CBSE Class 11 Chemistry Chapter-11 Important Questions - Free PDF Download
FAQs on Important Questions for CBSE Class 11 Chemistry Chapter 11 - The p-Block Elements
1. Which topics from Chapter 11, The p-Block Elements, are most important for the CBSE Class 11 exams 2025-26?
For the Class 11 final exams, certain topics from The p-Block Elements carry higher weightage. Students should focus on:
- Group 13 Elements: Anomalous properties of Boron, the structure of Diborane (B₂H₆), and the concept of the inert pair effect.
- Group 14 Elements: Anomalous behaviour of Carbon, catenation properties, and a detailed study of the allotropes of carbon (diamond and graphite).
- Important Compounds: Preparation, properties, and structures of Borax, Boric Acid, Silicones, and Silicates.
- Conceptual Questions: Reasoning for trends in atomic radii, ionisation enthalpy, and electronegativity down the groups.
These areas are frequently tested in 3-mark and 5-mark questions.
2. Explain the anomalous behaviour of Boron. Why is this an expected question in exams?
Boron, the first member of Group 13, shows significant differences from other elements in its group. This is considered an important question because it tests fundamental periodic trend concepts. The key reasons for its anomalous behaviour are:
- Extremely small atomic size.
- High ionisation enthalpy and high electronegativity compared to other group members.
- Absence of vacant d-orbitals in its valence shell.
As a result, Boron is non-metallic, forms only covalent compounds, and exhibits a maximum covalency of four (e.g., in [BF₄]⁻), unlike other members which can expand their octet.
3. What is the inert pair effect and how does it influence the properties of Group 13 and 14 elements?
The inert pair effect is the reluctance of the two s-electrons in the valence shell to participate in bond formation. This effect becomes more prominent as we move down the group.
- In Group 13: The common oxidation state is +3. However, due to the inert pair effect, the stability of the +1 oxidation state increases down the group. For example, Thallium (Tl) is more stable in the +1 state than the +3 state.
- In Group 14: The common oxidation state is +4. Similarly, the stability of the +2 oxidation state increases down the group. Lead (Pb) is more stable in the +2 state, making Pb⁴⁺ a strong oxidising agent.
This is a crucial concept for reasoning-based questions in the exam.
4. Why is Boron Trichloride (BCl₃) a Lewis acid? How can this concept be tested in a 2-mark question?
BCl₃ acts as a Lewis acid because the central Boron atom is electron-deficient. Although it forms three single covalent bonds with chlorine atoms, it only has six electrons in its valence shell, falling short of a stable octet. To complete its octet, it has a strong tendency to accept a pair of electrons from a donor molecule (a Lewis base) like ammonia (NH₃). In an exam, a typical question might ask you to explain why BCl₃ behaves as a Lewis acid or to show the reaction between BCl₃ and NH₃.
5. Compare the catenation properties of Carbon and Silicon. Why does this difference make it a frequently asked question?
Catenation is the ability of an atom to form long chains with itself. This is a classic comparative question because it highlights the unique nature of carbon.
- Carbon: Exhibits extensive catenation due to the very high strength of the C-C single bond (bond energy ~348 kJ/mol). This allows it to form stable, long chains and rings.
- Silicon: Shows very limited catenation (up to 7-8 atoms) because the Si-Si single bond is much weaker (bond energy ~297 kJ/mol) and more reactive.
For full marks, students must mention the difference in bond enthalpy as the primary reason.
6. Why does Carbon Tetrachloride (CCl₄) not undergo hydrolysis while Silicon Tetrachloride (SiCl₄) does? What is the core concept being tested here?
This is a high-order thinking skill (HOTS) question that tests your understanding of electronic configuration and bonding.
- CCl₄: Carbon has no vacant d-orbitals in its valence shell. Therefore, it cannot accept the lone pair of electrons from the oxygen atom of a water molecule to initiate hydrolysis.
- SiCl₄: Silicon has vacant 3d-orbitals. It can expand its coordination number by accepting a lone pair from a water molecule, which leads to its rapid hydrolysis to form silicic acid.
The key concept is the availability of vacant d-orbitals for expanding the octet.
7. From an exam perspective, what are the key structural differences between diamond and graphite to secure full marks?
To score well on questions about the allotropes of carbon, you must link their structure to their properties. The key differences are:
- Hybridisation and Structure: In diamond, each carbon is sp³ hybridised and tetrahedrally bonded to four other carbons, forming a rigid 3D network. In graphite, each carbon is sp² hybridised and bonded to three others in the same plane, forming hexagonal layers.
- Hardness: Diamond's rigid, covalent network makes it the hardest known substance. Graphite's layers are held by weak van der Waals forces, allowing them to slide, making it soft and slippery.
- Electrical Conductivity: In diamond, all four valence electrons are used in bonding, so there are no free electrons, making it an electrical insulator. In graphite, the fourth valence electron is delocalised within the layers, making it a good conductor of electricity.
8. What type of important questions can be expected from the topic of 'Silicones' in Chapter 11?
From the topic of Silicones, students can expect questions focusing on their preparation, structure, and uses. Important questions include:
- Preparation: A 2-mark question might ask for the reaction showing the preparation of silicones from alkyl chlorosilanes followed by hydrolysis.
- Structure: A question could ask you to explain why silicones are high-molecular-mass polymers with a repeating (R₂SiO) unit and a backbone of -Si-O-Si- linkages.
- Properties and Uses: A 3-mark question might ask to list the properties of silicones (like thermal stability, water repellency) and relate them to their uses in sealants, greases, and water-proofing materials.











