Relevant Titles
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CBSE Class 11 MCQs: p-Block Reactions & Bonding Patterns (Group 13 & 14)
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NCERT-Aligned 30 MCQs — p-Block Reactions for Class 11 Chemistry
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Practice Test: p-Block Reactions and Bonding (CBSE Class 11 Chemistry)
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Chapter 11 MCQs — Bonding Patterns & Reactions in Group 13 and 14
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CBSE Board Prep: p-Block Reactions MCQs with Answers & Explanations
Introduction
CBSE Class 11 Chemistry MCQs on p-Block reactions provide targeted, NCERT-aligned practice to master Chapter 11 (The p-Block Elements). This practice set focuses on the reaction types and bonding patterns characteristic of Group 13 and Group 14 elements — including electron-deficient boron chemistry (boranes like B<sub>2</sub>H<sub>6</sub>), Lewis acidity of BF<sub>3</sub>/BCl<sub>3</sub>, amphoterism of Al<sub>2</sub>O<sub>3</sub>, catenation and allotropy in carbon and silicon, and the behavior of oxides such as SiO<sub>2</sub> and CO<sub>2</sub>. Each MCQ is designed to test factual recall, conceptual understanding, and application under exam conditions; answers are followed by clear, concise explanations so learners can correct misconceptions immediately. Use this set for timed practice to simulate board pressure, identify weak topics, and reinforce NCERT examples and diagrams. The questions and explanations use properly formatted chemical notation (e.g., H<sub>2</sub>O, AlCl<sub>3</sub>) so they display correctly in WordPress. Regular practice will strengthen fundamentals for CBSE board exams and provide a strong foundation for competitive preparation.
Sample MCQs (with answers & explanations)
Q1. Which factor best explains why boron forms electron-deficient compounds such as B<sub>2</sub>H<sub>6</sub>?
A) Large atomic radius
B) Low electronegativity
C) Small size, high ionization enthalpy and high polarizing power
D) Presence of low-lying d-orbitals
Answer: C
Explanation: Boron’s small size and high charge density give strong polarizing power and limited valence electrons, causing electron-deficient multicenter bonding (3c–2e) in boranes.
Q2. Why does AlCl<sub>3</sub> (anhydrous) behave differently from AlCl<sub>3</sub> in water?
A) It is always ionic
B) In the gas/anhydrous state it dimerizes to Al<sub>2</sub>Cl<sub>6</sub> and shows covalent character; in water it forms solvated Al<sup>3+</sup>
C) It is inert in water
D) It becomes a noble gas
Answer: B
Explanation: Anhydrous AlCl<sub>3</sub> is covalent and dimeric (Al<sub>2</sub>Cl<sub>6</sub>); in aqueous solution it hydrolyses to give Al<sup>3+</sup>/complexes, showing different behavior depending on environment.
Q3. Which statement about BF<sub>3</sub> vs BCl<sub>3</sub> as Lewis acids is correct?
A) BF<sub>3</sub> is always stronger because F is more electronegative
B) BCl<sub>3</sub> is often the stronger Lewis acid because F→B back-bonding reduces BF<sub>3</sub>’s electron deficiency
C) Both are non-acidic
D) Only BF<sub>3</sub> reacts with bases
Answer: B
Explanation: Fluorine can engage in p(π)→p(π) back-bonding to boron, partially relieving electron deficiency in BF<sub>3</sub>, so BCl<sub>3</sub> is frequently the stronger Lewis acid in practice.
Q4. Which oxide is a network covalent solid and the main constituent of glass?
A) CO<sub>2</sub>
B) SiO<sub>2</sub>
C) Al<sub>2</sub>O<sub>3</sub>
D) B<sub>2</sub>O<sub>3</sub>
Answer: B
Explanation: SiO<sub>2</sub> forms strong, extended Si–O networks (silica) and is the principal component of glass and silicate minerals.
Q5. Which of the following reactions demonstrates aluminium’s amphoteric nature?
A) Al<sub>2</sub>O<sub>3</sub> + 6HCl → 2AlCl<sub>3</sub> + 3H<sub>2</sub>O
B) Al<sub>2</sub>O<sub>3</sub> + 2NaOH → 2NaAlO<sub>2</sub> + H<sub>2</sub>O
C) Both A and B
D) Neither A nor B
Answer: C
Explanation: Amphoterism means reacting with both acids and bases — Al<sub>2</sub>O<sub>3</sub> forms salts with acids and aluminates with bases, so both reactions show amphoteric behavior.