JEE (Advanced) · Chemistry, Class 12
Isolation of Metals: JEE Advanced PYQ, 2007–2026
Isolation of Metals (Chemistry, Class 12) has been asked 22 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.6 questions a paper, 3% of all Chemistry questions. It is the 17th most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 1in 2 papers |
| 2008 | 2in 2 papers |
| 2009 | 0in 2 papers |
| 2010 | 3in 2 papers |
| 2011 | 1in 2 papers |
| 2012 | 1in 2 papers |
| 2013 | 2in 2 papers |
| 2014 | 1in 2 papers |
| 2015 | 2in 2 papers |
| 2016 | 1in 2 papers |
| 2017 | 0in 2 papers |
| 2018 | 1in 2 papers |
| 2019 | 2in 2 papers |
| 2020 | 1in 2 papers |
| 2021 | 1in 2 papers |
| 2022 | 2in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 0in 2 papers |
| 2026 | 0in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Smelting and refining10
- Ores and concentration7
- Roasting and calcination5
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10 Isolation of Metals questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Isolation of Metals
Tin, which melts at about 505 K, is refined by heating its impure form on a sloping hearth so that the molten tin flows away from infusible impurities. Which property of tin makes this separation possible?
Answer: (A)
Principle: liquation is a physical method that separates a metal from impurities when the metal melts at a lower temperature than the impurities. The impure metal is heated on a sloping hearth. The molten metal flows downhill and leaves the infusible solids behind. No chemical reaction is needed. The key property is the difference in melting points, so check which option names it.
Q2 · Chemistry · Isolation of Metals
97 g of pure ZnS (Zn = 65, S = 32, O = 16) is roasted in air at a moderate temperature. Incomplete control of conditions results in half of the zinc (by moles) ending up as ZnO (with loss of sulphur as SO2) and the other half as ZnSO4. What is the mass of the solid residue?
Answer: (D)
The molar mass of ZnS is 97 g mol⁻¹, so 97 g is 1 mol and contains 1 mol Zn. Half goes to ZnO (M = 81 g mol⁻¹), giving 0.5 × 81 = 40.5 g. Half goes to ZnSO4 (M = 65 + 32 + 64 = 161 g mol⁻¹), giving 0.5 × 161 = 80.5 g. The total residue is 40.5 + 80.5 = 121 g. Zinc is conserved, so each mole of ZnS gives one mole of Zn-containing solid. A student who takes the all-ZnO or all-ZnSO4 masses has ignored the stated 1:1 split.
Q3 · Chemistry · Isolation of Metals
A finely powdered ore contains galena (PbS) and zinc blende (ZnS) together with siliceous gangue. It is concentrated by froth flotation using pine oil, and a small quantity of NaCN is added to the pulp. Which statement correctly describes the role of NaCN and the outcome?
Answer: (B)
In froth flotation, collectors make the sulphide mineral particles water-repelling so that they attach to air bubbles and rise, while the gangue is wetted by water and sinks. When two sulphides are present, a depressant can stop one of them from floating. NaCN forms the soluble complex Na2[Zn(CN)4] on the ZnS surface, which makes ZnS water-wet so it stays in the pulp. PbS is not affected in the same way, so it floats in the froth. The result is selective separation of PbS from ZnS.
Q4 · Chemistry · Isolation of Metals
A 50 g sample of pure calcium carbonate, CaCO3 (Ca = 40, C = 12, O = 16), is heated strongly until decomposition is complete. What mass of CO2 gas is evolved?
Answer: (A)
First convert the sample to moles: n = mass / molar mass = 50 g / 100 g mol⁻¹ = 0.5 mol. The decomposition CaCO3 → CaO + CO2 gives 1 mol CO2 per mol CaCO3. So 0.5 mol CO2 is produced. Its mass is 0.5 × 44 = 22 g. The other options come from skipping the mole step, using the wrong gas, or ignoring that the solid CaO stays behind.
Q5 · Chemistry · Isolation of Metals
Powdered bauxite (Al₂O₃·xH₂O containing Fe₂O₃, TiO₂ and SiO₂ as impurities) is digested with concentrated NaOH solution at 473–523 K and 35–36 bar. The solution is separated from the residue and CO₂ is passed through it. Which statement correctly describes the outcome?
Answer: (C)
Principle: leaching exploits the difference in acid–base character. Al₂O₃ (amphoteric) and SiO₂ (acidic) react with hot concentrated NaOH to give soluble Na[Al(OH)₄] and sodium silicate. Fe₂O₃ and TiO₂ (basic or inert) do not react, so they are filtered off as the residue. Passing CO₂ neutralises the alkali: Na[Al(OH)₄] is converted into hydrated Al₂O₃, which precipitates, while sodium silicate stays in solution. Only the third option fits this sequence.
Q6 · Chemistry · Isolation of Metals
Native gold (19.7 g, Au = 197) is leached from crushed ore using dilute aqueous NaCN with air bubbled through the mixture. All the gold dissolves as [Au(CN)2]⁻, and dissolved O2 is the only oxidant. The gold is then recovered completely from the solution by adding zinc dust, which is oxidised to [Zn(CN)4]²⁻. Taking Zn = 65, which option gives the amount of O2 consumed in the leaching step and the mass of Zn consumed in the precipitation step?
Answer: (A)
Cyanide leaching (Mac Arthur–Forrest process) relies on air oxidising the metal in the presence of CN⁻, which then forms a stable complex: 4Au + 8CN⁻ + O2 + 2H2O → 4[Au(CN)2]⁻ + 4OH⁻. Moles of Au = 19.7/197 = 0.100 mol. Each O2 takes 4 electrons and oxidises 4 Au, so O2 consumed = 0.100/4 = 0.025 mol. The complex is then displaced by the more electropositive zinc: 2[Au(CN)2]⁻ + Zn → [Zn(CN)4]²⁻ + 2Au. Each Zn supplies 2 electrons for 2 Au⁺, so Zn consumed = 0.100/2 = 0.050 mol, which is 0.050 × 65 = 3.25 g. The correct pair is 0.025 mol O2 and 3.25 g Zn.
Q7 · Chemistry · Isolation of Metals
Which statement correctly distinguishes calcination from roasting in metallurgy?
Answer: (A)
Principle: calcination is thermal decomposition of an ore below its melting point, usually with limited or no air, to drive off volatile matter such as CO2, H2O or SO2 from carbonates and hydrates. Roasting is heating in excess air, typically to oxidise sulphide ores to oxides, with sulphur leaving as SO2. Both are done below the melting point, so the air supply is the key difference.
Q8 · Chemistry · Isolation of Metals
In the extraction of many metals, a sulphide ore is roasted to the oxide before being reduced with carbon, rather than reduced directly. Which statement gives the correct thermodynamic reason?
Answer: (A)
A reducing agent works only if the overall ΔG of reduction is negative. For most sulphides, ΔfG° is more negative than that of CS2 and H2S, so carbon or hydrogen cannot reduce the sulphide to the metal. Roasting turns the sulphide into the oxide with release of SO2, and the oxide can then be reduced by carbon. The reason is therefore thermodynamic and not physical.
Q9 · Chemistry · Isolation of Metals
An ore concentrate is an equimolar mixture of calamine, ZnCO3, and zinc blende, ZnS. A 222 g sample is heated strongly in an excess of air until its mass is constant. The carbonate is calcined to ZnO and CO2, and the sulphide is roasted completely to ZnO and SO2. No ZnSO4 forms. (Zn = 65, S = 32, C = 12, O = 16) What percentage of the original mass is lost?
Answer: (B)
Calcination is ZnCO3 → ZnO + CO2. Roasting is 2ZnS + 3O2 → 2ZnO + 2SO2. The sample has 1 mol of each compound (125 + 97 = 222 g). It gives 2 mol ZnO, which is 2 × 81 = 162 g. Mass lost is 222 − 162 = 60 g, so the loss is 60/222 × 100 = 27.0%. During calcination the solid loses CO2 (44 g). During roasting the solid gains oxygen while losing sulphur, so ZnS (97 g) becomes ZnO (81 g), a net loss of only 16 g. Subtracting the gases' full masses would wrongly ignore the oxygen absorbed.
Q10 · Chemistry · Isolation of Metals
Which of the following is the role of a collector such as sodium ethyl xanthate in froth flotation?
Answer: (A)
A collector is a surfactant-type reagent that selectively adsorbs onto the surface of the desired sulphide mineral and makes it hydrophobic. The hydrophobic particles then attach to air bubbles and float. The collector acts on the ore, not the gangue, and it does not dissolve the ore.
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All 22 real JEE Advanced questions from Isolation of Metals.
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About Isolation of Metals in JEE Advanced.
How many questions come from Isolation of Metals in JEE Advanced?
Isolation of Metals (Chemistry, Class 12) has been asked 22 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.6 questions a paper, 3% of all Chemistry questions. It is the 17th most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it.
Which topics of Isolation of Metals are asked most in JEE Advanced?
Smelting and refining (10), Ores and concentration (7), Roasting and calcination (5), counted over the papers from 2007–2026.
Is Isolation of Metals important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 17 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Isolation of Metals JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Isolation of Metals with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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