Take timed practice tests on p-Block Elements for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.
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1. The most abundant metal in the Earth's crust is:
Explanation: Aluminium (Group 13) is the most abundant metal in Earth's crust (about 8%). It occurs mainly as bauxite (Al₂O₃·2H₂O) and aluminosilicates.
2. Boron differs from the rest of Group 13 elements because:
Explanation: Boron is a metalloid with a very high charge density (small size, +3 charge). It cannot form B³⁺ ions — the ionisation energy is too high. It forms covalent compounds (BCl₃, B₂H₆) instead.
3. The hybridisation of N in NH₃ is:
Explanation: N in NH₃ has 3 bonding pairs and 1 lone pair → total 4 electron pairs → sp³ hybridisation → pyramidal geometry.
4. Which oxide of nitrogen is used as a laughing gas (anaesthetic)?
Explanation: N₂O (nitrous oxide) is used as a mild anaesthetic (laughing gas). It decomposes at body temperature to release O₂. NO is a toxic radical, NO₂ is brown and toxic, N₂O₅ is the anhydride of HNO₃.
5. White phosphorus is more reactive than red phosphorus because:
Explanation: White phosphorus consists of P₄ tetrahedral units with 60° bond angles (highly strained). The strain weakens the P−P bonds, making it more reactive. Red phosphorus has a polymeric structure with less strain and is less reactive.
6. The structure of H₂SO₄ has sulphur with hybridisation:
Explanation: S in H₂SO₄ has 4 bonding pairs around it (2 OH + 2 =O), but with expanded octet using d-orbitals. Effectively sp³ hybridised with tetrahedral geometry.
7. The correct order of oxidising power of halogens is:
Explanation: Oxidising power of halogens decreases down Group 17 as electronegativity and reduction potential decrease: F₂ (E° = +2.87 V) > Cl₂ (+1.36 V) > Br₂ (+1.07 V) > I₂ (+0.54 V).
8. In ClF₃, the hybridisation of Cl is:
Explanation: Cl in ClF₃: 3 bonding pairs + 2 lone pairs = 5 electron pairs → sp³d hybridisation → T-shaped geometry (90° bond angles between axial and equatorial F).
9. Which of the following is NOT a property of noble gases?
Explanation: Noble gases (Group 18) have complete outer shells and very high ionisation energies. They rarely form compounds. XeF₂, XeF₄, XeO₃ exist for Xe (low IE in Group 18), but stable covalent compounds are NOT a general property.
10. Acid strength among oxychlorine acids follows:
Explanation: More oxygen atoms on central Cl = higher oxidation state of Cl = greater electron withdrawal from O−H bond = stronger acid. HClO₄ (Cl = +7) is the strongest of the oxychlorine acids.
11. The bond angle in NF₃ is less than in NH₃ because:
Explanation: In NF₃, F's high electronegativity pulls the N−F bonding electron pairs towards F, reducing electron density near N. This decreases bond-pair repulsions at N → bond angle 102.2° (less than NH₃'s 107°). In NH₃, bonding pairs are closer to N (H is less electronegative).
12. The number of lone pairs on the central Cl in ClF₃ is:
Explanation: Cl has 7 valence electrons. 3 are used for bonding with F → 4 non-bonding electrons = 2 lone pairs. T-shaped geometry.
13. Which allotrope of sulphur is stable at room temperature?
Explanation: Rhombic sulphur (α-sulphur, S₈ puckered rings) is stable below 96°C (transition temperature). Above 96°C, monoclinic sulphur (β-sulphur) is stable. Plastic sulphur is an amorphous form obtained by pouring molten S into cold water.
14. HF has higher boiling point than HCl despite lower molar mass because:
Explanation: F is the most electronegative element and forms very strong hydrogen bonds (F···H−F). These hydrogen bonds are stronger than the van der Waals forces in HCl, giving HF a much higher boiling point (19.5°C vs −85°C for HCl).