Ilmenite is a mineral made mainly of a mixture of iron(II) titanate, FeTiO3, and iron(III) titanate, Fe2(TiO3)3. Which row shows the oxidation state of titanium in the two compounds?
| FeTiO3 | Fe2(TiO3)3 | |
|---|---|---|
| A | +2 | +3 |
| B | +4 | +2 |
| C | +4 | +3 |
| D | +4 | +4 |
2024 AES O Prelim P1 Q22
The table shows the formulae of compounds formed by carbon and other elements.
| compound | chemical formula |
|---|---|
| potassium cyanide | KCN |
| iron(II) carbide | Fe2C |
| carbon dioxide | CO2 |
| calcium carbonate | CaCO3 |
In which two substances does carbon have the same oxidation state?
2024 BTYSS FYE O EOY P1 Q22
The compound Bi2Sr2Ca2Cu3O10 is a superconductor. In it, bismuth is +3, strontium and calcium are +2 and oxygen is −2. What are the possible oxidation numbers of the three copper atoms?
2023 HCI J2 Prelim P1 Q29
During the Contact process, sulfuric acid is produced from sulfur in a series of stages: S → SO2 → SO3 → H2S2O7 → H2SO4. Which statement correctly describes these stages?
2024 CHS O Prelim P1 Q18
Concentrated sulfuric acid reacts with solid potassium halides:
reaction 1: H2SO4 + 2KCl → 2HCl + K2SO4
reaction 2: 2H2SO4 + 2KBr → 2H2O + Br2 + K2SO4 + SO2
reaction 3: 5H2SO4 + 8KI → 4H2O + 4I2 + 4K2SO4 + H2S
What is the change in the oxidation state of the sulfur that reacts in each reaction?
| reaction 1 | reaction 2 | reaction 3 | |
|---|---|---|---|
| A | 0 | 0 | 4 |
| B | 0 | 2 | 4 |
| C | 0 | 2 | 8 |
| D | 2 | 4 | 8 |
2024 PLMGSS O Prelim P1 Q25
Nitrogen monoxide is removed from car exhaust in a catalytic converter: 2CO + 2NO → N2 + 2CO2. Which statement about the reaction is not correct?
2025 JYSS O Prelim P1 Q22
Sulfur dioxide is converted to sulfur trioxide in two stages.
stage 1: 2NO + O2 → 2NO2
stage 2: NO2 + SO2 → SO3 + NO
Which statement is correct?
2024 YSS O Prelim P1 Q22
The results of some experiments with sulfur dioxide are shown.
| experiment | description | result |
|---|---|---|
| 1 | mix with dilute hydrochloric acid | does not react |
| 2 | mix with concentrated sodium hydroxide | a salt is formed |
| 3 | add Universal Indicator | Universal Indicator turns violet |
| 4 | add acidified aqueous potassium manganate(VII) | purple solution turns colourless |
Which experiments show correct results?
2025 FHSS O Prelim P1 Q13
Aqueous iron(III) chloride reacts with aqueous potassium iodide: vFeCl3 + wKI → xFeCl2 + yKCl + I2. Which statements are correct?
1 In the balanced equation, v, w, x and y have the same value.
2 A purplish-black vapour is produced in the reaction.
3 Potassium iodide is used to test for the presence of an oxidising agent.
2023 Catholic High School O Prelim P1 Q17
NF3 can be made from ammonia and fluorine: 4NH3 + 3F2 → NF3 + 3NH4F. Which statements about the reaction are correct?
1 NH3 undergoes disproportionation.
2 The oxidation number of nitrogen in NF3 is +3.
3 NH3 is a stronger oxidising agent than F2.
4 One mole of F2 gains two moles of electrons.
2021 YIJC J1 MYE P1 Q8
50.0 cm3 of 0.100 mol dm−3 KClO4(aq) reacted exactly with 50.0 cm3 of 0.400 mol dm−3 TiCl3(aq). Ti(III) is oxidised to Ti(IV). What is the oxidation state of chlorine in the final product?
2018 RVHS J1 Common Test P1 Q2
The most common oxidation state of americium, Am, in aqueous solution is +3. Cu3+ oxidises Am3+(aq) in dilute HNO3 and is itself reduced to Cu2+. 20.0 cm3 of 0.0120 mol dm−3 Am3+(aq) required 24.00 cm3 of 0.0300 mol dm−3 Cu3+ for complete oxidation. What is the formula of the americium-containing species formed?
2019 ACJC J2 Prelim P1 Q2
25.0 cm3 of a 0.100 mol dm−3 solution of M2O5 is reduced by sulfur dioxide to a lower oxidation state. 50.0 cm3 of 0.0200 mol dm−3 acidified potassium manganate(VII) is needed to re-oxidise M back to its original oxidation state. What is the oxidation state of M after reaction with sulfur dioxide?
2019 RVHS J1 Common Test P1 Q4
0.84 g of an oxide MO of a metal M was dissolved in excess sulfuric acid. 25.0 cm3 of 0.12 mol dm−3 potassium manganate(VII) was required to oxidise M2+ to M3+. What is the relative atomic mass of M?
2019 SAJC J2 Prelim P1 Q2
0.02 mol of an iodine oxide reacts with 0.2 mol of acidified potassium iodide to give 0.12 mol of iodine, I2. What is the oxidation number of iodine in the oxide?
2024 DHS J2 Prelim P1 Q11
Consider the half-equations Fe2+ → Fe3+ + e− and C2O42− → 2CO2 + 2e−. 20 cm3 of an acidified 0.020 mol dm−3 solution of FeC2O4 is oxidised by 15 cm3 of 0.020 mol dm−3 KMnO4. Both Fe2+ and C2O42− are oxidised. What is the final oxidation state of manganese?
2024 HCI J2 Prelim P1 Q9
A comproportionation reaction is one where two reactants, each containing the same element in a different oxidation state, form a product in which that element has one oxidation state. 2 mol of hydrogen sulfide, H2S, react with 1 mol of another sulfur-containing compound to form 3 mol of sulfur, S. What is a possible identity of the sulfur-containing compound?
2021 CJC J2 Prelim P1 Q11
When potassium chlorate(V), KClO3, is heated at its melting point, it disproportionates to potassium chlorate(VII), KClO4, and potassium chloride. What is the maximum amount of potassium chlorate(VII) that could be produced from 0.200 mol of potassium chlorate(V)?
2021 RVHS J2 Prelim P1 Q1
Excess iodide ions react with 15 cm3 of a solution containing copper(II) ions: 4I− + 2Cu2+ → I2 + 2CuI. The iodine formed requires 20 cm3 of 0.10 mol/dm3 sodium thiosulfate: I2 + 2Na2S2O3 → 2NaI + Na2S4O6. What is the concentration of copper(II) ions in the solution?
2026 KSS O Prelim P1 Q15
The Winkler method finds the amount of dissolved oxygen in water.
step 1: 2Mn2+(aq) + O2(g) + 4OH−(aq) → 2MnO(OH)2(s)
step 2: MnO(OH)2(s) + 2I−(aq) + 4H+(aq) → I2(aq) + Mn2+(aq) + 3H2O(l)
step 3: I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O62−(aq)
A 5.00 dm3 sample of water needed a total of 4.00 × 10−3 mol of thiosulfate in step 3. What concentration of oxygen was present in the sample?
2024 ANDSS O Prelim P1 Q24
Excess chlorine was bubbled through 250 cm3 of aqueous sodium thiosulfate, Na2S2O3, oxidising thiosulfate to sulfate ions. 25.0 cm3 of the resulting solution gave 0.583 g of barium sulfate precipitate with excess aqueous barium nitrate. What is the concentration, in mol dm−3, of sodium thiosulfate in the 250 cm3 of solution?
2016 HCI J1 Promo P1 Q6
Iodate(V) ions react with iodide ions in acid: IO3− + 5I− + 6H+ → 3I2 + 3H2O. 25.0 cm3 of sulfuric acid is added to a solution containing excess iodate(V) and iodide ions, and the iodine produced is titrated with sodium thiosulfate. 22.00 cm3 of 0.100 mol dm−3 sodium thiosulfate is needed. What is the concentration of the sulfuric acid, in mol dm−3?
2020 VJC J1 Promo P1-P2 Q10
Carbon monoxide in air reacts with iodine pentoxide: 5CO + I2O5 → I2 + 5CO2. The iodine is titrated with sodium thiosulfate: I2 + 2S2O32− → S4O62− + 2I−. A 100 cm3 sample of polluted air gave iodine that needed 20.0 cm3 of 0.20 mol dm−3 sodium thiosulfate. What is the concentration, in g dm−3, of carbon monoxide in the sample?
2021 ASRJC J2 Prelim P1 Q7
Ethanedioate ions are oxidised by hot acidified potassium manganate(VII): 2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O. What volume of 0.020 mol dm−3 potassium manganate(VII) is needed to oxidise 1.0 × 10−3 mol of the salt KHC2O4·H2C2O4?
2022 EJC J1 MYE P1 Q9
25.0 cm3 of 0.01 mol dm−3 potassium manganate(VII) was added to 25.0 cm3 of 0.01 mol dm−3 ethanedioic acid. Z is the maximum volume of CO2 produced at s.t.p. What is the value of Z?
SJChO Mock 3 P1 Q19
Drow D[1]
FeTiO3: +2 + x − 6 = 0, so x = +4.
Fe2(TiO3)3: 2(+3) + 3x − 18 = 0, so x = +4.
Acarbon dioxide and calcium carbonate[1]
Carbon is +4 in CO2 and in CaCO3.
In KCN (K +1, N −3) carbon is +2, and in Fe2C (Fe +2) it is −4.
C+1, +2, +3[1]
Bi, Sr, Ca and O give 6 + 4 + 4 − 20 = −6, so the three copper atoms total +6.
Only +1, +2, +3 adds up to +6.
Dtwo oxidations, highest +6[1]
Sulfur goes 0, +4, +6, +6, +6: only the first two stages raise it.
In H2S2O7, 2 + 2x − 14 = 0 gives +6 for each sulfur, not +12.
Crow C[1]
Sulfur starts at +6. In reaction 1 it stays +6 (0). In reaction 2 it falls to +4 in SO2 (2).
In reaction 3 it falls to −2 in H2S (8).
Bnitrogen from +4 to 0[1]
In NO nitrogen is +2, not +4, so it changes from +2 to 0.
Carbon goes from +2 to +4 and oxygen stays −2, so the reaction is redox.
COxygen is reduced in stage 1.[1]
In stage 1 oxygen falls from 0 to −2, so it is reduced, while nitrogen rises from +2 to +4.
In stage 2 NO2 (N +4 to +2) is the oxidising agent and SO2 (S +4 to +6) the reducing agent.
C1, 2 and 4[1]
Sulfur dioxide is an acidic oxide: it does not react with an acid but forms a salt with an alkali. It is a reducing agent, so it decolourises acidified manganate(VII).
Its solution is acidic, so Universal Indicator turns red or orange, not violet.
B1 and 3 only[1]
2FeCl3 + 2KI → 2FeCl2 + 2KCl + I2, so v = w = x = y = 2, and KI is the test for an oxidising agent.
The iodine stays dissolved, giving a yellow-brown solution, not a vapour.
B2 and 4[1]
Nitrogen is +3 in NF3, and each F2 goes from 0 to two F−, gaining 2 electrons.
Nitrogen is only oxidised (it stays −3 in NH4F), so there is no disproportionation, and F2 is the oxidising agent.
C+3[1]
nClO₄⁻ = 5.00 × 10−3 mol, nTi³⁺ = 0.0200 mol, each losing 1 electron.
Each Cl gains 0.0200 ÷ 5.00 × 10−3 = 4 electrons: +7 − 4 = +3.
CAmO22+[1]
nAm³⁺ = 2.40 × 10−4 mol and nCu³⁺ = 7.20 × 10−4 mol, each gaining 1 electron, so each Am loses 3: +3 to +6.
In AmO22+, x − 4 = +2 gives +6. The others give +3, +4 and +3.
C+4[1]
M starts at +5 in M2O5, and nM = 2 × 2.50 × 10−3 = 5.00 × 10−3 mol.
nMnO₄⁻ = 1.00 × 10−3 mol takes 5.00 × 10−3 mol of electrons: 1 per M, so M was at +4.
B40.0[1]
nMnO₄⁻ = 3.00 × 10−3 mol takes 0.0150 mol of electrons, so nM²⁺ = 0.0150 mol = nMO.
Mr(MO) = 0.84 ÷ 0.0150 = 56.0, so Ar(M) = 56.0 − 16.0 = 40.0. Option D forgets the oxygen.
C+5[1]
0.2 mol of I− gives 0.10 mol of I2 and loses 0.2 mol of electrons. The other 0.02 mol of I2 (0.04 mol of I atoms) comes from the oxide.
0.04 mol of I atoms gain 0.2 mol of electrons: 5 each, so the oxide iodine is +5.
C+3[1]
nFeC₂O₄ = 4.0 × 10−4 mol, each giving 1 + 2 = 3 electrons: 1.2 × 10−3 mol.
nMnO₄⁻ = 3.0 × 10−4 mol, so each Mn gains 4: +7 − 4 = +3.
ASO2[1]
2 mol of H2S (S −2) lose 4 mol of electrons in reaching 0.
So 1 mol of the other sulfur must gain 4 electrons: it starts at +4, as in SO2.
B0.150 mol[1]
Cl +5 to +7 loses 2 electrons. Cl +5 to −1 gains 6. So three chlorate(VII) form for each chloride: 4KClO3 → 3KClO4 + KCl.
0.200 × 3/4 = 0.150 mol.
B0.133 mol/dm3[1]
nS₂O₃²⁻ = 2.0 × 10−3 mol, so nI₂ = 1.0 × 10−3 mol and nCu²⁺ = 2.0 × 10−3 mol.
c = 2.0 × 10−3 ÷ 0.015 = 0.133 mol/dm3.
B6.40 mg/dm3[1]
4.00 × 10−3 mol S2O32− needs 2.00 × 10−3 mol I2, from 2.00 × 10−3 mol MnO(OH)2, from 1.00 × 10−3 mol O2.
1.00 × 10−3 × 32.0 = 0.0320 g = 32.0 mg in 5.00 dm3, so 6.40 mg/dm3.
B0.050[1]
nBaSO₄ = 0.583 ÷ 233.4 = 2.50 × 10−3 mol in 25.0 cm3, so 0.0250 mol of sulfate in 250 cm3.
Each S2O32− gives 2SO42−, so nS₂O₃²⁻ = 0.0125 mol in 0.250 dm3: 0.050 mol dm−3.
A0.0440[1]
nS₂O₃²⁻ = 2.20 × 10−3 mol, so nI₂ = 1.10 × 10−3 mol. 3I2 needs 6H+, so nH⁺ = 2.20 × 10−3 mol.
H2SO4 gives 2H+: 1.10 × 10−3 mol in 0.0250 dm3 = 0.0440 mol dm−3.
C2.80[1]
nS₂O₃²⁻ = 4.0 × 10−3 mol, nI₂ = 2.0 × 10−3 mol, nCO = 0.010 mol.
0.010 × 28.0 = 0.280 g in 0.100 dm3 = 2.80 g dm−3.
B40 cm3[1]
Each formula unit contains two C2O4 groups, so 2.0 × 10−3 mol is oxidised.
nMnO₄⁻ = 2/5 × 2.0 × 10−3 = 8.0 × 10−4 mol, V = 8.0 × 10−4 ÷ 0.020 = 0.040 dm3.
B11.4 cm3[1]
2MnO4− + 5H2C2O4: 2.5 × 10−4 mol of each, so the acid is limiting (it needs only 1.0 × 10−4 mol of MnO4−).
nCO₂ = 2 × 2.5 × 10−4 = 5.0 × 10−4 mol × 22.7 dm3 = 11.4 cm3.