You are currently viewing CHM121 Past Questions and Answers: Comprehensive Study Guide for Exam Success

CHM121 Past Questions and Answers: Comprehensive Study Guide for Exam Success

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This study guide is tailored to support your CHM121 exam preparation by offering a thorough collection of past questions with detailed answers. Whether you’re revisiting key concepts or sharpening your problem-solving techniques, this guide will strengthen your understanding and build your confidence.

CHM121 past questions and answers 1

Question 1: The adoption of a particular isolation procedure to a large extent will depend on the ____
Options: (a) physical and chemical properties (b) boiling and melting point (c) solubility (d) density (e) all of the mentioned
Answer: (e) all of the mentioned
Explanation: Isolation procedures in chemistry often depend on various factors like the physical and chemical properties, boiling and melting points, solubility, and density of the compounds involved. Each of these properties can influence the choice of technique for separating and purifying compounds. For example, solubility is crucial in filtration and crystallization processes, boiling points are important in distillation, and density affects centrifugation. Therefore, all the mentioned factors are considered when selecting an isolation procedure.

CHM121 past questions and answers 2

Question 2: _____ are colored organic compounds
Options: (a) aldehydes (b) ethers (c) esters (d) amines
Answer: (d) amines
Explanation: Amines, particularly aromatic amines, can often exhibit color due to their ability to absorb light in the visible spectrum. This occurs because of the presence of conjugated systems in aromatic amines, which allow them to interact with light. Aldehydes, ethers, and esters are typically colorless unless they are part of a larger structure that includes chromophores.

CHM121 past questions and answers 3

Question 3: A fishlike odor indicates the presence of ______
Options: (a) complex amides (b) higher aliphatic amines (c) intermediate aliphatic amide (d) lower aliphatic amines
Answer: (d) lower aliphatic amines
Explanation: Lower aliphatic amines, such as methylamine and ethylamine, are known for their pungent, fishlike odor. This characteristic odor is due to the presence of nitrogen in the amine group, which contributes to the volatile nature of these compounds. Complex amides and higher aliphatic amines generally do not produce such a strong odor.

CHM121 past questions and answers 4

Question 4: Aromatic amines and amino-phenols have ____ color
Options: (a) brown (b) orange (c) blue (d) yellow (e) ethers
Answer: (a) brown
Explanation: Aromatic amines and amino-phenols typically exhibit a brown color due to the presence of conjugated double bonds and functional groups that can absorb light. This absorption shifts the perceived color to the brown region. The color can vary slightly depending on the exact structure and substituents present.

CHM121 past questions and answers 5

Question 5: Which of the following organic substances is colorless?
Options: (a) naphthols (b) nitro-compound (c) esters (d) ethers
Answer: (d) ethers
Explanation: Ethers are generally colorless because they lack chromophores, the part of the molecule responsible for absorbing visible light and producing color. Naphthols and nitro-compounds often have colors due to their complex aromatic structures and functional groups. Esters can be colorless, but many have slight coloration depending on their structure.

CHM121 past questions and answers 6

Question 6: An organic compound that burns with a non-sooty flame, which may or may not be clear, is most likely to be _____
Options: (a) aromatic compounds (b) amines (c) single aliphatic compound (d) an unsaturated aliphatic compound
Answer: (c) single aliphatic compound
Explanation: Single aliphatic compounds, particularly saturated hydrocarbons (alkanes), tend to burn with a non-sooty, sometimes clear flame because they combust completely to form carbon dioxide and water. Aromatic compounds and unsaturated aliphatic compounds (alkenes and alkynes) often produce a sooty flame due to incomplete combustion and higher carbon content.

CHM121 past questions and answers 7

Question 7: All the following react violently with bases and most give off very harmful vapors except _____
Options: (a) hydrobromic acid and hydrogen bromide (b) hydrochloric acid and hydrogen chloride (c) hydrofluoric acid and hydrogen fluoride (d) sulfuric acid and hydrogen sulfide
Answer: (d) sulfuric acid and hydrogen sulfide
Explanation: Sulfuric acid (H2SO4) reacts with bases in a neutralization reaction but is less likely to give off harmful vapors compared to hydrofluoric acid, which is extremely dangerous due to its ability to penetrate tissues and cause deep burns. Hydrogen sulfide, while toxic, does not have the same reactivity as hydrogen halides (e.g., HBr, HCl) in forming harmful vapors when reacting with bases.

CHM121 past questions and answers 8

Question 8: Which of the following cleaning mixtures has the corrosive properties of concentrated H2SO4?
Options: (a) chromic acid (b) hydrochloric acid (c) citric acid (d) nitric acid
Answer: (a) chromic acid
Explanation: Chromic acid is a potent oxidizing agent and is highly corrosive, similar to concentrated sulfuric acid. It is used for cleaning glassware in laboratories because of its strong ability to oxidize organic matter. Hydrochloric acid and nitric acid are also corrosive but are not as powerful as chromic acid. Citric acid is much weaker and not considered corrosive.

CHM121 past questions and answers 9

Question 9: Filtration of a mixture after extraction is important to ____
Options: (a) isolate a solid product that has crystallized out (b) remove insoluble impurities (c) remove insoluble reactants (d) all the mentioned
Answer: (d) all the mentioned
Explanation: Filtration is a crucial step in many chemical processes to separate solid materials from liquids. After extraction, filtration can isolate solid products, remove insoluble impurities, and remove insoluble reactants from the mixture, thereby purifying the desired substance.

CHM121 past questions and answers 10

Question 10: The filtration of substantial quantities of a solid from a suspension in a liquid ____ of convenient size is employed
Options: (a) separating funnel (b) Buchner funnel (c) filter paper (d) None of the mentioned
Answer: (b) Buchner funnel
Explanation: A Buchner funnel is commonly used for vacuum filtration of substantial quantities of solid from a suspension. It allows for efficient separation of the solid and liquid phases. The separating funnel is used for liquid-liquid extractions, while filter paper is used in combination with a funnel or a Buchner funnel for filtration.

CHM121 past questions and answers 11

Question 11: For the suction filtration of small quantities of solid less than 5g contained in a small volume of liquid, a small conical Buchner funnel known as _____ is used
Options: (a) Hirsch funnel (b) baby Buchner (c) micro Buchner (d) macro Buchner
Answer: (a) Hirsch funnel
Explanation: A Hirsch funnel is a small, conical version of the Buchner funnel designed for the filtration of small quantities of solid material. It is ideal for small-scale laboratory work where the amount of solid is minimal and the liquid volume is small.

CHM121 past questions and answers 12

Question 12: In crystallization, we assume the following except:
Options: (a) a substance is likely to be most soluble in a solvent to which it is closely related in chemical and physical properties (b) a polar substance dissolves in a polar solvent (c) a non-polar substance dissolves in a non-polar solvent (d) all the mentioned
Answer: (d) all the mentioned
Explanation: All the options provided are generally true assumptions in the process of crystallization. The concept of “like dissolves like” applies where polar substances dissolve in polar solvents, and non-polar substances dissolve in non-polar solvents. Therefore, no exception is found, making “all the mentioned” correct.

CHM121 past questions and answers 13

Question 13: Substances that decompose or undergo structural modification on exposure to air must be recrystallized in an inert atmosphere of nitrogen, hydrogen, or CO2 _______
Options: (a) True (b) False
Answer: (a) True
Explanation: Certain sensitive substances may decompose or change structure when exposed to air. To prevent this, recrystallization is carried out in an inert atmosphere, such as nitrogen, hydrogen, or carbon dioxide, which do not react with the substance, thus preserving its integrity during the recrystallization process.

CHM121 past questions and answers 14

Question 14: Tollen’s reagent is used to _____
Options: (a) distinguish acids from bases (b) confirm the presence of a nitro compound (c) distinguish aldehydes from ketones (d) none of the above
Answer: (c) distinguish aldehydes from ketones
Explanation: Tollen’s reagent is a chemical reagent used to differentiate between aldehydes and ketones. Aldehydes reduce Tollen’s reagent to produce a silver mirror on the surface of the reaction vessel, while ketones do not react in the same manner.

CHM121 past questions and answers 15

Question 15: The oxidizing agent in Tollen’s reagent is ____
Options: (a) Ag (b) Ag+ (c) Cu2+ (d) Cu2O
Answer: (b) Ag+
Explanation: In Tollen’s reagent, the active oxidizing agent is the silver ion (Ag+). This ion gets reduced to metallic silver (Ag) when it comes into contact with an aldehyde, leading to the formation of a silver mirror on the reaction vessel.

CHM121 past questions and answers 16

Question 16: Which of the laws of chemistry is used to establish the formula of a chemical compound?
Options: (a) Definite proportions (b) reciprocal proportions (c) conservation of mass (d) partial pressures
Answer: (a) Definite proportions
Explanation: The Law of Definite Proportions (or Constant Composition) states that a chemical compound always contains its component elements in fixed ratio by mass, regardless of the amount of the compound. This law is used to determine the formula of a chemical compound, as it helps in establishing the specific ratios of elements in the compound.

CHM121 past questions and answers 17

Question 17: The formula which shows the actual number of atoms of each element in a given compound is___
Options: (a) molecular formula (b) empirical formula (c) chemical formula (d) mass formula
Answer: (a) molecular formula
Explanation: The molecular formula of a compound shows the exact number of atoms of each element present in a molecule of the compound. It provides a complete picture of the molecule’s composition, unlike the empirical formula, which only gives the simplest whole-number ratio of the elements.

CHM121 past questions and answers 18

Question 18: The conditions for drying recrystallized material depend on the following EXCEPT
Options: (a) quantity of product (b) quality of product (c) sensitivity of product to heat (d) nature of solvent to be removed
Answer: (b) quality of product
Explanation: The conditions for drying recrystallized material are influenced by the quantity of the product, its sensitivity to heat, and the nature of the solvent being removed. The quality of the product, while important for the final analysis, does not directly affect the drying conditions.

CHM121 past questions and answers 19

Question 19: Substances which decompose or otherwise undergo structural modification on contact with air must be recrystallized in______ atmosphere
Options: (a) indifferent (b) normal (c) ambient (d) standard
Answer: (a) indifferent
Explanation: Substances that are sensitive to air or moisture must be recrystallized in an indifferent (inert) atmosphere, such as nitrogen or argon, to prevent decomposition or structural changes. The other options do not specifically refer to protective atmospheres.

CHM121 past questions and answers 20

Question 20: The actual sequence of bond breaking and formation during the course of a reaction is called ____
Options: (a) reaction kinetics (b) reaction stoichiometry (c) reaction product (d) reaction mechanism (e) activation energy
Answer: (d) reaction mechanism
Explanation: The reaction mechanism describes the step-by-step sequence of bond breaking and formation that occurs during a chemical reaction. It provides detailed information on the intermediate stages and how reactants are converted into products.

CHM121 past questions and answers 21

Question 21: The effect of attracting or repelling of electrons in sigma bonds is referred to as_____
Options: (a) resonance effect (b) positive inductive effect (c) inductive effect (d) steric effect (e) attraction/repulsion effect
Answer: (c) inductive effect
Explanation: The inductive effect refers to the transmission of charge through a chain of atoms in a molecule by electrostatic effects, leading to the attraction or repulsion of electrons in sigma bonds. It influences the distribution of electron density along the molecule.

CHM121 past questions and answers 22

Question 22: Heterolytic bond fission is favored by the following except_____
Options: (a) use of polar solvent (b) presence of resonance effect (c) presence of inductive effect (d) presence of free radical initiators (e) presence of solvent ethanol
Answer: (d) presence of free radical initiators
Explanation: Heterolytic bond fission involves the breaking of a bond where both electrons go to one of the fragments, typically favored by polar solvents, resonance effects, and inductive effects. Free radical initiators are more relevant to homolytic bond fission, where the bond breaks with each fragment receiving one electron.

CHM121 past questions and answers 23

Question 23: How many molecules of hydrogen are required for the conversion of alkynes to alkenes
Options: (a) 1 (b) 2 (c) 3 (d) 4 (e) 5
Answer: (b) 2
Explanation: The conversion of alkynes to alkenes involves the addition of 2 molecules of hydrogen (one molecule of hydrogen gas adds across each of the triple bonds in an alkyne to form a double bond, resulting in an alkene).

CHM121 past questions and answers 24

Question 24: In ozonolysis, the double bond of an alkene is broken and _____ and ______ are formed depending upon the structure of the parent alkene
Options: (a) alcohols and ketones (b) carboxylic acid and aldehydes (c) phenols and ketones (d) ketones and aldehydes (e) ozonides and ketones
Answer: (d) ketones and aldehydes
Explanation: Ozonolysis of alkenes involves the cleavage of the double bond to form two carbonyl compounds. Depending on the structure of the parent alkene, the products can be ketones, aldehydes, or a combination of both.

CHM121 past questions and answers 25

Question 25: The ozonolysis of an alkene yields 2 moles of ethanal as the major products; the alkene is likely to be
Options: (a) but-1-ene (b) but-2-ene (c) propene (d) 2-methylpropene
Answer: (c) propene
Explanation: The ozonolysis of propene (CH3-CH=CH2) produces 2 moles of ethanal (acetaldehyde, CH3CHO) as the major products. This occurs because the double bond in propene is cleaved to form two molecules of ethanal.

CHM121 past questions and answers 26

Question 26: The reaction of CH2=CH2 + Br2 → CH2BrCH2Br in carbon tetrachloride proceeds via_____ method
Options: (a) electrophilic addition (b) free radical addition (c) free radical substitution (d) ozonolysis
Answer: (a) electrophilic addition
Explanation: The reaction of ethene (CH2=CH2) with bromine (Br2) in carbon tetrachloride is an example of an electrophilic addition reaction. The double bond in ethene reacts with the bromine molecule, resulting in the addition of bromine atoms across the double bond.

CHM121 past questions and answers 27

Question 27: The predominant product in the reaction (CH3)2C=CH2 + HBr using Markovnikov’s rule is ____
Options: (a) CH3CH2Br (b) (CH3)CBrCH3 (c) (CH3)CBrCH3 (d) CH3CHBrCH2CH3
Answer: (b) (CH3)CBrCH3
Explanation: According to Markovnikov’s rule, in the addition of HBr to an alkene, the more substituted carbon atom gets the bromine atom while the hydrogen atom is added to the less substituted carbon. Thus, the predominant product is (CH3)CBrCH3.

CHM121 past questions and answers 28

Question 28: The major product formed when 2-methylbut-2-ene is reacted with hydrogen bromide is_____
Options: (a) 1-bromo-2-methylbutene (b) 2-bromo-3-methylbutane (c) 2-bromo-2-methylbutane (d) 3-bromo-2-methylbutane (e) 3-bromo-3-methylbutane
Answer: (c) 2-bromo-2-methylbutane
Explanation: The reaction of 2-methylbut-2-ene with HBr follows Markovnikov’s rule, where the bromine atom attaches to the more substituted carbon atom. Therefore, the major product is 2-bromo-2-methylbutane.

CHM121 past questions and answers 29

Question 29: 2-bromobutane reacts in an alcoholic solution of KOH to give mainly____
Options: (a) 1-butene (b) 2-butanol (c) 2-butene (d) 1-butanol
Answer: (c) 2-butene
Explanation: In an alcoholic solution of KOH, 2-bromobutane undergoes an elimination reaction (E2 mechanism) to produce 2-butene as the major product. This reaction favors the formation of the more substituted alkene (2-butene) over the less substituted alkene (1-butene).

CHM121 past questions and answers 30

Question 30: 1-butyne can be distinguished from 2-butyne by using____
Options: (a) Tollen’s reagent (b) potassium (c) bromine in CCl4 (d) Ammoniacal silver nitrate
Answer: (d) Ammoniacal silver nitrate
Explanation: 1-butyne and 2-butyne can be distinguished using ammoniacal silver nitrate (Tollen’s reagent). 1-butyne, an alkyne with a terminal hydrogen, will react with ammoniacal silver nitrate to form a silver acetylide precipitate, while 2-butyne will not react under these conditions.

CHM121 past questions and answers 31

31. Which of the following is used as a diagnostic test for the unit -C≡C-H?

  • Options:
    • (a) Cu(NH₃)₂OH
    • (b) BaSO₄/CaCO₃
    • (c) Butane/NaOH
    • (d) HB/H₂O₂
  • Answer: (a) Cu(NH₃)₂OH
  • Explanation: Cu(NH₃)₂OH, known as ammoniacal cuprous chloride, is used to test for terminal alkynes (-C≡C-H). When a terminal alkyne reacts with Cu(NH₃)₂OH, a red precipitate of copper acetylide is formed, indicating the presence of a terminal alkyne.

CHM121 past questions and answers 32

32. Ethylene reacts with HI to give:

  • Options:
    • (a) Iodoethane
    • (b) 2,3-Diiodoethane
    • (c) 1,1-Diiodoethane
    • (d) None
  • Answer: (a) Iodoethane
  • Explanation: Ethylene (C₂H₄) reacts with HI to give iodoethane (C₂H₅I) through an electrophilic addition reaction. The double bond of ethylene opens up and the iodine atom from HI attaches to one of the carbon atoms while the hydrogen attaches to the other, forming iodoethane.

CHM121 past questions and answers 33

33. Markownikoff’s rule is obeyed for which of the following?

  • Options:
    • (a) Symmetrical alkynes and alkenes
    • (b) Saturated hydrocarbons
    • (c) Symmetrical alkenes
    • (d) Unsymmetrical alkynes and alkenes
  • Answer: (d) Unsymmetrical alkynes and alkenes
  • Explanation: Markownikoff’s rule states that in the addition of HX to an unsymmetrical alkene or alkyne, the hydrogen atom attaches to the carbon with the greater number of hydrogen atoms (the carbon that is already more hydrogenated), and the halide (X) attaches to the carbon with fewer hydrogen atoms. This rule is applicable to unsymmetrical alkynes and alkenes.

CHM121 past questions and answers 34

34. Which of the following can be used as a diagnostic test for terminal alkynes?

  • Options:
    • (a) Lindlar catalyst
    • (b) Ammoniacal silver nitrate
    • (c) KMnO₄
    • (d) Cuprous ammonium nitrate
  • Answer: (b) Ammoniacal silver nitrate
  • Explanation: Ammoniacal silver nitrate is used to test for terminal alkynes. Terminal alkynes react with ammoniacal silver nitrate to form a white precipitate of silver acetylide, which confirms the presence of a terminal alkyne.

CHM121 past questions and answers 35

35. Alkynes are prepared from these compounds except:

  • Options:
    • (a) Alcohol
    • (b) Calcium carbide
    • (c) Gem-dihalides
    • (d) Vic-dihalides
  • Answer: (a) Alcohol
  • Explanation: Alkynes can be prepared from calcium carbide, gem-dihalides, and vic-dihalides through various chemical reactions. However, alcohols do not directly yield alkynes in common preparative reactions.

CHM121 past questions and answers 36

36. Which of the following organic compounds would you expect to observe a colour change when mixed with bromine water?

  • Options:
    • (a) Propanoic acid
    • (b) 2-Methylpropan-2-ol
    • (c) But-2-ene
    • (d) Propan-1-ol
  • Answer: (c) But-2-ene
  • Explanation: But-2-ene is an alkene and reacts with bromine water, decolorizing it due to the addition reaction that breaks the double bond. The other compounds listed are saturated or lack the necessary functional group to react with bromine water.

CHM121 past questions and answers 37

37. The product of the reaction between propyne and a mole of HBr in the presence of hydrogen peroxide is:

  • Options:
    • (a) 2-Bromopropene
    • (b) 2-Bromopropane
    • (c) 1-Bromopropene
    • (d) 1-Bromopropane
  • Answer: (c) 1-Bromopropene
  • Explanation: In the presence of hydrogen peroxide (H₂O₂), the anti-Markownikoff addition takes place, where the bromine atom adds to the carbon atom of the alkyne with fewer hydrogen atoms, resulting in the formation of 1-bromopropene.

CHM121 past questions and answers 38

38. Which of the following pairs of substances can be distinguished from each other by using Tollens silver mirror test alone?

  • Options:
    • (a) Propanal and Butanone
    • (b) Butanone and Propanoic acid
    • (c) Propanal and Propan-1-ol
    • (d) Butanone and Pentan-2-one
  • Answer: (a) Propanal and Butanone
  • Explanation: Tollens’ reagent (ammoniacal silver nitrate) is a test for aldehydes. Propanal, being an aldehyde, will reduce Tollens’ reagent to metallic silver, forming a silver mirror. Butanone, a ketone, does not react with Tollens’ reagent.

CHM121 past questions and answers 39

39. An unknown liquid Z was oxidized in the presence of excess sodium dichromate to produce liquid Y, which shows positive to the iodoform test. Z and Y are likely:

  • Options:
    • (a) Ethanol and Ethanoic acid
    • (b) 2-Propanol and Acetone
    • (c) Propan-1-ol and Propanoic acid
    • (d) 3-Pentanol and 3-Pentanone
  • Answer: (b) 2-Propanol and Acetone
  • Explanation: 2-Propanol can be oxidized to acetone, and acetone gives a positive iodoform test, indicating the presence of a methyl ketone.

CHM121 past questions and answers 40

40. Which of the following alkali metals react vigorously with water, forms nitrides M₃N on reaction with nitrogen, and forms only the normal oxide M₂O on heating in air?

  • Options:
    • (a) Li
    • (b) Na
    • (c) K
    • (d) Cs
  • Answer: (a) Li
  • Explanation: Lithium reacts vigorously with water, forms lithium nitride (Li₃N) on reaction with nitrogen, and forms only lithium oxide (Li₂O) when heated in air. Other alkali metals do not form nitrides as readily as lithium.

CHM121 past questions and answers 41

41. Of the following oxides, which is the most basic?

  • Options:
    • (a) MgO
    • (b) Na₂O
    • (c) P₂O₃
    • (d) BeO
    • (e) SO₂
  • Answer: (b) Na₂O
  • Explanation: Sodium oxide (Na₂O) is the most basic of the given oxides. It is a strong base and reacts with water to form sodium hydroxide (NaOH). MgO is less basic, BeO is amphoteric, P₂O₃ is acidic, and SO₂ is also acidic.

CHM121 past questions and answers 42

42. Halogen atoms in the ground state have____ number of electrons in their p-orbital.

  • Options:
    • (a) 5
    • (b) 6
    • (c) 7
    • (d) 8
    • (e) 2
  • Answer: (b) 6
  • Explanation: Halogen atoms in the ground state have 6 electrons in their p-orbital. Halogens belong to Group 17 of the periodic table, where the p-orbital is almost filled with 6 electrons, needing one more to complete the octet.

CHM121 past questions and answers 43

43. The oxidation number of oxygen in alkaline earth metal superoxides is:

  • Options:
    • (a) -2
    • (b) -1
    • (c) -1/2
    • (d) +1/2
    • (e) 2
  • Answer: (c) -1/2
  • Explanation: In superoxides, the oxidation state of oxygen is -1/2. This is because in superoxides, the oxygen atoms are in the O₂⁻ ion form, where the two oxygen atoms share the -1 charge, resulting in an oxidation state of -1/2 for each oxygen atom.

CHM121 past questions and answers 44

44. The oxidation state of oxygen in alkali metal superoxides is:

  • Options:
    • (a) -2
    • (b) -1
    • (c) -1/2
    • (d) +1/2
    • (e) +1
  • Answer: (c) -1/2
  • Explanation: In superoxides (e.g., KO₂), oxygen exists as the O₂⁻ ion. The charge on this ion is -1, but since there are two oxygen atoms sharing this charge, the oxidation state of each oxygen atom is -1/2.

CHM121 past questions and answers 45

45. Na₂O, Al₂O₃, SiO₂, and SO₃ when arranged in order of increasing basicity gives:

  • Options:
    • (a) SO₃<SiO₂ <Al₂O₃ <Na₂O
    • (b) Na₂O <Al₂O₃ <SO₃ <SiO₂
    • (c) Al₂O₃ <SO₃ <Na₂O <SiO₂
  • Answer: (a) SO₃<SiO₂ <Al₂O₃ <Na₂O
  • Explanation: Basicity decreases as you move from left to right across the periodic table. Na₂O (a strong base) > Al₂O₃ (amphoteric) > SiO₂ (weak acid) > SO₃ (strong acid). Therefore, the order of increasing basicity is <SiO₂ <Al₂O₃ <Na₂O.

CHM121 past questions and answers 46

46. Among pure metals, ______ has the highest electrical and thermal conductivity:

  • Options:
    • (a) Au
    • (b) Ag
    • (c) Cu
    • (d) Al
    • (e) Fe
  • Answer: (b) Ag
  • Explanation: Silver (Ag) has the highest electrical and thermal conductivity among all metals, making it an excellent conductor.

CHM121 past questions and answers 47

47. One of the following exhibits isomerism:

  • Options:
    • (a) Propane
    • (b) Propene
    • (c) Butane
    • (d) Benzene
  • Answer: (c) Butane
  • Explanation: Butane exhibits structural isomerism because it can exist as two different structures: n-butane and isobutane (methylpropane).

CHM121 past questions and answers 48

48. For a carbon atom to exhibit chirality, what feature must it possess?

  • Options:
    • (a) Multiple bonds attached to it
    • (b) A non-rotating C=C bond
    • (c) Have four different groups attached to it
    • (d) Colour
  • Answer: (c) Have four different groups attached to it
  • Explanation: A carbon atom is chiral if it is attached to four different groups, creating a non-superimposable mirror image, also known as an enantiomer.

CHM121 past questions and answers 49

49. The main classes of reagents commonly encountered in organic reactions are:

  • Options:
    • (a) Electrophiles, positrons, and free radicals
    • (b) Nucleophiles, positrons, and electrophiles
    • (c) Electrophiles, nucleophiles, and free radicals
    • (d) Electrophile, neutrons, and nucleophiles
    • (e) Nucleophiles, electrophiles, and carbon atom
  • Answer: (c) Electrophiles, nucleophiles, and free radicals
  • Explanation: The main types of reagents in organic chemistry are electrophiles (electron-loving species), nucleophiles (nucleus-loving species), and free radicals (species with unpaired electrons).

CHM121 past questions and answers 50

50. In alkali metals, the monoxide, peroxide, and superoxide ions are _____________

  • Answer:
  • Explanation: The oxides, peroxides, and superoxides of alkali metals are different forms of oxygen compounds. Monoxides contain O²⁻ ions, peroxides contain O₂²⁻ ions, and superoxides contain O₂⁻ ions.

CHM121 past questions and answers 51

51. The Solvay process is applied to the production of:

  • Options:
    • (a) Na₂CO₃
    • (b) NaHCO₃
    • (c) MgCO₃
    • (d) CaCO₃
  • Answer: (a) Na₂CO₃
  • Explanation: The Solvay process is an industrial method for producing sodium carbonate (Na₂CO₃) from sodium chloride, ammonia, and carbon dioxide.

CHM121 past questions and answers 52

52. Sodium reacts with a sufficient amount of oxygen in the air to produce:

  • Options:
    • (a) NaO
    • (b) Na₂O
    • (c) Na₂O₂
    • (d) NaO₂
  • Answer: (c) Na₂O₂
  • Explanation: Sodium reacts with oxygen to form sodium peroxide (Na₂O₂) when exposed to a sufficient amount of oxygen in the air.

CHM121 past questions and answers 53

53. In organic chemistry, electron-rich species are referred to as:

  • Options:
    • (a) Free radicals
    • (b) Electrophiles
    • (c) Cations
    • (d) Anions
    • (e) Nucleophiles
  • Answer: (e) Nucleophiles
  • Explanation: Nucleophiles are electron-rich species that are attracted to positively charged or electron-deficient centers (electrophiles) in a chemical reaction.

CHM121 past questions and answers 54

54. Which of the following is not a nucleophile?

  • Options:
    • (a) BF₃
    • (b) CH₃OH
    • (c) H₂O
    • (d) NH₃
    • (e) OH⁻
  • Answer: (a) BF₃
  • Explanation: BF₃ (boron trifluoride) is an electrophile, not a nucleophile. It is electron-deficient and seeks out electron-rich species.

CHM121 past questions and answers 55

55. Which of the following is not an electrophile?

  • Options:
    • (a) CH₃⁺
    • (b) Cl⁺
    • (c) BF₃
    • (d) NH₃
    • (e) FeBr₃
  • Answer: (d) NH₃
  • Explanation: NH₃ (ammonia) is a nucleophile because it has a lone pair of electrons that it can donate. The others (CH₃⁺, Cl⁺, BF₃, FeBr₃) are electrophiles, seeking electrons.

CHM121 past questions and answers 56

56. Which of the following is not true for the group 1A elements?

  • Options:
    • (a) Most of them are soft, silvery corrosive metals
    • (b) Their atomic radii increase with increasing molecular weight
    • (c) They are named the alkaline earth metals
    • (d) They are excellent conductors of heat and electricity
    • (e) They exhibit a +1 oxidation state in compounds
  • Answer: (c) They are named the alkaline earth metals
  • Explanation: Group 1A elements are called alkali metals, not alkaline earth metals (which are Group 2A elements). Alkali metals are soft, silvery metals, conduct electricity well, and always exhibit a +1 oxidation state.

CHM121 past questions and answers 57

57. Which element group is the most reactive of all the metallic elements?

  • Options:
    • (a) Alkali metals
    • (b) Alkaline earth metals
    • (c) Coinage metals
    • (d) Group 2B metals
  • Answer: (a) Alkali metals
  • Explanation: Alkali metals (Group 1A) are the most reactive metallic elements due to their single valence electron, which is easily lost to form positive ions.

CHM121 past questions and answers 58

58. In a surprisingly large number of their properties, Beryllium resembles Aluminium, and Boron resembles Silicon. Such a relationship is called:

  • Options:
    • (a) Diagonal relationship
    • (b) Periodic law
    • (c) Amphoterism
  • Answer: (a) Diagonal relationship
  • Explanation: The diagonal relationship refers to similarities between elements diagonally across the periodic table, such as Beryllium (Be) with Aluminium (Al) and Boron (B) with Silicon (Si).

CHM121 past questions and answers 59

59. Which of the following properties of the alkaline earth metals decreases with increasing atomic weight?

  • Options:
    • (a) Ionic radii
    • (b) Ionization energy
    • (c) Atomic radii
    • (d) Activity
    • (e) Atomic number
  • Answer: (b) Ionization energy
  • Explanation: As the atomic weight of alkaline earth metals increases, the ionization energy decreases because the valence electrons are farther from the nucleus and more easily removed.

CHM121 past questions and answers 60

60. The most abundant metal on the earth’s crust is:

  • Options:
    • (a) Cu
    • (b) Fe
    • (c) Na
    • (d) Al
  • Answer: (d) Al
  • Explanation: Aluminium (Al) is the most abundant metal in the Earth’s crust, comprising about 8.23% by weight. It is found primarily in the mineral bauxite.

CHM121 past questions and answers 61

61. Which of the following is the mineral of magnesium?

  • Options:
    • (a) Magnesite
    • (b) Magnetite
    • (c) Olivine
    • (d) Carnallite
  • Answer: (a) Magnesite
  • Explanation: Magnesite (MgCO₃) is a primary mineral source of magnesium. Magnetite and olivine are minerals, but they do not primarily contain magnesium, while carnallite (KCl·MgCl₂·6H₂O) is also a source of magnesium.

CHM121 past questions and answers 62

62. Carbonates of lithium are not stable like those of sodium due to:

  • Options:
    • (a) Low electronegativity
    • (b) Low electropositivity
    • (c) Low charge density
    • (d) All of the above
  • Answer: (c) Low charge density
  • Explanation: Lithium carbonate (Li₂CO₃) is less stable than sodium carbonate (Na₂CO₃) due to lithium’s smaller ionic size, which gives it a high charge density, making its carbonate less stable and more prone to decomposition.

CHM121 past questions and answers 63

63. The following is not a criterion for the purity of an organic compound:

  • Options:
    • (a) Sharp melting point for solids
    • (b) Sharp boiling point
    • (c) Distinctive color
    • (d) Distinctive odor
    • (e) Mass of substance
  • Answer: (e) Mass of substance
  • Explanation: The mass of a substance is not a criterion for purity. Purity is often determined by sharp melting or boiling points and sometimes by color and odor, though these are less reliable.

CHM121 past questions and answers 64

64. Which of the following is not a method of isolation of a compound?

  • Options:
    • (a) Sublimation
    • (b) Chromatography
    • (c) Boiling
    • (d) Fractional distillation
    • (e) Solvent extraction
  • Answer: (c) Boiling
  • Explanation: Boiling is a physical process to vaporize liquids but is not a method of isolation. Methods like sublimation, chromatography, fractional distillation, and solvent extraction are used to isolate and purify compounds.

CHM121 past questions and answers 65

65. The structural determination of an organic compound involves:

  • Options:
    • (a) Observation of physical properties
    • (b) Melting point analysis
    • (c) Boiling point analysis
    • (d) Isolation and purification
  • Answer: (d) Isolation and purification
  • Explanation: Structural determination involves isolating and purifying the compound, followed by analysis using techniques such as spectroscopy, chromatography, and X-ray crystallography to determine the structure.

CHM121 past questions and answers 66

66. Miscible liquids with large differences in boiling point can be separated by a technique called:

  • Options:
    • (a) Decantation
    • (b) Simple distillation
    • (c) Fractional crystallization
    • (d) Solvent extraction
  • Answer: (b) Simple distillation
  • Explanation: Simple distillation is used to separate miscible liquids that have significantly different boiling points. It works by heating the liquid mixture, causing the more volatile component (lower boiling point) to vaporize and then condense into a separate container.

CHM121 past questions and answers 67

67. The addition of HBr to alkenes and alkynes follows which reaction mechanism?

  • Options:
    • (a) Free radical substitution
    • (b) Nucleophilic substitution
    • (c) Electrophilic substitution
    • (d) Double displacement
  • Answer: (c) Electrophilic substitution
  • Explanation: The addition of HBr to alkenes and alkynes follows an electrophilic addition mechanism, where the π electrons of the double or triple bond attack the electrophilic hydrogen atom in HBr, leading to the formation of a carbocation intermediate.

CHM121 past questions and answers 68

68. Which of the following products is not obtained when methane undergoes chlorination in the presence of UV light?

  • Options:
    • (a) Chloroform
    • (b) Carbon tetrachloride
    • (c) Sand
    • (d) Dichloromethane
  • Answer: (c) Sand
  • Explanation: Chlorination of methane under UV light leads to the formation of chlorinated products such as chloromethane, dichloromethane, chloroform, and carbon tetrachloride. Sand is not a product of this reaction.

CHM121 past questions and answers 69

69. One of these is not a method of preparing alkanes:

  • Options:
    • (a) Rearrangement reaction
    • (b) From Grignard reagent
    • (c) Wurtz reaction
    • (d) Reduction of unsaturated HC
    • (e) Reduction of alkyl halides
  • Answer: (a) Rearrangement reaction
  • Explanation: Alkanes are not typically prepared through rearrangement reactions. Common methods for preparing alkanes include the use of Grignard reagents, the Wurtz reaction, and the reduction of unsaturated hydrocarbons and alkyl halides.

CHM121 past questions and answers 70

70. The fraction obtained during fractional distillation of petroleum containing C15–C20 is used for:

  • Options:
    • (a) Surface roads
    • (b) Motor car fuels
    • (c) Fuel for trucks and lorries
    • (d) Lubricating oil
  • Answer: (c) Fuel for trucks and lorries
  • Explanation: The fraction of petroleum with C15–C20 hydrocarbons is used primarily as diesel fuel, which is commonly used for trucks and lorries.

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