Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Electronic Configurations of the D-Block Elements”.

1. Which of the following are d-block elements, but not regarded as transition elements?
a) Ru, Ag, Au
b) Zn, Ru, Pd
c) Zn, Cd, Hg
d) Cd, Rh, Pd
Answer: c
Clarification: Transition elements refer to those d-block elements who have their penultimate shells incomplete. In the case of Zinc, cadmium and mercury, since their penultimate shell is completely occupied, they are referred to as d-block elements and not regarded as transition elements.

2. Why do transition elements form alloys so easily?
a) Atomic size
b) Orbital configuration
c) Very light
d) Hard elements
Answer: a
Clarification: Alloys are combinations of metals or elements in the form of a compound or a solution. Transition elements are highly capable in forming alloys because they have similar atomic size and can substitute each of their positions in a crystal lattice.

3. Which of the following has the highest atomic number?
a) Scandium
b) Titanium
c) Cadmium
d) Lanthanum
Answer: d
Clarification: In the periodic table, when you move down the group, the atomic number increases. In the given options, scandium and titanium belong to the 3d series, cadmium belongs to the 4d series and Lanthanum belongs to the 5d series. Therefore, Lanthanum will have the highest atomic number.

4. What is the oxidation state exhibited by actinium?
a) +4
b) +1
c) +2
d) +3
Answer: d
Clarification: Actinium is the first member of the actinoid series. Its atomic number is 89. The electronic configuration of actinium is [Rn]6d¹7s². It has only three electrons to lose. Therefore, actinium is generally only found in the +3-oxidation state.

5. Where does the last electron of d-block element go?
a) nd
b) (n-1)d
c) np
d) (n-1)s
Answer: b
Clarification: In these elements, the last electron enters the d orbital of the penultimate shell, i.e., the last electron goes to (n-1) d orbital. Hence, these elements are named as d-block elements. These elements have partly filled d-subshells in their elementary form or in their simple ions.

6. Why does Zinc not show variable valency?
a) Complete ‘d’ subshell
b) Inert pair effect
c) 4s² subshell
d) 4s³ subshell
Answer: a
Clarification: Both zinc and mercury have completely filled d-orbital. So, they tend to lose only 2 electrons from their outermost shell and have only one oxidation state. Hence, they only show a valency of +2 and do not show variable valency.

7. What is the common oxidation state of scandium? (At. No. of Sc = 21)
a) +4
b) +3
c) +1
d) +5
Answer: b
Clarification: The properties of scandium compounds are intermediate between those of aluminium and yttrium. A diagonal relationship exists between the behaviour of magnesium and scandium, just as there is between beryllium and aluminium. In the chemical compounds of the elements in group 3, the predominant oxidation state is +3.

8. Among d-block elements, the most abundant element belongs to the first transition state.
a) True
b) False
Answer: a
Clarification: Iron and titanium are the most abundantly found transition elements. The most abundant transition metal in Earth’s solid crust is iron, which is fourth among all elements and second (to aluminum) among metals in crustal abundance.

9. Which of the following statement is correct?
a) The properties of various actinoids are very similar
b) 4f and 5f orbitals are equally shielded
c) D-block elements show irregular and erratic chemical properties
d) 4d and 5d orbitals are equally shielded
Answer: c
Clarification: There is very little variance in the atomic size of the actinoids and so, they show similar properties. 4f and 5f orbitals belong to different energy levels and so are unequally shielded. The same applies to the 4d and 5d orbitals. The d-block elements are known to show irregular and erratic chemical properties.

10. The main factor for the cause of lanthanide contraction is poorer shielding of 5d electrons by 4f electrons.
a) True
b) False
Answer: b
Clarification: Lanthanide contraction is the steady decrease in the size of the atoms with increasing atomic number from lanthanum (atomic number 57) through lutetium (atomic number 71). The Lanthanide Contraction is caused by a poor shielding effect of the 4f electrons. It’s due to lanthanide contraction, as the atomic number increases, the size of the lanthanide atoms and their tripositive ions decreases.

Chemistry Multiple Choice Questions on “Stability of Coordination Compounds”.

1. The degree of ________ between the metal ion and the ligand directly influences the stability of the complex in solution.
a) association
b) dissociation
c) ionisation
d) freedom
Answer: a
Clarification: The magnitude of the equilibrium constant for the association quantitatively expresses the stability of the complex in solution.

2. How are the stepwise stability constants (K) related to the overall stability constant (β)?
a) β_n = K₁ + K₂ + …. + K_n
b) β_n = K₁ x K₂ x … x K_n
c) β_n = logK₁ + logK₂ + …. + logK_n
d) β_n = 1/(K₁ + K₂ + …. + K_n)
Answer: b
Clarification: The overall stability constant is the product of all the stepwise stability constants of each individual reaction.

3. If the values of stepwise stability constants for a series of successive reactions are logK₁ = 4, logK₂ = 3.2, logK₃ = 2.7 and logK₄ = 2, what is the value of β₄?
a) 11.9
b) 7.94 x 10¹¹
c) 1.47 x 10⁵
d) 27.4
Answer: b
Clarification: logβ₄ = logK₁ + logK₂ + logK₃ + logK₄
(Rightarrow)log₁₀β₄ = 4 + 3.2 + 2.7 + 2 = 11.9
(Rightarrow)β₄ = antilog(11.9)
(Rightarrow)β₄ = 7.94 x 10¹¹

4. If β is the formation constant, then the instability constant = ________
a) 1 – β
b) β – 1
c) (1/β)
d) log(β)
Answer: c
Clarification: The instability constant is also known as the dissociation constant. It is defined as the reciprocal of the formation or stability constant.

5. For a particular complex, the dissociation equilibrium constant is given as 2.3 x 10^-9. What will be the overall stability constant for this complex?
a) 4.35 x 10^-10
b) 4.35 x 10¹⁰
c) 4.35 x 10^-8
d) 4.35 x 10⁸
Answer: d
Clarification: Formation constant = 1/(dissociation constant)
(Rightarrow)β = 1/(2.3 x 10^-9)
(Rightarrow)β = 4.35 x 10⁸

6. Two complexes A and B have dissociation constants 1.0 x 10^-12 and 4.7 x 10^-14 respectively. Which complex will be more stable?
a) A
b) B
c) Both are equally stable
d) Cannot be determined
Answer: b
Clarification: The smaller the value of dissociation constant, more stable the complex will be in solution. Since 4.7 x 10^-14-12, complex B is more stable.

Chemistry Multiple Choice Questions on “Alcohols and Phenols – 1”.

1. How are alcohols prepared from haloalkanes?
a) By treating with concentrated H₂SO₄
b) By heating with aqueous NaOH
c) By treating with a strong reducing agent
d) By treating with Mg metal
Answer: b
Clarification: Haloalkanes when heated with aqueous NaOH or KOH give respective alcohols. This is a nucleophilic substitution reaction where the halide group is replaced by the OH nucleophile.

2. Which of the following process do not yield alcohols?
a) Acid catalysed hydration of alkenes
b) Hydroboration-oxidation of alkenes
c) Reduction of aldehydes
d) Free radical halogenation of alkanes
Answer: d
Clarification: Alkanes on free radical halogenation produce a mixture of haloalkanes and not alcohols. Alcohols can be prepared from alkenes by acid catalysed hydration and hydroboration-oxidation or from reduction of aldehydes.

3. Identify the catalyst in the hydration of alkenes to produce alcohols.
a) HCl
b) FeCl₃
c) Pt
d) Ni
Answer: a
Clarification: Alkenes react with water in the presence of a mineral acid as a catalyst to form alcohols. The H+ ion from the acid helps to form a carbocation for nucleophilic attack.

4. Propene when reacted with water in the presence of H₂SO₄ gives _________
a) Propan-1-ol
b) Propan-2-ol
c) 2-Methylpropan-1-ol
d) 2-Methylpropan-2-ol
Answer: b
Clarification: Since propene is an unsymmetrical alkene, the given hydration reaction takes place in accordance to Markovnikov’s rule, to form propan-2-ol. The double bond is broken and the OH group attaches at the second carbon.

5. The first step of the acid catalysed hydration of alkenes, involves the protonation of alkene to form a carbocation by electrophilic attack of _______
a) H⁺
b) H₂O
c) H₃O⁺
d) OH^–
Answer: c
Clarification: The water reacts with the H⁺ ion of the mineral acid to form a hydronium ion (H₃O⁺). This ion attacks the carbon double bond to form a carbocation and give a water molecule.

6. Identify the nucleophile that attacks the carbocation in the second step of acid catalysed hydration of alkenes?
a) OH^–
b) H₂O
c) H⁺
d) H₃O⁺
Answer: b
Clarification: Nucleophiles are electron rich species that attack the part of the structure that is electron deficient. In this step, the H₂O nucleophile attacks the carbocation forming a protonated alcohol.

7. Name the following step from the mechanism of acid catalysed hydration of ethene.

a) Protonation
b) Electrophilic attack
c) Nucleophilic attack
d) Deprotonation
Answer: d
Clarification: In this step, the electron pair of water attack the protonated alcohol, resulting in the loss of H⁺ from oxygen (deprotonation) to form an alcohol.

8. Which compound reacts with propene to form tripropyl borane?
a) Borane
b) Diborane
c) Boric acid
d) Sodium borohydride
Answer: b
Clarification: Diborane (B₂H₆) reacts with alkenes to give trialkyl boranes as a product of successive addition. Firstly, CH₃CH₂CH₂(BH₂) is formed which reacts with propene to give (CH₃CH₂CH₂)₂BH, which further reacts with propene to finally give (CH₃CH₂CH₂)B, which is tripropyl borane.

9. Which of the following is not required for the conversion of trialkyl borane to an alcohol?
a) Diborane
b) Water
c) Sodium hydroxide
d) Hydrogen peroxide
Answer: a
Clarification: Trialkyl boranes are oxidised by hydrogen peroxide in the presence of aqueous NaOH to form alcohols. Diborane is not required for this conversion but is essential in the production of trialkyl boranes.

10. Propan-2-ol can be prepared form the hydroboration-oxidation of propene.
a) True
b) False
Answer: b
Clarification: The oxidation of trialkyl boranes proceeds according to anti-Markovnikov’s rule, and hence the product formed will be propan-1-ol.

11. What happens when an aldehyde is treated with lithium aluminium hydride?
a) Primary alcohol is formed
b) Secondary alcohol is formed
c) Tertiary alcohol is formed
d) No reaction
Answer: a
Clarification: LiAlH₄ acts as a reducing agent which reduces an aldehyde by adding hydrogen atoms to it result in the formation of a primary alcohol.

12. Which carbonyl compound yields secondary alcohols when treated with LiAlH₄?
a) Aldehyde
b) Ketone
c) Carboxylic acid
d) Ester
Answer: b
Clarification: Ketones react with reducing agents LiAlH₄ or NaBH₄ to get reduced to alcohol where the OH group is formed at the C of the C-O group. This results in the C being bonded to two alkyl groups apart from the OH and H atom.

13. Esters on catalytic hydrogenation always give a mixture of two different alcohols.
a) True
b) False
Answer: b
Clarification: Esters (RCOOR’) give a mixture of two alcohols depending upon the acyl group (RCO^–) and the alkoxy group (-OR’). Methyl acetate (CH₃COOCH₃) gives a mixture of methanol and ethanol, whereas ethyl acetate (CH₃COOCH₂CH₃) gives only ethanol.

14. Hydrolysis of the adduct formed form the reaction of ________ with methyl magnesium bromide gives 2-Methylpropan-2-ol.
a) Methanal
b) Ethanal
c) Propanal
d) Propanone
Answer: d
Clarification: 2-Methylpropan-2-ol is a tertiary alcohol which is produced from the hydrolysis of the adduct formed between a ketone and a Grignard reagent.

15. Which of the following aldehydes can produce 1o alcohols when treated with Grignard reagent?
a) Methanal
b) Ethanal
c) Propanal
d) Butanal
Answer: a
Clarification: Methanal on treatment with Grignard reagent forms an adduct which has only one alkyl group attached to the C atom along with two hydrogens and one O-Mg-X (X=halogen) group. This on hydrolysis will form a primary alcohol where the OH group will replace the O-Mg-X group.

Chemistry Multiple Choice Questions on “Carboxylic Acids Chemical Reactions – 1”.

1. The reaction of carboxylic acids with NaHCO₃ produces ______ which helps it to differentiate it from phenols.
a) H₂O
b) CO
c) CO₂
d) NaCl
Answer: c
Clarification: Unlike phenols, carboxylic acids react with weak bases like hydrogen carbonates to give carbon dioxide, along with sodium carboxylate salt and water.

2. The pK_a value is equivalent to ________
a) logK_a
b) -logK_a
c) logK_eq
d) -logK_eq
Answer: b
Clarification: K_eq is the equilibrium constant of a reaction and K_a is the acid dissociation constant which is K_eq[H₂O]. The strength of an acid is generally by its pK_a value rather than its K_a value.

3. If the pK_a value of acetic acid is 4.76, predict the pK_a value of HCl.
a) -7
b) 4.2
c) 7
d) 10
Answer: a
Clarification: Generally, carboxylic acids are weaker acids than mineral acids like hydrochloric acid. Also, mineral acids are strong with pK_a values usually less than 1. Hence, the pK_a value of carboxylic acids will be higher.

4. What is the relation between the acidic strength of A and B?

a) A = B
b) A > B
c) A d) A >> B

Answer: c
Clarification: Compound A has a methoxy group (electron donating) and compound B has nitro group (electron withdrawing). In simple terms, the electron withdrawing groups stabilise the carboxylate ion and strengthen the acid, whereas electron donation groups weaken the acid.

5. Which of the following is the stronger acid?
a) Acetic acid
b) Propanoic acid
c) Isobutyric acid
d) 2,2-Dimethylpropanoic acid
Answer: a
Clarification: As the size of alkyl group (electron releasing) increases, the electron density on the O of OH group increases, and makes release of H⁺ ion more difficult. Thus, the release of proton from acetic acid will be easier compared to the higher acids, and is therefore the stronger acid.

6. Which of the following is the strongest acid?
a) CH₃COOH
b) CH₂ClCOOH
c) CHCl₂COOH
d) CCl₃COOH
Answer: d
Clarification: The -I inductive effect increases as the number of Cl atoms (electron withdrawing substituents) increases. The carboxylate ion becomes more stable as more negative charge is dispersed, thereby strengthening the acid.

7. Which of the following has the highest pK_a value?
a) Bromoacetic acid
b) Chloroacetic acid
c) Fluoroacetic acid
d) Iodoacetic acid
Answer: d
Clarification: Higher the electronegativity of a halogen, greater will be its electron withdrawing effect, and stronger will be the acid. Since electronegativity of halogens decrease in the order F > Cl > Br > I, iodoacetic acid will be the weakest and hence will have the highest pK_a value.

8. Which of the following will have the highest acidic strength?
a) Butanoic acid
b) 2-Chlorobutanoic acid
c) 3-Chlorobutanoic acid
d) 4-Chlorobutanoic acid
Answer: b
Clarification: Halogen substituted carboxylic acids are stronger acid than their respective parent acids, because of the electron withdrawing and stabilizing nature of the halogen. Furthermore, as the distance of the halogen form the COOH group increases, the electron withdrawing influence decreases, and therefore the acidic character decreases.

9. What is the correct order of acidity of the following?
a) Acetic acid > Acrylic acid > Propiolic acid
b) Acetic acid > Propiolic acid > Acrylic acid
c) Propiolic acid > Acrylic acid > Acetic acid
d) Acrylic acid > Propiolic acid > Acetic acid
Answer: c
Clarification: The electronegativity of carbon increases from sp³ to sp² to sp hybridisation. The carbon attached to the carbonyl carbon of propionic acid id sp hybridised (due to triple bond) ang has a higher inductive effect on the COOH group and as a result has higher acidity compared to acrylic acid (sp² carbon) or acetic acid (sp³ carbon).

10. It is given that the following compound has a higher pK_a value than benzoic acid. Which is the most probable substituent group X of the compound?

a) OH
b) Cl
c) CN
d) NO₂
Answer: a
Clarification: Given that the compound has a higher pK_a value than benzoic acid, it should have weaker acidic strength than benzoic acid. This means, that X should be an electron donating group that destabilises the compound and reduces its acidic strength. Since is an electron releasing group, the X group should be OH.

11. Acrylic acid is a stronger acid than benzoic acid.
a) True
b) False
Answer: b
Clarification: Both compounds contain a sp² hybridised carbon atom next to the COOH group, but the double bond of a benzene ring is less electron releasing in nature than the vinyl carbon double bond of acrylic acid. This is because delocalisation in the resonance structures of benzoic acid destroys its aromatic character. Hence, benzoic acid is stronger than acrylic acid.

12. What is the correct order of pK_a values of the following acids?
a) Benzoic acid > m-Nitrobenzoic acid > p-Nitrobenzoic acid
b) Benzoic acid > p-Nitrobenzoic acid > m-Nitrobenzoic acid
c) p-Nitrobenzoic acid > m-Nitrobenzoic acid > Benzoic acid
d) m-Nitrobenzoic acid > p-Nitrobenzoic acid > Benzoic acid
Answer: a
Clarification: Nitro group is electron withdrawing in nature and increases the acidic strength of acids. The electron withdrawing effect of NO₂ is pronounced at para position more than meta position, hence the p-isomer is more acidic and has a lower pK_a value than m-isomer.

13. Which of the following is the most acidic?
a) Benzoic acid
b) o-Toluic acid
c) m-Toluic acid
d) p-Toluic acid
Answer: b
Clarification: Although CH₃ is an electron releasing group and should reduce the acidic character of acids by destabilizing it, the presence of CH₃ group at ortho position has a reverse effect and increases the acidic strength compared to benzoic acid. This is due to a combination of stearic and electronic factors.

14. Which of the following has a higher acidic character than benzoic acid?
a) Acetic acid
b) p-Methoxybenzoic acid
c) p-Bromobenzoic acid
d) p-Aminobenzoic acid
Answer: c
Clarification: Bromine is an electron withdrawing group that disperse the negative charge on the carboxylate ion and strengthens its acidic character. p-Bromobenzoic acid has a pK_a value of 3.96 compared to 4.19, which is the pK_a value of benzoic acid.

15. The presence of CF₃ group in an acid gives a higher acidic strength compared to the presence of NO₂.
a) True
b) False
Answer: a
Clarification: CF₃ group is highly electronegative due to the presence of three F atoms and has a greater electron withdrawing and stabilizing influence on carboxylate ion than NO₂ group.

Chemistry Multiple Choice Questions on “Diazonium Salts Physical Properties”.

1. Aromatic diazonium salts are stable at _____ temperatures.
a) cold
b) room
c) warm
d) high
Answer: a
Clarification: Benzenediazonium salts are stable in cold environment, but readily react with water when the temperature increases even slightly.

2. When a blue litmus paper is dipped in an aqueous solution of benzenediazonium bromide, the litmus paper _____
a) turns red
b) turns orange
c) turns black
d) remains unchanged
Answer: d
Clarification: Benzenediazonium bromide is a salt, made of a positive diazonium group and a negative bromide ion. Thus, it remains neutral to litmus paper.

3. Benzenediazonium chloride is ______ in water.
a) slightly soluble
b) highly soluble
c) insoluble
d) highly insoluble
Answer: b
Clarification: Benzenediazonium chloride is readily soluble in polar solvents like water. However, it is less soluble in solvents like alcohol.

4. Which of the following best describes benzenediazonium chloride?
a) Yellow solid
b) Colourless crystals
c) White powder
d) Oily liquid
Answer: b
Clarification: Benzenediazonium chloride is usually obtained in crystal form. It is unstable and formed for a short period when aniline reacts with nitrous acid. It is a salt with chlorine as the negative group and diazo group as positive.

5. Which of the following diazonium salts is insoluble in water?
a) C₆H₅N₂⁺Cl^–
b) C₆H₅N₂⁺Br^–
c) C₆H₅N₂⁺HSO₄^–
d) C₆H₅N₂⁺BF₄^–
Answer: d
Clarification: Certain diazonium salts such as fluoroborates are insoluble in water. This makes them stable enough to be dried and stored.

6. Which of the following properties is suitable for all diazonium salts?
a) Water soluble
b) Coloured solids
c) Explosive
d) Unreactive
Answer: c
Clarification: Diazonium salts are unstable and may explode in dry state. Therefore, they are generally used in solution or aqueous state.

7. Benzenediazonium chloride is commercially available in packaged form.
a) True
b) False
Answer: b
Clarification: Because of the highly unstable nature of diazonium salts, they are not commercially available. Instead, they are prepared on demand.

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Chemistry Multiple Choice Questions on “Types of Polymerisation Reactions – 1”.

1. Addition polymerisation is also known as _________
a) copolymerisation
b) homopolymerisation
c) step growth polymerisation
d) chain growth polymerisation
View Answer

Answer: d
Clarification: Addition polymerisation takes place through steps leading to the increase in chain length and each step produces reactive intermediates for use in the next stage of the growth of the chain. It may take place between molecules of the same monomer or different monomers.

2. Which of the following cannot undergo addition polymerisation?
a) Ethane
b) Ethene
c) Propylene
d) Vinyl benzene
Answer: a
Clarification: The monomers involved in addition polymerisation are unsaturated compounds like alkenes and alkadienes. This is because it does not involve the removal of any molecule and hence requires the breaking of a double bond for it to progress.

3. Which of the following is not a type of addition polymerisation?
a) Free radical polymerisation
b) Polycondensation polymerisation
c) Anionic polymerisation
d) Cationic polymerisation
Answer: b
Clarification: Depending on the nature of the reactive species involved, addition polymerisation is classified as free radical, anionic and cationic based on whether the initiator is a free radical, anion or cation respectively.

4. Which of the following is not a suitable initiator for free radical addition polymerisation reaction?
a) Acetyl peroxide
b) Benzoyl peroxide
c) tert-Butyl peroxide
d) Benzoquinone
Answer: d
Clarification: Benzoquinone combines with the free radical intermediate to form a highly stable, non-reactive radical because of resonance. This inhibits the further progress of the chain growth and therefore, the reaction stops. Peroxides are very important free radical generating initiators.

5. In free radical mechanism, the step in which two very large free radicals combine with each other is called the _______ step.
a) chain initiating
b) chain propagating
c) chain growth
d) chain terminating
Answer: d
Clarification: The final step in which the product radical react with each other to form a polymerised product is called the chain terminating step. The other two steps involve reaction of one free radical with the unsaturated molecule.

6. Identify the chain initiation step of the polymerisation of ethene in the presence of benzoyl peroxide initiator, from the following?
a) C₆H₅• + C₂H₄ = C₆H₅-CH₂-CH₂•
b) C₆H₅-CH₂-CH₂• + C₂H₄ = C₆H₅-CH₂-CH₂-CH₂-CH₂•
c) C₆H₅-CH₂-CH₂-CH₂-CH₂• = C₆H₅-(-CH₂-CH₂-)_n-CH₂•
d) C₆H₅-(-CH₂-CH₂-)_n-CH₂• + C₆H₅-(-CH₂-CH₂-)_n-CH₂• = C₆H₅-(-CH₂-CH₂-)_n-C₆H₅
Answer: a
Clarification: Benzoyl peroxide undergoes homolytic fission to from phenyl free radical which acts an initiator. The first step is the addition of this radical to the ethene double bond, thus generating a new and larger free radical. This is called the chain initiating step.

7. Which of the following statements is correct regarding LDP and HDP?
a) Both have different monomers
b) Both have same structures
c) Both have similar preparation conditions
d) Both are chemically inert
Answer: d
Clarification: Low-density and high-density polythene have the same monomeric unit, ethene. Both are synthesized under very different pressure and temperatures conditions. LDP has a branches whereas HDP is a linear structure. Both are chemically inert and tough.

8. Identify the most suitable catalyst for the reaction shown.

a) Dioxygen
b) Ziegler-Natta catalyst
c) Persulphate
d) AlCl₃
Answer: c
Clarification: This is the addition polymerisation of tetrafluoroethene to give Teflon. It is carried out in the presence of a free radical or persulphate catalyst at high pressures.

9. LDP is used in the making of electrical wires.
a) True
b) False
Answer: a
Clarification: LDP is chemically inert and tough but flexible and a poor conductor of electricity. This makes it suitable for use in the insulation of electricity carrying wires.

10. Which of the following is not suitable for the polymerisation of ethene to form high-density polythene?
a) Presence of Ziegler-Natta catalyst
b) Temperature of 500K
c) Pressure of 7 atmospheres
d) Hydrocarbon solvent
Answer: b
Clarification: HDP is formed when polymerisation of C₂H₄ takes place through addition in a hydrocarbon solvent with Ziegler-Natta catalyst at temperature of 333-343K and under a pressure of 6-7 atmospheres. Temperatures and pressure higher than this will favour the formation of LDP.

11. Polymerisation of vinyl cyanide with peroxide catalyst forms _______
a) PVC
b) PAN
c) PET
d) HDP
Answer: b
Clarification: Polyacrylonitrile (PAN) is a polymerised product of acrylonitrile (vinyl cyanide) in the presence of a peroxide catalyst. Vinyl cyanide can be prepared by treating ethyne with HCN in the presence of Ba(CN)₂ catalyst. PAN is used for making orlon and acrilan.

12. Which of the following is used in non-stick pans?
a) HDP
b) LDP
c) Teflon
d) Orlon
Answer: c
Clarification: Teflon (polymer of tetrafluoroethene) is chemically inert and resistant to attacks by corrosive agents. This makes it suitable for use in making oil seals, gaskets and also in non-stick surface coated utensils.

13. What are the monomers of dacron?
a) Ethane-1,2-diol and terephthalic acid
b) Ethylene glycol and phthalic acid
c) 1,3-Butadiene and terephthalic acid
d) Ethylene glycol and 1,3-butadiene
Answer: a
Clarification: Dacron or terylene is a condensation polymer formed by the step growth polymerisation of ethylene glycol and terephthalic acid.

14. Terylene is a polyamide.
a) True
b) False
Answer: b
Clarification: Terylene is a polycondensation product of a dicarboxylic acid (terephthalic acid) and a diol (ethylene glycol) involving ester (COO) linkages. Hence, it is a polyester.