to remove carbonate, then add barium chloride solution; a white precipitate of $\mathrm{BaSO_4}$ confirms sulfate.
Method: Acid first converts carbonate to carbon dioxide and water: $\mathrm{CO_3^{2-}+2H^+\rightarrow CO_2+H_2O}$ . Adding $\mathrm{BaCl_2(aq)}$ then supplies $\mathrm{Ba^{2+}}$ , and sulfate gives the insoluble white solid by $\mathrm{Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)}$ .

Tier 3 · Hard

1. A flue-gas stream contains $1.28\times10^3\,\mathrm{mol}$ of $\mathrm{SO_2}$ . A power station uses limestone to capture $85.0\%$ of this amount in a $1{:}1$ mole ratio of $\mathrm{CaCO_3}$ to $\mathrm{SO_2}$ . The limestone is $92.0\%$ $\mathrm{CaCO_3}$ by mass. Calculate the mass of limestone required. Use $M_r(\mathrm{CaCO_3})=100.1$ .[5 marks]
Answer
- $1.18\times10^2\,\mathrm{kg}$ of limestone
Method: The amount captured is $(1.28\times10^3)(0.850)=1.088\times10^3\,\mathrm{mol}$ . The $1{:}1$ ratio requires the same amount of $\mathrm{CaCO_3}$ , whose mass is $(1.088\times10^3)(100.1)=1.089\times10^5\,\mathrm{g}$ . This is only $92.0\%$ of the limestone, so the required mass is $(1.089\times10^5)/0.920=1.18\times10^5\,\mathrm{g}=1.18\times10^2\,\mathrm{kg}$ .

3.2.3.1 · Trends in properties

Down Group 7, electronegativity decreases because radius and shielding increase, while boiling point increases because larger electron clouds give stronger London forces between $\mathrm{X_2}$ molecules.
Oxidising ability of the halogens decreases down the group, so a halogen displaces the ions of halogens below it; reducing ability of halide ions increases down the group.
Concentrated sulfuric acid gives only an acid-base reaction with chloride, but bromide reduces it mainly to $\mathrm{SO_2}$ and iodide can reduce it further to sulfur or $\mathrm{H_2S}$ .
For halide tests, use acidified $\mathrm{AgNO_3}$ : AgCl is white and dissolves in dilute ammonia, AgBr is cream and dissolves only in concentrated ammonia, and AgI is yellow and insoluble in ammonia; hydrochloric acid would introduce chloride ions.

Tier 1 · Easy

1. Chlorine water is added to aqueous potassium bromide. State the observation and write the ionic equation.[2 marks]
Answer
- The solution becomes orange; $\mathrm{Cl_2+2Br^-\rightarrow 2Cl^-+Br_2}$ .
Method: Chlorine is the stronger oxidising agent, so it gains electrons from bromide ions. Bromide is oxidised to orange bromine while chlorine is reduced to chloride; combining the half-equations gives the stated $1{:}2$ equation.

Tier 2 · Standard

1. Three colourless solutions contain chloride, bromide and iodide ions, one ion per solution. Describe a test that distinguishes all three, including every precipitate colour and its behaviour with ammonia.[6 marks]
Answer
- Acidify with dilute nitric acid and add silver nitrate: chloride gives white AgCl, bromide cream AgBr and iodide yellow AgI.
- AgCl dissolves in dilute ammonia, AgBr dissolves in concentrated ammonia, and AgI remains insoluble.
Method: Nitric acid removes interfering ions without adding a halide, and $\mathrm{Ag^+}$ precipitates each silver halide. Record white, cream and yellow in the order AgCl, AgBr and AgI. If colours are uncertain, add dilute ammonia first, then concentrated ammonia: successive dissolution identifies AgCl then AgBr, leaving AgI.

Tier 3 · Hard

1. Solid sodium iodide is treated with concentrated sulfuric acid. Give the initial acid-base equation, then an equation in which iodide reduces sulfuric acid to hydrogen sulfide. State two observations from the redox reaction and explain why iodide gives deeper reduction than bromide.[7 marks]
Answer
- $\mathrm{NaI+H_2SO_4\rightarrow NaHSO_4+HI}$ and $\mathrm{8HI+H_2SO_4\rightarrow 4I_2+H_2S+4H_2O}$ .
- Purple iodine vapour or a black solid and the rotten-egg smell of $\mathrm{H_2S}$ ; iodide is a stronger reducing agent because its outer electron is more shielded and farther from the nucleus.
Method: First transfer one proton from sulfuric acid to iodide to form HI. For the deepest reduction, sulfur changes from $+6$ in $\mathrm{H_2SO_4}$ to $-2$ in $\mathrm{H_2S}$ and gains eight electrons; eight iodide ions each lose one electron, forming four $\mathrm{I_2}$ . Balance H and O with four waters. Down the group the larger, more shielded halide ion loses an electron more readily, so iodide reduces sulfuric acid further than bromide.

3.2.3.2 · Uses of chlorine and chlorate(I)

Chlorine disproportionates in water to chloride and chlorate(I): $\mathrm{Cl_2+H_2O\rightleftharpoons 2H^++Cl^-+ClO^-}$ ; chlorate(I) species kill microorganisms by oxidation.
In sunlight, chlorine water can form chloride ions and oxygen overall: $\mathrm{2Cl_2+2H_2O\rightarrow 4H^++4Cl^-+O_2}$ .
Cold, dilute aqueous NaOH forms bleach by $\mathrm{Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O}$ ; the chlorate(I) solution is used for bleaching and disinfection.
A common evaluation error is to list only hazards: chlorine is toxic and chlorinated by-products may be harmful, but controlled treatment kills pathogens and provides a residual disinfectant, so the health benefit outweighs the risk.

Tier 1 · Easy

1. Write the equation for the reaction of chlorine with cold, dilute aqueous sodium hydroxide and state one use of the solution formed.[2 marks]
Answer
- $\mathrm{Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O}$ ; the solution is used as a bleach or disinfectant.
Method: One chlorine atom is reduced to chloride and the other is oxidised to chlorate(I), so equal amounts of NaCl and NaClO form. Balance sodium, hydrogen and oxygen with two NaOH and one water.

Tier 2 · Standard

1. Use oxidation states to show that the formation of chloride and chlorate(I) ions when chlorine reacts with water is disproportionation.[4 marks]
Answer
- Chlorine changes from $0$ in $\mathrm{Cl_2}$ to $-1$ in $\mathrm{Cl^-}$ and to $+1$ in $\mathrm{ClO^-}$ , so the same element is both reduced and oxidised.
Method: Assign chlorine oxidation state $0$ in the element. In chloride it is $-1$ , which is reduction, while in chlorate(I) the oxygen is $-2$ and the ion charge is $-1$ , making chlorine $+1$ , which is oxidation. Both changes occur in $\mathrm{Cl_2+H_2O\rightleftharpoons 2H^++Cl^-+ClO^-}$ , meeting the definition of disproportionation.

Tier 3 · Hard

1. $1.42\,\mathrm{g}$ of chlorine reacts completely with excess cold, dilute aqueous sodium hydroxide. The final solution has volume $250\,\mathrm{cm^3}$ . Calculate the concentration of $\mathrm{NaClO}$ formed. Use $M_r(\mathrm{Cl_2})=71.0$ .[4 marks]
Answer
- $0.0800\,\mathrm{mol\,dm^{-3}}$
Method: The amount of chlorine is $1.42/71.0=0.0200\,\mathrm{mol}$ . The equation $\mathrm{Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O}$ gives a $1{:}1$ ratio, so $0.0200\,\mathrm{mol}$ of NaClO forms. Convert the volume to $0.250\,\mathrm{dm^3}$ and calculate $c=n/V=0.0200/0.250=0.0800\,\mathrm{mol\,dm^{-3}}$ .

3.2.4 · Properties of Period 3 elements and their oxides (A-level only)

Na and Mg react with water, and the specified products of burning the elements are $\mathrm{Na_2O}$ , MgO, $\mathrm{Al_2O_3}$ , $\mathrm{SiO_2}$ , $\mathrm{P_4O_{10}}$ and $\mathrm{SO_2}$ ; $\mathrm{SO_3}$ is also studied as a Period 3 oxide, formed by further catalytic oxidation of $\mathrm{SO_2}$ rather than by burning sulfur directly.
The highest oxides change from ionic giant lattices through giant covalent $\mathrm{SiO_2}$ to molecular phosphorus and sulfur oxides, explaining the broad fall in melting point after the giant structures.
Basic $\mathrm{Na_2O}$ and MgO react with acids, amphoteric $\mathrm{Al_2O_3}$ reacts with acids and bases, and the acidic oxides from $\mathrm{SiO_2}$ onwards react with bases; $\mathrm{P_4O_{10}}$ , $\mathrm{SO_2}$ and $\mathrm{SO_3}$ form oxoacids in water.
A common error is to call every solid oxide a molecule or a precipitate; describe the actual structure and distinguish insolubility from failure to react.

Tier 1 · Easy

1. Write equations for the formation from the elements of magnesium oxide and phosphorus(V) oxide, $\mathrm{P_4O_{10}}$ .[2 marks]
Answer
- $\mathrm{2Mg+O_2\rightarrow 2MgO}$ and $\mathrm{P_4+5O_2\rightarrow P_4O_{10}}$ .
Method: Use the specified oxide formula in each case. Two Mg atoms balance one oxygen molecule for MgO, while ten oxygen atoms in $\mathrm{P_4O_{10}}$ require five $\mathrm{O_2}$ molecules.

Tier 2 · Standard

1. Explain why $\mathrm{Al_2O_3}$ and $\mathrm{SiO_2}$ have high melting points but $\mathrm{P_4O_{10}}$ has a much lower melting point.[5 marks]
Answer
- $\mathrm{Al_2O_3}$ has a giant ionic lattice with strong electrostatic attractions, $\mathrm{SiO_2}$ is giant covalent with many strong covalent bonds, and molecular $\mathrm{P_4O_{10}}$ has only intermolecular forces between molecules.
Method: Identify each structure before naming the force overcome. Melting $\mathrm{Al_2O_3}$ requires overcoming strong attraction between oppositely charged ions, and melting $\mathrm{SiO_2}$ requires breaking many Si-O covalent bonds. In $\mathrm{P_4O_{10}}$ , covalent bonds within each molecule remain intact and only the much weaker forces between molecules are overcome.

Tier 3 · Hard

1. For each oxide $\mathrm{Al_2O_3}$ , $\mathrm{SiO_2}$ and $\mathrm{SO_3}$ , state whether it is basic, amphoteric or acidic. Give one ionic equation showing the reaction of $\mathrm{Al_2O_3}$ with a base, one equation for $\mathrm{SiO_2}$ with a base, and the equation for $\mathrm{SO_3}$ with water.[7 marks]
Answer
- $\mathrm{Al_2O_3}$ is amphoteric, $\mathrm{SiO_2}$ and $\mathrm{SO_3}$ are acidic.
- $\mathrm{Al_2O_3+2OH^-+3H_2O\rightarrow 2[Al(OH)_4]^-}$ ; $\mathrm{SiO_2+2OH^-\rightarrow SiO_3^{2-}+H_2O}$ ; $\mathrm{SO_3+H_2O\rightarrow H_2SO_4}$ .
Method: Classify amphoteric aluminium oxide by its ability to react with both acids and bases. Balance its base reaction by forming two tetrahydroxoaluminate ions. Silicon dioxide is an acidic giant oxide and forms silicate with hydroxide. Sulfur trioxide is an acidic molecular oxide and hydrates directly to sulfuric acid.

3.2.5.1 · General properties of transition metals (A-level only)

A transition metal is a d-block element that forms at least one ion with an incomplete d sub-level; this definition excludes Sc, whose common ion is $d^0$ , and Zn, whose ion is $d^{10}$ .
Characteristic transition-metal behaviour includes complex formation, coloured ions, variable oxidation states and catalytic activity.
A ligand donates a lone pair to form a co-ordinate bond, a complex contains a central metal atom or ion surrounded by ligands, and co-ordination number counts co-ordinate bonds to the centre.
A common error is to count ligands rather than donor atoms: three bidentate ligands give co-ordination number six, not three.

Tier 1 · Easy

1. Define the term ligand.[2 marks]
Answer
- A molecule or ion that donates a pair of electrons to a central metal atom or ion to form a co-ordinate bond.
Method: Include both essential ideas: the ligand supplies an electron pair, and that pair forms a co-ordinate bond to the metal centre.

Tier 2 · Standard

1. For the complex ion $\mathrm{[Cr(C_2O_4)_3]^{3-}}$ , determine the oxidation state of chromium and the co-ordination number. Each ethanedioate ion is bidentate.[3 marks]
Answer
- Chromium is $+3$ ; co-ordination number $=6$ .
Method: Let the chromium oxidation state be $x$ . Three $\mathrm{C_2O_4^{2-}}$ ligands contribute $-6$ , so $x-6=-3$ and $x=+3$ . Each of the three ligands makes two co-ordinate bonds, giving $3\times2=6$ .

Tier 3 · Hard

1. Sc, Fe and Zn are d-block elements. Their common ions include $\mathrm{Sc^{3+}}$ with $d^0$ , $\mathrm{Fe^{2+}}$ with $d^6$ , and $\mathrm{Zn^{2+}}$ with $d^{10}$ . Use the definition of a transition metal to identify which of these three is a transition metal and justify every exclusion.[4 marks]
Answer
- Fe is a transition metal; $\mathrm{Fe^{2+}}$ has an incomplete d sub-level. Sc and Zn are excluded because their stated ions are respectively $d^0$ and $d^{10}$ .
Method: Apply the definition to an ion rather than merely checking that the element lies in the d block. An incomplete d sub-level contains between one and nine electrons, so $d^6$ qualifies. Empty $d^0$ and full $d^{10}$ sub-levels do not.

3.2.5.2 · Substitution reactions (A-level only)

$\mathrm{H_2O}$ , $\mathrm{NH_3}$ and $\mathrm{Cl^-}$ are monodentate ligands; water and ammonia are similarly sized and exchange without changing co-ordination number, while larger chloride can produce four-coordinate complexes.
Ethane-1,2-diamine and ethanedioate are bidentate, EDTA is multidentate, and replacing monodentate ligands with these chelating ligands is the chelate effect.
Chelate substitutions often have similar bond enthalpies on both sides but increase the number of particles, making $\Delta S$ positive and the substitution thermodynamically favourable.
Haem binds oxygen co-ordinately to Fe(II); carbon monoxide is toxic because it substitutes for the bonded oxygen and prevents effective oxygen transport.

Tier 1 · Easy

1. State whether each of $\mathrm{NH_3}$ , ethane-1,2-diamine and EDTA is a monodentate, bidentate or multidentate ligand.[3 marks]
Answer
- $\mathrm{NH_3}$ is monodentate; ethane-1,2-diamine is bidentate; EDTA is multidentate.
Method: Count the donor atoms that can bind to one metal centre. Ammonia donates through one nitrogen, ethane-1,2-diamine through two nitrogens, and EDTA through several donor atoms.

Tier 2 · Standard

1. Write an equation for the reaction of $\mathrm{[Cu(H_2O)_6]^{2+}}$ with excess ammonia and explain why the co-ordination number is unchanged.[4 marks]
Answer
- $\mathrm{[Cu(H_2O)_6]^{2+}+4NH_3\rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}+4H_2O}$ ; both sides have six donor atoms bonded to Cu.
Method: Excess ammonia replaces four water ligands but the product retains two waters. All ligands shown are monodentate, so the reactant has six Cu-O bonds and the product has four Cu-N plus two Cu-O bonds; both co-ordination numbers are six.

Tier 3 · Hard

1. The substitution $\mathrm{[M(H_2O)_6]^{2+}+3en\rightleftharpoons [M(en)_3]^{2+}+6H_2O}$ is strongly favoured, where en is neutral ethane-1,2-diamine. Explain the chelate effect using enthalpy, entropy and particle numbers.[5 marks]
Answer
- Similar numbers and types of metal-ligand bonds are broken and formed, so $\Delta H$ is small; four particles form seven particles, so $\Delta S$ is positive and the products are favoured.
Method: Count six M-O bonds broken and six M-N bonds formed, so the bond-enthalpy contributions broadly balance. Count independently moving species: one aqua complex plus three en molecules gives four particles, while one chelate complex plus six waters gives seven. The increase in disorder makes $T\Delta S$ favourable in $\Delta G=\Delta H-T\Delta S$ , shifting equilibrium toward the chelate.

3.2.5.3 · Shapes of complex ions (A-level only)

Small ligands commonly give octahedral six-coordinate complexes, whereas larger chloride ligands commonly give tetrahedral four-coordinate complexes.
Four-coordinate complexes can also be square planar, and $\mathrm{[Ag(NH_3)_2]^+}$ in Tollens' reagent is a linear two-coordinate complex.
Octahedral and square-planar complexes can show cis-trans isomerism; cisplatin is the cis square-planar isomer of $\mathrm{[Pt(NH_3)_2Cl_2]}$ .
Optical isomers are non-superimposable mirror images, often formed by octahedral complexes with bidentate ligands; a common error is to call any two different drawings optical isomers without checking mirror-image non-superimposability.

Tier 1 · Easy

1. State the shapes of $\mathrm{[Ag(NH_3)_2]^+}$ and $\mathrm{[CoCl_4]^{2-}}$ .[2 marks]
Answer
- $\mathrm{[Ag(NH_3)_2]^+}$ is linear; $\mathrm{[CoCl_4]^{2-}}$ is tetrahedral.
Method: The silver complex has two co-ordinate bonds arranged linearly. Four large chloride ligands around cobalt favour a tetrahedral arrangement.

Tier 2 · Standard

1. $\mathrm{[Pt(NH_3)_2Cl_2]}$ forms two stereoisomers. Name the type of stereoisomerism, describe the ligand arrangement in each isomer, and identify cisplatin.[4 marks]
Answer
- Cis-trans isomerism in a square-planar complex; like ligands are adjacent in cis and opposite in trans; cisplatin is the cis isomer.
Method: Place the four ligands in one plane around Pt. If the two chloride ligands occupy neighbouring positions the isomer is cis; if their positions are $180^\circ$ apart it is trans. The medicinal compound cisplatin is the adjacent-ligand form.

Tier 3 · Hard

1. An octahedral ion $\mathrm{[M(en)_3]^{3+}}$ contains three identical bidentate ligands. State its co-ordination number, the number and type of stereoisomers it forms, and explain the origin of the stereoisomerism.[5 marks]
Answer
- Co-ordination number $6$ ; it forms two optical isomers that are non-superimposable mirror images.
Method: Each en ligand uses two donor atoms, so three ligands make six co-ordinate bonds and an octahedral ion. The three chelate rings can wind around the metal centre in two opposite handed arrangements. These arrangements are mirror images but cannot be superimposed, so they are a pair of optical isomers.

3.2.5.4 · Formation of coloured ions (A-level only)

A transition-metal ion appears coloured because it absorbs some visible wavelengths and transmits or reflects the remaining wavelengths.
Absorbed light promotes a d electron from a ground state to an excited state, with the energy gap given by $\Delta E=h\nu=hc/\lambda$ .
Changing ligand, oxidation state or co-ordination number changes the d-orbital energy splitting and therefore changes the wavelengths absorbed and the observed colour.
In colorimetry, select a suitable filter, zero with a blank and use standards to make an absorbance-concentration calibration; a common error is to apply a dilution factor in the wrong direction.

Tier 1 · Easy

1. Explain why a transition-metal ion can appear coloured when white light passes through its solution.[3 marks]
Answer
- Some visible wavelengths are absorbed to promote d electrons to a higher energy level, and the remaining wavelengths are transmitted.
Method: Link the electronic change to the observation: a visible photon matching the d-level energy gap is absorbed, so the transmitted light lacks those wavelengths and is seen as a colour.

Tier 2 · Standard

1. The energy gap between two d-electron levels is $3.70\times10^{-19}\,\mathrm{J}$ . Calculate the wavelength of light absorbed. Take $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ and Planck's constant as $6.63\times10^{-34}\,\mathrm{J\,s}$ .[3 marks]
Answer
- $5.38\times10^{-7}\,\mathrm{m}$ or $538\,\mathrm{nm}$
Method: Rearrange $\Delta E=hc/\lambda$ to $\lambda=hc/\Delta E$ . Substitution gives $\lambda=(6.63\times10^{-34})(3.00\times10^8)/(3.70\times10^{-19})=5.38\times10^{-7}\,\mathrm{m}$ , which is $538\,\mathrm{nm}$ .

Tier 3 · Hard

1. A calibration is linear: solutions of a coloured ion at $1.00$ , $2.00$ , $3.00$ and $4.00\,\mathrm{mmol\,dm^{-3}}$ give absorbances $0.120$ , $0.240$ , $0.360$ and $0.480$ . A ten-fold diluted sample gives absorbance $0.330$ . Determine the concentration of the original sample in $\mathrm{mol\,dm^{-3}}$ .[4 marks]
Answer
- $2.75\times10^{-2}\,\mathrm{mol\,dm^{-3}}$
Method: The calibration gradient is $0.120$ absorbance per $1.00\,\mathrm{mmol\,dm^{-3}}$ . The diluted concentration is $0.330/0.120=2.75\,\mathrm{mmol\,dm^{-3}}$ . Undo the ten-fold dilution to obtain $27.5\,\mathrm{mmol\,dm^{-3}}=2.75\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ .

3.2.5.5 · Variable oxidation states (A-level only)

Transition elements show variable oxidation states; in acid, zinc reduces yellow $\mathrm{VO_2^+}$ through blue $\mathrm{VO^{2+}}$ and green $\mathrm{V^{3+}}$ to violet $\mathrm{V^{2+}}$ .
Redox potentials for a transition-metal couple depend on pH and ligand because these conditions change the relative stability of the oxidation states.
Tollens' reagent, $\mathrm{[Ag(NH_3)_2]^+}$ , is reduced to metallic silver by an aldehyde, distinguishing aldehydes from ketones.
In acidified manganate(VII) titrations, use $\mathrm{MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O}$ and preserve its $1{:}5$ ratio with $\mathrm{Fe^{2+}}$ or $2{:}5$ ratio with $\mathrm{C_2O_4^{2-}}$ .

Tier 1 · Easy

1. State the oxidation state and colour of vanadium in $\mathrm{VO^{2+}}$ , formed during reduction of acidified vanadate(V).[2 marks]
Answer
- Vanadium is in oxidation state $+4$ and the ion is blue.
Method: Oxygen contributes $-2$ and the ion has charge $+2$ , so $x-2=+2$ and $x=+4$ . The specified vanadium(IV) oxo ion is blue.

Tier 2 · Standard

1. Write the balanced ionic equation for the reaction of acidified manganate(VII) ions with ethanedioate ions, $\mathrm{C_2O_4^{2-}}$ .[4 marks]
Answer
- $\mathrm{2MnO_4^-+16H^++5C_2O_4^{2-}\rightarrow 2Mn^{2+}+8H_2O+10CO_2}$ .
Method: Two manganate(VII) ions gain ten electrons in total to form two $\mathrm{Mn^{2+}}$ . Five ethanedioate ions lose ten electrons while forming ten $\mathrm{CO_2}$ . Add $16\mathrm{H^+}$ and eight waters to balance hydrogen and oxygen; the charges are $-2+16-10=+4$ on the left and $+4$ on the right.

Tier 3 · Hard

1. A $0.455\,\mathrm{g}$ iron supplement is dissolved, then diluted to $200\,\mathrm{cm^3}$ . Titration of a $20.0\,\mathrm{cm^3}$ aliquot uses $17.30\,\mathrm{cm^3}$ of $0.00220\,\mathrm{mol\,dm^{-3}}$ acidified $\mathrm{KMnO_4}$ . Calculate the percentage by mass of iron in the supplement. Use $A_r(\mathrm{Fe})=55.8$ and $\mathrm{MnO_4^-{:}Fe^{2+}}=1{:}5$ .[6 marks]
Answer
- $23.3\%$ iron by mass
Method: Moles of $\mathrm{MnO_4^-}$ are $(0.00220)(17.30\times10^{-3})=3.806\times10^{-5}$ . The aliquot contains $5(3.806\times10^{-5})=1.903\times10^{-4}\,\mathrm{mol}$ of $\mathrm{Fe^{2+}}$ . The full solution contains ten times this amount, $1.903\times10^{-3}\,\mathrm{mol}$ , with iron mass $(1.903\times10^{-3})(55.8)=0.1062\,\mathrm{g}$ . Hence the percentage is $(0.1062/0.455)\times100=23.3\%$ .

3.2.5.6 · Catalysts (A-level only)

A heterogeneous catalyst is in a different phase from the reactants: reactants adsorb at active sites, react on the surface and desorb; a support increases surface area and a poison reduces efficiency by blocking sites.
Iron catalyses the Haber process and $\mathrm{V_2O_5}$ catalyses the Contact process; $\mathrm{V_2O_5}$ is reduced by $\mathrm{SO_2}$ and then regenerated by oxygen.
A homogeneous catalyst is in the same phase and forms an intermediate; variable oxidation states let $\mathrm{Fe^{2+}}$ catalyse the iodide-peroxodisulfate reaction through $\mathrm{Fe^{3+}}$ .
$\mathrm{Mn^{2+}}$ autocatalyses the acidified manganate(VII)-ethanedioate reaction, so the rate rises as product catalyst accumulates; a common error is to include a catalyst in the overall equation rather than cancel it.

Tier 1 · Easy

1. State the three surface stages in heterogeneous catalysis and explain how a catalyst poison lowers the rate.[4 marks]
Answer
- Adsorption, reaction at active sites and desorption; a poison occupies active sites so fewer reactant particles can adsorb and react.
Method: Follow a reactant through the catalytic cycle: it first binds to the surface, reacts by the lower-activation-energy route, then leaves so the site can be reused. A poison blocks these sites and reduces the available catalytic surface.

Tier 2 · Standard

1. Write two equations showing how $\mathrm{V_2O_5}$ catalyses oxidation of $\mathrm{SO_2}$ in the Contact process, and show that the catalyst is regenerated.[4 marks]
Answer
- $\mathrm{V_2O_5+SO_2\rightarrow V_2O_4+SO_3}$ and $\mathrm{V_2O_4+\tfrac{1}{2}O_2\rightarrow V_2O_5}$ .
Method: $\mathrm{SO_2}$ first removes an oxygen atom from $\mathrm{V_2O_5}$ , producing $\mathrm{SO_3}$ and $\mathrm{V_2O_4}$ . Oxygen then reoxidises $\mathrm{V_2O_4}$ to the original $\mathrm{V_2O_5}$ . Adding the equations cancels both vanadium oxides and gives $\mathrm{SO_2+\tfrac{1}{2}O_2\rightarrow SO_3}$ .

Tier 3 · Hard

1. The reaction $\mathrm{S_2O_8^{2-}+2I^-\rightarrow 2SO_4^{2-}+I_2}$ is catalysed by $\mathrm{Fe^{2+}}$ . Write two ionic equations for a catalytic route through $\mathrm{Fe^{3+}}$ , then show that $\mathrm{Fe^{2+}}$ is absent from the overall equation.[6 marks]
Answer
- $\mathrm{S_2O_8^{2-}+2Fe^{2+}\rightarrow 2SO_4^{2-}+2Fe^{3+}}$ and $\mathrm{2Fe^{3+}+2I^-\rightarrow 2Fe^{2+}+I_2}$ ; adding cancels both iron species.
Method: Peroxodisulfate first oxidises two $\mathrm{Fe^{2+}}$ ions to $\mathrm{Fe^{3+}}$ while it is reduced to sulfate. The resulting $\mathrm{Fe^{3+}}$ then oxidises iodide to iodine and is reduced back to $\mathrm{Fe^{2+}}$ . Add the steps and cancel $\mathrm{2Fe^{2+}}$ and $\mathrm{2Fe^{3+}}$ from opposite sides, leaving exactly the stated overall reaction.

3.2.6 · Reactions of ions in aqueous solution (A-level only)

The specified aqua ions are $\mathrm{[Fe(H_2O)_6]^{2+}}$ , $\mathrm{[Cu(H_2O)_6]^{2+}}$ , $\mathrm{[Al(H_2O)_6]^{3+}}$ and $\mathrm{[Fe(H_2O)_6]^{3+}}$ ; hydroxide precipitates are green Fe(II), blue Cu(II), white Al(III) and brown Fe(III).
$3+$ aqua ions are more acidic than $2+$ ions because their greater charge-to-size ratio polarises ligand O-H bonds more strongly, making loss of $\mathrm{H^+}$ easier.
Excess $\mathrm{OH^-}$ dissolves amphoteric $\mathrm{Al(OH)_3}$ to $\mathrm{[Al(OH)_4]^-}$ , while excess ammonia dissolves blue copper(II) hydroxide by forming deep-blue $\mathrm{[Cu(NH_3)_4(H_2O)_2]^{2+}}$ .
Carbonate precipitates $\mathrm{MCO_3}$ from the specified $2+$ aqua ions, but with $3+$ aqua ions it acts as a base, producing a hydroxide precipitate and $\mathrm{CO_2}$ ; a common error is to predict a metal(III) carbonate.

Tier 1 · Easy

1. State the observations when aqueous sodium hydroxide is added separately to solutions containing $\mathrm{Cu^{2+}}$ and $\mathrm{Fe^{3+}}$ ions.[2 marks]
Answer
- $\mathrm{Cu^{2+}}$ gives a blue precipitate; $\mathrm{Fe^{3+}}$ gives a brown precipitate.
Method: Hydroxide deprotonates water ligands and forms insoluble metal hydroxides. Copper(II) hydroxide is blue, whereas iron(III) hydroxide is brown.

Tier 2 · Standard

1. A blue precipitate forms when a small amount of ammonia is added to $\mathrm{[Cu(H_2O)_6]^{2+}}$ . State what happens in excess ammonia, name the product ion's colour, and write the ligand-substitution equation.[5 marks]
Answer
- The precipitate dissolves to give a deep-blue solution of $\mathrm{[Cu(NH_3)_4(H_2O)_2]^{2+}}$ .
- $\mathrm{[Cu(H_2O)_6]^{2+}+4NH_3\rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}+4H_2O}$ .
Method: A small amount of ammonia acts as a base and forms the blue hydroxide precipitate. With excess ammonia, ligand substitution becomes dominant: four water ligands are replaced by ammonia, the solid dissolves, and the soluble deep-blue tetraammine complex forms with charge still $2+$ .

Tier 3 · Hard

1. Solutions of $\mathrm{[Fe(H_2O)_6]^{2+}}$ and $\mathrm{[Fe(H_2O)_6]^{3+}}$ are treated with aqueous carbonate ions. Predict the different observations, write an ionic equation for each reaction, and explain why carbon dioxide forms with only one of the ions.[8 marks]
Answer
- $\mathrm{Fe^{2+}}$ gives a green $\mathrm{FeCO_3}$ precipitate with no gas: $\mathrm{[Fe(H_2O)_6]^{2+}+CO_3^{2-}\rightarrow FeCO_3(s)+6H_2O}$ .
- $\mathrm{Fe^{3+}}$ gives a brown hydroxide precipitate and effervescence: $\mathrm{2[Fe(H_2O)_6]^{3+}+3CO_3^{2-}\rightarrow 2[Fe(H_2O)_3(OH)_3](s)+3CO_2+3H_2O}$ .
- The $3+$ aqua ion is more acidic and transfers protons to carbonate, forming $\mathrm{CO_2}$ .
Method: For the $2+$ ion, carbonate acts as a precipitating ligand and forms green iron(II) carbonate. The higher charge-to-size ratio of $\mathrm{Fe^{3+}}$ strongly polarises its water ligands, so carbonate instead accepts protons; hydrolysis gives the brown hydroxide and carbonic acid decomposes to carbon dioxide and water. Atom and charge counts in the two equations confirm the different pathways.