Important Questions for CBSE Class 11 Chemistry Chapter 11 – The p-Block Elements

The P block Elements Important Questions Class 11 Chemistry Chapter 11

In this chapter, students will get in-depth knowledge about six groups of p block elements from the periodic table. This chapter covers concepts such as the general trend in p block, physical and chemical properties, behaviour and compound of boron, carbon and silicon and important use of elements and compounds. It carries significant weightage in the Chemistry syllabus. Students can easily access all this and more from the Extramarks website.

Students must go through all the chapters thoroughly to score good grades in examinations. . Extramarks understands the importance of solving questions. As a result, our science faculty experts have created a repository of resources, such as the NCERT Textbook, NCERT Exemplar, other reference books, past examination papers, and so on and compiled them into important questions of  Chemistry Chapter 11 Class 11. Students registered with Extramarks can access Important Questions Class 11 Chemistry Chapter 11 anytime. To enjoy the maximum benefit of these resources, students just need to register themselves at Extramarks’ official website and stay ahead of the competition. Why wait, be an early bird and make the most of it.

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CBSE Class 11 Chemistry Important Questions
Sr No Chapters Chapter Name
1 Chapter 1 Some Basic Concepts of Chemistry
2 Chapter 2 Structure of Atom
3 Chapter 3 Classification of Elements and Periodicity in Properties
4 Chapter 4 Chemical Bonding and Molecular Structure
5 Chapter 5 States of Matter
6 Chapter 6 Thermodynamics
7 Chapter 7 Equilibrium
8 Chapter 8 Redox Reactions
9 Chapter 9 Hydrogen
10 Chapter 10 The s-Block Elements
11 Chapter 11 The p block Elements
12 Chapter 12 Organic Chemistry – Some Basic Principles and Techniques
13 Chapter 13 Hydrocarbons
14 Chapter 14 Environmental Chemistry

P-block Elements Class 11 Important Questions with Answers

Extramarks subject experts have compiled important questions of p block elements class 11 from various sources. The question comprises a wide variety of topics, including the general trend in p block, physical and chemical properties, behaviour and compound of boron, carbon and silicon and important use of element and compound and so on. These questions and solutions help students to clear their concepts of the p block element and prepare thoroughly for the exams.

 The following are some of the questions from Class 11 Chemistry Chapter 11 Important Questions:

Question 1: Silicon dioxide is treated with hydrogen fluoride. Explain?

Answer 1: Silicon and hydrogen fluoride are treated to form silicon tetrafluoride. The bond is so strong that Halogens/Acids with Silicon dioxide do not allow the reaction. Still, they react at extreme temperatures, where the high electronegative fluorine substitutes silicon and Oxygen.


Question 2. Why does silicon not form a graphite structure? Explain.

Answer 2: C is sp2 hybridised in graphite, and each C is linked to three other C atoms and forms hexagonal rings. This forms two- Dimensional sheet compositions of graphite. 

Silicon does not form analogue C because of two reasons:

  1. Silicon has a much lesser tendency for catenation than C as Si-Si bonds are much weaker than C-C bonds.
  2. Because of its larger size than C, Silicon undergoes sp3 hybridisation.

Question 3. How is borax prepared from

(i) Colemanite ore (ii) Tincal (iii) Boric acid?

Answer 3: 

(i) Colemanite ore: The mineral is boiled with sodium carbonate; as a result, they form perceptions of calcium carbonate, the crystal of borax and sodium metaborate.

Ca2B6O11+ 2Na2Co3→ Na2B4O7 + 2NaBO2 + 2CaCO3.

when mother liquid passes through carbon dioxide, it gets converted to borax

4NaBO2 +CO2 → Na2B47 +Na2CO3

(ii)  Tincal: Naturally occurring borax is called Tincal; it is collected from dried lakes. We extract that Tincal from dried-up lakes, boil it with water, and filter it to remove impurities. The filter is then concentrated, and the crystal of borax separates.

(iii) Boric acid: Boric acid reacts with sodium carbonate; as a  result, it forms a crystal of borax, water, and carbon dioxide.

4H­3BO3 +Na2CO3 → Na2B4O7 +6H20 +CO2

Question 4. Rationalise the Given Statements and Chemical Reactions:

  1. Lead (II) chloride reacts with Cl2to give PbCl4
  2. Why Lead (IV) chloride is highly unstable toward heat.
  3. Why Lead is known not to form an Iodide, PBI4.

Answer (a): Due to the inert pair effect, PB shows an oxidation state of +2 and +4. Since Cl2. is a strong oxidising agent, it oxidises to Pb2 to Pb4, and hence PbClreacts with PbCl4.

PbCl2(s) + Cl2(g) → PbCl4(l) 

Group +2 is more stable on moving down, and +4 is less stable due to the inert pair effect. Hence pbcl4 is much more stable than PbCl2. However, the formation of PbCl4 takes place when Cl2 gas is bubbled through a saturated solution of PbCl2.

Answers (b): The greater stability of +2 over +4 oxidation state is because of the inert pair effect. 

Lead (IV) chloride on heating decomposes to give Lead (II) chloride and Cl2.

The reaction is given below:

PbCl4(l)→ Δ PbCl2(s) +  Cl2(g) 

Answer (c): PB4 is oxidising in nature. A combination of PB4 and iodide is not stable. Iodide ions are strongly reduced in nature.

The reaction is given below:

Pb(IV) oxidises I- to I2 and itself gets reduced to PB(II)

PBI4→ PBI2+ I2

Question 5. Mention some important properties of carbon monoxide.

Answers 5. Some important properties of carbon monoxide are as follows:

  1. It is a colourless, odourless gas, slightly soluble in water.
  2. When carbon monoxide is treated with chlorine in the presence of light or charcoal, a poisonous gas is formed. 

CO +Cl2  → COCl

  1. It burns in the air with a blue flame forming carbon dioxide 

2CO + O2→ 2CO2.

  1. Carbon monoxide acts as a strong agent when reacted with metals like iron, cobalt and nickel 

Fe2O3 + 3CO → 2Fe + 3CO2 .

Question 6: Explain the lower atomic radius of Gallium compared to Aluminium.

Answer 6: Aluminium size is 143 Pm, whereas Gallium size is 135 pm. Because of poor shielding of the Valence electron of Ga by the inner 3rd electron, the effective Nuclear charge of Ga is greater in magnitude than that of Al. As a result, the electron in Ga experiences a greater force of attraction by the Nucleus than in Al. Hence the atomic size of Ga is slightly less than that of Al.

Question 7: Explain the differences in properties of diamond and graphite based on their structures.

Answer 7: In a diamond, each carbon atom is sp3 hybridised and is bonded to four other carbon atoms through a sigma bond. In graphite, each carbon atom is sp2 hybridised and is bonded to three other carbon atoms through a sigma bond while the fourth electron forms a Π-bond.

The below table highlights the differences:

Sr no Diamond Graphite
1 It has crystalline lattice


It has a layered structure
2 It is made up of tetrahedral units


It has a planar geometry
3 The c-c bond of length in the diamond is 154 pm The c- c bond length in graphite is 141.5 pm


4 They have a rigid covalent bond network, which is difficult to break It is quite soft, and its layer can be represented easily.
5 It acts as an electrical insulator It is a good conductor of electricity

Question 8 Explain why there is a phenomenal decrease in ionisation enthalpy from Carbon to silicon?

Answer 8: The ionisation enthalpy of Carbon (the first element of group 14) is very high (1086 KJ/mol). Moving down in a group increases its size with some screening effects. The force of attraction in the Nucleus for the Valence electron decreases too much. This results in a phenomenal decrease in ionisation enthalpy from Carbon to silicon.

Question 9: Why does boron trifluoride behave as a Lewis acid?

Answer 9: The B atom in Bf3 has only six electrons in the valence shell and hence is an electron-deficient molecule. They can easily accept a pair of electrons from Nucleophiles to complete their activities and thus behave as Lewis Acid.

Question 10: How is excessive CO2 responsible for global warming?

Answer 10: CO2 is an essential gas for our survival. CO2 is produced during combustion. Plants utilise it during photosynthesis, and O2 is released into the atmosphere. As a result of this CO2 cycle, a constant percentage of 21% O2 is maintained in the atmosphere. The decomposition of limestone and decrease in the number of trees has led to increased levels of CO2.  which  has the property of trapping the heat provided by sun rays. A higher amount of heat trapped increases the atmospheric temperature, causing global warming.

Question 11: Explain (i) Inert pair effect, (ii) Allotropy, and (iii) Catenation.

Answer 11: 

(i) Inert pair effect: As we move down the group, the tendency of S – electrons of the valence shell to participate in a bond formation decreases. The reluctance of the S-electron to participate in bond formation is called the Inert-pair effect.

(ii) Allotropy: Allotropy is an element having more than one form. They have the same chemical properties but different physical properties. 

For example, Carbon exists in three allotropic forms Diamond, Graphite and fullerenes.

(iii) Catenation: Atoms of some elements (such as Carbon) can link to strong covalent bonds to form long chains. This is the reason they are called catenation. They can be found in Carbon and are quite significant in Si and S.

Question 12: Why is CO poisonous? Explain.

Answer 12: In the lungs, Haemoglobin in Red Blood cells combines with molecular Oxygen loosely and forms oxyhaemoglobin.

Haemoglobin + Oxygen provides oxyhaemoglobin.

Co is highly poisonous because of the ability to form a complex molecule with haemoglobin. The Co-Hb complex is more stable ( 300 times) than the O2 -Hb complex. 

Hb + CO → CO-Hb ( carboxyhemoglobin)

As a result, the oxygen-carrying capacity of Hb is destroyed, and the person dies of suffocation as it displaces oxygen in the blood and deprives the heart, brain and other vital organs of oxygen.

Question 13: Explain the higher stability of BCl3 as compared to TICl3.

Answer 13: Due to poor shielding of the s-electron of the Valence shell (6s) by 3d, 4d, 5d and 4F electrons. The inert-pair effect is maximum in Tl; as a result, only 6p1 electrons participate in Bond formation, and thus the most stable state of Tl is +1 and not +3. Therefore TICl is stable, but TICl3 is unstable. 

In contrast, due to the absence of d and f electrons, B does not show an inert pair effect. In other words, all three valence electrons (i.e. two 2s and one 2p) participate in Bond formation. Hence, B shows an oxidation state of +3 and thus forms BCl3. Thus BCl3 is more stable than TICl3.

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p-block elements are both a crucial l topic and a challenging one with varied concepts. Studying all these requires extra effort, and students must keep revising and practising Important Questions Class 11 Chemistry Chapter 11 to understand the topic thoroughly. Students don’t have sufficient time to prepare notes alone during the examination. Extramarks create study resources that will help them with a repository of important questions and hold their interest by clearing their doubts and strengthening their knowledge regarding the subject concepts. Students will get a sense of confidence by solving essential questions from all the chapters and assessing their preparation for the upcoming exams. 

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Q.1 (i) Why N(CH3)3 is pyramidal but N(SiH3)3 is planer. Give reason.
(ii) Why SnCl4 behaves as a covalent compound whereas SnCl2 as an ionic compound?
(iii) Which type of intermolecular forces exist among the following molecules?
(a) He atoms and HCl molecules
(b) SiH4 molecules
(c) Cl2 and CCl4 molecules
(d) H2S molecules



(i) This is because the lone pair of electrons present on N (present in the 2p-orbital) is transferred to the empty d-orbital of Si (


) overlapping) and hence is not available for protonation.

(ii) According to Fajans rule, a cation having small size and high charge exerts large polarising effect on the neighbouring anion resulting in the development of considerable amount of covalent character in the compound. The Sn4+ ion having high charge and small size as compared to those of Sn2+ ion polarizes Cl ion to a greater extent and because of this, SnCl4 behaves as a covalent compound but SnCl2 as an ionic compound.

(a) Dipole-induced dipole forces
(b) London dispersion forces
(c) London dispersion forces
(d) Dipole-dipole interactions

Q.2 Answer the following questions
(a) Highlight the products formed when orthoboric acid is heated.
(b) How is water gas produced?
(c) What happens when aluminium reacts with alkaline sodium hydroxide solution?



(a) On heating orthoboric acid above 370K, it forms metaboric acid, HBO2 which on further heating forms boric oxide, B2O3. The chemical reaction can be represented as

H3BO3Orthoboric acid+>370KΔHBO2Megabreccia+ΔB2O3Boric oxide

(b) Water gas is formed by passing steam over hot coke. Water gas is a mixture of carbon monoxide and hydrogen gas.


(c) Aluminium reacts with aqueous alkali solution to form sodium tetrahydroxoaluminate (III) ion and liberates dihydrogen gas. The reaction is represented as


Q.3 Account for the following:
(a) Sn4+ is more stable than Pb4+ but Pb2+ is more stable than Sn2+.
(b) BF3 acts as a Lewis acid.
(c) Octahedral complexes of boron such as [B(H2O)6]3+ do not exist in aqueous solution.



(a) In p-block elements, down the group from carbon to lead, the stability of +4 oxidation state decreases while the stability of +2 oxidation state increases owing to the inert pair effect. Consequently, the +2 oxidation state of Pb becomes more stable than the +2 oxidation state of Sn while the +4 oxidation state of Sn becomes more stable than the +4 oxidation state of Pb.
(b) In BF3, the electronic configuration of boron is [He]2s2 2p1. Boron has 3 electrons in its valence shell. It shares one electron each with 3 fluorine atoms due to which BF3 contains a total of 6 electrons. Since boron trifluoride remains electron-deficient, it can accept electrons and acts as a Lewis acid.
(c) Due to the non-availability of d orbitals, boron cannot expand its octet. Therefore, it can only form a maximum of four bonds. Therefore, octahedral complexes of boron such as [B(H2O)6]3+ do not exist in an aqueous solution.

Q.4 Compound X is used to remove temporary hardness. Identify the compound X.

(i) CaO

(ii) Ca(OH)2

(iii) CaCO3

(iv) CaCl2



(ii) Ca(OH)2

Explanation: Temporary hardness is removed by boiling and Clarks process. In Clarks process, calculated quantity of Ca(OH)2 is added. The bicarbonates present in water react with calcium hydroxide to form insoluble calcium and magnesium carbonates. These insoluble carbonates can be easily removed by filtration.

Q.5 Chemically, water gas is a mixture of
A. CO and H2
B. NO and H2
C. NO and H2O
D. CO2 and H2



A. CO and H2

An equimolar mixture of CO and H2 is called water gas.

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FAQs (Frequently Asked Questions)

1. What are p-block elements in Class 11?

In the p block element, the last electron enters the outermost p orbital. The periodic table has six groups of p block elements from 13 to 18. Bo2, co2, N2, O2, fl2 and hl2 are the most important elements to study. In the periodic table, metalloids and non-metal only exist in the p block element.

2. How many groups are there in p and d block elements as per the important questions in Class 11 Chemistry Chapter 11?

There are six groups in p- block elements as mentioned in the important questions in Class 11 Chemistry Chapter 11. The d-block elements are found in groups 3 to 12 in the periodic table, adding up to 10 groups in contrast to 6 in the p block.

3. Which are some good reference books for studying Chemistry in classes 11 and 12?

NCERT Organic Chemistry, Modern ABC, Pardeep books etc., are some of the decent reference books students can use for classes 11 and 12.