Category: Class 12 Chemistry Objective Questions

Chemistry Online Test for Class 12 on “Nomenclature of Coordination Compounds – 2”.

1. Identify the correct naming for K₃[Fe(CN)₆].
a) Tripotassium hexacyanidoferrate(III)
b) Potassium hexacyanoferrate(III)
c) Tripotassium hexacyanoferrate(III)
d) Potassium hexacyanidoferrate(III)

Answer: d
Clarification: The number of cations or anions is not denoted in the name of a coordination compound. In this case, the counter ion is named as potassium and not tripotassium. Also, CN is an anionic ligand and should end with -o, with the correct name being cyanido and not cyano.

2. Identify the correct naming for [Zn(OH)₄]^2-.
a) Zinc tetrahydroxide
b) Tetrahydroxozincate(II)
c) Tetrahydroxidozincate(II)
d) Tetrahydroxylzincate(II)
Answer: c
Clarification: The ligand OH is a negative group and should end with -o. The correct name for the OH group is hydroxide and the metal is mentioned after the ligands.

3. Identify the correct naming for [Co(ONO)(NH₃)₅]²⁺.
a) Pentaamminenitritocolbalt(III)
b) Pentaamminenitrito-O-cobalt(III)
c) Pentaamminenitrito-N-cobalt(III)
d) Pentaamminenitrosylcobalt(III)
Answer: b
Clarification: Nitrate is an ambidentate ligand and can bind through either O or N atom. In this case, it is written as (ONO) which implies that it binds through O atom, hence giving it the name nitrito-O.

4. Identify the correct formula for hexaaquamanganese(II) ion.
a) [Mn(H₂O)₆]²⁺
b) [Mn(H₂O)₆]^2-
c) [Mn₂(H₂O)₆]⁺
d) [Mn(H₂O)₆]⁺²
Answer: a
Clarification: The parenthesis after manganese represents its oxidation state (+2) which is indicated outside the square bracket with sign following the number. The H₂O molecule is a neutral ligand named as aqua.

5. Identify the correct formula for potassium tetracyanonickelate(II).
a) K[Ni(CN)₄]
b) K[Ni(CN)₄]₂
c) K₂[Ni(CN)₄]
d) K₂[Ni(CN)₄]₂
Answer: c
Clarification: CN is anionic with charge -1 and there are 4 of them. Ni is given to have oxidation number +2 in the name of the compound. So, the total charge on the compound will be -4 + 2 = -2. This must be balanced by potassium ions which have charge of +1, so two of them will be required. Hence, the formula is K₂[Ni(CN)₄].

6. The term inside the parenthesis in the formula of mercury(I) tetrathiocyanato-S-cobaltate(III) is _________
a) Hg
b) SCN
c) NCS
d) Co
Answer: b
Clarification: When the ligands are polyatomic, its formula is enclosed in parenthesis. The formula for the given compound is Hg[Co(SCN)₄], where the thiocyanate ligand binds through the S atom.

7. Which of the following representation of the complex ion is correct according to IUPAC?
a) [PtBrCl(NH₃)(NO₂)]^–
b) [PtBrCl(NO₂)(NH₃)]^–
c) [Pt(NH₃)BrCl(NO₂)]^–
d) [Pt(NO₂)BrCl(NH₃)]^–
Answer: c
Clarification: The ligands are bromine, chlorine, ammine and nitrite and according to IUPAC nomenclature they should be listed as per alphabetical order which is ammine (NH₃), bromine (Br), chlorine (Cl) and nitrite (NO₂).

8. An imaginary coordination entity named dipalidoqalidoralidometal(IV) according to the IUPAC norms, consists of ligands palide(P), qalide(Q) and ralide(R), all having charge of magnitude 1. The central atom is metal (M). Which of the following options best suits the complex described?
a)
b)
c)
d)
View Answer

Answer: c
Clarification: All the ligands end with -o, which implies they are anionic and have charge -1. The oxidation number of the central atom is given as +4. There are a total of 4 ligands (2 P, 1 Q and 1 R) in the entity. Hence, the overall charge on the complex is +4 – 4 = 0.

9. Which of the following is not enclosed in parenthesis?
a) Polyatomic ligands in the formula of coordination compounds
b) Ligand abbreviations in the formula of coordination compounds
c) Oxidation number of central metal in the naming of coordination compounds
d) The coordination entity in the formula of coordination compounds
Answer: d
Clarification: The coordination entity consisting of the metal atom/ion followed by ligands is enclosed within square brackets and not in parenthesis.

10. The prefixes bis, tris and tetrakis are used to indicate the number of individual ligands in the coordination entity.
a) True
b) False
Answer: a
Clarification: Mono, di, tri, etc. are used when name of ligands don’t have a numerical prefix. If not, then bis, tris and tetrakis are used as prefixes.

Chemistry Online Test for Class 12,

Chemistry Multiple Choice Questions on “Haloalkanes & Haloarenes – Chemical Reactions – 3”.

1. What is the correct order of reactivity of haloalkanes towards β-elimination reactions?
a) 1°>2°>3°
b) 3°>2°>1°
c) 1°>3°>2°
d) 3°>1°>2°
Answer: b
Clarification: The haloalkanes which will form more stable (highly substituted) alkenes are the ones that will react the fastest. Since tertiary alkyl halides will form the most stable alkenes, they are the most reactive.

2. How many β-carbon atoms does 2-Bromobutane have?
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: A beta carbon is the carbon atom that is adjacent to the carbon atom that is attached to the halogen. In 2-Bromobutane, the halogen is at position C-2, so the alpha carbon has one carbon on either side, making it two beta carbons

3. Which of the following is the suitable medium for preparing Grignard reagents?
a) Dry acetone
b) Dry ether
c) Concentrated HCl
d) Alcoholic KOH
Answer: b
Clarification: When haloalkanes react with magnesium metal in dry ether, special organo-metallic compounds called Grignard reagents are formed.

4. What is formed when ethyl magnesium bromide reacts with water?
a) Grignard reagent
b) Ethane
c) Ethanol
d) Magnesium hydroxide
Answer: b
Clarification: Ethyl magnesium bromide is a Grignard reagent and is highly reactive to any proton giving source because of the polar nature of bonding present in it. It forms hydrocarbons when reacted with water.

5. What will be the product of the following reaction?
2CH₃CH₂Br + 2Na + dry ether = ________
a) Ethane
b) 1-Bromoethane
c) Butane
d) 1-Bromobutane
Answer: c
Clarification: Alkyl halides react with sodium metal in dry ether to form hydrocarbons that have double the number of carbon atoms as that in the alkyl halide. This is called the Wurtz reaction.

6. Compared to haloalkanes, the reactivity of haloarenes towards nucleophilic substitution reactions is _________
a) low
b) high
c) very high
d) equal
Answer: a
Clarification: Haloarenes form resonating structures in which the carbon-halogen bond acquires a partial double bond character and its cleavage is difficult. Hence, they are less reactive towards nucleophilic substitution reactions.

7. The C-Cl bond length in chlorobenzene is less than the C-Cl bond length in chloromethane because of the ________
a) resonance effect in haloarenes
b) difference in hybridisation of carbon in C-Cl bond
c) instability of phenyl cation
d) possible repulsion between nucleophile and chlorobenzene
Answer: b
Clarification: In chlorobenzene, the C of C-Cl bond is sp² hybridised, whereas the C of C-Cl bond in chloromethane is sp³ hybridised. Thus, there is more s character in the carbon of the C-Cl bond in chlorobenzene and as a result holds the electron pair of C-Cl bond more tightly.

8. Which of the following compounds is most easily converted to a phenol when heated with aqueous NaOH solution?
a) Chlorobenzene
b) 4-Chloronitrobenzene
c) 2,4-Dinitrochlorobenzene
d) 2,4,6-Trinitrochlorobenzene
Answer: d
Clarification: The presence of electron withdrawing groups like NO₂ at ortho and para positions with respect to the halogen, increases the reactivity of the haloarene. Chlorobenzene has no such group and converts to phenol at the highest temperature and pressure.

9. 2-Chloronitrobenzene is more reactive than chlorobenzene.
a) True
b) False
Answer: b
Clarification: Although NO₂ is an electron withdrawing group, it is present at meta position with respect to chlorine in 2-Chloronitrobenzene and has no effect on the reactivity of the haloarene. This is because none of the resonating structures of m-nitrobenzene bear the negative charge on the carbon atom bearing the nitro group and as a result the presence of NO₂ at meta position does not stabilize the negative charge.

10. Identify the catalyst in Friedel-Crafts acylation reaction.
a) Anhydrous FeCl₃
b) Anhydrous AlCl₃
c) NaNO₂
d) Alcoholic KOH
Answer: b
Clarification: Friedel-Crafts acylation is carried out by treating a haloarene with and acetyl chloride in the presence of anhydrous aluminium chloride as a catalyst.

11. What is the major product formed when chlorobenzene reacts with nitric acid in concentrated sulphuric acid?
a) Nitrobenzene
b) 1-Chloro-2-nitrobenzene
c) 1-Chloro-3-nitrobenzene
d) 1-Chloro-4-nitrobenzene
Answer: d
Clarification: This is the nitration of chlorobenzene which is an electrophilic substitution reaction. The major product is that where the nitro group is present at a para position to Cl.

12. Identify ‘X’ from the reaction shown.

a) Dilute H₂SO₄
b) Concentrated H₂SO₄
c) Dilute HNO₃
d) Concentrated HNO₃
Answer: b
Clarification: This is the sulphonation of chlorobenzene which results in the formation of ortho and meta chlorobenzenesulfonic acid.

13. Which is the metal involved in Wurtz-Fittig reaction?
a) Iron
b) Magnesium
c) Aluminium
d) Sodium
Answer: d
Clarification: When an aryl halide and alkyl halide and together treated with sodium metal in dry ether, an alkylarene is formed. This is called Wurtz-Fittig reaction.

14. Predict the minor product formed when chlorobenzene reacts with chloromethane in the presence of anhydrous AlCl₃?
a) Toluene
b) m-Chlorotoluene
c) o-Chlorotoluene
d) p-Chlorotoluene
Answer: c
Clarification: This is the Friedel-Crafts alkylation reaction, where o-Chlorotoluene is the minor product and p-Chlorotoluene is the major product. This is because halogen group is ortho and para directing.

15. Fittig reaction results in the formation of a diphenyl.
a) True
b) False
Answer: a
Clarification: Aryl halides give analogous compounds in which two aryl groups are combined when treated with Na in dry ether. This is called Fittig reaction.

Chemistry Multiple Choice Questions on “Aldehydes and Ketones Chemical Reactions – 1”.

1. Aldehydes and ketones undergo __________ reactions.
a) electrophilic addition
b) electrophilic substitution
c) nucleophilic addition
d) nucleophilic substitution
Answer: c
Clarification: Aldehydes and ketones have a polar CO group which also has a double bond. The incoming nucleophile attacks the sp² hybridised carbon, thus breaking the double bond and converting it into sp³, and forming a tetrahedral alkoxide intermediate. This undergoes another fast step to form an addition product.

2. Which of the following is added to the tetrahedral intermediate when it reacts with a proton from the reaction medium?
a) Nu^–
b) H⁺
c) Both Nu^– and H⁺
d) OH^–
Answer: b
Clarification: A H+ proton is added to the negative O atom of the alkoxide intermediate to form an electrically neutral addition product. The Nu^– is added in the first step itself when it attacks the electrophilic carbon centre in a slow step.

3. The nucleophilic addition reactions of aldehydes are carried out in ________ medium.
a) neutral
b) acidic
c) weakly basic
d) extremely basic

Answer: b
Clarification: When the incoming nucleophile is weak, the acidic medium helps in protonation by increasing the positive charge on the carbonyl carbon so that it is more susceptible to attack by relatively weaker nucleophiles.

4. What is the correct order of reactivity of the following towards nucleophilic addition?
a) Methanal > Ethanal > Acetone
b) Acetone > Ethanal > Methanal
c) Methanal > Acetone > Ethanal
d) Ethanal > Methanal > Acetone
Answer: a
Clarification: Methanal is the most reactive among aldehydes and ketones due to electronic and stearic reasons. Methanal (HCHO) does not have any alkyl groups that release electrons toward carbonyl carbon, thus making it more electrophilic. Also, presence of only two H atoms do not provide any hinderance to the attack of incoming nucleophile.

5. Which of the following is least reactive towards a nucleophilic attack?
a) Acetaldehyde
b) Butanone
c) Diisopropyl ketone
d) Ditert-Butyl ketone
Answer: d
Clarification: The stearic effects come into consideration when comparing these compounds. Ditert-Butyl ketone has two very bulky tert-butyl groups, one on either side of the carbonyl carbon. This increases the hinderance to the incoming nucleophile and decreases the reactivity.

6. What is the correct order of reactivity towards nucleophilic addition?
a) Benzaldehyde > Benzophenone > Acetophenone
b) Benzophenone > Benzaldehyde > Acetophenone
c) Acetophenone > Benzaldehyde > Benzophenone
d) Benzaldehyde > Acetophenone > Benzophenone
Answer: d
Clarification: Generally, the aromatic carbonyl compounds are less reactive than corresponding aliphatic compounds. From benzaldehyde to acetophenone to benzophenone, the number of electron releasing groups increases and the magnitude of positive charge on carbonyl carbon reduces, hence decreasing its reactivity towards nucleophile attack.

7. Which of the following is the most reactive towards hydrogen cyanide?
a) Acetophenone
b) Benzaldehyde
c) p-Nitrobenzaldehyde
d) p-Tolualdehyde
Answer: c
Clarification: Addition of hydrogen cyanide is a nucleophilic reaction. CH₃ and COCH₃ are electron releasing groups which decrease the positive charge on carbonyl carbon, whereas NO₂ is an electron withdrawing group which increases the electrophilicity of carbonyl carbon, making it more reactive toward hydrogen cyanide than benzaldehyde.

8. Identify the catalyst in the nucleophilic addition of HCN to acetone.
a) NaOH
b) HCl
c) NaCl
d) NaCN
Answer: a
Clarification: Acetone reacts with HCN to give a cyanohydrin of acetone. But this reaction proceeds very slowly with pure HCN. So, the reaction is carried out in a basic medium which removes the proton from HCN and produces CN^– ion which is a stronger nucleophile and makes the reaction faster.

9. The hydrolysis of the addition product of acetone and methyl magnesium iodide gives _______
a) ethyl alcohol
b) isopropyl alcohol
c) tert-Butyl alcohol
d) phenol
Answer: c
Clarification: The methyl group of Grignard reagent is the nucleophile which initially get attached to the carbonyl carbon of acetone. This results in three methyl groups attached to the carbon atom and one OMgI group. This MgI is replaced by H⁺ on hydrolysis to give tert-Butyl alcohol. Similarly, methanal gives ethyl alcohol (1°) and other aldehydes give 2° alcohols.

10. The following is an addition product of the reaction between _______ and sodium bisulphite.

a) formaldehyde
b) acetaldehyde
c) ethanone
d) acetone
Answer: b
Clarification: Acetaldehyde reacts with NaHSO₃ to form a crystalline bisulphite addition compound. First, the nucleophile HSO₃^– attaches to the carbonyl carbon, after which the hydrogen gets attached to the oxygen atom on proton transfer.

11. The following acetal shown is formed by the reaction between ______ and ______ in the presence of dry HCl.

a) methanal; methanol
b) methanal; ethanol
c) ethanal; methanol
d) ethanal; ethanol
Answer: d
Clarification: Ethanal and ethanol react in the presence of dry HCl to initially form an unstable hemiacetal having one ethoxy (OC₂H₅) and one hydroxy (OH) group. This further reacts with ethanol to form the gem-diethoxy compound, acetaldehyde ethyl acetal as shown.

12. Acetone reacts with ethylene glycol in dry HCl gas to form _________
a) hemiacetals
b) acetals
c) cyclic acetals
d) cyclic ketals
Answer: d
Clarification: Dihydric alcohols like ethylene glycol react with aldehydes and ketones to directly form cyclic products known as cyclic acetals and cyclic ketals respectively. The cyclic ketals are also known as ethylene glycol ketals.

13. Which of the following products is formed when an aldehyde reacts with an amine in acidic medium?
a) Imine
b) Schiff’s base
c) Oxime
d) Hydrazone
Answer: b
Clarification: Amine is a nucleophile and a derivative of NH₃ which react with aldehydes and ketones followed by dehydration to form compounds with C-N double bond called Schiff’s base or substituted imines.

14. Identify the N-substituted derivative of carbonyl compounds that are coloured compounds and are useful in the identification of aldehydes and ketones.
a) Hydrazone
b) Phenylhydrazone
c) 2,4-Dinitrophenylhydrazone
d) Semicarbazone
Answer: c
Clarification: 2,4-DNP derivatives are formed from the reaction between aldehydes or ketones and 2,4-Dinitrohydrazine. These products may be yellow, orange or ed coloured and help in characterisation of aldehydes and ketones.

15. Semicarbazones undergo acidic nucleophilic addition followed by dehydration to give N-substituted carbonyl derivatives.
a) True
b) False
Answer: b
Clarification: Semicarbazone is the name of the product formed from the rapid dehydration of the product of the reaction between semicarbazide and aldehyde or ketone in acidic medium. Semicarbazide is the reagent and Semicarbazone is the N-substituted carbonyl derivative.

Chemistry Multiple Choice Questions on “Amines Physical Properties”.

1. Which of the following amines is not a gas?
a) Methylamine
b) Dimethylamine
c) Ethylamine
d) Trimethylamine
Answer: d
Clarification: Lower members of aliphatic amines exist as gases at ordinary temperature. The higher members with three or more carbon atoms are liquid and still higher ones are solid.

2. What is the characteristic odour of relatively lower aliphatic amines?
a) Fruity odour
b) Fishy odour
c) Rotten egg smell
d) Odourless
Answer: b
Clarification: Most of the amines have an unpleasant odour. The smell of lower amines is similar to that of ammonia with a fishy odour.

3. Which of the following best describes aniline in pure form?
a) Colourless liquid
b) White waxy solid
c) Brown gas
d) Yellowish gas
Answer: a
Clarification: Aniline and other aryl amines are usually colourless, but they become coloured on storage due to atmospheric oxygen. Aniline develops a yellow to brown colour dur to this reason.

4. The intermolecular hydrogen bonds, if any, in amines is formed between _______
a) N-H and N-H
b) N and N-H
c) Alkyl carbon and N-H
d) Alkyl H and N-H
Answer: b
Clarification: Primary and secondary amines are involved in intermolecular molecular hydrogen bonding between nitrogen of one and hydrogen of another molecule.

5. What is the correct order of boiling points of the isomeric amines where A=ethylmethylamine, B=propylamine and C=trimethylamine?
a) A > B > C
b) C > B > A
c) B > C > A
d) B > A > C
Answer: d
Clarification: A, B and C are respectively the 2°, 1° and 3° isomers of the compound C₃H₉N. Primary and secondary amines have two and one hydrogen atoms respectively for hydrogen bonding. Whereas, tertiary amines do not have intermolecular association due to absence of H for hydrogen bonding.

6. Which of the following has a lower boiling point than ethanamine?
a) Propane
b) Ethanal
c) Ethanol
d) Methanoic acid
View Answer

Answer: a
Clarification: All given compounds have similar molecular masses and can be compared. Ethanamine has a higher boiling point than propane because it is a polar molecule and forms intermolecular hydrogen bonds. However, it has a lower boiling point than ethanal, ethanol and methanoic acid as the O-H bonds in these compounds is more polar than the N-H bond in ethanamine. This makes the hydrogen bonds stronger.

7. If the boiling point of diethylamine is 329 K, predict the boiling point of ethyldimethylamine.
a) 310 K
b) 329 K
c) 340 K
d) 351 K
Answer: a
Clarification: Diethylamine is a 2° amine whereas, ethyldimethylamine is a 3° amine. In 3° amines, there are no hydrogen atoms available for intermolecular hydrogen bonding. Therefore, the tertiary amines have lowest boiling point among the primary and secondary counterparts.

8. If the boiling point of n-C₄H₉NH₂ is 351 K, what will be the boiling point of n-C₄H₉OH?
a) 329 K
b) 301 K
c) 351 K
d) 390 K
Answer: d
Clarification: Alcohols have higher boiling points than amines of comparable molecular masses. This is because the electronegativity of N is lower than that of O, which results in the OH bond in alcohols being more polar and hence stronger (in hydrogen bonding) than the N-H bond in amines.

9. If A is the boiling point of propanamine and B is the boiling point of butanamine, what is the correct relation between the two?
a) A > B
b) B > A
c) A = B
d) A >> B
Answer: b
Clarification: Butanamine is a larger molecule than ethanamine. As the carbon chain increase, the magnitude of van der Waals forces increases resulting in the increase in boiling point.

10. All aliphatic amines exist as associated molecules due to intermolecular hydrogen bonding.
a) True
b) False
Answer: b
Clarification: Only primary and secondary amines form intermolecular hydrogen bonds as they have individual at least one hydrogen atom to form bonds with. Tertiary amines do not have intermolecular association due to absence of hydrogen atom available for hydrogen bonding.

11. Which of the following amines are insoluble in water?
a) Methanamine
b) Ethanamine
c) Propanamine
d) Benzenamine
Answer: d
Clarification: Lower aliphatic amines are soluble in water as they easily form hydrogen bonds with water molecules. Higher amines containing six or more carbon atoms are insoluble in water due to a large hydrophobic alkyl/aryl part. Therefore, aniline is insoluble in water.

12. If ‘x’ ml of butan-1amine and ‘y’ ml of butan-1-ol is completely soluble in 100ml of water each, what is the relation between x and y?
a) x > y
b) x c) x = y
d) x + y = 100

Answer: b
Clarification: Alcohols are more polar than amines of comparable molecular masses and form stronger intermolecular hydrogen bonds with water. Therefore, the solubility of butan-1-ol is more than that of butan-1-amine.

13. Aniline is insoluble in which of the following solvents?
a) Ethanol
b) Ethoxyethane
c) Benzene
d) Water
View Answer

Answer: d
Clarification: Aromatic amines have a very large hydrocarbon part (benzene ring) which tends to retard the formation of hydrogen bonds with water. However, they are soluble in organic solvents like ethers, alcohols and benzene.

14. Which of the following is least soluble in water?
a) CH₃CH₂NH₂
b) CH₃CH₂OH
c) HCOOH
d) CH₃NH₂
Answer: a
Clarification: Lower amines are more soluble because of the smaller hydrophobic alkyl part. Also, alcohols and carboxylic acids have higher solubility than amines because of the greater polarity of OH bond than NH bond.

15. Primary amines are more soluble than tertiary amines of same formula.
a) True
b) False
Answer: a
Clarification: Solubility in water requires the presence of free hydrogen in the structure of a compound to form hydrogen bonds. Tertiary amines do not have any hydrogen attached to the N atom and are not soluble in water.

Chemistry MCQs for IIT JEE Exam on “Biomolecules – Proteins – 2”.

1. Proteins are formed primarily from ______ bonds.
a) glycosidic
b) peptide
c) phosphodiester
d) disulphide
Answer: b
Clarification: When many alpha amino acid units arranged themselves in a chain (or any other suitable structure), a protein is formed. These units are linked together by peptide bonds between NH₂ groups and COOH groups.

2. Identify the correct statement.
a) Peptide bond is formed by the loss of water molecule
b) A protein is made of only one type of amino acid
c) Dipeptides consists of different amino acids
d) Glycylalanine is a tripeptide
Answer: a
Clarification: The reaction between two molecules of same of different amino acids proceeds through the linking of COOH group of one and the NH₂ group of the other along with the loss of H₂O and formation of a peptide bond. The product is called a dipeptide.

3. How many peptide linkages does a hexapeptide have?
a) 4
b) 5
c) 6
d) 7
Answer: b
Clarification: A hexapeptide is a compound formed by the combination of six same or different amino acids with the help of the amino and carboxyl groups. The six amino acids are connected by five peptide bonds.

4. Proteins are _______
a) dipeptides
b) tripeptides
c) tetrapeptides
d) polypeptides
Answer: d
Clarification: When the number of amino acids in a peptide is more than ten, it is a polypeptide. But when a polypeptide has more than a 100 amino acid residues, with molecular mass higher than 10000u, it is called a protein. However, this is not true in all cases (like insulin).

5. Alanylglycyl phenylalanine is an example of a ______
a) dipeptide
b) tripeptide
c) tetrapeptide
d) polypeptide
Answer: b
Clarification: It is a tripeptide made from alanine, glycine and phenylalanine. It is abbreviated as Ala-Gly-Phe. The carboxyl group of alanine and the amino group of glycine combine to form one peptide bond. The second peptide linkage is formed between COOH of glycine and NH₂ of phenylalanine.

6. Which of the following bonds in not found in fibrous proteins?
a) Phosphodiester
b) Peptide
c) Hydrogen bonds
d) Disulphide
Answer: a
Clarification: Fibrous proteins are linear polypeptide chains that lie parallel to each other. Peptide binds are prevalent in the individual chains, whereas the threads are held together by hydrogen and disulphide bonds, to form a fibre-like structure.

7. Which of the following is not a fibrous protein?
a) Keratin
b) Myosin
c) Collagen
d) Albumin
Answer: d
Clarification: Keratin is a fibrous protein found in skin, nails, hair and wool. Myosin is present in muscles and collagen in tendons. Albumin is a globular protein.

8. The sequence in which amino acids are arranged in a protein is called ______ structure.
a) primary
b) secondary
c) fibrous
d) sheet
Answer: a
Clarification: Proteins have one or more polypeptide chains, in which each chain consists of a specific sequence of amino acids linked with each other. This is called the primary structure and is the most basic level. Any change in primary structure creates a different protein.

9. Which type of bonds govern the secondary structure of proteins?
a) Covalent
b) Hydrogen
c) Electrostatic
d) Peptide
Answer: b
Clarification: The secondary structure refers to the shape in which the polypeptide chain exists. There are two possible structures which arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between CO and NH groups of peptide bond.

10. Which of the following is soluble in water?
a) Insulin
b) Elastin
c) Fibroin
d) Collagen
Answer: a
Clarification: Fibroin (silk), collagen (tendons) and elastin (skin) are fibrous proteins which are insoluble in water. Insulin is an example of a globular protein which is water soluble.

11. The structure in which all peptide chains are stretched out to full extension and laid side by side through intermolecular hydrogen bonds is called ______
a) α-helix
b) β-pleated sheet
c) tertiary structure
d) quaternary structure
Answer: b
Clarification: β-pleated sheet is one of the secondary structures of proteins formed due to intermolecular hydrogen bonding. The structure resembles the pleated folds of drapery and hence the name.

12. Fibrous and globular proteins are classified on the basis of ______ structure.
a) primary
b) secondary
c) tertiary
d) quaternary

Answer: c
Clarification: Tertiary structure represents the overall folding of the polypeptide chains or the further folding of secondary structures. It gives rise to two major molecular shapes, i.e., fibrous and globular.

13. A protein ‘X’ was found in a biological system with a unique 3-D structure and biological activity. ‘X’ is known as _______
a) tertiary structure
b) quaternary structure
c) native protein
d) globular protein
Answer: c
Clarification: Native state of a protein is its most energetically stable state. When a native protein is subjected to physical change, the hydrogen bonds are disturbed, and the globules unfold.

14. The primary structure of protein is unaffected by denaturation.
a) True
b) False
Answer: a
Clarification: Denaturation is the process of altering the physical and biological properties of proteins without affecting chemical composition. It is caused by subjecting the protein to physical changes like temperature, pH, etc. During this, the secondary and tertiary structures are destroyed but the primary structure remains intact.

15. Boiling an egg is an example of reversible denaturation.
a) True
b) False
Answer: b
Clarification: The egg white gets coagulated on boiling. The globular proteins in egg white (albumin) change to a rubber like insoluble mass. This is irreversible denaturation as the protein cannot return to its original state.

Chemistry MCQs for IIT JEE Exam,