Inorganic (Year 2)

Period 3 Elements

Reaction of sodium with water, conditions and observations:

2Na (s) + 2H₂O (l) -> 2NaOH (aq) + H₂ (g)

The reaction can occur in cold, liquid water.

Observations: Sodium moves around on the surface of the water (reacts vigorously) + effervescence.

Sources

Not from a mark scheme, but useful content to know.

Reaction of magnesium with water, conditions and observations:

Mg (s) + H₂O (g) -> MgO (s) + H₂ (g)

Mg can react very slowly with cold water to form Mg(OH)₂, but this forms an insoluble layer around the Mg that usually prevents it from reacting further. It reacts readily with steam to form MgO.

Observations for Mg + Steam: White light and White powder (that’s it for mark schemes).

Sources

Not from a mark scheme, but useful content to know.

Reaction of sodium with oxygen + 2 observations:

4Na (s) + O₂ (g) -> 2Na₂O (s)

Observations: It burns with a yellow flame to produce a white solid.

Sources

Not from a mark scheme, but useful content to know.

Reaction of magnesium with oxygen + 2 observations:

2Mg (s) + O₂ (g) -> 2MgO (s)

Observations: It burns with a white flame to produce a white solid.

Sources

Not from a mark scheme, but useful content to know.

Reaction of aluminium with oxygen + 2 observations:

4Al (s) + 3O₂ (g) -> 2Al₂O₃ (s)

Observations: It burns with a white flame to produce a white solid.

Sources

Not from a mark scheme, but useful content to know.

Reaction of silicon with oxygen + 2 observations:

Si (s) + O₂ (g) -> SiO₂ (s)

Observations: It burns with a white flame to produce a white solid.

Sources

Not from a mark scheme, but useful content to know.

Reaction of phosphorous with oxygen + 2 observations:

P₄ (s) + 5O₂ (g) -> P₄O₁₀ (s)

Observations: It burns with a white flame to produce a white solid.

Note: P₄ is a solid at room temperature despite having a simple molecule structure.

Note 2: I believe it is preferred to write ‘P₄’ over ‘4P’ but I think both are accepted.

Sources

Not from a mark scheme, but useful content to know.

Reaction of sulfur with oxygen + 2 observations:

S (s) + O₂ (g) -> SO₂ (g)

Observations: It burns with a blue flame to produce an acidic choking gas.

SO₃ is formed instead when the reaction occurs in the presence of a catalyst at a very high temperature.

Sources

Not from a mark scheme, but useful content to know.

Reaction of Na₂O (s) in water + what pH solution does this form + 2-step explanation:

Na₂O (s) + H₂O (l) -> 2NaOH

A solution of pH 13 forms.

Explanation:

1. Sodium oxide contains O²⁻ ions.

2. These O²⁻ ions react with water to form OH⁻ ions by accepting H⁺ ions, acting as a Bronsted-Lowry base.

Note: If asked to explain why it forms an alkaline solution when added to water, also write the equation ‘O²⁻ + H₂O -> 2OH⁻’

Sources

Not from a mark scheme, but useful content to know.

Reaction of MgO (s) in water + what pH solution does this form + explanation:

MgO (s) + H₂O (l) -> Mg(OH)₂

A solution of pH 9 forms.

Explanation: Mg(OH)₂ forms a less alkaline solution than NaOH because Mg(OH)₂ is only sparingly soluble due to its stronger lattice, meaning it releases fewer free OH⁻ ions.

Sources

Not from a mark scheme, but useful content to know.

Reaction of Al₂O₃ (s) in water + what pH solution does this form + explanation:

No reaction as it doesn’t dissolve due to the high strength of its ionic lattice.

A solution of pH 7 forms.

Sources

Not from a mark scheme, but useful content to know.

What species forms when aluminium oxide is added to water?

Nothing, it remains as Al₂O₃ because aluminium oxide is insoluble due to its very high lattice enthalpy.

Sources

2021 A-Level Paper 1 Q3.4 (2 marks)

MS:

June 2010 A-Level Unit 5 Q2(d) (1 marks)

MS:

Reaction of SiO₂ (s) in water + what pH solution does this form + 3-step explanation:

No reaction as it doesn’t dissolve. This is because:

1. It has a macromolecular structure.

2. Held together by strong covalent bonds.

3. This means water cannot supply enough energy to break the many covalent bonds.

A solution of pH 7 forms.

Sources

Not from a mark scheme, but useful content to know.

Reaction of P₄O₁₀ (s) in water + what pH solution does this form + observation:

P₄O₁₀ (s) + 6H₂O (l) -> 4H₃PO₄ (aq)

A solution of pH 0 forms.Note: a mark scheme allowed -1 to 1 for 1 question.

Observation: A violent reaction that forms a colourless solution.

Sources

Not from a mark scheme, but useful content to know.

Reaction of SO₂ (g) in water + what pH solution does this form?

SO₂ (g) + H₂O (l) -> H₂SO₃ (aq)

A solution of pH 1 forms.Note: one mark scheme allowed a range of 1-3.

Note 2: If a question asks you to write an equation showing SO₂ forming a ‘solution of ions’, write the products as H⁺ + HSO₃⁻

Sources

Not from a mark scheme, but useful content to know.

Reaction of SO₃ (g) in water + what pH solution does this form?

SO₃ (g) + H₂O (l) -> H₂SO₄ (aq)

A solution of pH 0 forms.

Sources

Not from a mark scheme, but useful content to know.

Reaction of Na₂O with acids (in this case HCl):

Na₂O + 2HCl -> 2NaCl + H₂O

Note: It reacts in a standard ‘Acid + Base -> Salt + Water’ reaction.

Sources

Not from a mark scheme, but useful content to know.

Reaction of MgO with acids (in this case HCl):

MgO + 2HCl -> MgCl₂ + H₂O

Note: It reacts in a standard ‘Acid + Base -> Salt + Water’ reaction.

Sources

Not from a mark scheme, but useful content to know.

Reaction of Al₂O₃ with acids (in this case HCl):

Al₂O₃ (s) + 6HCl (aq) -> 2AlCl₃ (aq) + 3H₂O (l)

Sources

Not from a mark scheme, but useful content to know.

Reactions of Al₂O₃ with sodium hydroxide (2):

It can either react in a normal neutralisation reaction that forms this product:

Al₂O₃ + 2NaOH -> 2NaAlO₂ + H₂O

Or it can react with excess water to form this product:

Al₂O₃ + 2NaOH + 3H₂O -> 2NaAl(OH)₄

Note: you definitely need to learn the second reaction, and you should probably learn the first one too.

Sources

Not from a mark scheme, but useful content to know.

Reaction of SiO₂ with sodium hydroxide + condition:

SiO₂ + 2NaOH -> Na₂SiO₃ + H₂O

The NaOH must be concentrated.

Note: the product formed is sodium silicate. Think of the silicate ions like carbonate ions.

Sources

Not from a mark scheme, but useful content to know.

Reaction of P₄O₁₀ with bases (in this case NaOH):

P₄O₁₀ (s) + 12NaOH (aq) -> 4Na₃PO₄ (aq) + 6H₂O (l)

Note: think of it forming H₃PO₄, which then reacts with NaOH in a standard ‘Acid + Base -> Salt + Water’ reaction.

Sources

Not from a mark scheme, but useful content to know.

Reaction of SO₂ with bases (in this case NaOH):

SO₂ (g) + 2NaOH (aq) -> Na₂SO₃ (aq) + H₂O (l)

Note: think of it forming H₂SO₃, which then reacts with NaOH in a standard ‘Acid + Base -> Salt + Water’ reaction.

Note 2: SO₃ obviously acts the same way - just remember to form SO₄²⁻ ions instead of SO₃²⁻.

Sources

Not from a mark scheme, but useful content to know.

Why is MgO a more suitable agent for treating excess acid in the stomach and rivers than Na₂O?

Because it forms Mg(OH)₂, which is only sparingly soluble, meaning it doesn’t form an overly alkaline solution.

Sources

2025 A-Level Paper 1 Q1.3 (1 marks)

MS:

June 2012 A-Level Unit 5 Q1(e)(ii) (1 marks)

MS:

How can you show MgO is a base without reacting it with an acid?

You can dissolve it in water:

MgO + H₂O -> Mg(OH)₂

Note: this shows it is a basic oxide because it produces OH⁻ ions when added to water. However, if you’re asked to show MgO acting as a base, just react it with HCl.

Sources

June 2013 A-Level Unit 5 Q4(d) (2 marks)

MS:

Why is SiO₂ described as an acidic oxide even though it is insoluble in water?

It reacts with bases.Note: this explains why it can be described as ‘acidic’ despite not meeting the standard definition of an acid.

Sources

January 2013 A-Level Unit 5 Q4(e) (1 marks)

MS:

2015 A-Level Unit 5 Q5(c) (2 marks)

MS:

5 physical properties of SiO₂ that are a result of its macromolecular structure:

It has a high melting point.
It is insoluble.
It is non-conducting.
It is hard.
It is brittle.

Sources

2014 A-Level Unit 5 Q3(c) (4 marks)

MS:

2 reasons Al₂O₃ displays some covalent character:

There is a (fairly) small difference in electronegativity between Al and O.
The Al³⁺ ion has a high charge density, making it more able to polarise the O²⁻ ion by distorting its electron cloud.

Sources

Not from a mark scheme, but useful content to know.

Explain why P₄O₁₀ has a low boiling point in terms of its bonding and structure: (2 points)

It has a molecular structure with covalent bonding.Note: ‘covalent bonding’ was underlined, showing that if a question asks about a molecule's bonding, you should state that it is a simple molecule with covalent bonding, even though these aren’t the bonds/ forces being overcome.
There are weak Van der Waals forces between molecules, which don’t require much energy to overcome.

Sources

2021 A-Level Paper 1 Q3.3 (4 marks)

MS:

June 2013 A-Level Unit 5 Q4(c) (3 marks)

MS:

2014 A-Level Unit 5 Q3(d) (4 marks)

MS:

How are sodium and phosphorus stored and why?

Sodium is stored under oil and phosphorus is stored under water to prevent them reacting with oxygen in the air.

Sources

June 2012 A-Level Unit 5 Q1(a) (1 marks)

MS:

What term describes an oxide that reacts with both acids and bases?

Amphoteric.

Sources

2019 A-Level Paper 1 Q03.4 (1 marks)

MS:

How can you distinguish between aqueous solutions of SO₂ and SO₃ of the same concentration?

Determine which solution has a lower pH.

This could be done with universal indicator (SO2 would be orange-red, SO₃ would be red), a pH meter (SO2 would have a pH of 2-3, SO₃ would have a pH of 0-1), or by adding a metal carbonate/ reactive metal such as Mg (SO3 produces faster effervescence).

Sources

2022 A-Level Paper 1 Q05.3 (3 marks)

MS:

2016 As Paper 1 Q04.1 (5 marks)

MS:

Why is the melting point of sulfur (S₈) higher than that of phosphorus (P₄)? (2 points)

S₈ molecules are bigger than P₄ molecules and contain more electrons.
This means the van der Waals forces between molecules are stronger.

Sources

Specimen A-Level Paper 1 Q02.2 (2 marks)

MS:

Transition Metals + RP 11

Ligand definition:

A molecule or ion that forms a co-ordinate bond with a transition metal by donating a (lone) pair of electrons.

Note: a bidentate ligand donates two electron pairs from two different atoms.

Sources

January 2013 A-Level Unit 5 Q8(a) (2 marks)

MS:

Explain how a co-ordinate bond forms between a ligand and a central metal ion: (2 points)

An lone electron pair on the ligand.
Is donated from the ligand to the central metal ion.

Sources

Specimen A-Level Paper 1 Q8.1 (2 marks)

MS:

Why do Cl⁻ complexes have different shapes to hexaaqua ion complexes? (2 points)

Cl⁻ ions are bigger ligands than H₂O ligands.
This means only 4 can fit around the central metal ion (forming a tetrahedral structure) as opposed to 6 (octahedral structure) in the case of hexaaqua ions.

Sources

Not from a mark scheme, but useful content to know.

Why would the enthalpy change for a ligand substitution reaction where ethanedioate ions are substituted for iron(III) ions be approximately 0?

The same number and type of bonds are being broken and made.Note: use this exact wording.

Sources

2017 A-Level Paper 1 Q11.2 (2 marks)

MS:

Which metal ion forms complexes with a co-ordination number of 2?

Ag⁺.Note: also Au⁺.

Sources

Not from a mark scheme, but useful content to know.

What are the two types of isomerism that can occur in transition metal complexes and where does each occur?

Cis-trans isomerism. Occurs in both square planar complexes, which are formed with Pt²⁺ or Ni²⁺ ions, and octahedral complexes.Note: don’t forget the octahedral case, which usually happens when you have 4 of one ligand 2 of another.
Optical isomerism. Occurs in octahedral complexes with three bidentate ligands bonded to a central metal ion.Note: remember to always state it is bidentate ligands if asked to describe/ explain optical isomerism.

Sources

Not from a mark scheme, but useful content to know.

Explain why a ligand substitution reaction would occur despite the same number of the same type of bonds being broken and formed: (4 points)

Despite the enthalpy change of these reactions being negligible due to the same number of similar bonds being broken and formed.
Reactions where small, unidentate ligands are substituted out for larger, multidentate ligands still occur.
As entropy increases because there are more particles on the products side of the reaction than the reactants side.
Meaning free energy change is negative.

Sources

Specimen A-Level Paper 1 Q8.5 (2 marks)

MS:

Explanation of the chelate effect: (2 points)

When bidentate or multidentate ligands replace monodentate ligands to form a more stable complex.
Because there is an increase in entropy, so ΔG is negative despite ΔH being approximately 0.

Sources

2024 A-Level Paper 1 Q3.2 (2 marks)

MS:

Explanation for why the enthalpy change for the reaction between [Cu(H₂O)₆]²⁺ and 1-2 diaminoethane molecules is approximately 0: (2 points)

The Cu-N bonds formed have similar enthalpy to the Cu-O bonds broken.Note: the key point of this note is to show that you should always name the bonds being broken and formed and state that they have similar enthalpy.
The same number of bonds are broken and formed.

Sources

Not from a mark scheme, but useful content to know.

What is the function of haemoglobin in the body?

The transport of oxygen.Note: other facts about haemoglobin that may come in useful: Haem is a biological transition metal complex where a multidentate ligand (porphyrin) forms 4 coordinate bonds to a central Fe²⁺ ion. Oxygen forms a co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported in the blood. Carbon monoxide is toxic because it replaces oxygen co-ordinately bonded to Fe(II) in haemoglobin.

Sources

June 2010 A-Level Unit 5 Q4(e)(ii) (1 marks)

MS:

Which process does cisplatin disrupt during cell division?

DNA replication.

Sources

2020 A-Level Paper 3 Q4.1 (1 marks)

MS:

Explanation of why some transition metal complexes are coloured: (4 points)

Transition metal ions contain d orbitals of different energies (the d-orbitals are split).
Electrons get excited and move to higher energy d orbitals when.
They absorb a wavelength of visible light. The energy gap is related to the wavelength of light absorbed (∆E = hf)
The wavelengths not absorbed are reflected, and these are the wavelengths that create the colour of the complex.

Sources

2020 A-Level Paper 3 Q3.1 (3 marks)

MS:

Why do different complex ions have different colours? (3 points)

Different metals in complexes will have d orbitals with different energies, meaning d orbital splitting will be different.
This means when energy is absorbed, causing an electron to be excited,.
A different wavelength of light will be absorbed, so different wavelengths will therefore be reflected.Note: also mention oxidation state of metal, identity of ligands, and co-ordination number affecting colour, just in case.

Sources

January 2013 A-Level Unit 5 Q8(c) (3 marks)

MS:

2014 A-Level Unit 5 Q7(a) (6 marks)

MS:

4 factors that influence the colour of a transition metal complex:

Oxidation state of the metal.
Identity of the metal.
Identity of the ligands.
Co-ordination number.

Sources

2022 A-Level Paper 1 Q07.3 (3 marks)

MS:

Explanation for why a given transition metal complex appears yellow (or any other specific colour): (2 points)

Visible light is absorbed to excite electrons to higher energy orbitals.
Only yellow light is reflected.

Sources

2022 A-Level Paper 1 Q7.1 (2 marks)

MS:

Equation for relating energy change, planck’s constant and the wavelength of light absorbed:

∆E = h frequency = h speed of light/wavelengthNote: ‘h’ refers to Planck’s constant.

Sources

2024 A-Level Paper 3 Q5.2 (3 marks)

MS:

2020 A-Level Paper 3 Q3.2 (3 marks)

MS:

2022 A-Level Paper 1 Q7.2 (2 marks)

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June 2013 A-Level Unit 5 Q6(a) (2 marks)

MS:

Colour of [Fe(H₂O)₆]²⁺ ions in solution:

Green.

Sources

Not from a mark scheme, but useful content to know.

Colour of [Cu(H₂O)₆]²⁺ ions in solution:

Blue.

Sources

Not from a mark scheme, but useful content to know.

Colour of [Fe(H₂O)₆]³⁺ ions in solution:

Pale violet, but often appears orange due to hydrolysis to [Fe(H₂O)₅(OH)]²⁺.

Sources

Not from a mark scheme, but useful content to know.

Colour of [Al(H₂O)₆]³⁺ ions in solution:

It is colourless.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NaOH is added dropwise then in excess to [Fe(H₂O)₆]²⁺:

Green precipitate [Fe(H₂O)₄(OH)₂] formed, which darkens on standing due to oxidation to [Fe(H₂O)₃(OH)₃].

No further reaction when excess is added.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NaOH is added dropwise then in excess to [Cu(H₂O)₆]²⁺:

Blue precipitate [Cu(H₂O)₄(OH)₂] formed.

No further reaction when excess is added.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NaOH is added dropwise then in excess to [Fe(H₂O)₆]³⁺:

Brown precipitate [Fe(H₂O)₃(OH)₃] formed.

No further reaction when excess is added.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NaOH is added dropwise then in excess to [Al(H₂O)₆]³⁺:

A white precipitate [Al(H₂O)₃(OH)₃] is formed when NaOH is added dropwise.

The precipitate redissolves to form a colourless solution when excess NaOH is added, as the aqueous [Al(OH)₄]- ion forms.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NH₃ is added dropwise then in excess to [Fe(H₂O)₆]²⁺:

Green precipitate [Fe(H₂O)₄(OH)₂] formed, which darkens on standing due to oxidation to [Fe(H₂O)₃(OH)₃].

No further reaction when excess is added.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NH₃ is added dropwise then in excess to [Cu(H₂O)₆]²⁺:

Blue precipitate [Cu(H₂O)₄(OH)₂] formed.

Precipitate redissolves to form a deep blue solution when excess NH₃ is added, as the aqueous [Cu(H₂O)₂(NH₃)₄]²⁺ ion forms.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NH₃ is added dropwise then in excess to [Fe(H₂O)₆]³⁺:

Brown precipitate [Fe(H₂O)₃(OH)₃] formed.

No further reaction when excess is added.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when NH₃ is added dropwise then in excess to [Al(H₂O)₆]³⁺:

White precipitate [Al(H₂O)₃(OH)₃] formed.

No further reaction when excess is added.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when Na₂CO₃ is added to [Fe(H₂O)₆]²⁺:

Green precipitate FeCO₃ formed.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when Na₂CO₃ is added to [Cu(H₂O)₆]²⁺:

Blue-green precipitate CuCO₃ formed.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when Na₂CO₃ is added to [Fe(H₂O)₆]³⁺:

Brown precipitate [Fe(H₂O)₃(OH)₃] and CO₂ gas formed.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when Na₂CO₃ is added to [Al(H₂O)₆]³⁺:

White precipitate [Al(H₂O)₃(OH)₃] and CO₂ gas formed.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when conc HCl is added to [Fe(H₂O)₆]²⁺:

Yellow solution with the [FeCl₄]²⁻ ion forms.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when conc HCl is added to [Cu(H₂O)₆]²⁺:

Yellow-green solution with the [CuCl₄]²⁻ ion forms.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when conc HCl is added to [Fe(H₂O)₆]³⁺:

Yellow solution with the [FeCl₄]⁻ ion forms.

Sources

Not from a mark scheme, but useful content to know.

Observations and species formed when conc HCl is added to [Al(H₂O)₆]³⁺:

A colourless solution forms, containing the [AlCl₄]⁻ ion.

Sources

Not from a mark scheme, but useful content to know.

Why do [Fe(H₂O)₆]³⁺ ions form more acidic solutions than [Fe(H₂O)₆]²⁺ solutions: (3 points)

Fe³⁺ ions have a greater charge density.
Fe³⁺ ions are therefore more polarising, meaning they polarise the water molecules more.
Making the O-H bonds in the [Fe(H₂O)₆]³⁺ ion weaker and more likely to break, meaning more H⁺ ions dissociate and get released.

Sources

2021 A-Level Paper 1 Q4.7 (3 marks)

MS:

June 2011 A-Level Unit 5 Q8(b)(i) (3 marks)

MS:

Equation + 2 point explanation that explains why [Al(H₂O)₆]³⁺ ions form an acidic solution:

[Al(H₂O)₆]³⁺ ⇌ [Al(H₂O)₅(OH)]²⁺ + H⁺Note: remember it is a reversible reaction.

2 point explanation:

1. EITHER Al³⁺ ion is small and has a high charge OR Al³⁺ ion has a high charge density.Note: I’d always go with ‘high charge density’.

2. Al³⁺ weakens the O-H bonds in the water ligands by attracting the electrons from the O-H bond and donates H⁺ to water.

Sources

2023 A-Level Paper 1 Q7.3 (2 marks)

MS:

2020 A-Level Paper 1 Q5.4 (3 marks)

MS:

Why does Na₂CO₃ form CO₂ when added to [Fe(H₂O)]³⁺ but not [Fe(H₂O)]²⁺? (2 points)

Because [Fe(H₂O)]³⁺ is more acidic than [Fe(H₂O)]²⁺.
Because the Fe³⁺ ion has a greater charge density than the Fe²⁺ ion, making it more polarising, making the [Fe(H₂O)]³⁺ ion a better proton donor.

Sources

2021 A-Level Paper 1 Q4.7 (3 marks)

MS:

June 2011 A-Level Unit 5 Q8(b)(i) (3 marks)

MS:

If asked for a reagent that can hydrolyse hexaaqua ions, should you say NaOH or OH⁻ ions?

Always say the full ionic compound, NaOH.

Sources

2020 A-Level Paper 1 Q5.2 (3 marks)

MS:

What reducing agent can be used to change Iron III hexaaqua ions into Iron II hexaaqua ions?

Excess Zn in acid (H₂SO₄).Note: this means that if you wanted to determine the original concentration of an iron II hexaaqua ion solution you should add Zn (and acid) first to reduce the Iron III ions back to iron II ions.

Sources

Not from a mark scheme, but useful content to know.

What is a heterogeneous catalyst? (2 points)

A substance in a different phase from the reactants.
That speeds up the reaction by lowering the activation energy while remaining unchanged at the end of the reaction.Note: always include this in any definition of a homogeneous or heterogeneous catalyst - you need to define the ‘catalyst’ part as well as the ‘heterogeneous’ part.

Sources

2020 A-Level Paper 3 Q1.4 (2 marks)

MS:

How does a heterogeneous catalyst work? (3 points)

The reactants adsorb onto the active sites of the catalyst.
Bonds weaken, the activation energy is lowered, and the reaction takes place.
The products then desorb from the surface of the catalyst.

Sources

2018 A-Level Paper 1 Q3.4 (3 marks)

MS:

2 ways the effectiveness of a heterogeneous catalyst be maximised:

Using it as a powder.
Using it on a support mesh like a honeycombNote: both of these work by maximising the surface area, which you should mention in an answer to any question like this.

Sources

2015 As Unit 2 Q4(a)(iii) (2 marks)

MS:

How can a heterogeneous catalyst become ineffective? (2 points)

Catalyst poisoning.
This blocks the active sites and therefore prevents reactants adsorbing.

Sources

Specimen A-Level Paper 1 Q4.5 (1 marks)

MS:

January 2013 A-Level Unit 5 Q6(c)(ii) (1 marks)

MS:

2 examples of catalyst poisoning:

Sulfur poisoning the Iron catalyst in the Haber process.
Lead poisoning the Platinum/ Rhodium catalyst in catalytic conversion.

Sources

Not from a mark scheme, but useful content to know.

Why can’t Zn²⁺ act as a catalyst?

It only has one oxidation state as an ion, as Zn²⁺ is the only ion that forms.Note: although it is true that it also has a full d sub-shell, this is not the answer.Note 2: s block elements usually don’t act as catalysts for the same reason.

Sources

2021 A-Level Paper 1 Q4.3 (1 marks)

MS:

June 2010 A-Level Unit 5 Q3(b) (1 marks)

MS:

Explain what an autocatalyst is and why the potassium manganate + sodium ethanedioate reaction is slow at first but speeds up over time: (3 points)

An autocatalyst is a product of a reaction that then catalyses the reaction.
The reaction would be very slow as two negatively charged reactants are reacting with each other, meaning they repel each other so the activation energy is very high.
However, once the Mn₂+ has formed, it acts as a catalyst by reducing the MnO₄⁻ ions to create Mn³⁺ ions (and water), which then react with the ethanedioate ions to regenerate (by reduction) the Mn²⁺ ions and produce the CO₂.

Sources

2022 A-Level Paper 3 Q5.2 (6 marks)

MS:

2 equations that show how the potassium manganate + sodium ethanedioate reaction works: The overall reaction is: 2MnO₄⁻ (aq) + 16H⁺ (aq) + 5C₂O₄²⁻ (aq) -> 2Mn²⁺ (aq) + 10CO₂ (g) + 8H₂O (l)

8Mn²⁺ (aq) + 2MnO₄⁻ (aq) + 16H⁺ (aq) → 10Mn³⁺ (aq) + 8H₂O (l)

10Mn³⁺ (aq) + 5C₂O₄²⁻ (aq) → 10Mn²⁺ (aq) + 10CO₂ (g)

Note: you must learn these. I remember the 10 Mn³⁺ being formed from 8 Mn²⁺ with all molar ratios the same as the overall reaction and go from there.

Sources

2022 A-Level Paper 3 Q5.2 (6 marks)

MS:

Describe the shape of a graph that shows the volume of CO₂ formed in the potassium manganate + sodium ethanedioate autocatalysis reaction and explain what causes this shape: (2 points)

The gradient/ slope increases as the Mn²⁺ catalyst forms.
The gradient/ slope then decreases again as the reactants are used up and the rate slows.

Sources

2015 A-Level Unit 5 Q9(b) (4 marks)

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Fe²⁺ catalyses the reaction between S₂O₈²⁻ and I⁻ ions. Why can Fe²⁺ catalyse this reaction? Why is it needed to catalyse this reaction? What are the reactions that show Fe²⁺ being used as a catalyst + what is the overall equation?

Fe²⁺ can catalyse this reaction as it has variable oxidation states, as it can get oxidised to Fe³⁺.

It is needed to catalyse this reaction as both of the ions in the reaction are negative, meaning they’d repel each other, making the activation energy very high. Therefore, they need Fe²⁺ to lower the activation energy, which it can do by attracting and reacting with the negatively charged ions.

It is able to do this because it has variable oxidation states and can be oxidised and reduced:

Fe²⁺ -> Fe³⁺ + e⁻

How it is used as a catalyst:

2 Fe²⁺ + S₂O₈²⁻ → 2 SO₄²⁻+ 2 Fe³⁺

2 Fe³⁺ + 2 I⁻ → 2 Fe²⁺ + I₂

Overall equation:

S₂O₈²⁻ + 2 I⁻ → 2 SO₄²⁻ + I₂.

Sources

2021 A-Level Paper 1 Q4.2 (4 marks)

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2021 A-Level Paper 1 Q4.3 (1 marks)

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June 2010 A-Level Unit 5 Q3(e) (3 marks)

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2014 A-Level Unit 6 EMPA Q9(b) (1 marks)

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2 equations that show how vanadium oxide (V₂O₅) is used in the contact process to form SO₃ from SO₂ and O₂:

V₂O₅ + SO₂ -> V₂O₄ + SO₃

V₂O₄ + ½ O₂ -> V₂O₅.

Sources

2025 A-Level Paper 1 Q7.3 (2 marks)

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2020 A-Level Paper 3 Q2.3 (2 marks)

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2018 A-Level Paper 1 Q9.5 (1 marks)

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2016 A-Level Unit 5 Q6(f)(ii) (2 marks)

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What is the formula for ammonium vanadate and what oxidation state is vanadium in?

NH₄VO₃.

V is in a +5 oxidation state.

Sources

Not from a mark scheme, but useful content to know.

In what conditions is the vanadium in ammonium vanadate reduced? What apparatus are used and why?

It can be reduced by Zinc in acidic conditions.

A cotton wool plug is inserted into the top of the flask to keep air out to stop V(+2) instantly forming.

Sources

Not from a mark scheme, but useful content to know.

What species are formed in each stage of the reduction of ammonium vanadate and what colour solutions do these form?

V(+ 5): VO₂⁺ Yellow

V(+ 4): VO²⁺ Blue

V(+ 3): V³⁺ Green

V(+ 2): V²⁺ Violet.

Sources

Not from a mark scheme, but useful content to know.

What happens in a Potassium Manganate (VII) titration where Potassium Manganate is added to a Fe²⁺ solution? What are the conditions, what species are added and produced, and what colour are they?

The Manganate ions (MnO₄⁻) are purple, and they are added to the reducing agent Fe²⁺ and H₂SO₄ (which provides H⁺ for the reaction).

At first, the purple MnO₄⁻ is reduced by the Fe²⁺, producing colourless Mn²⁺ ions and Fe³⁺ ions.

This means the solution in the conical flask remains colourless until all of the Fe²⁺ has been used up, at which point it turns pink/ pale purple due to the excess MnO₄⁻.

Sources

2019 A-Level Paper 3 Q5.3 (1 marks)

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2021 A-Level Paper 3 Q1.6 (1 marks)

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Why is an indicator not used for the Potassium Manganate (VII) titration?

KMnO₄ is self-indicating as it changes the solution from colourless to pale pink at the end point.

Sources

2022 A-Level Paper 1 Q3.2 (1 marks)

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3 steps in using a colorimeter to identify the concentrations of unknown hexaaqua ion solutions:

Make some solutions with known concentrations then find the absorbance of each using a colorimeter.
Plot a graph of concentration vs. absorbance.
Find the absorbance of the unknown solutions, then use the calibration curve to find their concentration.

Sources

2020 A-Level Paper 3 Q3.3 (3 marks)

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2016 A-Level Unit 5 Q9(e) (3 marks)

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Why should the container for each sample be the same when using colorimetry to identify the concentration of hexaaqua ion solutions? (2 points)

Absorption depends on the distance travelled through the solution.
This means this needs to be controlled for by using the same container.

Sources

Not from a mark scheme, but useful content to know.

Why is a coloured filter sometimes used in colorimetry? (3 points)

A complementary filter is used.Note: so an orange filter would be used for a blue solution.
This maximises absorbance for the wavelength that is most strongly absorbed by the sample.
This maximises sensitivity, allowing for a greater range in results and the ability to detect colour differences at low concentrations, creating a more accurate calibration curve.

Sources

Not from a mark scheme, but useful content to know.

What is another way the sensitivity of colorimetry readings can be increased?

Adding a ligand to the solution that increases the intensity of the solutions, again making it easier to detect low concentration solutions.

Sources

Not from a mark scheme, but useful content to know.

Explain why copper(I) iodide is white: (2 points)

The Cu⁺ ion has a full 3 subshell.Note: remember that Cu has the electron configuration of 4s1 3d10, so loses the 4s electron to form a 1+ ion.
This means it cannot absorb visible light for a d-d transition/ to promote an electron to a higher energy d orbital.

Sources

2019 A-Level Paper 1 Q06.5 (2 marks)

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If an aqueous salt forms a green precipitate when NH₃ is added to it, and a white precipitate when AgNO₃ is added to it, what is its identity?

FeCl₂.Note: I’ve included this example to show that in the questions where you have to identify a species based on the products it forms with different reactants, you need to remember that it doesn’t have to be the hexaaqua ion to form the standard ppts with NaOH, NH₃ etc., but instead can be normal a normal salt like FeCl₂ or FeSO₄.

Sources

Not from a mark scheme, but useful content to know.

Test to distinguish between a solution of MgCl₂ and AlCl₃ ions:

Add NaOH.

MgCl₂ forms a white precipitate.

AlCl₃ forms a white precipitate that redissolves in excess NaOH.

Sources

2017 A-Level Paper 1 Q8.2 (3 marks)

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