3.2 Inorganic chemistry — 7405 spec coverage

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 mark

ORIGINAL

An element has the electron configuration $\mathrm{[Ne]3s^2}$ . State its block in the Periodic Table.

Tier 2 · Standard

2 marks

ORIGINAL

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.

Tier 3 · Hard

3 marks

ORIGINAL

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.

3.2.1.2

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

3 marks

ORIGINAL

Arrange Na, Mg and Cl in order of decreasing atomic radius and explain the order.

Tier 2 · Standard

4 marks

ORIGINAL

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.

Tier 3 · Hard

6 marks

ORIGINAL

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.

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

2 marks

ORIGINAL

Steam is passed over heated magnesium ribbon. Give the chemical equation and the visible change.

Tier 2 · Standard

4 marks

ORIGINAL

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.

Tier 3 · Hard

5 marks

ORIGINAL

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 $M_r(\mathrm{CaCO_3})=100.1$ .

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

2 marks

ORIGINAL

Chlorine water is added to aqueous potassium bromide. State the observation and write the ionic equation.

Tier 2 · Standard

6 marks

ORIGINAL

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.

Tier 3 · Hard

7 marks

ORIGINAL

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.

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

2 marks

ORIGINAL

Write the equation for the reaction of chlorine with cold, dilute aqueous sodium hydroxide and state one use of the solution formed.

Tier 2 · Standard

4 marks

ORIGINAL

Use oxidation states to show that the formation of chloride and chlorate(I) ions when chlorine reacts with water is disproportionation.

Tier 3 · Hard

4 marks

ORIGINAL

$1.42\,\mathrm{g}$ 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 $M_r(\mathrm{Cl_2})=71.0$ .

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

2 marks

ORIGINAL

Write equations for the formation from the elements of magnesium oxide and phosphorus(V) oxide, $\mathrm{P_4O_{10}}$ .

Tier 2 · Standard

5 marks

ORIGINAL

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.

Tier 3 · Hard

7 marks

ORIGINAL

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.

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

2 marks

ORIGINAL

Define the term ligand.

Tier 2 · Standard

3 marks

ORIGINAL

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.

Tier 3 · Hard

4 marks

ORIGINAL

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.

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

3 marks

ORIGINAL

State whether each of $\mathrm{NH_3}$ , ethane-1,2-diamine and EDTA is a monodentate, bidentate or multidentate ligand.

Tier 2 · Standard

4 marks

ORIGINAL

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.

Tier 3 · Hard

5 marks

ORIGINAL

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.

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

2 marks

ORIGINAL

State the shapes of $\mathrm{[Ag(NH_3)_2]^+}$ and $\mathrm{[CoCl_4]^{2-}}$ .

Tier 2 · Standard

4 marks

ORIGINAL

$\mathrm{[Pt(NH_3)_2Cl_2]}$ forms two stereoisomers. Name the type of stereoisomerism, describe the ligand arrangement in each isomer, and identify cisplatin.

Tier 3 · Hard

5 marks

ORIGINAL

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.

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

3 marks

ORIGINAL

Explain why a transition-metal ion can appear coloured when white light passes through its solution.

Tier 2 · Standard

3 marks

ORIGINAL

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 $6.63\times10^{-34}\,\mathrm{J\,s}$ .

Tier 3 · Hard

4 marks

ORIGINAL

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 $\mathrm{mol\,dm^{-3}}$ .

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

2 marks

ORIGINAL

State the oxidation state and colour of vanadium in $\mathrm{VO^{2+}}$ , formed during reduction of acidified vanadate(V).

Tier 2 · Standard

4 marks

ORIGINAL

Write the balanced ionic equation for the reaction of acidified manganate(VII) ions with ethanedioate ions, $\mathrm{C_2O_4^{2-}}$ .

Tier 3 · Hard

6 marks

ORIGINAL

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 $\mathrm{MnO_4^-{:}Fe^{2+}}=1{:}5$ .

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

4 marks

ORIGINAL

State the three surface stages in heterogeneous catalysis and explain how a catalyst poison lowers the rate.

Tier 2 · Standard

4 marks

ORIGINAL

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.

Tier 3 · Hard

6 marks

ORIGINAL

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.

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

2 marks

ORIGINAL

State the observations when aqueous sodium hydroxide is added separately to solutions containing $\mathrm{Cu^{2+}}$ and $\mathrm{Fe^{3+}}$ ions.

Tier 2 · Standard

5 marks

ORIGINAL

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.

Tier 3 · Hard

8 marks

ORIGINAL

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.