Questions & explanations
1. Which of the following orders of Lewis acid strength towards trimethylamine is correct?
- BF3 < BCl3 < BBr3 < BI3 < AlCl3 < Al2Cl6 < Tl(NO3)3
- BF3 < BCl3 < BBr3 < BI3 < Al2Cl6 < AlCl3 < Tl(NO3)3
- BI3 < BBr3 < BCl3 < BF3 < AlCl3 < Al2Cl6 < Tl(NO3)3
- BF3 < BCl3 < BBr3 < BI3 < AlCl3 < Tl(NO3)3 < Al2Cl6
Answer: BF3 < BCl3 < BBr3 < BI3 < AlCl3 < Al2Cl6 < Tl(NO3)3
Lewis acidity of boron trihalides increases down the group due to decreasing pπ-pπ back-bonding: BF3 < BCl3 < BBr3 < BI3. For aluminium halides, monomeric AlCl3 is a stronger Lewis acid than its dimer Al2Cl6 because dimerization reduces electron deficiency. Tl(NO3)3 is the weakest due to the inert pair effect. Hence the correct order is BF3 < BCl3 < BBr3 < BI3 < AlCl3 < Al2Cl6 < Tl(NO3)3.
2. A white precipitate formed by adding dilute HCl to Pb(NO3)2 solution is:
- soluble in hot water
- insoluble in hot water
- soluble in cold water
- insoluble in concentrated HCl
Answer: insoluble in concentrated HCl
Pb(NO3)2 + 2HCl → PbCl2↓ (white) + 2HNO3. PbCl2 is sparingly soluble in cold water but dissolves in hot water due to increased hydration energy. However, it is insoluble in concentrated HCl because the common ion effect suppresses dissociation, and complex formation [PbCl4]2- requires excess Cl- which is not present in dilute HCl. Thus, the precipitate is insoluble in concentrated HCl.
3. Which of the following statements about producer gas and water gas is correct?
- Water gas has a higher calorific value than producer gas.
- Water gas is obtained by passing air over red-hot coke at 1100°C.
- Producer gas contains about 49% H₂ and 45% CO.
- Producer gas is used as a source of hydrogen for methanol synthesis.
Answer: Water gas has a higher calorific value than producer gas.
Water gas (CO + H₂) has a higher calorific value than producer gas (CO + N₂) because both components are combustible, while producer gas is diluted with non-combustible N₂. Option a describes water gas composition, not producer gas. Option b describes producer gas preparation, not water gas. Option d is incorrect: water gas is used for methanol synthesis, not producer gas.
4. Which of the following statements about carbonates and bicarbonates is correct?
- NaHCO₃ is thermally stable up to 1000°C.
- Na₂CO₃ decomposes on gentle heating to give Na₂O and CO₂.
- Both Na₂CO₃ and NaHCO₃ give a white precipitate with MgSO₄ in cold solution.
- Addition of dilute HCl to NaHCO₃ produces CO₂ gas that turns lime water milky.
Answer: Addition of dilute HCl to NaHCO₃ produces CO₂ gas that turns lime water milky.
Both carbonates and bicarbonates react with dilute acids to release CO₂, which turns lime water milky due to formation of CaCO₃. NaHCO₃ decomposes on gentle heating (2NaHCO₃ → Na₂CO₃ + H₂O + CO₂), not stable up to 1000°C. Na₂CO₃ is thermally stable and does not decompose on gentle heating. Only carbonate (CO₃²⁻) gives a white precipitate with MgSO₄; bicarbonate does not.
5. Which of the following statements correctly compares CO and CO₂?
- CO is weakly acidic, while CO₂ is neutral.
- CO has a bond order of 2, while CO₂ has a bond order of 3.
- CO has a dipole moment of 0.11 D, while CO₂ has zero dipole moment.
- CO₂ is a stronger reducing agent than CO.
Answer: CO has a dipole moment of 0.11 D, while CO₂ has zero dipole moment.
CO has a triple bond (C≡O) with sp hybridisation, giving a small dipole moment of 0.11 D due to unsymmetrical electron distribution. CO₂ is linear and symmetrical (O=C=O), so its dipole moment is zero. The other options are incorrect: CO has bond order 3, CO₂ has bond order 2 per C=O; CO is neutral, CO₂ is weakly acidic; CO is a strong reducing agent, while CO₂ is not.
6. Which property of fluorine is the root cause of its low F-F bond enthalpy, less negative electron gain enthalpy than chlorine, and weak acidity of HF?
- Highest electronegativity in the periodic table
- Small atomic size leading to high inter-electronic repulsion in 2p subshell
- Absence of d-orbitals in its valence shell
- Very high ionization enthalpy
Answer: Small atomic size leading to high inter-electronic repulsion in 2p subshell
Fluorine's small atomic size causes high inter-electronic repulsion in its compact 2p subshell. This repulsion weakens the F-F bond (lone-pair repulsion), reduces the energy released on adding an electron (less negative EGE than Cl), and makes the H-F bond very strong, so HF is a weak acid. All three anomalies trace back to this single underlying property.
7. Which property of carbon is primarily responsible for its ability to form long chains and three different allotropes?
- Presence of d-orbitals
- High electronegativity and small size
- Low ionization enthalpy
- Large covalent radius
Answer: High electronegativity and small size
Carbon's small size and high electronegativity lead to strong C–C bonds (348 kJ/mol), the highest among Group 14. This strong bonding enables catenation—the formation of long chains. Additionally, carbon's ability to adopt sp, sp2, and sp3 hybridisations (without d-orbitals) gives rise to allotropes: diamond (sp3), graphite (sp2), and fullerenes (sp2).
8. In the brown ring test for nitrate, what is the oxidation state of iron in the brown complex [Fe(H2O)5(NO)]2+?
- 0
- +3
- +2
- +1
Answer: +1
The complex [Fe(H2O)5(NO)]2+ has overall charge +2. H2O is neutral, NO is neutral (NO+ would be nitrosyl, but here NO is neutral). Let Fe oxidation state be x. Then x + 0*5 + 0 = +2 => x = +2. However, in the brown ring complex, NO is actually NO+ (nitrosyl cation) with charge +1. Then x + 0*5 + (+1) = +2 => x = +1. Thus iron is in +1 oxidation state.
9. Which of the following is an interhalogen compound of the type XX'7?
- F7I
- ClF7
- BrF7
- IF7
Answer: IF7
Interhalogen compounds of type XX'7 are formed when the central halogen is the heaviest and the terminal halogen is the lightest. Only iodine (largest) can accommodate seven fluorine atoms, giving IF7. Chlorine and bromine are too small to hold seven fluorines. The formula is written with the heavier atom first, so IF7 is correct, not F7I.
10. Which of the following is NOT a major use of sulphuric acid?
- Manufacture of ammonium sulphate fertiliser
- Electrolyte in lead-acid storage batteries
- Pickling of steel to remove rust before electroplating
- Production of hydrochloric acid by the salt-cake process
Answer: Production of hydrochloric acid by the salt-cake process
Sulphuric acid is a key industrial chemical used in fertiliser production (ammonium sulphate), as battery electrolyte, and for steel pickling. The salt-cake process (NaCl + H2SO4) is a laboratory method for HCl; industrially, HCl is produced via the Hargreaves process or as a by-product of chlorination. Thus, option c is not a major use.
11. Which statement correctly compares HNO3 and H3PO4?
- HNO3 is a moderate tribasic acid and a strong oxidising agent; H3PO4 is a strong monobasic acid and not oxidising.
- HNO3 is a strong tribasic acid and a strong oxidising agent; H3PO4 is a moderate monobasic acid and not oxidising at room temperature.
- HNO3 is a strong monobasic acid and not oxidising; H3PO4 is a moderate tribasic acid and a strong oxidising agent.
- HNO3 is a strong monobasic acid and a strong oxidising agent; H3PO4 is a moderate tribasic acid and not oxidising at room temperature.
Answer: HNO3 is a strong monobasic acid and a strong oxidising agent; H3PO4 is a moderate tribasic acid and not oxidising at room temperature.
HNO3 is a strong monobasic acid (Ka ~24) and a powerful oxidising agent because N in +5 state is small and electronegative, making it easy to reduce. H3PO4 is a moderate tribasic acid (pKa1 = 2.15) and not oxidising at room temperature because P in +5 state is larger and less electronegative, making reduction difficult.
12. Which statement correctly compares PCl3 and PCl5 regarding their structure and hydrolysis?
- PCl3 has sp3 hybridization, trigonal pyramidal geometry, and on hydrolysis gives H3PO3; PCl5 has sp3d hybridization, trigonal bipyramidal geometry with equatorial bonds 202 pm and axial bonds 240 pm, and on hydrolysis gives H3PO4.
- PCl3 has sp3d hybridization, trigonal bipyramidal geometry, and on hydrolysis gives H3PO4; PCl5 has sp3 hybridization, trigonal pyramidal geometry, and on hydrolysis gives H3PO3.
- PCl3 has sp3 hybridization, trigonal planar geometry, and on hydrolysis gives H3PO4; PCl5 has sp3d hybridization, square pyramidal geometry, and on hydrolysis gives H3PO3.
- PCl3 has sp3 hybridization, trigonal pyramidal geometry, and on hydrolysis gives H3PO4; PCl5 has sp3d hybridization, trigonal bipyramidal geometry with all bonds equal at 220 pm, and on hydrolysis gives H3PO3.
Answer: PCl3 has sp3 hybridization, trigonal pyramidal geometry, and on hydrolysis gives H3PO3; PCl5 has sp3d hybridization, trigonal bipyramidal geometry with equatorial bonds 202 pm and axial bonds 240 pm, and on hydrolysis gives H3PO4.
PCl3 has sp3 hybridization with one lone pair, giving trigonal pyramidal geometry. Hydrolysis yields H3PO3 (dibasic). PCl5 in gas phase has sp3d hybridization, trigonal bipyramidal geometry; equatorial bonds are 202 pm and axial bonds are 240 pm due to greater repulsion. Hydrolysis yields H3PO4 (tribasic).