IOE entrance Chemistry · Chapter 2
Inorganic Chemistry
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57 questions in 3 syllabus topics · 20 are 2-mark questions.
2.1 Non-metals and environmental pollution
25 questions
1. Temporary hardness of water is caused by the presence of dissolved:
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Which salts are removed simply by boiling?
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Answer: A. Bicarbonates of calcium and magnesium
Temporary hardness is due to Ca(HCO₃)₂ and Mg(HCO₃)₂, which decompose on boiling to insoluble carbonates. Chlorides and sulphates of Ca and Mg cause permanent hardness; sodium salts do not cause hardness.
2. In Clark's method, temporary hardness of water is removed by adding a calculated amount of:
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Recall which reagent reacts with bicarbonate to precipitate CaCO₃.
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Answer: A. Slaked lime, Ca(OH)₂
Clark's method: Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃↓ + 2H₂O. Washing soda, Calgon and permutit are used mainly for permanent hardness.
3. One litre of a water sample contains 324 mg of Ca(HCO₃)₂ as the only hardness-causing salt. Its hardness in ppm (expressed as CaCO₃) is:
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Convert moles of the salt to the same moles of CaCO₃.
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Answer: B. 200 ppm
Ca(HCO₃)₂ (M = 162) ≡ CaCO₃ (M = 100). 324 mg = 2 mmol, equivalent to 2 × 100 = 200 mg CaCO₃ per litre (10⁶ mg of water), i.e. 200 ppm.
4. The loss of meniscus and sticking of mercury to glass ("tailing of mercury") is observed when mercury comes in contact with:
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It is a classic test of an allotrope of oxygen.
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Answer: B. Ozone
Ozone oxidises mercury to Hg₂O, which dissolves in mercury and makes it stick to glass: 2Hg + O₃ → Hg₂O + O₂.
5. Chlorofluorocarbons deplete the stratospheric ozone layer mainly because UV light releases from them:
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Which bond in CF₂Cl₂ is weakest?
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Answer: D. Chlorine free radicals
UV breaks the C–Cl bond giving Cl•, which catalytically destroys O₃: Cl• + O₃ → ClO• + O₂, ClO• + O → Cl• + O₂. The C–F bond is too strong to break.
6. When 100 mL of pure oxygen is passed through an ozoniser, the volume decreases by 10 mL (same T and P). The volume of ozone formed is:
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Write 3O₂ → 2O₃ and compare the volume change with the ozone formed.
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Answer: C. 20 mL
3O₂ → 2O₃: 3 volumes give 2 volumes, a decrease of 1 volume for every 2 volumes of O₃ formed. A 10 mL decrease means 20 mL of O₃.
7. Which isotope of hydrogen is radioactive?
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Think of the isotope with the highest neutron-to-proton ratio.
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Answer: B. Tritium
Tritium (³H, one proton and two neutrons) is a β-emitter with half-life about 12.3 years; protium and deuterium are stable.
8. Hydrogen can be placed both with the alkali metals and with the halogens in the periodic table because it:
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Compare 1s¹ with ns¹ and with ns²np⁵.
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Answer: C. Has one valence electron and also needs only one electron to complete its shell
Like alkali metals hydrogen has 1s¹ configuration and can form H⁺; like halogens it is one electron short of a stable (He) configuration and can form H⁻.
9. In the Haber process for ammonia, molybdenum is added to the iron catalyst. Molybdenum acts as a:
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A substance that boosts the activity of a catalyst is called…
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Answer: D. Promoter
Molybdenum increases the efficiency of the iron catalyst without being a catalyst itself, i.e. it is a promoter.
10. In Ostwald's process, ammonia is oxidised to nitric oxide by air in the presence of:
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Match each catalyst to its industrial process.
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Answer: C. Platinum–rhodium gauze
4NH₃ + 5O₂ → 4NO + 6H₂O over Pt–Rh gauze at about 800 °C. Fe/Mo is for Haber's process, V₂O₅ for the contact process and Ni for hydrogenation.
11. In the Ostwald process, assume every mole of NH₃ is finally converted into one mole of HNO₃ (NO being recycled). The mass of HNO₃ obtained from 34 g of NH₃ is:
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Find moles of NH₃ first; M(HNO₃) = 63.
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Answer: A. 126 g
34 g NH₃ = 34/17 = 2 mol. With a 1 : 1 conversion, 2 mol HNO₃ = 2 × 63 = 126 g.
12. Copper reacts with dilute nitric acid to give mainly:
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The product depends on acid concentration; dilute acid is reduced less far than you might think.
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Answer: A. Nitric oxide (NO)
3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 2NO + 4H₂O. Concentrated acid gives NO₂; copper never liberates H₂ from nitric acid.
13. Which metal is rendered passive by concentrated nitric acid?
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A protective oxide layer forms on certain metals.
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Answer: B. Iron
Concentrated HNO₃ forms a thin protective oxide film on Fe (also Al and Cr), making it passive. Cu, Zn and Ag dissolve in it.
14. 20 L of N₂ and 30 L of H₂ (same T and P) are allowed to react to the maximum possible extent to form ammonia. The volume of NH₃ formed is:
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Identify the limiting reactant using Gay-Lussac's law of volumes.
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Answer: B. 20 L
N₂ + 3H₂ → 2NH₃. 30 L H₂ needs only 10 L N₂, so H₂ is limiting; 30 L H₂ gives (2/3) × 30 = 20 L NH₃ (10 L N₂ left over).
15. Which hydrogen halide is the strongest acid in aqueous solution?
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Consider H–X bond strength down the group.
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Answer: B. HI
Acid strength increases HF < HCl < HBr < HI because the H–X bond becomes longer and weaker down the group, so HI ionises most easily.
16. Chlorine reacts with hot, concentrated sodium hydroxide to give:
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Temperature decides which oxoanion forms.
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Answer: D. NaCl and NaClO₃
3Cl₂ + 6NaOH(hot, conc) → 5NaCl + NaClO₃ + 3H₂O (disproportionation). Cold dilute NaOH gives NaCl and NaOCl.
17. 8.7 g of MnO₂ is heated with excess concentrated HCl. The volume of chlorine evolved at NTP is (Mn = 55):
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One mole of MnO₂ gives one mole of Cl₂.
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Answer: C. 2.24 L
MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O. M(MnO₂) = 87, so 8.7 g = 0.1 mol, giving 0.1 mol Cl₂ = 0.1 × 22.4 = 2.24 L.
18. Graphite is a good conductor of electricity because:
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Count how many of carbon's four valence electrons are used in σ bonds.
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Answer: B. Each carbon is sp² hybridised and has one delocalised electron
In graphite each C uses three sp² orbitals for σ bonds in a layer; the fourth electron is delocalised over the layer and carries current.
19. 10.7 g of ammonium chloride is heated with excess slaked lime. The volume of ammonia obtained at NTP is:
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Moles of NH₃ equal moles of NH₄Cl.
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Answer: D. 4.48 L
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O. M(NH₄Cl) = 53.5, so 10.7 g = 0.2 mol, giving 0.2 mol NH₃ = 4.48 L.
20. The number of P–P single bonds in one molecule of white phosphorus (P₄) is:
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Count the edges of a tetrahedron.
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Answer: B. 6
P₄ is a tetrahedron with a P atom at each corner; a tetrahedron has 6 edges, so there are 6 P–P bonds (each P bonded to 3 others: 4 × 3/2 = 6).
21. White phosphorus reacts with hot NaOH solution: P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂. The oxidation states of phosphorus in PH₃ and NaH₂PO₂ respectively are:
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Use H = +1, O = −2, Na = +1.
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Answer: A. −3 and +1
In PH₃: x + 3 = 0, x = −3. In NaH₂PO₂: +1 + 2(+1) + x + 2(−2) = 0, x = +1. P goes from 0 to −3 and +1: disproportionation.
22. In the contact process, SO₃ is absorbed in concentrated H₂SO₄ rather than directly in water because:
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Think about the heat released and what the product looks like.
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Answer: B. With water it forms a dense mist of acid that is hard to condense
SO₃ + H₂O is very exothermic and produces a fog of H₂SO₄ droplets. So SO₃ is absorbed in conc. H₂SO₄ to form oleum (H₂S₂O₇), which is then diluted.
23. Assuming 100% conversion, the mass of H₂SO₄ that can be manufactured from 16 g of sulphur is:
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Each sulphur atom ends up in one H₂SO₄ molecule.
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Answer: D. 49 g
S → SO₂ → SO₃ → H₂SO₄, one S per H₂SO₄. 16 g S = 0.5 mol, giving 0.5 × 98 = 49 g H₂SO₄.
24. Which pair of xenon compound and molecular shape is correctly matched?
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Count bond pairs and lone pairs on Xe using VSEPR.
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Answer: D. XeF₂ – linear
XeF₂: sp³d with 3 lone pairs, linear. XeF₄ is square planar, XeO₃ is pyramidal (one lone pair) and XeF₆ is a distorted octahedron (one lone pair).
25. Acid rain is mainly caused by the presence in the atmosphere of:
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Which gases form strong acids with water and oxygen?
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Answer: D. Oxides of sulphur and nitrogen
SO₂ and NO/NO₂ are oxidised and dissolve in rain to form H₂SO₄ and HNO₃, lowering the pH of rain below about 5.6.
2.2 Metallurgy, alkali, alkaline earth and coinage metals
20 questions
26. Which of the following statements is correct?
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Ore = mineral + economic profitability.
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Answer: B. All ores are minerals, but all minerals are not ores
A mineral is any naturally occurring compound of a metal; an ore is a mineral from which the metal can be extracted economically. E.g. clay is a mineral of Al, but bauxite is its ore.
27. The froth flotation process is mainly used for the concentration of:
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Which ore particles are preferentially wetted by oil?
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Answer: D. Sulphide ores
Sulphide ore particles are preferentially wetted by oil (pine oil, with collectors like xanthates) and rise with the froth, while gangue is wetted by water and sinks.
28. Which of the following changes is an example of calcination?
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Calcination expels volatile matter without needing oxygen.
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Answer: B. ZnCO₃ → ZnO + CO₂
Calcination is heating an ore in a limited supply or absence of air to remove moisture or volatile matter like CO₂. Heating ZnS in air is roasting; the others are reductions.
29. During smelting, an ore contains silica (SiO₂) as gangue. The flux used to remove it is:
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An acidic impurity needs a basic flux.
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Answer: A. CaO (from limestone)
SiO₂ is an acidic impurity, so a basic flux is needed: CaO + SiO₂ → CaSiO₃ (slag). SiO₂ and P₄O₁₀ are acidic fluxes used for basic impurities.
30. In the electrolytic refining of copper, the anode is made of:
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The metal to be purified must dissolve into the electrolyte.
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Answer: C. Impure copper
Impure copper is the anode and a thin sheet of pure copper is the cathode in acidified CuSO₄. Cu dissolves at the anode and deposits at the cathode; Ag and Au collect as anode mud.
31. A current of 2 A is passed for 9650 s through acidified CuSO₄ solution during electrolytic refining. The mass of copper deposited at the cathode is (Cu = 63.5):
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Use Q = It and remember Cu²⁺ needs two electrons.
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Answer: D. 6.35 g
Q = 2 × 9650 = 19300 C = 0.2 F. Cu²⁺ + 2e⁻ → Cu needs 2 F per mole, so 0.1 mol Cu = 6.35 g.
32. Which ore can be concentrated by magnetic separation?
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The ore's name itself is a clue.
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Answer: B. Magnetite
Magnetite (Fe₃O₄) is magnetic and is separated from non-magnetic gangue using a magnetic roller. Bauxite (Al₂O₃·2H₂O), galena (PbS) and calamine (ZnCO₃) are non-magnetic.
33. When potassium burns in excess oxygen, the main product is:
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Larger alkali metal cations stabilise larger anions.
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Answer: C. KO₂
Li forms the normal oxide Li₂O, Na forms the peroxide Na₂O₂, while larger K, Rb and Cs form superoxides MO₂. A large cation stabilises the large O₂⁻ ion.
34. Potassium carbonate cannot be prepared by the Solvay (ammonia–soda) process because:
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Which step of the Solvay process needs a bicarbonate to come out of solution?
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Answer: D. KHCO₃ is too soluble in water to be precipitated
The Solvay process depends on precipitation of NaHCO₃ from the solution. KHCO₃ is much more soluble and does not precipitate, so the process fails for potassium.
35. The percentage of water of crystallisation in washing soda, Na₂CO₃·10H₂O, is approximately:
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Find the molar mass of the whole hydrate first.
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Answer: C. 62.9%
M = 106 + 180 = 286. % water = (180/286) × 100 ≈ 62.9%. (37.1% is the share of Na₂CO₃.)
36. A salt gives a brick-red colour in the flame test. The metal present is most likely:
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Recall flame colours of group 1 and 2 metals.
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Answer: A. Calcium
Flame colours: Ca brick red, Na golden yellow, Ba apple green, K lilac (violet).
37. Among the sulphates of alkaline earth metals, the least soluble in water is:
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Hydration energy decreases sharply as cation size increases.
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Answer: C. BaSO₄
Solubility of sulphates decreases down the group (BeSO₄ > MgSO₄ > CaSO₄ > SrSO₄ > BaSO₄) because hydration energy falls faster than lattice energy.
38. Gypsum, CaSO₄·2H₂O, is heated at about 393 K to form plaster of Paris. The mass of water lost from 172 g of gypsum is:
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Plaster of Paris is the hemihydrate.
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Answer: B. 27 g
CaSO₄·2H₂O (M = 172) → CaSO₄·½H₂O + 1½H₂O. 1 mol gypsum loses 1.5 mol water = 1.5 × 18 = 27 g.
39. Magnesium carbonate is heated strongly till there is no further change. The percentage loss in mass is about:
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The mass lost is the CO₂ given off.
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Answer: D. 52.4%
MgCO₃ (84) → MgO (40) + CO₂ (44). Loss = 44/84 × 100 ≈ 52.4%. (47.6% is the residue MgO.)
40. Copper obtained after the self-reduction step (Cu₂S + 2Cu₂O → 6Cu + SO₂) is called blister copper because:
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Look at the gaseous product of the self-reduction.
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Answer: D. Escaping SO₂ gas produces blisters on its surface
As the molten copper solidifies, dissolved SO₂ escapes and leaves blister-like marks. Blister copper is about 98% pure.
41. In the cyanide process for extraction of silver, the metal is recovered from the soluble complex Na[Ag(CN)₂] by adding:
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A more reactive metal displaces silver from its complex.
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Answer: A. Zinc
Zinc, being more electropositive, displaces silver: 2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag.
42. Pure gold is 24 carat. The mass of pure gold present in a 12 g ornament made of 22-carat gold is:
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Carat is parts of gold per 24 parts of alloy.
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Answer: A. 11 g
Fraction of gold = 22/24. Mass = (22/24) × 12 = 11 g. (2.64 g comes from wrongly using 22/100.)
43. Gold dissolves in aqua regia mainly forming:
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The chloride ions help by forming a complex.
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Answer: C. Chloroauric acid, H[AuCl₄]
Aqua regia (3 HCl : 1 HNO₃) produces nascent chlorine/NOCl; gold forms AuCl₃, which with HCl gives the stable complex H[AuCl₄].
44. In blue vitriol, CuSO₄·5H₂O, the number of water molecules directly coordinated to the Cu²⁺ ion is:
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One water molecule is held differently from the others.
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Answer: B. 4
Four H₂O molecules are coordinated to Cu²⁺ in a square plane; the fifth is held by hydrogen bonding between the coordinated water and the sulphate ion.
45. In photography, unexposed silver bromide is removed from the film by sodium thiosulphate (hypo) by forming:
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Fixing works by forming a soluble complex.
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Answer: A. Na₃[Ag(S₂O₃)₂]
AgBr + 2Na₂S₂O₃ → Na₃[Ag(S₂O₃)₂] + NaBr. The soluble complex washes away, fixing the image.
2.3 Extraction of zinc and mercury; iron
12 questions
46. In the extraction of zinc from zinc blende, the roasted ore (ZnO) is reduced by heating with:
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The cheapest reducing agent used in metallurgy.
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Answer: C. Coke
ZnO + C → Zn + CO at about 1400 °C (Belgian/horizontal retort process). Coke is the cheap reducing agent used.
47. In the extraction of zinc by carbon reduction, zinc is obtained as vapour and collected by condensation because:
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Compare the boiling point of zinc with the furnace temperature.
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Answer: B. Its boiling point (about 907 °C) is lower than the reduction temperature
Reduction of ZnO by carbon needs about 1400 °C, well above zinc's boiling point (≈907 °C), so zinc distils off and is condensed; this also separates it from non-volatile impurities.
48. The principal ore of mercury is:
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It is a red sulphide ore.
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Answer: B. Cinnabar
Cinnabar is HgS. Calamine is ZnCO₃, siderite is FeCO₃ and cuprite is Cu₂O.
49. Mercury is extracted by roasting cinnabar: HgS + O₂ → Hg + SO₂. The mass of mercury obtained from 46.4 g of pure HgS is (Hg = 200, S = 32):
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One mole of HgS gives one mole of Hg.
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Answer: C. 40 g
M(HgS) = 232. 46.4 g = 0.2 mol HgS → 0.2 mol Hg = 0.2 × 200 = 40 g.
50. Galvanised iron continues to be protected from rusting even if its zinc coating is scratched because:
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Compare the standard reduction potentials of Zn and Fe.
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Answer: B. Zinc is more electropositive and corrodes in preference to iron
Zinc has a more negative electrode potential than iron, so it becomes the anode and oxidises instead of iron (sacrificial/cathodic protection).
51. In the upper (cooler) part of the blast furnace, haematite is reduced mainly by:
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The gas produced from burning coke at the bottom rises upwards.
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Answer: D. Carbon monoxide
At about 500–900 °C: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Direct reduction by coke takes place only in the hotter lower region.
52. The purest commercial form of iron is:
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Which form has the least carbon?
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Answer: A. Wrought iron
Wrought iron contains only about 0.1–0.25% carbon. Steel has about 0.2–1.5% carbon and cast/pig iron about 2.5–4.5% carbon.
53. Which statement about rusting of iron is NOT correct?
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Rusting is electrochemical, so it needs an electrolyte medium.
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Answer: D. Iron rusts readily in dry oxygen
Rusting is an electrochemical process that requires both O₂ and moisture; electrolytes like NaCl speed it up. Iron does not rust in dry oxygen.
54. During rusting of iron, the reaction occurring at the anodic region is:
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Oxidation always occurs at the anode.
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Answer: D. Fe → Fe²⁺ + 2e⁻
At the anode iron is oxidised to Fe²⁺; the electrons travel to the cathodic region where O₂ is reduced. Fe²⁺ is later oxidised to Fe³⁺ forming rust.
55. A haematite ore contains 80% Fe₂O₃ by mass. The maximum mass of iron that can be extracted from 1000 kg of this ore is:
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First find the mass of Fe₂O₃, then the mass fraction of Fe in it.
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Answer: D. 560 kg
Fe₂O₃ present = 800 kg. Fraction of Fe in Fe₂O₃ = 112/160 = 0.7, so Fe = 800 × 0.7 = 560 kg.
56. When green vitriol is heated strongly, the products obtained are:
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Green vitriol is FeSO₄·7H₂O; Fe²⁺ is oxidised while some S⁶⁺ is reduced.
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Answer: D. Fe₂O₃, SO₂ and SO₃
FeSO₄·7H₂O first loses water to give anhydrous FeSO₄, which then decomposes: 2FeSO₄ → Fe₂O₃ + SO₂ + SO₃.
57. Hardened (quenched) steel is reheated to a temperature well below red heat and then cooled slowly to reduce its brittleness. This process is called:
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Quenching makes steel hard; which treatment then moderates it?
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Answer: C. Tempering
Tempering reheats quenched steel to about 200–300 °C and cools it slowly, reducing brittleness while keeping hardness. Annealing heats to a high temperature and cools slowly to make steel soft.