Category: Class 12 Chemistry Objective Questions

Chemistry Online Quiz for Schools on “Colligative Properties and Determination of Molar Mass – 2”.

1. Why is ‘raising of viscosity’ of a solution after addition of solute, not considered to be a colligative property?
a) The resultant viscosity depends on the nature of the solute
b) The resultant viscosity depends on the amount of solute
c) The resultant viscosity depends on the nature of solvent
d) The resultant viscosity depends on the amount of solvent
Answer: a
Clarification: A colligative property is identified by the fact that it has no dependence on the nature of the particles of solute. However, in the case of viscosity it really depends on the solute that is added to the solvent. Thus, the change in viscosity after addition of a non-volatile solute cannot be considered to be a colligative property.

2. At 70°C the vapor pressure of pure water is 31 kPa. Which of the following is most likely the vapor pressure of a 2.0 molal aq. glucose solution at 70°C?
a) 30.001 kPa
b) 29.915 kPa
c) 28.226 kPa
d) 32.392 kPa
Answer: b
Clarification: Given, P⁰_water = 31 kPa
Concentration of solution, c = 2 molal = 2 moles of glucose/kg of water
From law of relative lowering of vapor pressure, ΔP/P⁰ = X₂, where X₂ is the mole fraction of glucose in the solution.
Mass of water = 1 kg = 1000 g
Molecular weight of water = 18 g/mole
Moles of water = 1000/18 = 55.556 moles
X₂ = 2/(2 + 55.556) = 0.035
ΔP = 31 kPa x 0.035 = 1.085 kPa
Final pressure = 31 kPa – 1.085 kPa = 29.915 kPa.

3. 117 g of NaCl is added to 222 g of water in a saucepan. At what does temperature does water boil at 101.325 kPa? Ebullioscopy constant for water = 0.52 K kg mol^-1 and b.p. = 100°C
a) 98.3°C
b) 102.8°C
c) 104.7°C
d) 101.5°C
Answer: c
Clarification: Given,
Weight of solvent, w₁ = 222 g
Weight of solute, w₂ = 117 g
K_b = 0.53 K kg mol^-1
Now, addition of a non-volatile solute causes elevation in boiling point, ΔT_b
ΔT_b = (k_b x 1000 x w₂)/(M₂ x w₁)
On substituting, ΔT_b = (0.52 x 1000 x 117)/(58.5 x 222) = 4.7°C
New boiling temperature = 100 + 4.7 = 104.7°C.

4. Boiling point of chloroform is 61°C. After addition of 5.0 g of a non-volatile solute to 20 g chloroform boils at 64.63°C. If k_b = 3.63 K kg mol^-1, what is the molecular weight of the solute?
a) 320 g/mol
b) 100 g/mol
c) 400 g/mol
d) 250 g/mol
Answer: d
Clarification: Given,
b.p. of chloroform = 61°C
New b.p. after addition = 64.63°C
Mass of solute, w₂ = 5.0 g
Mass of solvent, w₁ = 20 g
K_b = 3.63 K kg mol^-1
From these, ∆T_b = 64.63 – 61 = 3.63°C
Using ΔT_b = (k_b x 1000 x w₂)/(M₂ x w₁)
M₂ = (k_b x 1000 x w₂)/(ΔT_b x w₁)
M₂ = (3.63 x 1000 x 5)/(3.63 x 20) = 250 g/mol.

5. Pure CS₂ melts at -112°C. 228 grams of propylene glycol crystals is mixed with 500 grams of CS₂. If k_f of CS₂ = -3.83 K kg mol^-1 what is the depression in freezing point?
a) -23°C
b) -135°C
c) -20°C
d) -100°C
Answer: a
Clarification: Given,
k_f = -3.83 k kg mol^-1
Mass of solute, w₂ = 228 g
Mass of solvent, w₁ = 500 g
Molar mass of solute, M₂ = 76 g/mole
Moles of solute = w₂/M₂ = 228/76 = 3 moles
Molality of the solution, m = Number of moles of solute/mass of solvent (kg)
m = 3 moles/0.5 kg = 6 molal
We know, ΔT_f = k_f x m
ΔT_f = -3.83 x 6 = -23°C.

6. What are colligative properties useful for?
a) Determining boiling and melting temperature
b) Determining molar mass
c) Determining equivalent weight
d) Determining van’t Hoff factor
Answer: b
Clarification: Colligative properties serve the purpose for determining molar masses of unknown compounds. Colligative properties are not used to determine boiling and melting temperatures as it would result in an incorrect value upon the addition of a solute. Equivalent weight can only be determined if the molar mass is known and the van’t Hoff factor is determined in a similar manner.

7. A pair of solution bears the same osmotic pressure. What is this pair of solutions called?
a) Hypertonic
b) Hypotonic
c) Isotonic
d) Osmolarity
Answer: c
Clarification: Two solutions having the same osmotic pressure at a given temperature are called isotonic solutions. Hypertonic solutions are those in which the concentration of solute outside the membrane and a lesser concentration within the membrane. Osmolarity is a type of concentration expressed in number of solute particles per liter.

8. A cell with lots of salt inside it is placed in a vessel containing just water. Which process takes place?
a) Dialysis
b) Filtration
c) Shriveling
d) Osmosis
Answer: d
Clarification: Osmosis is a mass transfer process due to which water molecules move from a region of higher water potential to lower water potential down the potential gradient. Dialysis and filtration are processes using the concept of diffusion of impurities. Shriveling is the shrinking of a cell. In this case, the cell swells up.

9. What is a necessary condition for osmosis to take place?
a) Semi-permeable membrane
b) Same concentration of solvent
c) High temperature
d) Pressure greater than osmotic pressure
Answer: a
Clarification: A semi permeable membrane is a must condition since it facilitates the blocking of solute particles from diffusing through and allows only water molecules to pass through. Obviously, water is the solvent in osmosis hence the concentration of solvent cannot be same if osmosis has to occur. If pressure greater than osmotic pressure is applied, reverse osmosis takes place. Meaning, water molecules will flow from region of lower water potential to higher water potential.

10. Which is the most appropriate method for determining the molar masses of biomolecules?
a) Relative lowering of vapor pressure
b) Elevation of boiling point
c) Depression in freezing point
d) Osmosis
Answer: d
Clarification: Osmotic pressure method has the greatest advantage over other methods that even for very dilute concentrations it gives a large magnitude. This would not be true in case other methods. It is highly useful for biomolecules since they are unstable at extremely high and low temperatures thus eradicating the method using boiling and freezing points.

Chemistry Online Quiz for Schools,

Chemistry Multiple Choice Questions on “Surface Chemistry – Adsorption”.

1. A finely divided substance is more effective as an adsorbent.
a) True
b) False
Answer: a
Clarification: Adsorption is a surface phenomenon. It is dependent on the amount of the adsorbent exposed to the adsorbate. A finely divided substance provides a large surface area and hence, provides more sites where adsorption can take place.

2. What is the process called when the molecules of a substance are retained at the surface of a solid or a liquid?
a) Absorption
b) Adsorption
c) Sorption
d) Desorption
Answer: b
Clarification: Adsorption is the process which involves the accumulation of the molecules of a substance in higher concentration on the surface of a solid or a liquid. For example, gasses are adsorbed on the surface of charcoal.

3. Which of the following forces is involved in physical adsorption?
a) Gravitational force
b) Magnetic force
c) Van der Waals force
d) Electromagnetic force
Answer: c
Clarification: In physical adsorption, the molecules of the adsorbate stick to the surface of the adsorbent due to very weak forces which is known as Van der Waals force. Van der Waals force is similar to the forces that cause condensation of gas into liquid.

4. In physisorption, the adsorbent does not show specificity towards a particular gas.
a) True
b) False
Answer: a
Clarification: In physisorption, there is no chemical bond between the adsorbent and adsorbate. They are held together by Van der Waals forces. Since Van der Waals forces are universal and almost same for all gases, the adsorbent does not show specificity to a particular gas.

5. Which of the following statements is true with respect to the extent of physisorption?
a) Increases with increase in temperature
b) Decreases with increase in surface area
c) Decreases with increase in the strength of Van der Waals forces
d) Decreases with increase in temperature
Answer: d
Clarification: Physisorption is an exothermic process. According to Le-Chatelier’s principle, an exothermic reaction is favoured by a decrease in temperature. Therefore, the extent of physisorption decreases on increasing temperature.

6. Which of the following can result in a transition from physisorption to chemisorption?
a) Decrease in temperature
b) Increase in temperature
c) Decrease in pressure
d) Increase in surface area
Answer: b
Clarification: Physisorption occurs due to Van der Waals forces. On increasing the temperature, the adsorbate can split into atoms and form chemical bonds with the adsorbent and hence, cause chemisorption.

7. Which of the following statements is incorrect with respect to physisorption?
a) It is reversible
b) It is spontaneous
c) ΔH d) ΔS > 0
Answer: d
Clarification: Physisorption is reversible as the molecules of the adsorbate are held to the adsorbent by weak Van der Waals forces which can be broken easily. Physisorption is exothermic and so, ΔH is negative. Adsorption is a spontaneous process. In adsorption, the movement of the adsorbed molecule is restricted. As a result the entropy change (ΔS) is negative.

8. Which of the following statements is true with respect to the types of adsorption?
a) Chemisorption is stronger than physisorption
b) Physisorption is stronger than chemisorption
c) They are both equal
d) They cannot be compared
Answer: a
Clarification: Chemisorption is stronger than physisorption. Chemisorption involves the forming of a chemical bond between the adsorbent and the adsorbate whereas in physisorption, the molecules are held together by weak Van der Waal’s forces. Therefore, chemisorption is stronger.

9. Which of the following is an example of sorption?
a) Sponge in water
b) Cotton dipped in ink
c) Water on silica gel
d) Oxygen on metal surface
Answer: b
Clarification: Sorption refers to the process where adsorption and absorption occur at the same time. Cotton dipped in ink is one of the cases where both adsorption and absorption, that is, sorption occurs.

10. Which of the following statements is not true with respect to chemisorption?
a) Depends on nature of adsorbate and adsorbent
b) Has a large heat of adsorption
c) Forms a unimolecular layer
d) Occurs at low temperature
Answer: d
Clarification: Chemisorption is a kind of adsorption that involves a chemical reaction between the adsorbent and adsorbate, resulting in the formation of a chemical bond between the two. Since a bond is to be formed, chemisorption is highly dependent on the reactants(adsorbate and adsorbent) and has a large heat of adsorption. Also, since it involves bond formation, it cannot be multi-layered. Chemisorption is favoured to occur at high temperature.

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

1. What is the chemical formula of ammonia?
a) NH₂
b) NH₃
c) NH₄⁺
d) NH₅
View Answer

Answer: a
Clarification: Ammonia is a tri-hydride of nitrogen. It is a colorless gas with a highly pungent smell that causes nasal irritation. It is the most stable hydride of any element in group 15.

2. What kind of smell is ammonia recognized by?
a) Acidic
b) Sweet
c) Rotten
d) Pungent

Answer: d
Clarification: Ammonia has a highly pungent smell. With a sharp, pinching smell, it causes severe irritation in the nose and throat. This pungent smell of ammonia is used as a means to identify the presence of ammonium cation, NH₄⁺ in salt analysis.

3. How many unshared pair of electrons does an ammonia molecule have?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: The nitrogen atom in NH₃ (ammonia) bears one lone pair of electron. Each single electron of hydrogen is bonded to the nitrogen atom. Three of out five electrons of nitrogen are involved in bonding whereas the unbounded two electrons make up the single unshared pair of electron.

4. What happens when sodium is put in a solution of ammonia?
a) It generates a lot of heat
b) It does not dissolve
c) It produces deep blue color
d) Ammonia liquid evaporates due to heat
Answer: c
Clarification: When sodium is introduced to liquid ammonia, it separates into sodium ion and a single electron (Na → Na⁺ + e^–). The single electron produces the deep blue color due to solvation effect producing e^–[NH₃]_n.

5. What is one method of qualitatively analyzing a given salt for presence of ammonia?
a) Solution turns blue litmus red
b) Heating the salt causing decrepitation
c) Using a reagent to obtain dirty brown precipitate
d) Addition of NaOH causing white gelatinous precipitate
Answer: c
Clarification: Typically, Nessler’s reagent is used to test for ammonia. Nessler’s reagent is an alkaline solution of potassium tetraiodomercurate (II). When it combines with ammonia it gives dirty brown precipitate due to the formation of [OHg₂.NH₂]I. Another method is to add a strong base and heat the mixture. The gas given off is pungent and turns red litmus blue.

6. What is the most acidic of all?
a) NH₃
b) NaOH
c) KOH
d) Alkaline KMnO₄
Answer: a
Clarification: In the listed options, except ammonia, every other compound is a strong base which automatically makes ammonia the most acidic relative to the mentioned compounds. Although it is a weak base, compared to NaOH, KOH and alkaline KMnO₄ it is the most acidic amongst the four options.

7. In the Haber-Bosch process, what is formed by the reaction of natural gas and steam?
a) Ammonia
b) Nitrogen
c) Oxygen
d) Hydrogen
View Answer

Answer: d
Clarification: The reaction of natural gas and steam is known as the steam reformation of methane. It is an utmost important step in the Bosch-Haber (industrial process to produce ammonia), which produces hydrogen gas, H₂, one of the two reactants alongside nitrogen to produce ammonia in presence of iron catalyst.

8. Which of the following conditions would improve the yield of ammonia production from Bosch-Haber process?
a) High temperature, high pressure
b) High temperature, low pressure
c) Low temperature, low pressure
d) Low temperature, high pressure
Answer: d
Clarification: Following the Le Chatelier’s principle, since the forward reaction is exothermic (i.e. N₂ + 3H₂ → 2NH₃), low temperatures would increase the rate of reaction in turn producing more ammonia. In addition to this, since the total number of molecules on the product side lesser, high pressures will increase the rate of reaction.

9. Nitrogen in plants is taken in what form?
a) Ammonia
b) Amide
c) Nitrate
d) Nitrite
Answer: c
Clarification: Plants take up nitrogen predominantly in the form of nitrate ions, along side ammonium ions. This is made available to them through fertilizers, whose typical examples usually include Ammonium Phosphate, calcium ammonium nitrate.

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

1. What is the molecular mass of chlorine?
a) 71.0 kg/kmole
b) 35.5 g/kmole
c) 71.0 g/kmole
d) 35.5 kg/mole
Answer: a
Clarification: In the periodic table, the atomic mass of chlorine atom is 35.5 AMU. Since chlorine exists in diatomic form, the molecular mass = 35.5g/mole x 2 = 71.0 g/mole. On multiplying and dividing by 1000, 71.0 kg/kmole.

2. Which of the following would instantly react with chlorine?
a) Copper
b) Magnesium
c) Iron
d) Aluminum
Answer: b
Clarification: Out of the four metals given, magnesium will readily lose electrons due to its relatively higher reactivity. Copper and iron are not affected by gaseous or liquid chlorine. Lastly, aluminum forms a tough, passive oxide layer which prevents it from reacting with external reagents.

3. What is the color of chlorine?
a) Yellow grey
b) Greenish brown
c) Green yellow
d) Yellow
Answer: c
Clarification: Every halogen bears a characteristic color which is used in identifying it. Fluorine is a bright yellow gas. Chlorine is greenish yellow, followed by bromine which is brown and iodine which is purple/violet. Astatine is a black solid which is radioactive in nature.

4. When excess chlorine reacts with methane in presence of sunlight, which is the major product?
a) Refrigerant – 40
b) Methylene chloride
c) Chloroform
d) Carbon tetrachloride
Answer: d
Clarification: In presence of sunlight, a free radical substitution reaction takes place between the reacting methane and chlorine molecules where in each elementary reaction a hydrogen atom is replaced by a chlorine atom in the new product. First, methane converts to 1-chloromethane which in turn produces dichloromethane followed by tri-chloromethane and tetra-chloromethane, respectively.

5. Which of the following is a suitable test to identify presence of chlorine atoms or ions?
a) Acidify the given compound and add silver nitrate
b) Add silver nitrate followed by warming the solution
c) Heat the compound strongly
d) Smell the compound
Answer: a
Clarification: The standard, universal qualitative test for detecting presence of chloride is carried out by acidifying the solution and then adding silver nitrate. If chloride ions are present, a thick white ppt. is produced.

6. Why is chlorine bubbled through water during the latter’s treatment?
a) To remove foul odor
b) To kill microorganisms
c) To improve taste
d) To remove hardness
Answer: b
Clarification: Chlorine is a very popular disinfecting agent and is widely used in containers to kill microorganisms. In water treatment plants, chlorine gas may be bubbled through to remove trace quantities of microorganisms.

7. What is the electronic configuration of chloride?
a) 1s²2s²2p⁶3s²3p⁵
b) 1s²2s²2p⁶3s²3p⁶
c) 1s²2s²2p⁵
d) [Ar]4s²4p³
View Answer

Answer: b
Clarification: Chlorine has the atomic number 17. However, chloride has 18 electrons. Consequently, it attains the electronic configuration of argon i.e.1s²2s²2p⁶3s²3p⁶.

8. How is chlorine gas manufactured industrially?
a) Deacon process
b) Ostwald process
c) Chlor-alkali process
d) Haber process
Answer: c
Clarification: Although Deacon process involves release of chlorine gas as a byproduct, the chief purpose of the deacon process is used to manufacture alkalis. Chlorine is best produced by the chloralkali process, which is the electrolysis of brine, producing hydrogen and chlorine gases as by products.

9. Hydrogen chloride as an acid is much stronger than hydrogen fluoride under the same conditions. True or False?
a) True
b) False
Answer: a
Clarification: Fluorine atom is extremely small compared to chlorine. This means that the bond of H – F is much more polar than that of H – Cl, causing the H atom to be more tightly bonded in case of H – F than in H – Cl. Since hydrogen chloride is able to release the H+ much easily, the latter is a stronger acid.

10. Which of the following is a result of bubbling hydrogen chloride gas through gas?
a) H^– + Cl^–
b) H⁺ + Cl^2-
c) H₃O⁺ + Cl^–
d) H⁺ + Cl^–
Answer: c
Clarification: The hydrogen chloride molecule, HCl dissociates in water giving rise to one hydrogen ion and a chloride ion. The extremely small hydrogen ion bearing the positive charge is instantly attracted by the unshared pair of electrons of the oxygen atom in water molecules, consequently giving rise to the hydronium ion, H₃O⁺.

11. What will be the result of subjecting a blue litmus to chlorine gas?
a) Turn to red only
b) Remain blue
c) Turn to white finally
d) Turn white in the first go
Answer: c
Clarification: Chlorine bears bleaching properties. When blue litmus is subjected to chlorine gas, it first turns red (due to the acidic nature) then it turns white (due to chlorine’s ability to bleach materials).

12. To which group does chlorine element belong?
a) Halides
b) Pre inert
c) Chalcogens
d) Halogens
Answer: d
Clarification: Fluorine, chlorine, bromine, iodine and astatine belong to Group 17 of the periodic table. This group is commonly named as Halogens because of their ability to produce salt on reacting with metals.

13. Which of the following is the displays the most basic nature in water?
a) Fluoride
b) Chloride
c) Bromide
d) Iodide
Answer: a
Clarification: The fluoride ion is actually not an oxidizing agent. Due to its high electronegativity and high reactivity, it converts water to form hydroxide ions. The reaction is given by 2F^– + H₂O → H – F – H^– + OH^–.

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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.