Top 5 Hardest A Level AQA Inorganic Chemistry Questions with Fully Worked Answers (2026)

Practise the hardest A Level AQA Inorganic Chemistry questions with step-by-step solutions. Perfect for Grade A/A* revision covering Periodicity, Group 2, Group 7, Transition Metals and more.

Master the Hardest AQA A Level Inorganic Chemistry Questions

Introduction

Preparing for your A Level Chemistry exams? This guide contains the most challenging AQA Inorganic Chemistry questions, carefully selected to reflect real exam difficulty. Every question includes detailed, examiner-style answers to help you understand the reasoning behind each mark. Whether you’re aiming for an A or A*.This resource will strengthen your understanding of Periodicity, Group 2, Group 7, Transition Metals, Redox Reactions, Equilibria, and Inorganic Analysis.

Ques1) This question is about vanadium compounds and ions.

a)Use data from Table 4 to identify the species that can be used to reduce VO₂⁺  ions to VO²⁺ in aqueous solution and no further. Explain your answer. 

b)Give the oxidation state of vanadium in [VO(H₂O)₅]²⁺

c)The [V(H₂O)₄Cl₂] ⁺ ion exists as two isomers. One isomer is shown. Draw the structure of the other isomer and state the type of isomerism.

d)Heating NH4VO3 produces vanadium(V) oxide, water and one other product. Give an equation for the reaction.

e)Vanadium(V) oxide is the catalyst used in the manufacture of sulfur trioxide. Give two equations to show how the catalyst is used and regenerated.

This is repeatedly flagged in examiner reports and only 11.3% of students are able to score full on this question.

Solution – 
1. Fe²⁺ because 

Which means iron(II) can reduce VO₂⁺  ions to VO²⁺   but no further reduction is possible. 

1. Let assume V to be x 

Oxygen =  -2  ; H₂O = 0(neutral)

x+(-2)+0 = +2

X = +4

2. In the question they have actually given cis isomer, we simply have to draw trans isomer. Just put 2 Cl directly opposite to each other.

3. In the question they have actually given cis isomer, we simply have to draw trans isomer. Just put 2 Cl directly opposite to each other.
   
   

1. 2 NH₄VO₃ → V₂O₅ + H₂O + 2NH₃ 
2. V2O5 + SO2  →   V2O4 + SO3 

V2O4 + ½ O2  →  V2O5 

Ques2)The ClF₂⁺ ion contains two different Group 7 elements. Use your understanding of the electron pair repulsion theory to draw the shape of this ion. Include any lone pairs of electrons that influence the shape. Explain why the ion has the shape you have drawn. Suggest a value for the bond angle in the ion. 

Ions with unusual central charges ClF₂⁺ regularly stump students. Most candidates miscounted the lone pairs or mistakenly treated the compound as a linear/trigonal bipyramidal structure like its neutral relative ClF₃.

Solution –

Explanation – chlorine has 6 electrons since there is a +1 charge on the molecule hence 2 fluorine will combine to form 2 B.P and 2 lone pairs giving it a bent / V- shape appearance.

Ion gets this particular shape because of the extent of repulsions between lone pair and lone pair followed by L.P and B.P , so it tries to occupy a shape where these repulsions are minimum. The original Bond angle is 109.5 for tetrahedral but since its bent it will be less the original angle and lies between 104 to 106 


Trick –  You can actually relate this molecule to water and draw bent shape & a comparable bond angle (104

Ques3)The following ques are based on equations.

a)Write an equation for the reaction of sulfur trioxide with potassium hydroxide solution. 
b)Write an equation for the reaction of an excess of magnesium oxide with phosphoric acid. 
c)Draw the displayed formula of the undissociated acid formed when sulfur dioxide reacts with water.

While basic oxide + acid reactions are fundamental, this specific combination tripped up most candidates. Students struggled with the balancing ratio required to form magnesium phosphate and often incorrectly treated phosphoric acid as monoprotic or diprotic.

Solution – 

1. SO₃ + 2KOH →  K₂SO₄ + H₂O (its a normal acid base reaction, acidic oxides are basic in nature and they reacts with alkali metals to form salt and water)
2. 3 MgO + 2 H₃PO₄ → Mg₃(PO₄)₂ + 3 H₂O (its again a neutralization reaction ; metallic oxides are basic in nature and they react with acids to make salt and water )

Tutor Tip – Make sure to do correct criss cross of all compounds in this equation & balance it nicely as it is a little tricky to balance.

Tutor Tip – Remember SO₂(Sulfur dioxide ) reacts with water to form Sulfurous acid and SO₃(Sulfur trioxide) reacts with water to form Sulfuric acid.

Ques4)Complete the equation to show the formation of one complex that contains chromium in its +3 oxidation state. 

                             C_rCl₃ + 5H₂O →

Only 14% of students wrote the correct equation. Most student make a standard hexa-aqua complex or failed to balance the chlorine atoms between ligands and counter-ion.

Solution –       CrCl₃ + 5H₂O →   [Cr(H₂O)5Cl]Cl₂

Ques5)Hydrated aluminium sulfate, Al₂(SO₄)3.xH₂O, is soluble in water. The relative formula mass and value of x can be found from a titration experiment. Aqueous [Al(H₂O)₆] ³⁺ ions react to form a stable complex when treated with an excess of EDTA^{4 –} ions. The excess of  EDTA^{4 –}ions is determined by titration with ZnSO₄ solution. 

Method –

Dissolve 1.036 g of Al₂(SO₄)3.xH₂O in distilled water and make up to 250 cm³ 

Add 25.0 cm³ of this solution to 50.0 cm3 of a solution containing  EDTA^{4 –} ions of    concentration 0.0100 mol dm^–3

  Determine the excess of  EDTA^{4 –} ions by titrating with ZnSO₄  solution in the presence of an indicator. 

The excess of  EDTA^{4 –} ions requires 18.00 cm³ of 0.0105 mol dm^–3 ZnSO₄ solution to react completely. The equations for the reactions are

 [Al(H2O)₆] ³⁺ + EDTA^{4 –} → [AlEDTA]– + 6H2O

 [Zn(H2O)₆] ²⁺  EDTA^{4 –} → [ZnEDTA]^2– + 6H₂O

 For Al₂(SO₄)₃ M_r = 342.3

It is one of the hardest Calculation based question with good number of multi – step where students have to calculate various moles and find unreacted moles and so on.

Solution : moles of  ZnSO₄  = concentration x volume(dm³)

                                            = 0.0105 x 0.018(dm³) = 0.000189 

                Moles of EDTA taken at starting = 0.01 x 0.050 = 0.0005

Each 1 mole of EDTA needs 1 mole of  ZnSO₄ , hence 0.000189 needs 0.000189 

  Hence from here we can find moles of EDTA reacted with Al³⁺ 

{Remember unlike other back titrations Here EDTA is just reacting with Al³⁺ ion.}   

      Moles of EDTA reacted with Al³⁺ ion = 0.0005 – 0.000189 = 0.000311

             Moles of Al³⁺  ion = 1 mole of EDTA requires 1 mole of Al³⁺  ion

                                         = 0.000311 EDTA requires 0.000311 molesAl³⁺ ion

Remember these moles of Al³⁺ ion are in 25.0 cm³ of solution but we have to find in 250 cm³ of solution.

{This is the most common mistake students make,  they always forget to find moles in actual sample}

Moles of  Al³⁺ ion in 250 cm³ of solution = 0.000311 x 10 = 3.11 x 10^{– 3} mol  

Al³⁺ ion comes from Al₂(SO₄)3.xH₂O

Each 1 mole of Al₂(SO₄)3.xH₂O release 2 mole of Al³⁺ ion 

Therefore we can find moles of original sample = 3.11 x 10^{– 3} mol  2 = 1.555 x10^{– 3} mol 

Now we have moles of  Al₂(SO₄)3.xH₂O , we can find Mr of this molecule  

                             = given mass moles 

                             = 1.036  1.555 x10^{– 3} 

                             = 666.2 

Now lets find water of crystallization 

                      342.3 + 18 x = 666(.2)   [Al₂(SO₄)₃ M_r = 342.3 ]

                       18x  = 666.2 – 342.3

                          x   = 18