Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Electrochemistry – Batteries”.

1. A battery is an arrangement of electrolytic cells.
a) True
b) False
Answer: b
Clarification: A battery is not an arrangement of electrolytic cells, but an arrangement of electrochemical cells. An electrochemical cell is one which converts chemical energy into electrical energy whereas an electrolytic cell is one which converts electrical energy into chemical energy. Since batteries convert chemical energy to electrical energy, it is an arrangement of electrochemical cells.

2. Which of the following is not a requirement for a useful battery?
a) It should be light and compact
b) It should have a reasonable life span
c) It should ideally have a constant voltage throughout its lifespan
d) It should supply Alternating Current(AC)
Answer: d
Clarification: A useful battery is expected to be light and compact to be easily transported. It is expected to have a reasonable lifespan to justify its usage. Its voltage should not vary appreciably during usage so that it doesn’t adversely affect the circuit it is used in. A battery supplies Direct Current(DC) and not Alternating Current(AC).

3. Which of the following statements is true regarding a primary cell?
a) The electrode reactions can be reversed
b) It can be recharged
c) An example of a primary cell is a mercury cell
d) An example of a primary cell is a nickel-cadmium storage cell
Answer: c
Clarification: A primary cell is one in which the electrode reactions occur only once and cannot be reversed by applying electrical energy. Therefore, primary cells cannot be recharged. A mercury cell is an example of a primary cell, whereas a nickel-cadmium storage cell is an example of a secondary cell.

4. Secondary cells are also called storage cells.
a) True
b) False
Answer: a
Clarification: Secondary cells are those cells in which the electrode reaction can be reversed by applying an electrical energy. Therefore, they can be used to store electrical energy. So, they are also known as storage cells.

5. Which of the following is used as an anode in a dry cell?
a) Zinc
b) Graphite
c) Mercury(II) oxide
d) Nickel
Answer: c
Clarification: A dry cell is constructed using zinc and graphite. It consists of a zinc cylinder through whose centre passes a graphite rod. The zinc cylinder acts as an anode, whereas the graphite rod acts as a cathode.

6. Why do leak proof dry cells have an iron or steel sheet covering the zinc cylinder?
a) It increases the potential difference between the anode and cathode
b) It acts as a barrier around the zinc cylinder which can develop holes during use
c) It makes it waterproof
d) It prevents the leakage of current
Answer: b
Clarification: In a dry cell, zinc loses electrons and the zinc ions dissolve into the electrolyte. As a result, the zinc cylinder of the dry cell develops holes as it is used. To prevent the leakage of electrolyte through these holes, an iron or steel sheet is used to cover the cylinder.

7. Which of the following is the electrolyte used in a dry cell?
a) Ammonium chloride
b) Manganese dioxide
c) Potassium hydroxide
d) Sulphuric acid
Answer: a
Clarification: The electrolyte in a dry cell is ammonium chloride in the form of a moist paste placed next to the zinc anode. In some dry cells marketed as “heavy-duty”, the ammonium chloride is replaced by zinc chloride.

8. What is the role of manganese dioxide in a dry cell?
a) It acts as an electrolyte
b) It acts as the cathode
c) It acts as an anode
d) It acts as a depolariser
Answer: d
Clarification: In a dry cell, in the remaining space between the electrolyte and the graphite cathode, a second paste consisting of ammonium chloride and manganese dioxide is applied. The manganese dioxide acts as a depolariser as it helps to prevent the build-up of hydrogen gas bubbles.

9. What is the final oxidation state of manganese after the electrochemical reactions in a dry cell?
a) +4
b) +3
c) +2
d) +1
Answer: b
Clarification: In a dry cell, in the cathode reaction, manganese dioxide(MnO₂) is reduced to form manganese oxide-hydroxide(MnO(OH)). In this process, the oxidation state of manganese changes from +4 to +3. Hence the final oxidation state of manganese is +3.

10. Which of the following scientists invented the first dry cell?
a) Carl Gassner
b) Nikola Tesla
c) Antione Lavoisier
d) Georges Leclanché
Answer: a
Clarification: In the year 1886, Carl Gassner obtained a German patent on a variant of the wet Leclanché cell, which can be known as the dry cell because it did not have a liquid electrolyte. Instead, a mixture of ammonium chloride and plaster of paris was used.

11. What is the final product that zinc forms during the functioning of a mercury cell?
a) ZnO
b) ZnO₂
c) Zn
d) Zn(OH)₂
Answer: a
Clarification: A mercury cell consists of a zinc anode and a mercury(II) oxide cathode. Potassium hydroxide is used as the electrolyte. In the electrochemical reaction, zinc is oxidised to become zinc oxide(ZnO) whereas mercury(II) oxide is reduced to elemental mercury.

12. Which of the following appliances would not use sodium hydroxide as an electrolyte in their mercury cells?
a) Calculators
b) Hearing aids
c) Electronic watches
d) Photographic cameras with a flash
Answer: d
Clarification: Sodium hydroxide cells have nearly constant voltage at low discharge currents whereas potassium hydroxide cells provide a constant voltage at high discharge currents. Therefore, sodium hydroxide cells are ideal for calculators, hearing aids and electronic watches whereas potassium hydroxide cells are ideal for photographic cameras with a flash.

13. Which of the following is the voltage output of a mercury cell?
a) 1.55V
b) 1.35V
c) 2.55V
d) 1V
Answer: b
Clarification: Mercury batteries use a reaction between mercuric oxide and zinc in an alkaline electrolyte to produce electricity. Its voltage during discharge is a constant 1.35V. The common dry cell, on the other hand, provides a voltage of 1.5V.

14. Which of the following statements is not true with respect to a lead storage cell (or a lead-acid battery)?
a) The electrolyte used is an aqueous solution of sulphuric acid
b) The anode is made up of lead
c) The cathode is made up of lead(IV) oxide
d) It is a primary cell
Answer: d
Clarification: A lead storage cell is a secondary cell which has a grid of lead packed with finely divided spongy lead for an anode and a grid of lead packed with lead(IV) oxide for a cathode. The electrolytic solution used in a lead-acid battery is an aqueous solution of sulphuric acid.

15. Which of the following products are formed when a lead storage battery is discharged?
a) SO₂
b) Pb
c) PbO₂
d) PbSO₄
Answer: d
Clarification: During the working of the lead storage battery, PbSO₄ is formed at both the electrodes and sulphuric acid is used up. At the anode, Pb is oxidised to form PbSO₄ and at the cathode, PbO₂ is reduced to form PbSO₄.

Chemistry Multiple Choice Questions on “Concentration of Ores”.

1. Concentration of ores is also known as ore-dressing.
a) True
b) False
Answer: a
Clarification: The removal of unwanted earthy and silicious impurities (that is, gangue or matrix) from an ore is called concentration of ores. Concentration of ores is also known as ore-dressing.

2. Which of the cases is suitable for concentration by hand picking?
a) When either the ore or the impurities are magnetic
b) When the ore particles are heavier than the impurities
c) When the ores are good conductors of electricity
d) When the impurities can be distinguished from the ore by the naked eye
Answer: d
Clarification: In the case of concentration by hand picking, it is necessary that the impurities be easily distinguishable from the particles of the ore. Only if these impurities can be distinguished by the naked eye, is the ore said to be suitable for concentration by hand picking.

3. Which of the following equipment is used for concentration by hydraulic washing?
a) Stamp mill
b) Ball mill
c) Wilfley tables
d) Magnetic roller
Answer: c
Clarification: Hydraulic washing is usually used for concentration of ores when the particles of the ore have a higher mass compared to the earthy or rocky gangue particles. The process is carried out in specially designed tables called Wilfley tables.

4. Ores obtained from the Earth’s crust are always pure.
a) True
b) False
Answer: b
Clarification: Ores, as they are obtained from the Earth’s crust, are never pure. They are usually associated with earthy and silicious impurities (along with impurities of other minerals) called gangue or matrix.

5. Which of the following ores cannot be concentrated by hydraulic washing?
a) Haematite
b) Pyrolusite
c) Tinstone
d) Native ores of silver
Answer: b
Clarification: Concentration by hydraulic washing requires the ore particles to have a higher mass compared to its gangue particles. Hydraulic washing can be used for oxide ores of iron and tin such as haematite and tinstone and also for native ores of silver and gold. Pyrolusite is concentrated by magnetic separation.

6. Which of the following cannot be concentrated by electromagnetic separation?
a) Chromite
b) Magnetite
c) Pyrolusite
d) Zinc blende
Answer: d
Clarification: Electromagnetic separation is used to concentrate an ore when either the particles of the ore or the impurities present in it are magnetic in nature. Chromite, magnetite and pyrolusite have magnetic nature and are separated from their non-magnetic gangue particles using this method. Zinc blende is concentrated using electrostatic separation or froth floatation.

7. Where are the magnetic particles collected in concentration by electromagnetic separation?
a) Below the magnetic roller
b) Above the magnetic roller
c) Away from the magnetic roller
d) On the conveyer belt
Answer: a
Clarification: In electromagnetic separation, the powdered ore is dropped over a conveyer belt moving around two rollers, one of which has an electromagnet in it. As the ore particles roll over the belt, the magnetic particles are attracted by the magnetic roller. As a result, two heaps are formed separately. The heap collected below the magnetic roller contains magnetic particles while the heap away from the roller consists of non-magnetic particles.

8. What is the property of necessary for an ore to be concentrated by electrostatic separation?
a) The ore particles should be heavier than the gangue particles
b) The ore particles should be magnetic in nature
c) The ores should conduct electricity
d) It should be a sulphide ore
Answer: c
Clarification: Electrostatic separation is a method used for concentration or separation of ores when they are good conductors of electricity. It is based on the principle that when an electrostatic field is applied the ore particles, which conduct electricity, get charged and get repelled by electrodes having same charge and all thrown away.

9. Which of the following are collectors used in the froth floatation process?
a) Aniline
b) Pine oil
c) Ethyl xanthate
d) Potassium ethyl xanthates
Answer: a
Clarification: Froth floatation is generally used for sulphide ores. It is based on the fact that sulphide ores are preferentially wetted by oils while that of the gangue is wetted by water. To a suspension of the crushed ore in water, collectors are added to enhance the non-wettability of the ore particles. Pine oil, ethyl xanthate and potassium ethyl xanthates are examples of collectors.

10. What is the role of the rotating paddle in froth floatation?
a) Enhances wettability of gangue particles
b) Stabilizes the froth
c) Draws in air causing frothing
d) Enhances non-wettability of the ore particles
View Answer

Answer: c
Clarification: In froth floatation process, the suspension of the powdered ore in water along with the collectors and froth stabilizers is violently agitated by the rotating paddle which draws in air, causing frothing.

Chemistry Multiple Choice Questions on “P-Block – Group 16 Elements”.

1. Group 16 elements are also called Chalcogens.
a) True
b) False
View Answer

Answer: a
Clarification: Group 16 elements, also known as the Oxygen family, are also called Chalcogens (pronounced as kal’-ke-jens) (meaning ore forming elements) because many metals occur as oxides and sulphides.

2. Which of the following elements does not belong to group 16 of the periodic table?
a) Oxygen
b) Phosphorus
c) Sulphur
d) Selenium
Answer: b
Clarification: Group 16 of the periodic table consists of five elements viz., Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te) and Polonium (Po). The elements of this group are commonly known as the oxygen family after the name of its first member.

3. Which is the most abundant group 16 element?
a) Oxygen
b) Sulphur
c) Selenium
d) Tellurium
Answer: a
Clarification: Oxygen is the most abundant of all elements. It occurs in the free state as dioxygen (O₂) and makes up 20.46% by volume and 23% by mass of the atmosphere. It also occurs in the form of ozone (O₃).

4. Ozone is an allotrope of Oxygen.
a) True
b) False
Answer: a
Clarification: Oxygen also occurs as ozone (O₃), an allotrope of oxygen, in the upper atmosphere which protects us from the harmful radiations of the sun. Ozone is a pale blue gas with a distinctively pungent smell.

5. What is the general outer electronic configuration of the Oxygen family?
a) ns²np⁴
b) ns²np³
c) ns²np⁵
d) ns²np²
Answer: a
Clarification: The elements of group 16 have six electrons in the valence shell and hence their general outer electronic configuration is ns²np⁴. The four p-electrons are arranged in three p-orbitals as p_x²p_y¹p_z¹ in accordance with Hund’s rule.

6. Which of the following statements regarding group 16 elements is not true?
a) The electronic configuration of Oxygen is [He]2s²2p⁴
b) The electronic configuration of Sulphur is [Ne]3s²3p⁴
c) The atomic radii of the elements of group 16 are larger than those of the corresponding elements of group 15
d) The electronic configuration of Tellurium is [Kr]4d¹⁰5s²5p⁴
Answer: c
Clarification: When you move across the periodic table, from left to right, the atomic size of the respective elements decreases due to increase in nuclear charge. Therefore, the atomic radii of the elements of group 16 are smaller than those of the corresponding elements of group 15.

7. Which group 16 element is the most electronegative?
a) Sulphur
b) Polonium
c) Oxygen
d) Selenium
Answer: c
Clarification: The elements of group 16 have higher values of electronegativity than the corresponding elements of group 15. Actually, oxygen is the second most electronegative element (EN=3.5), the first being fluorine (EN=4.0).

8. Which of the following is radioactive in nature?
a) Oxygen
b) Sulphur
c) Tellurium
d) Polonium
Answer: d
Clarification: Oxygen is the most non-metallic element of group 16. Sulphur is also a typical non-metal and Tellurium is a metalloid. Polonium is, however, metallic in nature but is radioactive with a short half-life (13.8 days).

9. Which of the following does not exist as an octatomic solid?
a) Sulphur
b) Tellurium
c) Selenium
d) Oxygen
Answer: d
Clarification: Oxygen exists as a diatomic gas at room temperature while other elements (S, Se and Te) exist as octatomic solids. Due to small size and high electronegativity, oxygen atom forms pπ – pπ double bond with other oxygen atom to form O = O molecule. The intermolecular forces of attraction between oxygen molecules are weak van der Waals forces and hence oxygen exists as a diatomic gas at room temperature.

10. Which group 16 element has the highest tendency for catenation?
a) Oxygen
b) Sulphur
c) Selenium
d) Polonium
Answer: b
Clarification: Sulphur has a stronger tendency for catenation than Oxygen. Due to small size, the lone pairs on the Oxygen atoms repel the bond pair of the O-O bond to a greater extent than the lone pairs of electrons on the Sulphur atoms in S-S bond. As a result, S-S bond is much stronger than O-O bond and hence Sulphur has a much stronger tendency for catenation than Oxygen.

11. Which group 16 element has 8 allotropic forms?
a) Sulphur
b) Oxygen
c) Selenium
d) Polonium
Answer: c
Clarification: Selenium has eight allotropic forms, of which three are red monoclinic forms containing Se₈ rings. The thermodynamically most stable form is grey hexagonal metallic Selenium which consists of polymeric helical chains.

12. What is the most reactive element of group 16?
a) Oxygen
b) Sulphur
c) Tellurium
d) Selenium
Answer: a
Clarification: Oxygen is the most reactive element of group 16. It is the second most electronegative element in the periodic table (EN=3.5), the first being fluorine (EN=4.0), making it more reactive than the other elements of the group.

13. Which of the following is a photosensitive element?
a) S
b) Se
c) O
d) Po
Answer: b
Clarification: The grey form of Selenium (metallic) and Tellurium consist of parallel chains held by weak metallic bonds. In the presence of light, the weak metallic bonds are excited and as a result, the number of free electrons increases and so does the conductivity. Thus, these elements conduct electricity significantly only in the presence of light. That is why Se and Te are called photosensitive elements.

14. What is the correct order of reactivity of group 16 elements?
a) O > Se > S > Te > Po
b) S > O > Te > Po > Se
c) S > O > Se > Te > Po
d) O > S > Se > Te > Po
Answer: d
Clarification: Oxygen is the most reactive group 16 element. Its reactivity is only slightly less than the most reactive elements, halogens. Sulphur is also very reactive particularly at high temperatures which helps in the cleavage of S-S bonds present in S₈ molecules. However, as we move down the group, the reactivity decreases, i.e., O > S > Se > Te > Po.

15. Which of the following is not poisonous?
a) H₂O
b) H₂S
c) H₂Se
d) H₂Te
Answer: a
Clarification: The hydride of Oxygen, i.e., H₂O is a colourless, odourless liquid while the hydrides of all the other group 16 elements are unpleasant, foul smelling, poisonous gases. H₂O has the highest boiling point of 373 K amongst the hydrides of group 16 elements.

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

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

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

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

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

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

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

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

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

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

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

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

Chemistry Multiple Choice Questions on “Haloalkanes Classification”.

1. Monohalo, dihalo, trihalo and tetrahalo are types of haloalkanes and haloarenes based on the ______
a) type of halogen atom
b) number of halogen atoms
c) nature of carbon atom
d) hybridisation of C atom to which halogen is bonded
Answer: b
Clarification: Haloalkanes may be classified as mono, di, tri, tetra and so on depending on the number of halogen atoms present in their structure.

2. A monohaloarene is an example of a/an __________
a) aliphatic halogen compound
b) aromatic halogen compound
c) alkyl halide
d) side chain substituted aryl halide
Answer: b
Clarification: A monohaloarene is a compound in which the halogen is directly attached to the benzene ring. Side chain substituted aryl halides are also aromatic halogen compounds with the halogen not directly attached to the benzene ring.

3. What is the general formula for haloalkanes? (X=halogen atom, n = 1, 2, 3…)
a) C_nH_2nX
b) C_nH_2n+1X
c) C_nH_2n-1X
d) C_nH_2n-3X
Answer: b
Clarification: The general formula for haloalkanes is C_nH_2n+1X where X is a halogen atom and n = 1, 2, 3… The formulae C_nH_2n-1X and C_nH_2n-3X are that of haloalkenes and haloalkynes respectively.

4. Which of the following compounds contains an allylic carbon?

a) A
b) B
c) C
d) D
Answer: a
Clarification: A sp³ hybridised C atom present adjacent to a C-C double bond is called an allylic carbon, and when the halogen atom is bonded to this carbon, it is called an allylic halide.

5. Which of the following categories does the compound shown belong to?

a) Primary haloalkane
b) Secondary haloalkane
c) Tertiary haloalkane
d) Haloarene
Answer: b
Clarification: The compound shown is cyclohexyl iodide, in which the halogen atom is bonded to an alkyl group that is alicyclic in nature. Since there are effectively two alkyl groups attached to the carbon bonded to the halogen atom, it is classified as secondary or 2° cyclo alkyl halide.

6. What is the nature of the circled C atom in the following compound?

a) sp² hybridised
b) allylic
c) benzylic
d) vinylic
Answer: c
Clarification: The circled C atom is sp³ hybridised and is attached directly to an aromatic ring, hence it is a benzylic carbon and the compound is a 1o benzylic halide.

7. In which of the following cases will the compound be a tertiary (3°) halogen compound? (X=halogen atom)

a) R’=R”=H
b) R’=CH₃, R”=H
c) R’=H, R”=CH₃
d) R’=R”=CH₃
Answer: d
Clarification: When R and R’ both are CH₃ groups, then the carbon atom bonded to the halogen will have three alkyl groups attached to it including the benzene ring.

8. Which of the following is a vinylic halide?
a) CH₂=CHCHCl₂
b) CH₃CHClCH₃
c) (CH₃)₂C=CHCH₂Cl
d) CH₃CH=CClCH₂CH₃
Answer: d
Clarification: In CH₃CH=CClCH₂CH₃, the Cl is bonded directly to the C atom of a C-C double bond, and hence it is a vinylic halide. CH₃CHClCH₃ is an alkyl halide whereas CH₂=CHCHCl₂ and (CH₃)₂C=CHCH₂Cl are allylic halides.

9. The compound C6H5F is an example of a ________ halide.
a) allylic
b) benzylic
c) vinylic
d) aryl
Answer: d
Clarification: In C₆H₅F, the F atom is directly attached to the sp² hybridised carbon atom of an aromatic ring, i.e., benzene.

10. The compound in which a CH₂Br group is attached to a benzene ring is an aryl halide.
a) True
b) False
Answer: b
Clarification: The Br atom is bonded to a carbon atom which is attached to sp² hybridised carbon of benzene. This is an example of a side chain substituted aryl halide.

Chemistry Exam Questions and Answers for Class 12 on “Alcohols and Phenols – 3”.

1. When the bond between O and H of the hydroxyl group is broken, alcohols react as _________
a) nucleophiles
b) electrophile
c) protonated molecules
d) electron seeking compounds
Answer: a
Clarification: Alcohols may react as nucleophiles or electrophiles depending upon whether the O-H bond or the C-O bond is broken respectively. Carbocations are formed in the nucleophile case and protonated alcohols are formed in the electrophile case.

2. Alcohols and phenols are ________
a) Lewis acids
b) Lewis bases
c) Bronsted acids
d) Bronsted bases
Answer: c
Clarification: Alcohols and phenols are known to be acidic in nature based on their interaction with metals. They are able to donate a proton to a stronger base. Additionally, alcohols also act as Bronsted bases due to the presence of unshared electron pairs on oxygen.

3. What is the correct relation between acidic strength of primary, secondary and tertiary alcohols?
a) 1°>2°>3°
b) 1°c) 1°=2°=3°
d) 1°>3°>2°
Answer: a
Clarification: Alkyl groups are electron releasing groups and tend to increase the electron density on oxygen and reduce the polarity of O-H bond. So, the greater the number of alkyl groups present, lesser will be the tendency to release protons, and therefore weaker will be the acidic strength.

4. Water is a ______ than alcohol.
a) better proton acceptor
b) stronger base
c) weaker acid
d) better proton donor
View Answer

Answer: d
Clarification: The reaction between water and an alkoxide illustrates that water is a better proton donor (stronger acid) than alcohol.

5. Phenols are stronger acids than alcohols.
a) True
b) False
Answer: a
Clarification: The aryl carbon in phenols is more electronegative than the sp³ hybridized carbon in alcohols due to the presence of double bond. This extra negative charge is also delocalized at positions across the benzene, resulting in resonating structure and hence stability.

6. Which of the following compounds is the best proton acceptor?
a) Ethanol
b) Phenol
c) p-Cresol
d) p-Nitrophenol
Answer: a
Clarification: Proton acceptor indicates the basic nature of the compound. Ethanol is the only alcohol in the given compounds, and it is the least acidic, because phenols are more acidic than alcohols due to their stability through resonance structures.

7. Which of the following phenols is the most acidic?
a) o-Cresol
b) m-Cresol
c) o-Nitrophenol
d) m-Nitrophenol
Answer: c
Clarification: Alkyl groups are electron releasing, and contribute towards a reduction in acidic strength of phenols. Electron withdrawing groups like nitro, enhances the acidic strength especially at ortho and para positions, due to effective delocalisation of negative charge in the phenoxide ion.

8. What is the role of concentrated H₂SO₄ in the reaction between alcohol and carboxylic acid?
a) Deprotonating agent
b) Dehydrating agent
c) Reducing agent
d) Nucleophile
Answer: b
Clarification: Concentrated H₂SO₄ is the catalyst in the esterification of alcohols, which acts a protonating and dehydrating agent. The reaction is reversible and is shifted in the forward direction by the removal of water as soon as it is formed.

9. Which of the following alcohols is the most reactive towards esterification reaction?
a) CH₃OH
b) CH₃CH₂OH
c) (CH₃)₂CHOH
d) (CH₃)₃COH
Answer: a
Clarification: Presence of bulky groups reduces the reactivity of alcohols and carboxylic acid towards esterification due to effect of stearic hinderance. Therefore, the tertiary alcohols are least reactive.

10. Identify the products of the following reaction:
Methyl alcohol + Ethyl magnesium bromide = ?
a) CH₄ and CH₃OMgBr
b) CH₄ and CH₃CH₂OMgBr
c) C₂H₆ and CH₃OMgBr
d) C₂H₆ and CH₃CH₂OMgBr
Answer: c
Clarification: Alcohols reacts with Grignard reagents to form hydrocarbons. Also, the alkane formed from this reaction corresponds to the alkyl group of the Grignard reagent and not the alcohol.

11. What is the catalyst used in the acetylation of alcohols?
a) Palladium
b) Hydrogen peroxide
c) Pyridine
d) Aluminium chloride
Answer: c
Clarification: The reaction of alcohols with acid chlorides and acid anhydride is carried out in the presence of a base so as to neutralise the acid formed during the reaction. This facilitates the shift f equilibrium in the forward direction.

12. Aspirin is produced from the reaction between salicylic acid and which compound?
a) Acetyl chloride
b) Acetic anhydride
c) Phenyl acetate
d) Benzoyl chloride
Answer: b
Clarification: Salicylic acid, which is a phenolic acid when treated with acetic anhydride produces acetyl salicylic acid, also known as aspirin.

13. What is the correct order of reactivity of alcohols toward reactions involving cleavage of C-O bond?
a) 1°>2°>3°
b) 3°>2°>1°
c) 1°>3°>2°
d) 3°>1°>2°
Answer: b
Clarification: Alkyl groups are electron releasing groups and increase the electron density towards oxygen making the C-O more polar. This makes the cleavage of the bond between C and O a lot easier.

14. What is the catalyst used in the following reaction?

a) Anhydrous AlCl₃
b) Anhydrous FeCl₃
c) Anhydrous ZnCl₂
d) No catalyst required
Answer: d
Clarification: The given alcohol is a tertiary alcohol and can easily form alkyl halides with HCl and does not require any catalyst. However, primary and secondary alcohols require ZnCl₂ as catalyst in the same reaction for it to proceed.

15. Propan-2-ol undergoes acidic dehydration relatively at a lower temperature than propan-1-ol.
a) True
b) False
Answer: a
Clarification: Acidic dehydration of alcohols occurs through the formation of a carbocation. Since tertiary alcohol produce the most stable carbocations and primary alcohols produce the least stable carbocation, the dehydration of tertiary and secondary alcohols is relatively easier.

Chemistry Exam Questions and Answers for Class 12,