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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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.

                      SiO2+4HF→SiO4+2H2O

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 → Na2B4O­7 +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 PbCl2  reacts 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  → COCl2 

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

2CO + O2→ 2CO2.

4. 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:

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.

Benefits of Solving Important Questions of P-block Elements Class 11

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. 

Mentioned below are some benefits of solving Important Questions Class 11 Chemistry Chapter 11:

• Students will benefit by practising similar examination questions to get high scores in their examinations and stay ahead of the competition.
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• The questions and answers follow the CBSE pattern, and the topics are directly picked from the CBSE syllabus by the experienced faculty to provide authentic, up-to-date and accurate information to students to help them achieve their best scores. No wonder students have absolute faith and trust in Extramarks.
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