Ncert Solutions Class 12 Chemistry Chapter 7 | Download Free PDF

Ncert Solutions Class 12 Chemistry Chapter 7 in English

Class 12 Chemistry Chapter 7 NCERT Solutions: Adda247 provides NCERT Solutions for class 12 Chemistry chapter 7. These solutions will not only help students to boost their board exams and score brilliant marks but for competitive exams. The Solutions are according to the NCERT guidelines.

Complete 16 chapters wise solutions were provided for the benefit of students.12^th class sets the base for higher education for every student. This makes it the most crucial class for any student who is aiming for his/her dream of quality education. Scoring good marks in the 12^th class is equivalent to a quality high education. So, it becomes extremely important for students to give a boost to their class 12 Chemistry preparation with Adda247 NCERT solutions.

Students can access the solution anywhere while browsing the web easily. The solutions are very precise and accurate.

Advantages of NCERT Solutions for class 12 Chemistry

• In-depth explanation of all the questions with a logical reason.
• Numericals are solved with step by step process.
• Free PDF download option.
• All the answers are given to the point of theoretical questions.

The detailed, explanative NCERT solutions are widely known for being extremely student-friendly and easy to digest. Solutions to relatively complex questions are always broken down into simpler one in order to help students focus on the ‘method to solve’ rather than the solution itself. This concept-centered approach is what enables students that use the Adda247 NCERT Solutions to solve all similar questions with ease.

NCERT Solution for class 12 Chemistry Chapter 7: the p-Block Elements

NCERT Solution Class 12 Chapter 7 is exclusively written for CBSE students of Class 12. These solutions provide an excellent approach to master the subject. These solutions assist you in understanding the concept deeply by giving P block elements Class 12 questions and answers in the textbook.

P block elements are in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons therefore in all there six groups of p-block elements.

The p-block elements contain groups 13, 14, 15, 16, 17, 18 with the expectation of Helium. The principal quantum number ‘n’ fills the p-orbital.

Uses of p-block Elements

• A compound of boron called borax is used in the glass making industry and pottery.
• Boron is also used in the soap or detergent industry.
• Boron is used in aircrafts and bullet proof vests.
• Boron is used in steel to increase its hardness.

One of the most interesting facts about the p-block elements is that it contains both non-metals and metalloids.

Chapter 7 Chemistry class 12 is a continuation of class 11 syllabus. While the course of grade 11 deals with the first two groups of p-block, 13 and 14; class 12 syllabus deals with group 16 to 18 groups. Group 15 is no longer a part of the syllabus as per the latest changes made by CBSE.

With numerous subtopics, it is one of the lengthiest chapter5 in the book. Hence, a solution book becomes beneficial for a student to score well in their exams.

Key features of NCERT Solution of Chemistry Class 12 Chapter 7

• The NCERT Solution provide clear and precise answer.
• The columns are used wherever necessary.
• Covers the basic of a topic.
• Adheres to the pattern of CBSE.

Students referring to the solutions book will get to know about the following things.

• Physical, chemical and atomic properties of the elements of group 16, 17 and 18.
• Properties of interhalogens.
• Allotropic forms of sulphur.
• Uses of noble gases.
• Structure of sulphur’s oxoacids.
• Preparation and properties of chlorine and hydrochloric acids.

Important questions of NCERT Solution of Class 12 Chemistry Chapter 7

Question: 1 Why does nitrogen show catenation properties less than phosphorus?

Answer: The extent of catenation depends on strength of the element-element bond. As N-N bond strength is weaker than P-P bond hence nitrogen shows less catenation properties than phosphorous.

Question: 2 Give the disproportionation reaction of H₃PO₃?

Answer: On heating, orthophosphorus acid disproportionates to give orthophosphoric acid and phosphine. The oxidation state of P in various species involved in the reaction are mentioned below:

4H₃PO₃ –à   3H₃PO₄ + PH₃

Question: 3 Can PCl₅ act as an oxidising as well as a reducing agent? Justify.

Answer:

PCl₅ cannot act as reducing agent because it cannot increase its oxidation state beyond +5 and it can easily act as oxidising agent because it can decrease its oxidation state from +5 to +3.

Question: 4 Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Answer:  The elements of group 16 are collectively called chalcogens.

• Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns² np⁴, where n varies from 2 to 6.
• Oxidation state: As these elements have six valence electrons (ns² np⁴), they should display an oxidation state of −2. However, only oxygen predominantly shows the oxidation state of −2 owing to its high electronegativity. It also exhibits the oxidation state of −1 (H₂O₂), zero (O₂), and +2 (OF₂). However, the stability of the −2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.
• Formation of hydrides: These elements form hydrides of formula H₂M, where M = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H₂M₂. These hydrides are quite volatile in nature.

Question: 5 Why is dioxygen a gas but sulphur a solid?

Answer: The intermolecular forces in oxygen are weak van der Waals forces, which causes it to exist as gas. On the other hand, sulphur does not form strong S=S double bonds hence exists as a puckered structure held together by strong covalent bonds and exists as a polyatomic molecule. So, it exists as a solid.

Question: 6 Which aerosols deplete ozone?

Answer:

Chlorofluoro carbons, sulphur dioxide, nitrous oxide are some examples of ozone depleting substances. The aerosols that depletes ozone layer are chlorofluorocarbons such as freon. The chemical formula of freon is CCl₂F₂.

Question: 7 Describe the manufacture of H₂SO₄ by contact process?

Answer: Sulphur acid is manufactured by the contact process. It involves the following steps:

Step 1: Sulphur or sulphide ores are burnt in air to form SO₂.

Step 2: By a reaction with oxygen, SO₂ is converted in SO₃ in the presence of V₂O₅ as a catalyst.

2SO_2(g) + O_2(g)     ^V2O5       2SO_3(g)

Step 3: SO₃ produced is absorbed on H₂SO₄ of the desired concentration.

In practice, the plant is operated at 2 bar(pressure) and 720 K (temperature). The sulphur acid thus obtained is 96 – 98 % pure.

Question: 8 How is SO₂ an air pollutant?

Answer: Sulphur dioxide (SO₂) is an air pollutant made up of sulphur and oxygen atoms and is harmful to both plants and people. On dissolution in rain water, SO₂ produces acid rain. This SO₃ gets converted into H₂SO₄ in the presence of moisture, which comes down in the form acid rain.

Even in very low concentrations, SO₂ causes irritation in the respiratory track. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.

Question: 9 Why are halogens strong oxidising agents?

Answer: Halogens act as strong oxidising agents because they have high tendency to accept electron, i.e., they have high electron affinity values. The reduction potentials are high (positive) and decreased from F to I.

Question: 10 Explain why fluorine forms only one oxoacids. HOF.

Answer:

Fluorine forms only one oxoacid, HOF due to small atomic size and high electronegativity. Fluorine cannot act as central atom in higher oxoacids.

Question: 11 Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.

Answer: Atomic size of oxygen is smaller than the atomic size of chlorine. Small atomic size favours hydrogen bond formation.

Hence, inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Question: 12 Write two uses of ClO₂.

Answer: Uses:

• It is used for purifying water.
• It is used as a bleaching agent.

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