300+ REAL TIME Engineering Chemistry Objective Questions & Answers 2023

250+ TOP MCQs on Adhesives, their Advantages and Limitations and Answers

Engineering Chemistry Multiple Choice Questions on “Adhesives, their Advantages and Limitations”.

1. The substances which are capable of uniting two other materials together by adhering strongly to the surface of both are called ____________
a) Organic conducting polymers
b) Biodegradable polymers
c) Adhesives
d) Composites
Answer: c
Clarification: The substances which are capable of uniting two other materials together by adhering strongly to the surface of both are called adhesives. Adhesives may be found naturally or produced synthetically.

2. Select the incorrect statement from the following option.
a) Two surfaces are rapidly, economically and easily joined to each other by adhesives
b) The process of applying adhesives requires highly specialized workers or high heat
c) Adhesives bonding requires less after-finishing
d) Bonding by adhesives gives smoother finished surfaces
Answer: b
Clarification: The process of applying adhesives does not require highly specialized workers or high heat. All the other options are correct.

3. Corrosion of metals at the joints is minimized by joining dissimilar metals with adhesives because ____________
a) Direct metal to metal contact is avoided
b) Adhesive act as a lubricator at the joints
c) Metals are covered with adhesives
d) Adhesives prevent the exposure to environment
Answer: a
Clarification: Corrosion of metals at the joints is minimized by joining dissimilar metals with adhesives because direct metal to metal contact is avoided so galvanic action is reduced.

4. Adhesives joints are leak-proof for ____________
a) Solids only
b) Liquids only
c) Gases only
d) Both liquids and gases
Answer: d
Clarification: Adhesives joints are leak-proof for liquids and gases. There is no leaking of solids.

5. Joining of labels to plastic cans or bottles is not possible without the use of adhesives.
a) True
b) False
Answer: a
Clarification: Joining of labels to plastic cans or bottles is not possible without the use of adhesives. Adhesives are typically organized by the method of adhesion.

6. The strength of adhesive joint is _____________ than mechanical joint obtained by welding or riveting.
a) More
b) Less
c) Equals to
d) Very less
Answer: b
Clarification: The strength of adhesive joint is less than mechanical joint obtained by welding or riveting. Mechanical joints are permanent and stronger as they are joined with the help of nuts and bolts.

7. On increasing the temperature, the strength of adhesive ____________
a) Remain same
b) Increases
c) Decreases
d) First increases than decreases
Answer: c
Clarification: On increasing the temperature, the strength of adhesive decreases due to relative weakness in bonding large objects with a small bonding surface area and greater difficulty in separating objects during testing.

8. The bond energy of a metallic bond is ____________
a) 600-1200 kJ/mol
b) 60 kJ/mol
c) 20 kJ/mol
d) 100-350 kJ/mol
Answer: d
Clarification: The bond energy of a metallic bond is 100-350 kJ/mol. Metallic bonding is the force of attraction between valence electrons and metal ions. Therefore, it is very strong in nature.

9. Which of the following is not an example of thermosetting synthetic resins?
a) Epoxy
b) Acrylic polyester
c) Cellulose derivatives
d) Silicones
Answer: c
Clarification: Epoxy, silicones, acrylic polyester, polyurethane etc are some of the examples of thermosetting synthetic resins.

10. Which of the following adhesive is prone to brittle failure?
a) Acrylic polyester
b) PF resin
c) UF resin
d) Epoxy
Answer: b
Clarification: PF resins are prone to brittle failure due to its tough structure.

11. The epoxy resins are the _______________ products of epichlorohydrin and bisphenol.
a) Free radical
b) Addition
c) Ion-exchange
d) Condensation
Answer: d
Clarification: The epoxy resins are the condensation products of epichlorohydrin and bisphenol. They are formed by the condensation reaction.

12. Epoxies are used in wide diversity of applications because of their strength, durability and great versatility.
a) True
b) False
Answer: a
Clarification: Epoxies are used in a wide diversity of applications because of their strength, durability and great versatility. Epoxy resins are low molecular weight pre-polymers or higher molecular weight polymers which normally contain at least two epoxide groups. The epoxide group is also sometimes referred to as a glycidyl or oxirane group.

13. Which of the following is used as curtain-wall sealants?
a) Polyurethane
b) Silicones
c) Polyvinyls
d) Cellulose derivatives
Answer: b
Clarification: Although expensive, silicones are used as curtain-wall sealants because of their excellent aging characteristics.

14. Emulsion adhesives are based on ____________
a) Polyurethane
b) Collodion
c) Urea-formaldehyde
d) Polyvinyl acetate
Answer: d
Clarification: Emulsion adhesives are based on polyvinyl acetate. It is formed by the emulsification and polymerization of acetic vinyl, EVA and acrylic resin in water.

15. Collodion is a mixture of cellulose nitrate with ____________
a) Alcohol
b) Ketone
c) Carboxylic acid
d) Phenol
Answer: a
Clarification: Collodion is a mixture of cellulose nitrate with alcohol. It is a flammable, syrupy solution.

250+ TOP MCQs on Coal Technology and Answers

Engineering Chemistry Multiple Choice Questions on “Coal Technology”.

1. How do the non-coking coals are subjected to consumers in India?
a) By washing
b) By sizing
c) By sorting
d) By blending
Answer: b
Clarification: Sizing is done for non coking coal where as for coking coal, coal are beneficiated to washing before their distribution to consumers. Sorting is done for both type of coal.

2. How is the sizing of coal processed?
a) By the use of measuring instruments
b) By the use of computer software
c) By crushing and screening
d) By the mass of the coal
Answer: c
Clarification: Crushing makes the top surface of a coal furnish due to which it becomes easy to handle (by reducing its overall top size). Screening is used to group process particles into ranges by size. By appropriate sizing, transportation of coal becomes easier.

3. By which process does the impurities of coal are generally removed?
a) Screening process
b) Sorting
c) Blending
d) Wet washing
Answer: d
Clarification: Jig washers and water cyclones are used for the removal of impurities of coal by wet washing. Wet washing is affected with the help of water or a dense medium as the sorting liquid. This process is used for making the density of the clean coal and impurities same.

4. Which of the following is used for the removal of impurities by heavy media separation in industries?
a) Magnetite
b) Zinc chloride
c) Bromoform
d) Carbon Tetrachloride
Answer: a
Clarification: For laboratory we can use zinc chloride but for industrial process we have to use magnetite or fine grains of sand. This is due to cost factor and properties of both of them at large scale.

5. For coarse fractions of dewatering of coal, which of the following process is used?
a) Filtration
b) Thickening
c) Flocculation
d) Centrifugation
Answer: d
Clarification: Centrifugation is a separation process that uses centrifugal force to settle the particles of a solid-liquid mixture. Since at the coarse fraction there is some coal reduction during dewatering due to which centrifugation process is used. Filtration, thickening and flocculation processes are used for finer fractions.

6. Froth flotation process is used for ____________________
a) Screening of coal
b) Beneficiation of coal
c) Dewatering of coal
d) Mining of coal
Answer: b
Clarification: Froth floatation is a process for selectively separating a hydrophobic substance. Before froth flotation the coal is crushed and grinded. Selective agglomeration process is also used for beneficiation of coal.

7. Which of the following is not an advantage of the cleaning of coal?
a) It increases the cost of the coal
b) It increases the efficiency of coal
c) It removes the impurities like phosphorous and sulphur
d) It reduces its ash content
Answer: a
Clarification: The cleaning of coal reduces impurities due to which it becomes easy to transport and store, which reduces the cost of the coal. It also increase the calorific value of the coal.

8. What is the main purpose for a blending of coal?
a) To produce more amount of coal
b) To produce good quality of coal
c) To decrease the cost of coal
d) To produce different types of coal at same time
Answer: b
Clarification: The process to mix different types of coal at the same time to increase the efficiency of coal is called blending of coal. This is due to the fact that different types of coal are produced during the coal processing.

9. Which of the following is a disadvantage of storing coal for a long period of time?
a) Increase in the friability of coal
b) Decrease in its ignition temperature
c) Increase in its calorific value
d) Increase in the proportion of fine
Answer: d
Clarification: When the coal is stored for a long period of time, it leads to slow oxidation which causes deterioration of the coal. This process causes an increase in the proportion of fine which are not easy to cure. Also the oxidation of coal can lead to its spontaneous combustion.

10. How do the chances of spontaneous combustion of coal can decrease?
a) Decrease in the maturity of coal
b) By washing the coal time to time
c) Increase in the maturity of coal
d) By the reducing the quantity of coal
Answer: c
Clarification: The coal of high maturity (anthracite) contains less impurities due to which it does not gets oxidise easily, also the volatile matter present in them is less. Washing of coal can cause an increase in moisture content due to which the coal produced will not be of good quality.

250+ TOP MCQs on Octane and Cetane Number and Answers

Engineering Chemistry Multiple Choice Questions on “Octane and Cetane Number”.

1. Which of the following compound is considered for calculating the octane number?
a) n-heptane
b) n-hexane
c) iso-octane
d) iso-butane
Answer: c
Clarification: The octane number of a fuel is numerically taken as the percentage of iso-octane in a mixture of iso-octane and n-heptane. The octane number of iso-octane is 100 and of n-heptane is 0.

2. Which of the following compound readily goes under the process of knocking?
a) 2-ethly butane
b) n-heptane
c) benzene
d) toluene
Answer: b
Clarification: n-heptane is a straight chain hydrocarbon which knocks readily and hence its octane number is zero. Presence of straight chain hydrocarbon in gasoline reduces its rate of combustion.

3. Which of the following compound is considered for calculating the cetane number?
a) α-methyl naphthalene
b) n-hexane
c) iso-octane
d) cetane molecule
Answer: d
Clarification: Cetane, C₁₆H₃₄, is a saturated hydrocarbon that has a very short ignition lag as compared to any commercial diesel fuel. Its cetane number is 100. Cetane number is the percentile of cetane in a mixture of cetane and α-methyl naphthalene.

4. What should be the cetane number of middle speed diesel engine?
a) At least 35
b) At least 25
c) At least 65
d) At least 45
Answer: a
Clarification: The cetane number of high speed diesel engine should be at least 45, that of middle speed diesel engine should be at least 35 and that of low speed diesel engine should be at least 25.

5. The knocking characteristic of gasoline fuel are expressed in terms of cetane number.
a) True
b) False
Answer: b
Clarification: The knocking characteristics of diesel fuel are expressed in terms of cetane number and that of gasoline fuel is expressed in terms of octane number. The molecule with the highest octane number has a least cetane number.

6. Which of the following substance is used to decrease knocking in diesel fuel?
a) Tetra ethyl lead
b) Benzene
c) Sodium hydroxide
d) Acetone peroxide
Answer: d
Clarification: The knocking in diesel fuel can be reduced by adding substances like ethyl nitrite, ethyl nitrate, isoamyl nitrate etc. This process also increases the cetane number of diesel fuel.

7. Which of the following is used to reduce the surface tension in diesel fuel?
a) Chemical additives
b) Inhibitors
c) Poly hydrocarbon
d) Dopes
Answer: a
Clarification: Chemical additives are used to reduce surface tension thus promoting the formation of finer spray. Inhibitors are used to delay or prevent gum formation. Poly hydrocarbons are used to preserve the fluid properties.

8. By which process does the knocking starts in diesel engine?
a) Due to sudden spontaneous combustion of last portion of fuel
b) Due to delay in spontaneous combustion of last portion of fuel
c) Due to the rise in temperature of diesel engine
d) Due to the parts of diesel engine
Answer: b
Clarification: Due to the presence of impurities in diesel, its spontaneous combustion of the last portion of fuel takes some time. Due to this their produces an uneven sound in the engine which is called knocking.

9. Which of the following exhaust gas produces least harmless components after its combustion?
a) Carbon monoxide
b) Ethyne
c) NO_x
d) N₂
Answer: d
Clarification: Gases produced by the compression ignition engines are said to be exhaust gases. This engines operate with air-fuel ratios of 14:1 on load and 70:1 on idling.

10. Which of the following has the highest cetane number?
a) n-heptane
b) n-hexane
c) n-pentane
d) n-butane
Answer: a
Clarification: n-heptane has maximum branched atom due to which it has the highest cetane number of all. This fuel is good for diesel engines.

250+ TOP MCQs on Wet Theory of Corrosion and Answers

Engineering Chemistry Multiple Choice Questions on “Wet Theory of Corrosion”.

1. In wet corrosion _________________ are formed at the cathodic areas.
a) Organic compounds
b) Metallic ions
c) Non-metallic ions
d) Inorganic compounds
Answer: c
Clarification: In wet corrosion, non-metallic ions are formed at the cathodic areas. Cathodes are negatively charged electrodes and attract positive charges or non-metallic ions.

2. Which type of reaction occurs in anodic areas?
a) Oxidation
b) Reduction
c) Displacement
d) Addition
Answer: a
Clarification: Oxidation occurs in anodic areas. Oxidation means the addition of oxygen or removal of hydrogen.

3. Rusting of iron in neutral aqueous solution of electrolyte occurs in the presence of oxygen with the evolution of ___________
a) Nitrogen
b) Chloride
c) Sulphide
d) Hydrogen
Answer: d
Clarification: Rusting of iron in neutral aqueous solution of electrolyte occurs in the presence of oxygen with the evolution of hydrogen. Only hydrogen gas is evolved in this process.

4. Where does corrosion occurs in the rusting of iron?
a) At cathode
b) At anode
c) In electrolytic solution
d) Outside the solution
Answer: b
Clarification: Corrosion occurs at anode but rust is deposited near cathode.

5. Which of the following cathodic reaction does not occur due to release of electrons at the anode?
a) Oxygen absorption
b) Hydrogen evolution
c) Electrodialysis
d) Electroplating
Answer: c
Clarification: Oxygen absorption, hydrogen evolution and electroplating occur due to the release of electrons at the anode.

6. Select the incorrect statement about the wet corrosion from the following option.
a) It involves the setting up of large number of galvanic cells
b) It is explained by absorption mechanism
c) It occurs only on heterogeneous metal surface
d) It is a fast process
Answer: b
Clarification: Wet corrosion is explained by the mechanism of electrochemical reaction. All the other options are correct.

7. Which of the following factor does not contribute to the rusting of iron?
a) Presence of acids and electrolytes
b) Contact with less reactive metal
c) Presence of water and oxygen
d) Contact with more reactive metal
Answer: d
Clarification: Contact with the more reactive metal does not contribute to the rusting of iron whereas all the other given factors contribute to the rusting of iron.

8. Concentration cell corrosion occurs when a metallic surface is partially immersed in an electrolyte and partially exposed to air.
a) True
b) False
Answer: a
Clarification: Concentration cell corrosion occurs when a metallic surface is partially immersed in an electrolyte and partially exposed to air. This is due to the half part which is exposed to air. It gets in contact with air molecules and gets corroded.

9. Which of the following does not promote the differential aeration corrosion?
a) Accumulation of dirt
b) Partially covering metals
c) Wire fence kind of structures
d) Accumulation of oxygen
Answer: d
Clarification: Accumulation of dirt, partially covering metals and wire fence kind of structures are the factors which promote the differential aeration corrosion.

10. Poorly oxygenated part becomes cathode whereas well oxygenated part becomes anode in the differential aeration corrosion.
a) True
b) False
Answer: b
Clarification: Poorly oxygenated part becomes anode and undergoes oxidation whereas well oxygenated part becomes cathode in the differential aeration corrosion.

250+ TOP MCQs on Catalysts and Answers

Engineering Chemistry Multiple Choice Questions on “Catalysts”.

1. Which of the following is not a characteristic of a catalyst?
a) It participates in the reaction
b) It enhances the equilibrium rate
c) It activates equilibrium
d) It initialises the reaction
Answer: a
Clarification: The main purpose of using a catalyst is only to increase the rate of achieving the equilibrium state in case of slow reaction. Other than that it will not initiate or involve in the reaction.

2. Which of the following will decrease the rate of reaction?
a) Catalytic poison
b) Positive catalyst
c) Negative catalyst
d) Catalytic promoters
Answer: a
Clarification: Catalytic poisons are the once which are used to slow down the reaction inorder to avoid explosion when the temperature or pressure increase the critical value. Hence catalytic poisons are used such as aluminium oxide.

3. Which of the following is a catalytic poison?
a) Potassium nitrate
b) Aluminium nitrate
c) Aluminium oxide
d) Chlorine
Answer: c
Clarification: Aluminium oxide is known for its sophisticated properties and is mainly used in the process of corrosion preventing it absorbs moisture and has high retentivity and hence is used as a catalytic poison.

4. Which of the following is used as a catalytic promoter?
a) Chlorine
b) Nitrate
c) Bromine
d) Helium
Answer: b
Clarification: The use of a promoter is to increase the rate of equilibrium. We know that nitrogen has promoting properties in the case of nitrate molecules. Thus the can act as good promoters in reaction.

5. Which of the following is used as a catalyst for the following reaction?
2SO₂ + O₂ ⟶ 2SO₃
a) Chlorine
b) Nitrous oxide
c) Sulphur di oxide
d) Potassium chloride
Answer: b
Clarification: Since sulphur and nitrogen components have the same equilibrium valencies, we can see that one will promote the other and vice versa. Thus nitrogen molecule is the best for sulphur molecules.

6. Which of the following is used as a catalyst for the following reaction?
N₂ + 3H₂ ⟶ 2NH₃
a) Zinc
b) Chlorine
c) Platinum
d) Water
Answer: c
Clarification: Since nitrogen and noble metals components have the same equilibrium valencies, we can see that one will promote the other and vice versa. Thus nitrogen molecule is the best for platinum molecules.

7. Which of the following is used as a catalyst for the following reaction?
H₂O₂ ⟶ 2H₂O + O₂
a) Chlorine
b) Bromine
c) Platinum
d) Noble solutions
Answer: c
Clarification: Since oxygen and noble metals components have same equilibrium valencies, we can see that one will promote the other and vice versa. Thus oxygen molecule is the best for platinum molecules.

8. Which of the following is used as a catalytic poison for the same reaction:
SO₂ + O₂ ⟶ 2SO₃
a) Platinum
b) Bromine
c) Magnesium
d) Silver
Answer: a
Clarification: Though nitrogen is used as a promoter, we can see that platinum reduces its activity and hence decreases the rate of reaction of equilibrium in the above reaction.

9. Where is the intermediate theory used?
a) Heterogeneous components
b) Homogeneous components
c) Miscible components
d) Immiscible components
Answer: a
Clarification: Intermediate theory is the process in which an intermediate product is formed and hence it is suitable for homogeneous components and not for heterogeneous components.

10. Calculate the amount of intermediate components formed from the given data.
S=12
T=10
C=12
a) 1
b) 0
c) 2
d) 10
Answer: d
Clarification: we know that the amount of intermediate components formed is given by the formula s*c/12. So here we have 12*10/12=10.

250+ TOP MCQs on Molecular Orbital Theory and Answers

Engineering Chemistry Multiple Choice Questions on “Molecular Orbital Theory”.

1. If the sign of the wave function is unchanged when the orbital is reflected about its centre, the orbital is ______________
a) Gerade
b) Ungerade
c) Gerade as well as Ungerade
d) None of the mentioned
Answer: a
Clarification: If the sign of the wave function is unchanged when the orbital is reflected about its center (i.e., x, y and z are replaced by –x, -y and –z), the orbital is gerade.

2. The filling of molecular orbital takes place according to __________
a) The Aufbau Principle
b) Pauli Exclusion Principle
c) Hund’s rule of maximum multiplicity
d) All of the mentioned
Answer: d
Clarification: According to these principles, molecular orbitals are filled in order of increasing energy. Two electrons in the same orbital cannot have the same set of all four quantum numbers identical.

3. Bond Order of O₂, F₂, N₂ respectively are __________
a) +1, +2, +3
b) +2, +3, +1
c) +2, +1, +3
d) +3, +2, +1
Answer: c
Clarification: Bond Order=[[Number of electrons in bonding molecular orbital]-[Number of electrons in antibonding molecular orbital]]/2.

4. Arrange the following molecules in decreasing bond length.
a) O₂ > O₂^– > O₂⁺ > O₂^2-
b) O₂^2- > O₂^– > O₂ > O₂⁺
c) O₂^2- > O₂^– > O₂⁺ > O₂
d) O₂^– > O₂⁺ > O₂^2- > O₂
Answer: b
Clarification: The bond length is inversely proportional to the bond order. Therefore, the correct is:
O₂^2- > O₂^– > O₂ > O₂⁺.

5. Arrange the following molecules in the order of increasing stability.
a) N₂⁺ < N₂ < N₂^– < N₂^2-
b) N₂^2- < N₂^– < N₂ < N₂⁺
c) N₂^2- < N₂^– = N₂⁺ < N₂
d) N₂ < N₂⁺ = N₂^– < N₂^2-
Answer: c
Clarification: The order of stability is directly proportional to the bond order. Therefore, the correct order of stability is N₂^2- < N₂^– = N₂⁺ < N₂.

6. On the basis of molecular orbital theory, select the most appropriate option.
a) The bond order of O₂ is 2.5 and it is paramagnetic
b) The bond order of O₂ is 1.5 and it is paramagnetic
c) The bond order of O₂ is 2 and it is diamagnetic
d) The bond order of O₂ is 2 and it is paramagnetic
Answer: d
Clarification: Oxygen is paramagnetic in nature and its bond order is 2. It is an exceptional case.

7. Which of the following molecule does not exist due to its zero bond order?
a) H₂⁺
b) He₂⁺
c) He₂
d) H₂^–
Answer: c
Clarification: Molecular orbital electronic configuration of He₂ molecule = (σ_1s)² (σ*_1s)². Bond order = 0, so He₂ molecule does not exist.

8. The relative energies of molecular orbitals in increasing order have been found to be as follows.
(σ_1s) < (σ*_1s) < (σ_2s) < (σ*_2s) < [(π_2py)(π_2pz)] < (σ _2px) < [(π*_2py)(π*_2pz)] < (σ*_2px)
a) For O₂ to Ne₂
b) For H₂ to N₂
c) For H₂ to Ne₂
d) For N₂ to Ne₂
Answer: b
Clarification: The relative energies of molecular orbitals in increasing order have been found to be (σ_1s) < (σ*_1s) < (σ_2s) < (σ*_2s) < [(π_2py)(π_2pz)] < (σ_2px) < [(π*_2py)(π*_2pz)] < (σ*_2px) for H₂ to N₂ only.

9. According to Molecular Orbital Theory, the shape and size of a molecular orbital depends upon _________
a) Shape and size of the combining atomic orbitals
b) Numbers of the combining atomic orbitals
c) Orientation of the combining atomic orbitals
d) All of the mentioned
Answer: d
Clarification: The shape and size of a molecular orbital depend upon the shape, size, number and orientation of the combining atomic orbitals.

10. Choose the incorrect statement from the following options.
a) In bonding molecular orbital, electron density is low in the region between the nuclei of bonded atoms
b) The energy of antibonding molecular orbital is higher than that of atomic orbitals from which it is formed
c) Every electron in bonding molecular orbital contributes toward stability of the molecule
d) Antibonding takes place when lobes of atomic orbitals have different signs
Answer: a
Clarification: In bonding molecular orbital, the electron density is high in the region between the nuclei of bonded atoms. All the other options are correct. The energy of antibonding molecular orbital is higher than that of atomic orbitals from which it is formed. Every electron in bonding molecular orbital contributes to the stability of the molecule. Antibonding takes place when lobes of atomic orbitals have different signs.