NCERT Solutions for Class 12 Chemistry Chapter 15: Polymers

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Chapter15. Polymers
TextbookChemistry II
CategoryNCERT Solutions for Class 12

NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 15: Polymers.

NCERT Solutions for Class 12 Chemistry Chapter 15


Q1) Explain the terms polymer and monomer.

Answer) Polymers are high molecular mass macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (103 − 107 u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6, 6 are examples of polymers.

Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example, ethene, propene, styrene, vinyl chloride.

Q2) What are natural and synthetic polymers? Give two examples of each type.

Answer) Natural polymers are polymers that are found in nature. They are formed by plants and animals. Examples include protein, cellulose, starch, etc.

Synthetic polymers are polymers made by human beings. Examples include plastic (polythene), synthetic fibres (nylon 6, 6), synthetic rubbers (Buna − S).

Q3) Distinguish between the terms homopolymer and copolymer and give an example of each.


The polymers that are formed by the polymerization of a single monomer are known as homopolymers. In other words, the repeating units of homopolymers are derived only from one monomer. For example, polythene is a homopolymer of ethene.The polymers whose repeating units are derived from two types of monomers are known as copolymers. For example, Buna−S is a copolymer of 1, 3-butadiene and styrene.

Q4) How do you explain the functionality of a monomer?

Answer) The functionality of a monomer is the number of binding sites that is/are present in that monomer.

For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene and adipic acid is two.

Q5) Define the term polymerisation.

Answer) Polymerization is the process of forming high molecular mass (103 − 107 u) macromolecules, which consist of repeating structural units derived from monomers. In a polymer, various monomer units are joined by strong covalent bonds.

Q6) Is (-NH-CHR-CO-)n, a homopolymer or copolymer?

Answer) (-NH-CHR-CO-)n is a homo-polymer, the reason being that it is derived from a single monomer unit, NH2—CHR—COOH.

Q7) Determine the groups where the polymers are graded according to molecular forces?

Answer) On the basis of magnitude of intermolecular forces present in polymers, they are classified into the following groups:

  1. Elastomers
  2. Fibres
  3. Thermoplastic polymers
  4. Thermosetting polymers

Q8) How can you differentiate between addition and condensation polymerisation?

Answer) Addition polymerization is the process of repeated addition of monomers, possessing double or triple bonds to form polymers. For example, polythene is formed by addition polymerization of ethene.

Condensation polymerization is the process of formation of polymers by repeated condensation reactions between two different bi-functional or tri-functional monomers. A small molecule such as water or hydrochloric acid is eliminated in each condensation. For example, nylon 6, 6 is formed by condensation polymerization of hexamethylenediamine and adipic acid.

Q9) Explain the term copolymerisation and give two examples.

Answer) The process of forming polymers from two or more different monomeric units is called copolymerization. Multiple units of each monomer are present in a copolymer. The process of forming polymer Buna−S from 1, 3-butadiene and styrene is an example of copolymerization

Q10) Write the free radical mechanism for the polymerisation of ethene.

Answer) Polymerization of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide as the initiator.

The reaction involved in this process is given below:

Chain initiation step

Q11) Define thermoplastics and thermosetting polymers with two examples of each.

Answer) Thermoplastic polymers are linear (slightly branched) long chain polymers, which can be repeatedly softened and hardened on heating. Hence, they can be modified again and again. Examples include polythene, polystyrene.

Thermosetting polymers are cross-linked or heavily branched polymers which get hardened during the molding process. These plastics cannot be softened again on heating. Examples of thermosetting plastics include bakelite, urea-formaldehyde resins.

Q12) Write the monomers used for getting the following polymers.

(i) Polyvinyl chloride (ii) Teflon (iii) Bakelite


(i) Vinyl chloride (CH2=CHCl)
(ii) Tetrafluoroethylene (CF2 = CF2)
(iii) Formaldehyde (HCHO) and phenol (C6H5OH)

Q13) Write the name and structure of one of the common initiators used in free radical addition polymerisation.

Answer) One common initiator used in free radical addition polymerization is benzoyl peroxide. Its structure is given below.

Q14) How does the presence of double bonds in rubber molecules influence their structure and reactivity?

Answer) Natural rubber is a linear cis-polyisoprene in which the double bonds are present between C2 and C3 of the isoprene units.

Because of this cis-configuration, intermolecular interactions between the various strands of isoprene are quite weak. As a result, various strands in natural rubber are arranged randomly. Hence, it shows elasticity.

Q15) Discuss the main purpose of vulcanisation of rubber.

Answer) Natural rubber though useful has some problems associated with its use. These limitations are discussed below:

  • Natural rubber is quite soft and sticky at room temperature. At elevated temperatures (> 335 K), it becomes even softer. At low temperatures (< 283 K), it becomes brittle. Thus, to maintain its elasticity, natural rubber is generally used in the temperature range of 283 K-335 K.
  • It has the capacity to absorb large amounts of water.
  • It has low tensile strength and low resistance to abrasion.
  • It is soluble in non-polar solvents.
  • It is easily attacked by oxidizing agents.

Vulcanization of natural rubber is done to improve upon all these properties. In this process, a mixture of raw rubber with sulphur and appropriate additive is heated at a temperature range between 373K and 415K.

Q16) What are the monomeric repeating units of Nylon-6 and Nylon-6, 6?

Nylon 6 monomeric repeat unit is [NH- (CH)2 )5 -CO, derived from Caprolactam. The nylon 6, 6

monomeric repeat complex is [NH- (CH)2 )6 – NH – CO – (CH)2 )4 – CO], which is derived from diamine hexamethylene and adipic acid.

Q17) Write the names and structures of the monomers of the following polymers:

(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene


Q18) Identify the monomer in the following polymeric structures.



Answer i) The monomers of the given polymeric structure are decanoic acid

[HOOC – (CH)2)8 – COOH] and hexamethylene diamine [H2N (CH)2)6 – NH2]

Answer ii) The monomers of the given polymeric structure are

Q19) How is dacron obtained from ethylene glycol and terephthalic acid?

Answer) The condensation polymerisation of ethylene glycol and terephthalic acid leads to the formation of dacron.

Q20) What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester.

Answer) A polymer that can be decomposed by bacteria is called a biodegradable polymer.

Poly-β-hydroxybutyrate-CO-β- hydroxyvalerate (PHBV) is a biodegradable aliphatic polyester.

That’s it. These were the solutions of NCERT Class 12 Chemistry Chapter 15 – Polymers. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.

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