Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “General Properties of the Transition Elements (D-Block)”.

1. The d-electrons affect the properties of the transition elements to a great extent.
a) True
b) False
View Answer

Answer: a
Clarification: The transition elements differ from one another only in the number of electrons in the d-orbitals of the penultimate shell. The d-orbitals of the transition elements project to the periphery of the atom more than s and p-orbitals. Therefore, the d-electrons affect the properties of transition elements to a great extent.

2. Why is there an increase in the atomic radius of transition elements at the end of the period?
a) Increased electron-electron repulsions
b) Decreased electron-electron repulsions
c) Increase in nuclear charge
d) Increase in atomic mass
Answer: a
Clarification: Near the end of the period, the increased electron-electron repulsions between added electrons in the same orbitals are greater than the attractive forces due to increased nuclear charge. This results in the expansion of the electron cloud and therefore, increases the atomic size.

3.Which of the following iswhy the atomic radii of the second and third transition series are almost same?
a) Actinoid contraction
b) Radioactive nature
c) Lanthanoid contraction
d) Filled d-orbital
Answer: c
Clarification: In the atoms of the second transition series, the number of shells increases and so, their atomic radii is greater than that of the first transition series. The atomic radii of the second and third transition series are almost same due to lanthanoid contraction.

4. The decrease in atomic size of the d-block elements in a series is small after midway.
a) True
b) False
Answer: a
Clarification: In the beginning of the series, the atomic radius decreases with increase in atomic number as the nuclear charge increases and the shielding effect of the d-electrons is small. After midway, the increased number of d-electrons show an increase in shielding effect which counterbalances the increase in nuclear charge. Therefore, the decrease in atomic radius post midway is minimal.

5. Which of the following is not a property of a transition metal?
a) They are lustrous
b) They are malleable
c) They are ductile
d) They have low boiling points
Answer: d
Clarification: The transition metals exhibit all the characteristics of metals. They are hard, lustrous, malleable and ductile, have high melting and boiling points, high thermal and electrical conductivity and high tensile strength.

6. What type of bond do the transition elements form with themselves?
a) Ionic bond
b) Covalent bond
c) Coordinate bond
d) Metallic bond
Answer: d
Clarification: Transition elements have relatively low ionization energies and have one or two electrons in their outermost energy level. As a result, they form metallic bonds. This is also the reason behind the metallic properties of transition elements.

7. Which of the following is not a very hard metal?
a) Chromium
b) Molybdenum
c) Tungsten
d) Zinc
Answer: d
Clarification: Greater the number of unpaired electrons, stronger is the bonding due to overlapping of unpaired electrons between different metal atoms. Cr, Mo and W have maximum unpaired d-electrons and are harder metals whereas Zn, Cd and Hg are not very hard due to the absence of unpaired electrons.

8. Which of the following statements is incorrect?
a) Manganese has an abnormally low boiling point
b) Transition elements have low enthalpies of atomisation
c) Transition elements exhibit metallic bonding
d) Transition metals generally have a high boiling point
Answer: b
Clarification: Transition metals generally have high melting and boiling points. Magnesium and technetium have abnormally low boiling points. This is due to the strong metallic bonds between the atoms. They have high enthalpies of atomization.

9. Which of the following element has the highest density among transition metals?
a) Iridium
b) Osmium
c) Scandium
d) Chromium
Answer: a
Clarification: Among the d-block elements, iridium has the highest density (22.61 g cm^-3) whereas scandium has the lowest density (3.43 g cm^-3). Osmium has slightly lesser density (22.59 g cm^-3) than iridium.

10. Which of the following element has the highest ionisation enthalpy?
a) Scandium
b) Vanadium
c) Copper
d) Zinc
Answer: d
Clarification: Ionisation enthalpy shows only a little variation on moving along the period of d-block elements. The first ionisation enthalpy of zinc, cadmium and mercury are higher than the other elements in their respective periods because of the fully filled (n-1)d¹⁰ns² configuration.

Chemistry Multiple Choice Questions on “Coordination Compounds Importance and Applications”.

1. Which reagent is used for detecting Ni²⁺ ions in solution?
a) EDTA
b) Dimethylglyoxime
c) α-nitroso-β-naphthol
d) Cupron
View Answer

Answer: b
Clarification: When dimethylglyoxime is added to a solution along with ammonium hydroxide, the formation of a bright red precipitate indicates the presence of Ni²⁺ ions in solution. This is due to the formation of nickel dimethylglyoxime complex.

2. Hardness of water is estimated by simple titration with which compound?
a) Na₂(EDTA)
b) Fe(EDTA)
c) Mg(EDTA)
d) Co(EDTA)
Answer: a
Clarification: The Ca²⁺ and Mg²⁺ ions form stable complexes with EDTA, and the difference in the stability constants of Ca and Mg complexes helps in the estimation of hardness of water.

3. Given below is the incomplete equation for the extraction process of gold using potassium cyanide. Identify the missing reagent.

a) Water
b) Oxygen
c) Water + air
d) Zinc dust
Answer: c
Clarification: Gold combines with cyanide in the presence of air and water to form [Au(CN)₂]^– in aqueous solution. This is then treated with zinc dust to obtain gold in metallic form.

4. Haemoglobin is a complex compound of which metal ion?
a) Fe²⁺
b) Fe³⁺
c) Co²⁺
d) Co³⁺
Answer: a
Clarification: Haemoglobin is the red pigment of blood which acts as the oxygen carrier and it a coordination compound of iron(II).

5. Which of the following biologically important coordination compounds has a magnesium central atom?
a) Chlorophyll
b) Haemoglobin
c) Vitamin B₁₂
d) Carboxypeptidase-A
Answer: a
Clarification: Chlorophyll is the green pigment in plants, responsible for photosynthesis, is a coordination compound of Mg.

6. The coordination complex chloridotris (triphenylphosphine) rhodium(I) is used in the hydrogenation of alkenes. It is also known as __________ catalyst.
a) Grubb
b) Pearlman
c) Wilkinson
d) Ziegler-Natta
Answer: c
Clarification: Coordination compounds may be used as homogeneous or heterogeneous catalysts for many reactions. Wilkinson catalyst is used as a homogeneous catalyst in hydrogenation of alkenes, whereas Ziegler¬-Natta catalyst is used as a heterogeneous catalyst for the polymerisation of olefins.

7. The platinum complex, trans-platin has been used in cancer treatment.
a) True
b) False
Answer: b
Clarification: Cis-platin is a coordination complex of platinum which is effective in inhibiting the growth of tumours.

8. The complex ion [Ag(S₂O₃)₂]^3- is associated with which field?
a) Electroplating
b) Medicine
c) Water treatment
d) Photography
Answer: d
Clarification: In black and white photography, the film is fixed by washing with hypo solution which dissolves the undecomposed silver bromide to form [Ag(S₂O₃)₂]^3-.

Chemistry Multiple Choice Questions on “Alcohols and Phenols – 2”.

1. Phenol can be obtained by ________ of sodium phenoxide.
a) acidification
b) oxidation
c) sulphonation
d) hydrolysis
Answer: a
Clarification: When halobenzene is fused with NaOH at 623K and 320atm with a copper salt as catalyst, sodium phenoxide is produced which on treatment with dilute HCl yields phenol. This is also known as Dow’s process.

2. Cumene hydroperoxide on hydrolysis with dilute H₂SO₄ gives _________
a) alcohol and phenol
b) only phenol
c) phenol and acetone
d) alcohol and acetone
Answer: c
Clarification: When cumene hydroperoxide is hydrolysed with a solution of dilute acid, it results in the formation of phenol and acetone, of which the latter is obtained in large quantities.

3. How is carbolic acid prepared from benzene diazonium chloride?
a) Treating it with nitrous acid at 275K
b) Preparing an aqueous solution and warming it
c) Treating it with sodium hydroxide
d) Freezing it
Answer: b
Clarification: Carbolic acid is another name for phenol. Simply warming an aqueous solution of benzene diazonium chloride salt or treating it with a dilute acid, will result in the formation of phenol.

4. Identify the compound ‘X’.

a) Oleum
b) Sulphuric acid
c) Nitrous acid
d) Sodium sulphate
Answer: a
Clarification: Oleum is a mixture of H₂SO₄ and SO₃ that is added to benzene to form benzene sulphonic acid, which on subsequent treatment as shown, gives phenol.

5. Which of the following isomeric alcohols is the most soluble in water?
a) n-Butyl alcohol
b) Isobutyl alcohol
c) sec-Butyl alcohol
d) tert-Butyl alcohol
Answer: d
Clarification: tert-Butyl alcohol is the most heavily branched isomer and is tightly packed compared to the others. This decreases the effective surface area of the alkyl part which is hydrophobic in nature.

6. Which of the following is the least soluble in water?
a) n-Butyl alcohol
b) n-Pentyl alcohol
c) n-Hexyl alcohol
d) n-Heptyl alcohol
Answer: d
Clarification: As the size of the alkyl group becomes larger, it prevents the formation of hydrogen bonds with water molecules and hence the solubility goes on decreasing with the increase in the size or mass of the compound.

7. Which of the following is necessary for the bromination of phenol?
a) CS₂
b) FeBr₃
c) AlCl₃
d) BF₃
Answer: a
Clarification: FeBr₃, AlCl₃ and BF₃ are Lewis acids, that are required for the substitution in benzene. Phenol has an OH group that activates the aromatic ring and requires the presence of only slightly or non-polar solvents for its halogenation.

8. Phenol is approximately how much times acidic than ethanol?
a) 10
b) 25
c) 100
d) million
Answer: d
Clarification: Phenol is more acidic than alcohols because of their better stability and resonant structures. The pK_a values of phenol and ethanol are 10 and 15.9 respectively, which when converted to K_a values gives a ratio of about a million.

9. Phenols are not soluble in which of the following?
a) Water
b) Alcohol
c) Ether
d) Sodium hydroxide
Answer: a
Clarification: Phenols are almost insoluble in water because of the presence of a very large non-polar aryl group which overcomes the polar character of the OH group.

10. What is the correct order of boiling points of alcohols having the same number of carbon atoms?
a) 1°>2°>3°
b) 3°>2°>1°
c) 3°>1°>2°
d) 2°>1°>3°
Answer: a
Clarification: As the branching in the structure increases, the surface area decreases and hence, the van der Waal forces decrease. This results in the reduction of boiling points of isomeric alcohols from primary to tertiary structures.

11. Four compounds A, B, C and D having similar molecular masses were tested for their boiling points. It was found that compound C has the highest boiling point of all four. What is the compound C most likely to be?
a) Hydrocarbon
b) Haloalkane
c) Alcohol
d) Ether
Answer: c
Clarification: Of compounds having similar molecular masses, alcohols will certainly have the highest boiling point due to the presence of intermolecular hydrogen bonding which is lacking in hydrocarbons, haloalkanes and ethers.

12. If the boiling point of methoxymethane is 248K, predict the boiling point of ethanol.
a) 231K
b) 248K
c) 351K
d) 455K
Answer: c
Clarification: Methoxymethane and ethanol are of comparable masses and it can be easily pointed out that ethanol will have a higher boiling point than it because of the presence of intermolecular hydrogen bonding. 455K is very high and is actually the boiling point of phenol.

13. Which of the following has the highest boiling point?
a) Propan-1-ol
b) Butan-1-ol
c) Butan-2-ol
d) Pentan-1-ol
Answer: d
Clarification: As the size the molecule increases, the number of carbon atoms increases and the van der Waal forces increase, resulting in an increase in boiling points.

14. Propan-1-ol boils at a higher temperature than propan-2-ol.
a) True
b) False
Answer: a
Clarification: Propan-2-ol is a secondary alcohol and has some degree of branching compared to propan-1-ol. This results in the decreases in surface area and van der Waals forces and hence reduction in boiling point.

15. Toluene and phenol have similar boiling points.
a) True
b) False
Answer: b
Clarification: Phenols have much higher boiling points than aromatic compounds of similar molecular mass. This is because they exist as associated molecules due to intermolecular hydrogen bonding.

Chemistry Objective Questions and Answers for Class 12 on “Carboxylic Acids Chemical Reactions – 2”.

1. How many molecules of acetic acid react with H₂SO₄ on heating to give one molecule of acetic anhydride?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Two molecules combine with the loss of one water molecule (H atom from one molecule and OH group from another molecule) when heated with dehydrating agents like H₂SO₄ to give acetic anhydride.

2. Benzoic acid reacts with phosphorus pentaoxide in the presence of heat to give ________
a) C₆H₅OC₆H₅
b) C₆H₅COC₆H₅
c) C₆H₅COOC₆H₅
d) C₆H₅COOCOC₆H₅
Answer: d
Clarification: Benzoic acid on heating with a strong dehydrating agent like P₂O₅ forms its corresponding anhydride, i.e., benzoic anhydride, which consists of two C₆H₅C=O groups bridged by a (-O-) group.

3. Esterification of carboxylic acids is a/an __________ reaction.
a) irreversible
b) nucleophilic addition
c) electrophilic substitution
d) nucleophilic substitution
Answer: d
Clarification: When carboxylic acids are heated with alcohols with suitable catalyst, esters are formed. This is a reversible reaction due to all the individual steps involved in the mechanism of this reaction are reversible in nature. It is a type of nucleophilic acyl substitution reaction.

4. Which of the following is the most suitable catalyst for the esterification of acetic acid with ethanol?
a) P₂O₅
b) HCl gas
c) KMnO₄-KOH
d) Pyridine
Answer: b
Clarification: HCl gas and concentrated H₂SO₄ are catalyst which provide a proton for the protonic attack on the carbonyl oxygen of acetic acid, which further activates the carbonyl groups towards nucleophilic addition of ethanol, which ultimately results in the formation of ethyl acetate, which is an ester.

5. Which of the following is used to shift the esterification reaction of carboxylic acids towards the right?
a) Using excess of carboxylic acid
b) Using excess of acid catalyst
c) Removal of water by distillation
d) Removal of ester by distillation
Answer: c
Clarification: Since the esterification of carboxylic acids is a reversible reaction, it can be shifted towards the right by using excess of alcohol or by the removal the water formed by distillation.

6. Which of the following carboxylic acids is most reactive towards esterification with given alcohol?
a) HCOOH
b) CH₃COOH
c) (CH₃)₂CHCOOH
d) (CH₃)₃CCOOH
Answer: a
Clarification: The rate at which carboxylic acids are esterified depends upon the stearic hinderance and to some extent the acidic strength of the acid, as it is a nucleophilic substitution reaction. Since formic acid has the lowest molecular mass and smallest group, it is the most reactive.

7. Which is the most preferred reagent in the product of acetyl chloride from acetic acid?
a) Cl₂
b) PCl₃
c) PCl₅
d) SOCl₂
Answer: d
Clarification: Thionyl chloride reacts with acetic acid to form acetyl chloride along with two gaseous compounds (SO₂ and HCl) that escape the reaction mixture, making the purification of the product easier.

8. Benzoic acid reacts with phosphorus pentachloride to give ________
a) chlorobenzene
b) benzyl chloride
c) benzoyl chloride
d) chlorobenzoic acid
Answer: c
Clarification: The hydroxyl group (OH) of benzoic acid id easily replaced by chlorine atom on treatment with PCl₅. This results in the formation of C₆H₅COCl, or benzoyl chloride.

9. Other than ethanoyl chloride, what are the by-products formed when ethanoic acid reacts with PCl₃?
a) H₃PO₃
b) H₃PO₃ and HCl
c) H₃PO₂ and HCl
d) POCl₃ and HCl
Answer: a
Clarification: Three molecules of ethanoic acid react with one molecule of PCl₃ to give three molecules of ethanoyl chloride and one molecule of phosphorous acid (H₃PO₃).

10. Acetic acid behaves exactly like ethanol on reaction with thionyl chloride.
a) True
b) False
Answer: a
Clarification: Acetic acid as well as ethanol consist of an OH group and reaction with SOCl₂ results in the cleavage of C-OH bond with the hydroxyl group to be replaced by Cl atom. The only difference is that acetic acid forms acetyl chloride, whereas ethanal forms ethyl chloride.

Chemistry Objective Questions and Answers for Class 12,

Chemistry Multiple Choice Questions on “Diazonium Salts Chemical Reactions – 1”.

1. In reactions of diazonium salts involving the displacement of diazonium group, the nitrogen escapes as _____
a) precipitate
b) liquid
c) gas
d) crystals
Answer: c
Clarification: Nitrogen gas is liberated when diazonium salts are treated with suitable reagents so as to displace the N₂⁺X^– (X=anion) with an anionic group. This may or may not involve the production of another by-product.

2. Identify the most suitable reagent for converting benzene diazonium chloride to chlorobenzene.
a) CuCl₂/HCl
b) Cu₂Cl₂/HCl
c) CuSO₄/HCl
d) CuSO₂/H₂SO₄
Answer: b
Clarification: The Cl^– nucleophile can be introduced in the benzene ring in the presence of Cu(I) ion which is produced by Cu₂Cl₂ in hydrochloric acid. This on warming with the diazonium salt will give chlorobenzene.

3. Which of the following cannot be formed from Sandmeyer reaction on benzenediazonium chloride?
a) Chlorobenzene
b) Bromobenzene
c) Iodobenzene
d) Benzonitrile
Answer: c
Clarification: Sandmeyer reaction is used for the introduction of chloride, bromide and cyanide ions only in the benzene ring of diazonium salts, by replacing the diazo group. Iodobenzene is produced by another reaction.

4. Identify Y in the following reaction.

a) Cu₂Cl₂/KCN
b) CuCN/KCN
c) Cu₂Cl₂/HCN
d) CuCN/HCN
Answer: b
Clarification: When benzenediazonium chloride is treated with copper cyanide in aqueous potassium cyanide solution, benzonitrile is formed. This is also categorized under Sandmeyer reaction.

5. Which of the following can be produced by Gatterman reaction of diazonium salts?
a) Bromobenzene
b) Fluorobenzene
c) Nitrobenzene
d) Cyanobenzene
Answer: a
Clarification: Gatterman reaction is a modification of Sandmeyer reaction. It is used for introducing chlorine or bromine in the benzene ring by treating diazonium salt (aqueous) with corresponding halogen acid (HCl or HBr) in the presence of Cu powder.

6. The yield of chlorobenzene from Sandmeyer reaction is better than that from Gatterman reaction.
a) True
b) False
Answer: a
Clarification: In Gatterman reaction, there is an additional by product along with chlorobenzene and nitrogen gas, which is a copper salt (with the anion of diazonium salt). The contribution towards formation of this product results in the yield on chlorobenzene to be slightly less.

7. Which of the following compounds has to be used in Gatterman reaction?
a) Chloroform
b) HCl
c) Copper powder
d) Copper chloride
Answer: c
Clarification: Gatterman reaction is used for the preparation of either chlorobenzene or bromobenzene by treating a diazonium salt solution with either HCl or HBr respectively, in the presence of copper powder.

8. Benzenediazonium bromide on Gatterman reaction with HCl does not form which of the following?
a) Chlorobenzene
b) Nitrogen gas
c) Copper bromide
d) Copper chloride
Answer: d
Clarification: The anion from the diazonium salt (Br^–) combines with Cu⁺ from copper powder (in halogen acid), to form CuBr. The chlorobenzene formed gets the chlorine from HCl in solution.

9. When a diazonium salt is treated with ______ iodobenzene is formed.
a) potassium iodide
b) copper iodide
c) ethyl iodide
d) iodoform
Answer: a
Clarification: Iodine is not easily introduced in the benzene ring directly. When an aqueous solution of benzene diazonium salt is warmed with excess potassium iodide, iodobenzene is formed.

10. Which of the following is not produced when C₆H₅N₂⁺Br^– is warmed with KI?
a) C₆H₅I
b) N₂ gas
c) KBr
d) KNO₂
Answer: d
Clarification: Replacement of diazo group by an iodide ion in benzenediazonium bromide is accompanied by the formation of potassium bromide and liberation of nitrogen gas.

11. Benzenediazonium chloride on treatment with _____ gives a product which on heating forms fluorobenzene.
a) hydrofluoric acid
b) fluoroboric acid
c) potassium fluoride
d) boron trifluoride
Answer: b
Clarification: When benzenediazonium chloride is treated with fluoroboric acid, diazonium fluoroborate is formed, which on heating gets decomposed to give fluorobenzene. This is known as Balz-Schiemann reaction.

12. Aryl fluoride can be obtained directly by heating which arenediazonium salt?
a) Benzenediazonium fluoride
b) Benzenediazonium chloride
c) Benzenediazonium hydrogensulphate
d) Benzenediazonium fluoroborate
Answer: d
Clarification: Arenediazonium fluoroborate is obtained as a precipitate when arenediazonium chloride is treated with HBF₄. This salt on heating decomposes to give aryl fluoride.

13. Which of the following is formed as a by product of decomposition of arenediazonium fluoroborate to give fluorobenzene?
a) BF₃
b) BH₃
c) H₃BO₃
d) BN
Answer: a
Clarification: Arenediazonium fluoroborate has four fluorine atoms put of which one forms fluorobenzene. The other three fluorine atoms combine with boron from BF₄^– to form boron trifluoride along with the liberation of nitrogen gas.

14. Alkyl fluorides and iodides cannot be formed from Sandmeyer reaction.
a) True
b) False
Answer: a
Clarification: It is difficult to introduced iodide and fluoride ions into the benzene ring and are carried out by separate reactions involving potassium iodide and fluoroboric acid respectively.

Chemistry Multiple Choice Questions for IIT JEE Exam on “Types of Polymerisation Reactions – 2”.

1. What is the polymer obtained from the condensation of NH₂-(CH₂)₆-NH₂ and COOH-(CH₂)₈-COOH?
a) Nylon-6,6
b) Nylon-6
c) Nylon-6,10
d) Terylene
Answer: c
Clarification: Nylons are polyamides obtained from the condensation between amines and dicarboxylic acids. The have amide linkages (NHCO). Hexamethylenediamine and sebacic acid from the polymer nylon-6,10.

2. Cyclohexanone on treatment with hydroxylamine gives P, which on treatment with sulphuric acid gives Q. When Q is heated with water at high temperature, R is obtained. Identify R.
a) Nylon-6
b) Caprolactam
c) Amino caproic acid
d) Nylon-6,6
Answer: a
Clarification: Nylon-6 is the polymer of caprolactam (Q), which is obtained from the oxidation of cyclohexane to give cyclohexanone(P), followed by treatment with H₂SO₄. Caprolactam on heating with water gives amino caproic acid which polymerises to nylon-6.

3. Glyptal is obtained from the condensation polymerisation of ethylene glycol with _______
a) terephthalic acid
b) phthalic acid
c) isophthalic acid
d) trimesic acid
Answer: b
Clarification: Glyptal is a polyester (containing COO linkages) obtained from the polycondensation reaction of ethane-1,2-diol and benzene-1,2-dicarboxylic acid (phthalic acid). It is cross-linked polymer.

4. Which of the following is not required for the production of dacron?
a) 450K temperature
b) Ethylene glycol
c) Terephthalic acid
d) Triethylaluminium-titanium tetrachloride
Answer: d
Clarification: Dacron or terylene is a polyester of ethylene glycol and terephthalic acid. It is produced by heating the two at 420-460K and in the presence of zinc acetate-antimony trioxide, which acts as a catalyst. On the other hand, triethylaluminium-tetrachloride is used in the synthesis of HDPE.

5. Identify the polymer generated by the reaction of methanal with the shown compound.

a) Melamine-formaldehyde
b) Novolac
c) Bakelite
d) Phenol-formaldehyde
Answer: a
Clarification: The compound shown is a heterocyclic triamine called melamine. It reacts with methanal to form a resin intermediate which polymerises to form a melamine-formaldehyde polymer.

6. Bakelite is a cross-linked polymer.
a) True
b) False
Answer: a
Clarification: Phenol-formaldehyde condensation initially forms a linear polymer, Novolac. This on further heating with formaldehyde undergoes cross-linking through CH₂ groups to form bakelite.

7. Which of the following phenols are able to form Novolac?
a) m-Hydroxybenzyl alcohol
b) o-Hydroxybenzyl alcohol and p-hydroxybenzyl alcohol
c) 2,4-(Dihydroxymethyl)phenol
d) 2,4,6-(Trihydroxymethyl)phenol
Answer: b
Clarification: Novolac is a linear chain phenol-formaldehyde polymer. The ortho and para hydroxy derivatives of benzyl alcohol, initially formed by the reaction of phenol with HCHO, further react with phenol to form compounds having benzene rings joined together by -CH₂– links (Novolac).

8. The compound shown is a ________

a) chain growth copolymer
b) step growth copolymer
c) chain growth homopolymer
d) step growth homopolymer
Answer: a
Clarification: The polymer shown is butadiene-styrene copolymer which is formed by the addition polymerisation of two different unsaturated compounds, 1,3-butadiene and styrene. It is tough and used as a substitute for natural rubber.

9. The polymer of type [-AAAABBBAAAABBB-]_n, where A and B are two different monomers, is called a ______ copolymer.
a) alternating
b) block
c) graft
d) random
Answer: b
Clarification: Bock copolymers are those in which blocks of each type of monomer alternate with each other. In this case, the block AAAA and block BBB are alternatively repeated after each other.

10. Which of the following is the monomer of natural rubber?
a) Neoprene
b) Isoprene
c) Chloroprene
d) Butadiene
Answer: b
Clarification: Isoprene (or 2-methyl-1,3-butadiene) repeats itself in a linear chain to form natural rubber. This is also called as cis-1,4-polyisoprene, in which the monomers are linked by weak van der Waals forces, thus making it elastic in nature.

11. Which of the following is incorrect regarding natural rubber?
a) Shows thermoplastic behaviour
b) Soluble in benzene
c) Resistant to attack by oxidising agents
d) Shows high water absorption capacity
Answer: c
Clarification: Natural rubber becomes soft at temperatures above 335K and shows brittle character at temperatures below 283K. Hence, it shows thermoplastic behaviour. It is soluble in non-polar solvents like benzene, toluene, etc. and is not susceptible to attack by oxidising agents.

12. Vulcanization is carried out with which element?
a) Sulphur
b) Phosphorous
c) Oxygen
d) Chlorine
Answer: a
Clarification: Vulcanization is a process of improving the physical properties of natural rubber by introducing sulphur cross-links at the reactive sites of double bonds. This is done by heating the rubber with sulphur and an additive.

13. X on polymerisation gives neoprene. Identify X.
a) 1-Chloro-1,3-butadiene
b) 2-Chloro-1,3-butadiene
c) 1-Methyl-1,3-butadiene
d) 2-Methyl-1,3-butadiene
Answer: b
Clarification: Neoprene or polyisoprene is a synthetic rubber obtained from the addition polymerisation of chloroprene (2-chloro-1,3-butadiene) through free radical mechanism. It has a better stability towards oxidation than natural rubber.

14. Buna-N involves acrylonitrile as one of the monomers.
a) True
b) False
Answer: a
Clarification: Buna-N is a nitrile rubber formed by the copolymerisation of 1,3-butadiene and acrylonitrile. On the other hand, buna-S is a styrene rubber, with vinylbenzene as one monomer in place of acrylonitrile.

Chemistry Multiple Choice Questions for IIT JEE Exam,