Category: Class 12 Chemistry Objective Questions

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 “Coordination Compounds Importance and Applications”.

1. Which reagent is used for detecting Ni²⁺ ions in solution?
a) EDTA
b) Dimethylglyoxime
c) α-nitroso-β-naphthol
d) Cupron
View Answer

Answer: b
Clarification: When dimethylglyoxime is added to a solution along with ammonium hydroxide, the formation of a bright red precipitate indicates the presence of Ni²⁺ ions in solution. This is due to the formation of nickel dimethylglyoxime complex.

2. Hardness of water is estimated by simple titration with which compound?
a) Na₂(EDTA)
b) Fe(EDTA)
c) Mg(EDTA)
d) Co(EDTA)
Answer: a
Clarification: The Ca²⁺ and Mg²⁺ ions form stable complexes with EDTA, and the difference in the stability constants of Ca and Mg complexes helps in the estimation of hardness of water.

3. Given below is the incomplete equation for the extraction process of gold using potassium cyanide. Identify the missing reagent.

a) Water
b) Oxygen
c) Water + air
d) Zinc dust
Answer: c
Clarification: Gold combines with cyanide in the presence of air and water to form [Au(CN)₂]^– in aqueous solution. This is then treated with zinc dust to obtain gold in metallic form.

4. Haemoglobin is a complex compound of which metal ion?
a) Fe²⁺
b) Fe³⁺
c) Co²⁺
d) Co³⁺
Answer: a
Clarification: Haemoglobin is the red pigment of blood which acts as the oxygen carrier and it a coordination compound of iron(II).

5. Which of the following biologically important coordination compounds has a magnesium central atom?
a) Chlorophyll
b) Haemoglobin
c) Vitamin B₁₂
d) Carboxypeptidase-A
Answer: a
Clarification: Chlorophyll is the green pigment in plants, responsible for photosynthesis, is a coordination compound of Mg.

6. The coordination complex chloridotris (triphenylphosphine) rhodium(I) is used in the hydrogenation of alkenes. It is also known as __________ catalyst.
a) Grubb
b) Pearlman
c) Wilkinson
d) Ziegler-Natta
Answer: c
Clarification: Coordination compounds may be used as homogeneous or heterogeneous catalysts for many reactions. Wilkinson catalyst is used as a homogeneous catalyst in hydrogenation of alkenes, whereas Ziegler¬-Natta catalyst is used as a heterogeneous catalyst for the polymerisation of olefins.

7. The platinum complex, trans-platin has been used in cancer treatment.
a) True
b) False
Answer: b
Clarification: Cis-platin is a coordination complex of platinum which is effective in inhibiting the growth of tumours.

8. The complex ion [Ag(S₂O₃)₂]^3- is associated with which field?
a) Electroplating
b) Medicine
c) Water treatment
d) Photography
Answer: d
Clarification: In black and white photography, the film is fixed by washing with hypo solution which dissolves the undecomposed silver bromide to form [Ag(S₂O₃)₂]^3-.

Chemistry Multiple Choice Questions on “Alcohols and Phenols – 2”.

1. Phenol can be obtained by ________ of sodium phenoxide.
a) acidification
b) oxidation
c) sulphonation
d) hydrolysis
Answer: a
Clarification: When halobenzene is fused with NaOH at 623K and 320atm with a copper salt as catalyst, sodium phenoxide is produced which on treatment with dilute HCl yields phenol. This is also known as Dow’s process.

2. Cumene hydroperoxide on hydrolysis with dilute H₂SO₄ gives _________
a) alcohol and phenol
b) only phenol
c) phenol and acetone
d) alcohol and acetone
Answer: c
Clarification: When cumene hydroperoxide is hydrolysed with a solution of dilute acid, it results in the formation of phenol and acetone, of which the latter is obtained in large quantities.

3. How is carbolic acid prepared from benzene diazonium chloride?
a) Treating it with nitrous acid at 275K
b) Preparing an aqueous solution and warming it
c) Treating it with sodium hydroxide
d) Freezing it
Answer: b
Clarification: Carbolic acid is another name for phenol. Simply warming an aqueous solution of benzene diazonium chloride salt or treating it with a dilute acid, will result in the formation of phenol.

4. Identify the compound ‘X’.

a) Oleum
b) Sulphuric acid
c) Nitrous acid
d) Sodium sulphate
Answer: a
Clarification: Oleum is a mixture of H₂SO₄ and SO₃ that is added to benzene to form benzene sulphonic acid, which on subsequent treatment as shown, gives phenol.

5. Which of the following isomeric alcohols is the most soluble in water?
a) n-Butyl alcohol
b) Isobutyl alcohol
c) sec-Butyl alcohol
d) tert-Butyl alcohol
Answer: d
Clarification: tert-Butyl alcohol is the most heavily branched isomer and is tightly packed compared to the others. This decreases the effective surface area of the alkyl part which is hydrophobic in nature.

6. Which of the following is the least soluble in water?
a) n-Butyl alcohol
b) n-Pentyl alcohol
c) n-Hexyl alcohol
d) n-Heptyl alcohol
Answer: d
Clarification: As the size of the alkyl group becomes larger, it prevents the formation of hydrogen bonds with water molecules and hence the solubility goes on decreasing with the increase in the size or mass of the compound.

7. Which of the following is necessary for the bromination of phenol?
a) CS₂
b) FeBr₃
c) AlCl₃
d) BF₃
Answer: a
Clarification: FeBr₃, AlCl₃ and BF₃ are Lewis acids, that are required for the substitution in benzene. Phenol has an OH group that activates the aromatic ring and requires the presence of only slightly or non-polar solvents for its halogenation.

8. Phenol is approximately how much times acidic than ethanol?
a) 10
b) 25
c) 100
d) million
Answer: d
Clarification: Phenol is more acidic than alcohols because of their better stability and resonant structures. The pK_a values of phenol and ethanol are 10 and 15.9 respectively, which when converted to K_a values gives a ratio of about a million.

9. Phenols are not soluble in which of the following?
a) Water
b) Alcohol
c) Ether
d) Sodium hydroxide
Answer: a
Clarification: Phenols are almost insoluble in water because of the presence of a very large non-polar aryl group which overcomes the polar character of the OH group.

10. What is the correct order of boiling points of alcohols having the same number of carbon atoms?
a) 1°>2°>3°
b) 3°>2°>1°
c) 3°>1°>2°
d) 2°>1°>3°
Answer: a
Clarification: As the branching in the structure increases, the surface area decreases and hence, the van der Waal forces decrease. This results in the reduction of boiling points of isomeric alcohols from primary to tertiary structures.

11. Four compounds A, B, C and D having similar molecular masses were tested for their boiling points. It was found that compound C has the highest boiling point of all four. What is the compound C most likely to be?
a) Hydrocarbon
b) Haloalkane
c) Alcohol
d) Ether
Answer: c
Clarification: Of compounds having similar molecular masses, alcohols will certainly have the highest boiling point due to the presence of intermolecular hydrogen bonding which is lacking in hydrocarbons, haloalkanes and ethers.

12. If the boiling point of methoxymethane is 248K, predict the boiling point of ethanol.
a) 231K
b) 248K
c) 351K
d) 455K
Answer: c
Clarification: Methoxymethane and ethanol are of comparable masses and it can be easily pointed out that ethanol will have a higher boiling point than it because of the presence of intermolecular hydrogen bonding. 455K is very high and is actually the boiling point of phenol.

13. Which of the following has the highest boiling point?
a) Propan-1-ol
b) Butan-1-ol
c) Butan-2-ol
d) Pentan-1-ol
Answer: d
Clarification: As the size the molecule increases, the number of carbon atoms increases and the van der Waal forces increase, resulting in an increase in boiling points.

14. Propan-1-ol boils at a higher temperature than propan-2-ol.
a) True
b) False
Answer: a
Clarification: Propan-2-ol is a secondary alcohol and has some degree of branching compared to propan-1-ol. This results in the decreases in surface area and van der Waals forces and hence reduction in boiling point.

15. Toluene and phenol have similar boiling points.
a) True
b) False
Answer: b
Clarification: Phenols have much higher boiling points than aromatic compounds of similar molecular mass. This is because they exist as associated molecules due to intermolecular hydrogen bonding.

Chemistry Objective Questions and Answers for Class 12 on “Carboxylic Acids Chemical Reactions – 2”.

1. How many molecules of acetic acid react with H₂SO₄ on heating to give one molecule of acetic anhydride?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Two molecules combine with the loss of one water molecule (H atom from one molecule and OH group from another molecule) when heated with dehydrating agents like H₂SO₄ to give acetic anhydride.

2. Benzoic acid reacts with phosphorus pentaoxide in the presence of heat to give ________
a) C₆H₅OC₆H₅
b) C₆H₅COC₆H₅
c) C₆H₅COOC₆H₅
d) C₆H₅COOCOC₆H₅
Answer: d
Clarification: Benzoic acid on heating with a strong dehydrating agent like P₂O₅ forms its corresponding anhydride, i.e., benzoic anhydride, which consists of two C₆H₅C=O groups bridged by a (-O-) group.

3. Esterification of carboxylic acids is a/an __________ reaction.
a) irreversible
b) nucleophilic addition
c) electrophilic substitution
d) nucleophilic substitution
Answer: d
Clarification: When carboxylic acids are heated with alcohols with suitable catalyst, esters are formed. This is a reversible reaction due to all the individual steps involved in the mechanism of this reaction are reversible in nature. It is a type of nucleophilic acyl substitution reaction.

4. Which of the following is the most suitable catalyst for the esterification of acetic acid with ethanol?
a) P₂O₅
b) HCl gas
c) KMnO₄-KOH
d) Pyridine
Answer: b
Clarification: HCl gas and concentrated H₂SO₄ are catalyst which provide a proton for the protonic attack on the carbonyl oxygen of acetic acid, which further activates the carbonyl groups towards nucleophilic addition of ethanol, which ultimately results in the formation of ethyl acetate, which is an ester.

5. Which of the following is used to shift the esterification reaction of carboxylic acids towards the right?
a) Using excess of carboxylic acid
b) Using excess of acid catalyst
c) Removal of water by distillation
d) Removal of ester by distillation
Answer: c
Clarification: Since the esterification of carboxylic acids is a reversible reaction, it can be shifted towards the right by using excess of alcohol or by the removal the water formed by distillation.

6. Which of the following carboxylic acids is most reactive towards esterification with given alcohol?
a) HCOOH
b) CH₃COOH
c) (CH₃)₂CHCOOH
d) (CH₃)₃CCOOH
Answer: a
Clarification: The rate at which carboxylic acids are esterified depends upon the stearic hinderance and to some extent the acidic strength of the acid, as it is a nucleophilic substitution reaction. Since formic acid has the lowest molecular mass and smallest group, it is the most reactive.

7. Which is the most preferred reagent in the product of acetyl chloride from acetic acid?
a) Cl₂
b) PCl₃
c) PCl₅
d) SOCl₂
Answer: d
Clarification: Thionyl chloride reacts with acetic acid to form acetyl chloride along with two gaseous compounds (SO₂ and HCl) that escape the reaction mixture, making the purification of the product easier.

8. Benzoic acid reacts with phosphorus pentachloride to give ________
a) chlorobenzene
b) benzyl chloride
c) benzoyl chloride
d) chlorobenzoic acid
Answer: c
Clarification: The hydroxyl group (OH) of benzoic acid id easily replaced by chlorine atom on treatment with PCl₅. This results in the formation of C₆H₅COCl, or benzoyl chloride.

9. Other than ethanoyl chloride, what are the by-products formed when ethanoic acid reacts with PCl₃?
a) H₃PO₃
b) H₃PO₃ and HCl
c) H₃PO₂ and HCl
d) POCl₃ and HCl
Answer: a
Clarification: Three molecules of ethanoic acid react with one molecule of PCl₃ to give three molecules of ethanoyl chloride and one molecule of phosphorous acid (H₃PO₃).

10. Acetic acid behaves exactly like ethanol on reaction with thionyl chloride.
a) True
b) False
Answer: a
Clarification: Acetic acid as well as ethanol consist of an OH group and reaction with SOCl₂ results in the cleavage of C-OH bond with the hydroxyl group to be replaced by Cl atom. The only difference is that acetic acid forms acetyl chloride, whereas ethanal forms ethyl chloride.

Chemistry Objective Questions and Answers for Class 12,