Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “P-Block – Group 18 Elements”.

1. Helium is the second most abundant gas in the universe.
a) True
b) False
View Answer

Answer: a
Clarification: Helium is the second element in the periodic table and has an atomic number of 2. Helium is formed as the result of the fusion of two hydrogen atoms. Helium is the most abundant gas in the universe (23% as compared to 76% of hydrogen).

2. Which of the following noble gases do not occur in the elemental state in the atmosphere?
a) Helium
b) Neon
c) Argon
d) Radon
Answer: d
Clarification: Radon is the last-named member of the 18^th group of elements in the periodic table. It is radioactive in natureand as a result, all other noble gases except radon occur in the elemental state in the atmosphere.

3. Which of the following is the second most abundant noble gas in the atmosphere?
a) Helium
b) Argon
c) Neon
d) Krypton
Answer: c
Clarification: Argon is the most abundant noble gas in the atmosphere, accounting for about 0.934% by volume. Neon is the second most abundant noble gas in the atmosphere (about 0.00182%) followed by helium as the third most abundant noble gas (about 0.000524%).

4. The main commercial source of helium is natural gas.
a) True
b) False
Answer: a
Clarification: The main commercial source of helium is natural gas which mainly contains hydrocarbons along with varying amounts of carbon dioxide, nitrogen, hydrogen sulphide and helium (2-7%).

5. Which of the following scientists first discovered helium in the solar spectrum during a total solar eclipse?
a) Lord Rayleigh
b) J N Lockyer
c) William Ramsay
d) Antoine Lavoisier
Answer: b
Clarification: J N Lockyer first discovered helium in the solar spectrum during a total solar eclipse on 18^th August, 1868. Later on, in 1895, it was discovered on the planet Earth by a scientist named William Ramsay.

6. What is the half-life of radon?
a) 10 days
b) 4.56 days
c) 3.82 days
d) 5.46 days
Answer: c
Clarification: Half-life of a radioactive substance refers to the amount of time that it takes for half of the radioactive atoms in a sample to decay into a more stable form. Radon has a half-life of 3.82 days.

7. What are the decay products of radium?
a) Radon and oxygen
b) Radon and nitrogen
c) Lanthanum and oxygen
d) Radon and Helium
Answer: d
Clarification: Radium is a radioactive element with the atomic number 88. It is the 6^th element in group 2 of the periodic table. Radium (Ra) on radioactive decay, gives Radon (Rn) and Helium (He) as the products.

8. Which of the following noble gases is not obtained via fractional distillation?
a) Krypton
b) Argon
c) Neon
d) Helium
Answer: d
Clarification: Neon, argon, krypton and xenon are obtained by fractional distillation of air. Fractional distillation of liquid air gives oxygen, nitrogen and a mixture of noble gases. The individual noble gases are then separated by adsorption over coconut charcoal which adsorbs different gases at different temperatures.

9. Which of the following statements is incorrect about noble gases?
a) They are monoatomic
b) They are colourless
c) They are odourless
d) They all have an outer electronic configuration of ns²np⁶
Answer: d
Clarification: The general outer electronic configuration for noble gases is ns²np⁶. But, for helium, the outer electronic configuration is 1s² as it its atomic number is only 2. All noble gases are monoatomic, odourless and colourless.

10. What is the ratio of the molar heat at constant pressure to the molar heat at constant volume for a noble gas?
a) 1.5
b) 1.67
c) 1.73
d) 1.37
Answer: b
Clarification: All noble gases are monoatomic in nature due to their stable outer electronic configuration. The ratio of their specific or molar heat at constant pressure to the specific or molar heat at constant volume is equal to 1.67, which is the same for all monoatomic gases.

11. What is the electron gain enthalpy for noble gases?
a) > 0
b) c) = 0
d) It is not defined for noble gases
Answer: a
Clarification: Noble gases have completely filled subshells. As a result, there is no vacant room in their valence shell and hence the additional electron has to be placed in an orbital of the next higher shell. Therefore, energy has to be supplied in order to add an electron and so, the electron gain enthalpy of noble gases is positive (> 0).

12. What is the boiling point of Helium?
a) 7.8 K
b) 0 K
c) 4.2 K
d) 3.7 K
Answer: c
Clarification: The melting and boiling points of noble gases are very low. This is because the atoms are held together by very weak Van der Waals forces of attraction in both liquid and solid states. Helium has the lowest boiling point, equal to 4.2 K, among all known substances.

13. Which of the following noble gases is most soluble in water?
a) Helium
b) Radon
c) Krypton
d) Neon
Answer: b
Clarification: Of all the noble gases, radon is the most soluble in water. Noble gases are non-polar substance, however, when dissolved in water, they are slightly soluble. This is due to dipole induced dipole interactions.

14. Which of the following noble gases can diffuse through rubber?
a) Helium
b) Xenon
c) Argon
d) Krypton
Answer: a
Clarification: Hydrogen is the lightest and the smallest noble gas, having an atomic number of 2. Owing to its small size and inert nature, it has the unusual property of diffusing through most commonly used laboratory materials such as rubber, glass or plastic.

15. Which among the following noble gases does not form clathrates?
a) Argon
b) Xenon
c) Krypton
d) Helium
Answer: d
Clarification: Noble gases can form compounds in which the gases are entrapped in the cavities of crystal lattices. Such compounds are called clathrates. Only Argon, Krypton, Xenon and Radon are known to form clathrates among the noble gases.

Chemistry Problems for IIT JEE Exam on “Bonding in Coordination Compounds – 2”.

1. The crystal field theory considers the metal-ligand bond to be a _______ bond.
a) covalent
b) ionic
c) polar
d) hydrogen
Answer: b
Clarification: The CFT is an electrostatic model which considers the bond between metal ion and the ligand to be ionic arising due to the electrostatic interactions between them.

2. In CFT, which of the following ligands will be treated as point dipoles?
a) Cl
b) NO₂
c) CN
d) NO
Answer: d
Clarification: Anionic ligands are treated as point charges and neutral ligands are treated as point dipoles in crystal field theory.

3. Which of the following orbitals does not belong to t_2g orbitals?
a) d_xy
b) d_yz
c) d_xz
d) d_z²
Answer: d
Clarification: t_2g is a set of three d orbitals of lower energy in octahedral complexes. The two remaining d orbitals of higher energy form the e_g set.

4. Identify the correct order of increasing field strength of the ligands.
a) SCN^–, NH₃, NCS^–, CN^–
b) NCS^–, SCN^–, NH₃, CN^–
c) SCN^–, NCS^–, NH₃, CN^–
d) NCS^–, NH₃, SCN^–, CN^–
Answer: c
Clarification: According to the spectrochemical series, thiocyanate is weaker than isothiocyanate which is in turn weaker than amine and cyanide ligands.

5. By how much percentage does the energy of e_g orbitals increase by from the average field splitting energy in octahedral complexes?
a) 40%
b) 45%
c) 50%
d) 60%
Answer: d
Clarification: The energy of the two e_g orbitals will increase by (3/5)Δ_o, that is 60% of the average energy that separates the field in octahedral complexes.

6. Identify the correct order of decreasing field strength of the neutral ligands.
a) CO, en, NH₃, H₂O
b) CO, en, H₂O, NH₃
c) en, CO, NH₃, H₂O
d) en, CO, H₂O, NH₃
Answer: a
Clarification: According to the spectrochemical series, CO is the strongest ligand. The ligands en, amine and water are weaker than CO and their field strength decreases in that order.

7. Identify the correct relation between Δ_o and Δ_t, where Δ_o denotes crystal field splitting in octahedral complexes and Δ_t denotes crystal field splitting in tetrahedral complexes.
a) Δ_ot
b) Δ_o > Δ_t
c) Δ_o = Δ_t
d) Δ_o ≥ Δ_t
View Answer

Answer: b
Clarification: It has been observed that splitting energy in tetrahedral fields is considerably less than in octahedral complexes. To be precise, it has been found that Δ_t = (4/9) Δ_o.

8. Which of the following ligands is most likely to form high spin complexes in octahedral fields?
a) Cl^–
b) OH^–
c) C₂O₄^–
d) CN^–
Answer: a
Clarification: Cl- is the weakest field ligand in the given options. The splitting field Δ_o is small and the energy separation between the e_g and t_2g orbitals are narrower. Hence, the electrons tend to occupy the eg orbitals easily and are likely to have a greater number of half-filled orbitals.

9. What will be the electronic configuration of d⁵ in terms of t_2g and e_g in an octahedral field when Δ_o

a) t_2g⁵ e_g⁰
b) t_2g² e_g³
c) t_2g³ e_g²
d) t_2g⁰ e_g⁵
Answer: c
Clarification: When the field splitting is less than the pairing energy, the energy gap between the two sets of d orbitals is small. This allows the electrons to occupy the higher energy orbitals once the lower orbitals are singly occupied. Thus, the configuration will be t_2g³ e_g², with one electron in each orbital.

10. Tetrahedral fields can easily form low spin complexes.
a) True
b) False
Answer: b
Clarification: The splitting energy in tetrahedral fields is almost half of that of octahedral fields. Due to this, the orbital splitting energies are not very large, which makes it difficult for electron pairing and therefore low spin complexes are rarely formed.

11. Observe the orbital splitting diagram of a tetrahedral field and find the value of x.

a) 0.178
b) 0.267
c) 0.4
d) 0.44
Answer: a
Clarification: The value of x is asked in terms of Δ_o and not Δ_t. The t₂ orbitals is at an energy of (2/5) Δ_t above the average energy. This is equal to (2/5) times (4/9) Δ_o = 0.178Δ_o.

12. As the crystal field splitting energy in octahedral field increases, the wavelength of light absorbed _________
a) Increases
b) Decreases
c) Remains the same
d) May increase or decrease
Answer: b
Clarification: If Δ_o is less (as in weak field ligands), the excitation energy will be the smallest and the largest wavelength will be absorbed. On the other hand, Δ_o is high (as in strong field ligands), the excitation energy will be the largest and the smallest wavelength will be absorbed.

13. The complex [Co(H₂O)₆]³⁺ absorbs the wavelength of light corresponding to orange colour. Predict the colour of the coordination compound based on this information.
a) Red
b) Yellow
c) Blue
d) Colourless
Answer: c
Clarification: This is because of the concept of complementary colours. The colour of the complex is complementary to that which is absorbed and is generated from the wavelength that is left over.

14. Hexaaquatitanium(III) chloride is a coloured compound that loses its colour on heating.
a) True
b) False
Answer: a
Clarification: Heating the compound removes the water which is the ligand. In the absence of a ligand, crystal field splitting does not occur and hence the compound becomes colourless.

Chemistry Problems for IIT JEE Exam,

Chemistry Written Test Questions for IIT JEE Exam on “Alcohols, Phenols and Ethers Nomenclature – 2”.

1. What is the correct name of the following compound?

a) But-3-enol
b) But-3-en-2-ol
c) But-1-en-3-ol
d) But-2-en-3-ol
Answer: b
Clarification: The numbering preference is given to the hydroxyl group over the double bond and hence, the OH lies on the second carbon and double bond on the third.

2. What is the correct IUPAC name of the following phenol?

a) Benzene-1,2,4-triol
b) 1,2,4-Benzenetriol
c) Benzene-1,2,4-triphenol
d) 1,2,4-Triphenol
Answer: a
Clarification: Polyhydric phenols are named as hydroxy derivatives of benzene in the IUPAC system. The positions 1,2,4 comply with the lowest set of locants rule.

3. What is the correct common name of 2-Bromophenol?
a) 2-Bromophenol
b) m-Bromophenol
c) o-Bromophenol
d) 2-Hydroxybromobenzene
Answer: c
Clarification: Phenol is used as the common as well as the IUPAC name and haloarenes with hydroxy groups are named as halo derivatives of phenols.

4. Which of the following is given highest preference while naming when present in the same structure?
a) Hydroxyl group
b) Nitro group
c) Triple bond
d) Halogen group
Answer: a
Clarification: Functional groups are given the highest preference while naming compounds according to the IUPAC system.

5. What is the correct common name for CH₃OC₂H₅?
a) Methyl methyl ether
b) Methyl ethyl ether
c) Ethyl ethyl ether
d) Ethyl methyl ether
Answer: d
Clarification: It is an unsymmetrical ether with one methyl group and one ethyl group. It is named such that the alkyl group are in alphabetical order.

6. Which of the following compounds is ethoxyethane?
a) CH₃OCH₃
b) CH₃OC₂H₅
c) C2H₅OC₂H₅
d) C2H₅OC₃H₇
Answer: c
Clarification: Ethoxyethane is an ether which is a derivative of ethane with one hydrogen atom replaced by ethoxy group (OC₂H₅).

7. Identify the incorrect name for CH₃-O-C₆H₅.
a) Methyl phenyl ether
b) Methoxybenzene
c) Anisole
d) Phenetole
Answer: d
Clarification: This is an aromatic ether with a methoxy substituent and has the common name anisole or methyl phenyl ether.

8. What is the IUPAC name of CH₃-O-CH₂-CH₂-OCH₃?
a) 1,2-Dimethoxyethane
b) 1,3-Dimethoxybutane
c) 2,3-Dimethoxyethane
d) 1,4-Dimethoxypropane
Answer: a
Clarification: The base hydrocarbon is ethane, in which one hydrogen of each carbon atom is replaced by a methoxy group, making it a dimethoxy ether.

9. What is the IUPAC name of CH₃(CH₂)₆-OC₆H₅?
a) Heptyl phenyl ether
b) 1-Phenoxyheptane
c) 1-Hepoxybenzene
d) Phenyl heptyl ether
Answer: b
Clarification: Since heptane is a larger group than benzene, it is chosen as the parent hydrocarbon and the compound is named as an aryloxy derivative of heptane.

10. The name 1-Ethoxy-2, 2-dimethylcyclohexane is correct according to IUPAC nomenclature.
a) True
b) False
Answer: b
Clarification: The methyl group should be given higher preference over the alkoxy group in ethers. The correct naming will be 2-Ethoxy-1, 1-dimethylcyclohexane.

11. Phenetole is an aromatic ether.
a) True
b) False
Answer: a
Clarification: Phenetole is also known as ethoxybenzene or ethyl phenyl ether and has an aromatic ring on one side.

Chemistry Written Test Questions for IIT JEE Exam,

Chemistry Exam Questions for IIT JEE Exam on “Methods of Carboxylic Acids Preparation”.

1. Which of the following is known as Jones reagent?
a) KMnO₄ in alkaline medium
b) CrO₃ in H₂SO₄
c) K₂Cr₂O₇ in acidic medium
d) KMnO₄ in H₂SO₄
View Answer

Answer: b
Clarification: Chromium trioxide in an aqueous solution with sulphuric acid in known as Jones reagent. It Is an important compound in the preparation of carboxylic acids from alcohols.

2. Benzoic acid is obtained from the oxidation of _______ with alkaline KMnO₄ followed by treatment with mineral acid.
a) phenol
b) benzaldehyde
c) acetophenone
d) benzyl alcohol
Answer: d
Clarification: Only primary alcohols undergo oxidation in the presence of common oxidising agents followed by reaction with H₃O⁺ to give the respective carboxylic acids. This reaction is proceeded by the removal of both hydrogen atoms from the alpha carbon and formation of a double bond between C and O.

3. Identify the most suitable reagent for the conversion of ethanal to acetic acid.
a) Alkaline KMnO₄; H₃O⁺
b) Jones reagent
c) Tollen’s reagent
d) LiAlH⁴
Answer: c
Clarification: Aldehydes are easily oxidized to carboxylic acids having the same number of carbon atoms as the parent aldehyde, when reacted even with mild oxidising agents like Tollen’s reagent.

4. Identify the product B in the reaction chain shown.

a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) 2-Methylpropanoic acid
Answer: b
Clarification: Ethyl bromide on reaction with alc. KCN gives ethyl cyanide by nucleophilic substitution. Nitriles can be hydrolysed to give amides and on further heating give a carboxylic acid which contains one more carbon than the original ethyl bromide. Hence, the acid contains three C atoms, making it propanoic acid.

5. Identify X in the following conversion.

a) Alkaline KMnO₄; H₃O+
b) KOH; heat
c) H₃O⁺; heat
d) KMnO₄-KOH; heat
Answer: c
Clarification: Amides undergo hydrolysis in the presence of heat to give corresponding carboxylic acids along with ammonia gas. In the reaction shown, benzamide undergoes oxidation to form benzoic acid.

6. Benzoic acid can be prepared by oxidation of tert-Butyl benzene.
a) True
b) False
Answer: b
Clarification: tert-Butyl benzene consists of a tertiary group with no benzylic H. It is highly stable and does not undergo oxidation even under drastic conditions, to give aromatic carboxylic acids.

7. Which of the following pairs do not give the same compound on heating with alkaline potassium permanganate?
a) Toluene and propyl benzene
b) Toluene and n-Butyl benzene
c) Propyl benzene and isopropyl benzene
d) o-Xylene and n-Butyl benzene
Answer: d
Clarification: All mono-substituted alkyl benzenes, with primary or secondary alkyl groups, on vigorous oxidation give benzoic acid. The entire side chain is oxidised irrespective of the length. In case of o-xylene, there are two methyl groups on the benzene ring and both of them are oxidised to carboxyl group, resulting in phthalic acid.

8. p-Xylene on reaction with acidified potassium dichromate at high temperature gives ________
a) benzoic acid
b) phthalic acid
c) terephthalic acid
d) no reaction
Answer: c
Clarification: p-Xylene is a dimethyl substituted benzene which on vigorous oxidation, gets oxidised to an aromatic dicarboxylic acid, with the two carboxyl groups at para positions with respect to each other. This compound is called terephthalic acid.

9. Which of the following cannot be converted to benzoic acid on reaction with KMnO₄-KOH followed by H₃O⁺?
a) Ethyl benzene
b) Acetophenone
c) 4-Methylacetophenone
d) Styrene
Answer: c
Clarification: The ethyl group, acetyl group and ethene group are all oxidised to potassium carboxylate groups which are further oxidised to carboxyl groups. In the case of 4-Methylacetophenone, there are two groups that will be oxidised to carboxyl groups, hence forming terephthalic acid instead of benzoic acid.

10. How can methyl magnesium bromide be converted to propanoic acid?
a) Jones reagent
b) KMnO₄-KOH; heat
c) H₃O⁺; heat
d) CO₂-dry ether; H₃o⁺
Answer: d
Clarification: Methyl magnesium bromide (Grignard reagent) on reaction with CO₂ forms an addition product containing an additional C carbon atom. This is decomposed in the presence of mineral acid to form propanoic acid.

11. 3-Chlorophenyl magnesium bromide on reaction with dry ice followed by acidification in mineral acid gives _________
a) 3-Chlorophenol
b) 3-Chlorophenylethanoic acid
c) 3-Chlorobenzaldehyde
d) 3-Chlorobenzoic acid
Answer: d
Clarification: The MgBr group of 3-Chlorophenyl magnesium bromide (Grignard reagent) will be substituted by COOH group in the above reaction, to give a halogen substituted aromatic carboxylic acid.

12. Benzoic ethanoic anhydride on hydrolysis gives __________
a) benzoic acid and methanoic acid
b) benzoic acid and ethanoic acid
c) phenylethanoic acid and methanoic acid
d) no products
Answer: b
Clarification: Benzoic ethanoic anhydride (C₆H₅COOCOCH₃) is easily hydrolysed with water to give its corresponding acids, benzoic acid (by adding H to C₆H₅COO) and ethanoic acid (by adding OH to COCH₃).

13. Ethanoyl chloride on hydrolysis with aqueous NaOH gives _______
a) acetate ion
b) acetic acid
c) propanoic acid
d) no reaction
Answer: a
Clarification: Acid chlorides are readily hydrolysed with aqueous base to give carboxylate ions which further give corresponding carboxylic acids on acidification. It can be directly obtained by direct hydrolysis with water.

14. The final product(s) of basic hydrolysis followed by acidification of ethyl butanoate is _______
a) ethanoic acid
b) butanoic acid
c) ethanoic acid and butanoic acid
d) butanoic acid and ethanol
Answer: b
Clarification: Ethyl butanoate (CH₃CH₂CH₂COOC₂H₅) on basic hydrolysis forms CH₃CH₂CH₂COONa and ethanol. Then this sodium carboxylate compound gets acidified to give butanoic acid.

15. Acidic hydrolysis of ethyl benzoate directly gives benzoic acid.
a) True
b) False
Answer: a
Clarification: Acidic hydrolysis of esters directly gives carboxylic acids. In this case, the ethyl group of the ester is separated out as ethanol along with the main product, benzoic acid.

Chemistry Assessment Questions for Class 12 on “Amines Chemical Reactions – 4”.

1. Which gas is produced when ethanamine reacts with nitrous acid?
a) N₂
b) H₂
c) HCl
d) O₂
Answer: a
Clarification: Nitrogen gas is evolved due to the decomposition of diazonium salts which are formed from the reaction of 1° aliphatic amines with HNO₂. This is used in the estimation of amino acids and proteins.

2. What is the main product of the following reaction?

a) Benzene diazonium chloride
b) Phenol
c) o-Nitrosoaniline
d) p-Nitrosoaniline
Answer: b
Clarification: Primary aromatic amines react with nitrous acid at cold temperatures (273-278K) to form diazonium salts. However, if the temperature is more than 278K, phenol is formed along with the evolution of nitrogen gas.

3. Tertiary amines dissolve in nitrous acid to form corresponding salts.
a) True
b) False
Answer: a
Clarification: Tertiary amines on reaction with cold HNO₂ remain dissolved, forming amine nitrite salts, which decompose to nitrosoamines and alcohol on warming.

4. Which of the following is not a final product of the reaction between propylamine and nitrous acid?
a) CH₃CH₂CH₂N₂Cl
b) CH₃CH₂CH₂OH
c) N₂ gas
d) H₂O
Answer: a
Clarification: CH₃CH₂CH₂N₂Cl is a diazonium salt derivative of propanamine and is first formed when it reacts with HNO₂. This is a very unstable compound and immediately decomposes to give propanol and evolves nitrogen gas.

5. Diethylamine reacts with nitrous acid in the cold to form _______
a) diazonium salt
b) alcohol
c) imine
d) nitrosoamine
Answer: d
Clarification: Secondary amines react slowly with nitrous acid at low temperatures to give yellow oily nitrosoamines. Diethylamine specifically produces N,N-diethylnitrosoamine.

6. Hinsberg’s reagent is _______
a) benzenesulfonic acid
b) benzenesulphonyl chloride
c) p-toluenesulphonyl chloride
d) chlorosulphuric acid
Answer: b
Clarification: Benzenesulphonyl chloride (C₆H₅SO₂Cl), or arylsulphonyl chloride, is known as Hinsberg’s reagent. It is an important compound used in the distinction of different classes of amines.

7. Which of the following amines, on reaction with benzenesulphonyl chloride, will give a sulphonamide that is insoluble in alkali?
a) Ethylamine
b) Ethylmethylamine
c) Trimethylamine
d) Aniline
Answer: b
Clarification: Secondary amines on reaction with Hinsberg’s reagent produce sulphonamides without any hydrogen atom attached to the nitrogen atom. Hence, it is not acidic and therefore insoluble in alkali. The amide produced in this case is N-ethyl-N-methylbenzenesulphonamide.

8. Which of the following amines will form a product that is soluble in KOH, on reaction with Hinsberg’s reagent?
a) Isopropylamine
b) Diethylamine
c) N,N-Dimethylpropylamine
d) N,N-Dimethylaniline
Answer: a
Clarification: The reaction of Hinsberg’s reagent with Isopropylamine gives N-isopropylbenzene sulphonamide. The lone hydrogen attached to the N atom is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, is it soluble in alkali like KOH.

9. Which of the following reactions/tests does not help in the distinction between ethylamine and diethylamine?
a) Carbylamine test
b) Hinsberg’s test
c) Reaction with HNO₂
d) Reaction with CH₃CH₂Br
Answer: d
Clarification: Both ethylamine and diethylamine on reaction with CH₃CH₂Br eventually gives quaternary ammonium salt. Hence, the alkylation of primary amines cannot be used as a distinction method.

10. Aniline is a _______ directing compound.
a) ortho
b) meta
c) ortho and para
d) ortho, meta and para
Answer: c
Clarification: The ortho and para positions with respect to NH₂ group in aniline are centres of high electron density. This is due to the resonance structures of aniline. Thus, NH₂ is ortho and para directing and a powerful activating group.

11. The activating effect of -NHCOCH₃ group is ______ as compared to -NH₂ group.
a) less
b) same
c) more
d) very high
Answer: a
Clarification: The -NHCOCH₃ group forms two resonance structures, where the lone pair of nitrogen interacts with the lone pairs on oxygen atom. This makes the lone electron pair on N less available for donation to benzene ring.

12. What is the major product formed when aniline reacts with bromine water at room temperature?
a) 2-Bromoaniline
b) 4-Bromoanline
c) 2,6-Dibromoaniline
d) 2,4,6-Tribromoaniline
Answer: d
Clarification: Since amino group is highly ortho and para directing, it substitutes the Br group at the para as well as both ortho positions (2,4,6) to give a white precipitate of a tribromo substituted aniline.

13. What is the order of quantities of all isomers of nitroaniline formed on the reaction of aniline with nitric acid and sulphuric acid at 288K?
a) ortho > meta > para
b) para > ortho > meta
c) para > meta > ortho
d) meta > para > ortho
Answer: c
Clarification: Nitration of aniline produces nitro derivatives. Under controlled conditions, aniline is protonated (by strong acidic medium) to form anilinium ion, which is meta directing. Therefore, apart from p-nitroaniline (51%) and o-nitroaniline (2%), a significant amount of m-nitroaniline (47%) is also formed.

14. It is possible to obtain para isomers of anilines as the major product in electrophilic ring substitution of aniline.
a) True
b) False
Answer: a
Clarification: This is possible by protecting the amino group of aniline by reacting it with acetic anhydride (acetylation) to give acetanilide. This compound on desired substitution reaction followed by hydrolysis gives para substituted aniline as major product.

15. A compound ‘P’ on treating with concentrated H₂SO₄ forms ‘R’. The product ‘R’ on heating at 460K forms a zwitter ionic compound. Identify P, R respectively.
a) Aniline; anilinium hydrogensulphate
b) Aniline; sulphanilic acid
c) Anilinium hydrogensulphate; sulphanilic acid
d) Sulphanilic acid; anilinium hydrogensulphate
Answer: a
Clarification: Aniline (P) reacts with conc. H₂SO₄ to form anilinium hydrogensulphate (R) which on heating at 453-473K produces sulphanilic acid (zwitter ionic compound).

Chemistry Assessment Questions for Class 12,

Chemistry Multiple Choice Questions on “Biomolecules – Hormones”.

1. Which of the following hormone is a polypeptide?
a) Estrogen
b) Insulin
c) Androgen
d) Epinephrine
Answer: b
Clarification: Insulin consists of a chain of 51 amino acids and hence is a polypeptide. Estrogen and androgen are steroids, whereas epinephrine is an amine.

2. Hormones are ______
a) messengers
b) catalysts
c) enzymes
d) inhibitors
Answer: a
Clarification: Hormones are chemical substances produced by the endocrine glands in the human body. They are carried to different parts of the body through the blood stream. Because of the action of hormones as intercellular communicators, they are called chemical messengers.

3. Which of the following is not an amine hormone?
a) Norepinephrine
b) Adrenaline
c) Thyroxine
d) Oxytocin
Answer: d
Clarification: Amine hormones are water-soluble compounds which have amino groups and are structurally derived from amino acids. Oxytocin is a peptide hormone.

4. Identify the hormone that increases the glucose level in blood.
a) Insulin
b) Glucagon
c) Oxytocin
d) Vasopressin
Answer: b
Clarification: Insulin is released in response to the rapid rise in blood glucose level. On the other hand, the hormone glucagon tends to increase the glucose level in blood. These two hormones together regulate the blood glucose level.

5. Which of the following is known as fight or flight hormone?
a) Epinephrine
b) Norepinephrine
c) Insulin
d) Thyroxine
View Answer

Answer: a
Clarification: Epinephrine and norepinephrine are amine hormones that mediate response to external stimuli. Adrenaline plays an important role in fight or flight situations by increasing blood flow to muscles, blood sugar level and pulse rate.

6. Which hormone plays an important role during child birth and post it?
a) Estrogen
b) Progesterone
c) Cortisone
d) Oxytocin
Answer: d
Clarification: Oxytocin controls the contraction of the uterus during child birth and helps in the release of milk from mammary glands. Progesterone is responsible for the preparation of uterus for implantation of fertilised egg.

7. The condition goitre is associated with which hormone?
a) Insulin
b) Thyroxine
c) Adrenaline
d) Cortisone
Answer: b
Clarification: Thyroxine is an iodinated derivative of tyrosine. Abnormally low levels of thyroxine leads to hypothyroidism, which causes enlargement of thyroid gland (goitre). Increases level of thyroxine causes hyperthyroidism.

8. Lack of which component in diet causes hypothyroidism?
a) Potassium
b) Vitamin C
c) Iodine
d) Water
Answer: c
Clarification: Low levels of iodine in the diet may lead to hypothyroidism and enlargement of the thyroid gland. This can be controlled by adding sodium iodide to table salt to form iodised salt.

9. Which of the following does not release steroid hormones?
a) Testes
b) Ovary
c) Adrenal cortex
d) Pancreas
Answer: d
Clarification: Steroid hormones are produced by adrenal cortex and gonads. Hormones released by the adrenal cortex play a very important role in the functions of the bod.

10. Which hormone controls the balance of water and minerals in the body?
a) Vasopressin
b) Mineralocorticoids
c) Testosterone
d) Thyroxine
Answer: b
Clarification: Mineralocorticoids are steroid hormones which control the level of excretion of water and salt by the kidney, thus balancing the water and mineral levels as required.

11. Lack of which hormone causes Addison’s disease?
a) Glucocorticoids
b) Oxytocin
c) Insulin
d) Norepinephrine
Answer: a
Clarification: If adrenal cortex does not function properly then one of the results may be Addison’s disease characterized by hypoglycaemia, weakness and increased susceptibility to stress. This may be fatal unless treated by glucocorticoids and mineralocorticoids.

12. All hormones are proteins.
a) True
b) False
Answer: b
Clarification: Hormones are compounds having varied chemical structures. They may be polypeptide chains or amino acids or contain a steroid nucleus.

13. Estradiol is the main sex hormone in females.
a) True
b) False
Answer: a
Clarification: Estradiol is a steroid hormone which is majorly responsible for the development of secondary sex characteristics in females and participates in the control of menstrual cycle.

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