Category: Engineering Chemistry Objective Questions

Engineering Chemistry Multiple Choice Questions on “Homonuclear and Heteronuclear Molecules”.

1. Which of the following molecule is not homonuclear?
a) H₂
b) N₂
c) NO
d) O₂
Answer: c
Clarification: NO is Heteronuclear diatomic molecule in which atomic number of nitrogen is 7 and that of oxygen is 8 i.e., total number of electrons = 15.

2. Bond order of NO⁺ molecule is _________
a) 2
b) 3
c) 2.5
d) 4
Answer: b
Clarification: Bond order = (10-4)/2 = 3.

3. Which of the following molecule is least stable?
a) NO⁺
b) N₂
c) NO
d) O₂
Answer: c
Clarification: In NO molecule there is a significant difference of about 250 kJ/mol in the energy of atomic orbitals involved.

4. The molecular orbital electronic configuration of HF molecule is _________
a) 1s² 2s² σ_spx² [2_py² 2_pz²] σ_spx*
b) 1s² 2s² σ_spx² [2_px² 2_py²] σ_spx*
c) 1s² 2s² σ_spx² [2_px² 2_pz²] σ_spx*
d) 1s² 2s² σ_spx² [2_px⁴] σ_spx*
Answer: a
Clarification: HF molecule has 10 electrons and its electronic configuration is 1s² 2s² σ_spx² [2_py² 2_pz²] σ_spx*.

5. From the following options, choose the heteronuclear diatomic molecules which are paramagnetic in nature?
a) HF and NO
b) HF and O₂
c) NO and O₂
d) Only NO
Answer: d
Clarification: HF is diamagnetic in nature and O₂ is homonuclear molecule. So, only NO is a heteronuclear diatomic molecule that is paramagnetic in nature.

6. Calculate the number of σ and π bonds in cyclohexadiene. (C₆H₈)
a) σ bond = 2, π bond = 14
b) σ bond = 14, π bond = 2
c) σ bond = 12, π bond = 2
d) σ bond = 12, π bond = 4
Answer: b
Clarification: Cyclohexadiene molecule has two double bonds(hence two π-bonds) and one ring.

7. The combination of H (1s¹) and F (2px¹) gives _________
a) Bonding orbital
b) Antibonding orbital
c) Both bonding and antibonding orbital
d) None of the mentioned
Answer: c
Clarification: The combination of H (1s1) and F (2px1) gives both bonding (σ_spx) and antibonding (σ_spx*) orbitals.

8. Choose the incorrect option from the following.
a) Valence bond theory does not explain the paramagnetic nature of O₂
b) Molecular orbital theory explains the extra stability of O₂⁺ cation over O₂
c) Valence bond theory explains the ionization or gain of electrons, giving O₂⁺ and O₂^– ions, if O₂ has the stable octet
d) Resonance has no role in Molecular orbital theory
Answer: c
Clarification: Valence bond theory does not explain the ionization or gain of electrons, giving O₂⁺ and O₂^– ions, if O₂ has the stable octet.

9. Which of the following molecules have bond order equal to 1?
a) NO, HF, HCl, Li₂, CO
b) H₂, Li₂, B₂, HF, HCl
c) Li₂, B₂, CO, NO,He₂⁺
d) B₂, CO, He₂⁺, NO, HF
Answer: b
Clarification: He₂⁺ = 0.5, H₂, Li₂, B₂, HF, HCl = 1, NO = 2.5, CO = 3.

10. The interaction between a pair of orbitals of the same type is _________
a) Attractive
b) Repulsive
c) There is no interaction
d) None of the mentioned
Answer: b
Clarification: The interaction between pair of orbitals of the same type is repulsive like between σ_2s and σ_2p orbitals.

Engineering Chemistry Multiple Choice Questions on “Distillation”.

1. The process of heating a liquid mixture to form vapours and then cooling the vapours to get pure component is called ____________
a) Crystallisation
b) Distillation
c) Chromatography
d) Sublimation
Answer: b
Clarification: The process of heating a liquid mixture to form vapours and then cooling the vapours to get pure component is called distillation. Distillation is a process of separating the component substances from a liquid mixture by selective evaporation and condensation.

2. Porcelain pieces are put into the distillation flask to avoid ____________
a) Overheating
b) Uniform boiling
c) Bumping of the solution
d) None of the mentioned options
Answer: c
Clarification: Porcelain pieces are put into the distillation flask to avoid bumping of the solution due to uneven heating.

3. The boiling point of chloroform is ____________
a) 334 K
b) 286 K
c) 350 K
d) 298 K
Answer: a
Clarification: The boiling point of chloroform is 334 K. Chloroform is an organic compound with formula CHCl₃. It is a colorless, sweet-smelling, dense liquid that is produced on a large scale as a precursor to PTFE and refrigerants.

4. The boiling point of aniline is ____________
a) 438 K
b) 370 K
c) 338 K
d) 457 K
Answer: d
Clarification: The boiling point of aniline is 457 K. Aniline is a toxic organic compound with the formula C₆H₅NH₂.

5. How aniline and chloroform can be separated?
a) Sublimation
b) Condensation
c) Distillation
d) Evaporation
Answer: c
Clarification: Aniline and chloroform can be separated through the distillation process. Aniline is a toxic organic compound with the formula C₆H₅NH₂. Chloroform is an organic compound with formula CHCl₃. It is a colorless, sweet-smelling, dense liquid that is produced on a large scale as a precursor to PTFE and refrigerants.

6. Which of the following is not separated through distillation process?
a) Acetone and water
b) Aniline and chloroform
c) Impurities in Sea water
d) Milk and water
Answer: d
Clarification: Milk and water are not separated through the distillation process. All the other options can be separated through a distillation process.

7. Which of the following will vaporize faster?
a) Aniline
b) Chloroform
c) Water
d) Kerosene
Answer: b
Clarification: Chloroform will vaporize faster than aniline and water. Chloroform is an organic compound with formula CHCl₃. It is a colorless, sweet-smelling, dense liquid that is produced on a large scale as a precursor to PTFE and refrigerants.

8. The distilled water is collected in ____________
a) Receiver
b) Adapter
c) Condenser
d) Round bottom flask
Answer: a
Clarification: The distilled water is collected in the receiver. An installation for distillation, especially of alcohol, is a distillery. The distillation equipment is still.

9. The process of distillation is used for the liquids having ____________
a) Sufficient difference in their boiling point
b) Sufficient difference in their melting point
c) Sufficient difference in their solubility
d) None of the mentioned
Answer: a
Clarification: The process of distillation is used for the liquids having a sufficient difference in their boiling point. Distillation also permits the separation of air into its components.

10. The residue in the round bottom flask is ____________
a) Volatile
b) Non volatile
c) None of the mentioned
d) Volatile & Non volatile
Answer: b
Clarification: The residue salt int he round bottom flask is non volatile in nature. The process of distillation is used for the liquids having a sufficient difference in their boiling point. Distillation also permits the separation of air into its components.

Engineering Chemistry Online test focuses on “Vibrational Spectroscopy”.

1. Vibrational spectroscopy involves the transitions falling in the spectral range of ____________
a) 100-1000 cm^-1
b) 300-3000 cm^-1
c) 400-4000 cm^-1
d) 500-5000 cm^-1
Answer: c
Clarification: Vibrational spectroscopy involves the transitions falling in the spectral range of 400-4000 cm^-1 (infrared region).

2. Which of the region of IR spectra appears between (1400-600) cm^-1?
a) Functional group region
b) Fingerprint region
c) Low-frequency region
d) None of the mentioned
Answer: b
Clarification: Fingerprint region of IR spectra appears between (1400-600) cm^-1.

3. Select the correct statement from the following option.
a) Infrared spectra can identify the unknown materials
b) It can determine the amount of components in a mixture
c) It can also determine the quality of a sample
d) All of the mentioned
Answer: d
Clarification: Infrared spectra can identify the unknown materials. It can determine the amount of components in a mixture and can also determine the quality of a sample.

4. Which of the following molecule have infrared active vibrations?
a) NO
b) CH₄
c) H₂
d) All of the mentioned
Answer: a
Clarification: NO molecule have infrared active vibrations. Infrared spectra can identify unknown materials. It can determine the amount of components in a mixture and can also determine the quality of a sample.

5. Which of the following cannot show a vibrational absorption spectrum?
a) OCS
b) H₂O
c) CO₂
d) C H₂ = C H₂
Answer: d
Clarification: C H₂ = C H₂ cannot show vibrational absorption spectrum. OCS, H₂O and CO₂ can show a vibrational absorption spectrum.

6. Which of the following is not a type of bending molecular vibration?
a) Twisting
b) Stretching
c) Wagging
d) Rocking
Answer: b
Clarification: Stretching is not a type of bending molecular vibration. Twisting, wagging and rocking are types of bending molecular vibration.

7. What is the absorption frequency(cm^-1) of –C=N functional group?
a) 3610-3640
b) 1690-1760
c) 1180-1360
d) 2850-2960
Answer: c
Clarification: The absorption frequency for –C=N group is 1180-1360 cm^-1. Absorption spectroscopy refers to spectroscopic techniques that measure the absorption of radiation, as a function of frequency or wavelength, due to its interaction with a sample.

8. Presence of a functional group in a compound can be established by using _____________
a) Chromatography
b) IR spectroscopy
c) Mass spectroscopy
d) X-ray diffraction
Answer: b
Clarification: Presence of a functional group in a compound can be established by using IR spectroscopy. It is the spectroscopy that deals with the infrared region of the electromagnetic spectrum, that is light with a longer wavelength and lower frequency than visible light.

9. Select the incorrect option from the following option.
a) IR spectroscopy helps in the determination of purity
b) IR spectroscopy helps in the determination of force constant from vibrational spectrum
c) IR spectroscopy helps in identifying an unknown compound
d) None of the mentioned
Answer: d
Clarification: All the options are correct. IR spectroscopy helps in the determination of purity, force constant from the vibrational spectrum and identifying an unknown compound.

10. IR spectroscopy helps in detecting the presence of hydrogen bonding.
a) True
b) False
Answer: a
Clarification: IR spectroscopy helps in detecting the presence of hydrogen bonding. It is the spectroscopy that deals with the infrared region of the electromagnetic spectrum, that is light with a longer wavelength and lower frequency than visible light.

To practice all areas of Engineering Chemistry for online tests,

Engineering Chemistry MCQs focuses on “Effect of Temperature on Reaction Rate”.

1. Temperature dependence of reaction rates can be studied by plotting a graph between ___________
a) Concentration of reactants and temperature
b) Concentration of products and temperature
c) Rate constant and temperature
d) Rate of catalysis and temperature
Answer: c
Clarification: Temperature dependence of reaction rates can be studied by plotting a graph between rate constant and temperature for different reactions.

2. The ratio of the rate constant of a reaction at two temperatures differing by __________⁰C is called temperature coefficient of reaction.
a) 2
b) 10
c) 100
d) 50
Answer: b
Clarification: The ratio of the rate constant of a reaction at two temperatures differing by 10⁰C is called temperature coefficient of reaction.

3. According to Arrhenius equation, rate constant(k) is proportional to ___________
a) Activation Energy (E)
b) e^E
c) e^1/E
d) e^-E
Answer: d
Clarification: According to Arrhenius equation, k = Ae^-Ea/RT. So, rate constant(k) is proportional to e^-E.

4. How does half life period of a first order reaction vary with temperature?
a) It increases
b) It decreases
c) It remains the same
d) Both increases as well as decrease
Answer: a
Clarification: Half-life period of a first order reaction is directly proportional to the rate constant. So, it increases with increase in temperature.

5. How many times the rate of reaction increases at 20⁰C for a reaction having the activation energies in the presence and absence of catalyst as 50 kJ/mol and 75 kJ/mol?
a) 1000
b) 10000
c) 30000
d) 50000
Answer: c
Clarification: The rate of reaction will increase 28,592 times, i.e. 30,000 times.

6. Which of the following method is used to determine the order of the reaction in which two or more reactants take part?
a) Integration method
b) Half life period method
c) Graphical method
d) Ostwald’s isolation method
Answer: d
Clarification: Ostwald’s isolation method is used to determine the order of the reaction in which two or more reactants take part. Kinetic isolation conditions were identified that enabled determination of the reaction order for interfacial charge recombination.

7. Which of the following method is satisfactory only for a simple homogeneous reaction?
a) Integration method
b) Half life period method
c) Graphical method
d) Ostwald’s isolation method
Answer: a
Clarification: Integration method is satisfactory only for simple homogeneous reaction and lead to wrong conclusion for complex reactions.

8. A zero order reaction is one ___________
a) In which rate is independent of reactants concentration
b) In which one of the reactant is in large excess
c) Whose rate is not affected by time
d) Whose rate increases with time
Answer: a
Clarification: A zero order reaction is one in which rate is independent of reactants concentration. It only depends upon the rate constant.

To practice MCQs on all areas of Engineering Chemistry,

Engineering Chemistry Multiple Choice Questions on “Polymer Blends”.

1. The physical mixture of two or more polymers that are not linked by covalent bonds is called ____________
a) Polymerisation
b) Dendrimer
c) Polyblend
d) Multiblend
Answer: c
Clarification: The physical mixture of two or more polymers that are not linked by covalent bonds is called polyblend or polymer blend. They are blended together to create a new material with different physical properties.

2. Polymers are dissolved in common solvent to give ____________
a) Mechanical blends
b) Solution-cast blends
c) Latex blends
d) Chemical blends
Answer: b
Clarification: Polymers are dissolved in common solvent to give solution-cast blends. If the blend is made of two polymers, two glass transition temperatures will be observed.

3. For amorphous and semi-crystalline polymers, mixing leads to the formation of ____________
a) Mechanical blends
b) Solution-cast blends
c) Latex blends
d) Chemical blends
Answer: a
Clarification: For amorphous and semi-crystalline polymers, mixing leads to the formation of mechanical blends. The macroscopically uniform properties are usually caused by sufficiently strong interactions between the component polymers.

4. The immiscible blends exhibit ____________
a) Excellent mechanical properties
b) Poor mechanical properties
c) Poor chemical properties
d) Intermediate chemical properties
Answer: b
Clarification: The immiscible blends exhibit poor mechanical properties. This is by far the most populous group. If the blend is made of two polymers, two glass transition temperatures will be observed.

5. Cross-linked polymers are swollen with a different monomer, which results in ____________
a) Mechanical blends
b) Solution-cast blends
c) Latex blends
d) Chemical blends
Answer: d
Clarification: Cross-linked polymers is swollen with a different monomer, which results in chemical blends. They have different chemical properties.

6. Select the incorrect option from the following option.
a) Immiscible blends are usually opaque
b) Immiscible blends exhibit a single glass transition temperature (T_g) intermediate between those of individual components
c) Miscible blends are usually clear
d) Immiscible blends exhibit separate T_g’s characteristics of each component
Answer: b
Clarification: Miscible blends exhibit a single glass transition temperature (T_g) intermediate between those of individual components. All the other options are correct.

7. (PPO + PS) polyblend is commercially available engineering plastic Noryl (General Electric).
a) True
b) False
Answer: a
Clarification: (PPO + PS) polyblend is commercially available engineering plastic Noryl (General Electric). The use of the term polymer alloy for a polymer blend is discouraged, as the former term includes multi-phase co-polymers but excludes incompatible polymer blends.

8. Fine dispersions of polymers in water are mixed and coagulated to give ____________
a) Mechanical blends
b) Solution-cast blends
c) Latex blends
d) Chemical blends
Answer: c
Clarification: Fine dispersions of polymers in water are mixed and coagulated to give latex blends. The use of the term polymer alloy for a polymer blend is discouraged, as the former term includes multi-phase co-polymers but excludes incompatible polymer blends.

9. The blend is said to be synergistic for the mechanical property if the term (I) in the empirical formula is?
a) Zero
b) Positive
c) Negative
d) Imaginary
Answer: b
Clarification: The blend is said to be synergistic for the mechanical property if the term (I) in the empirical formula is positive. For amorphous and semi-crystalline polymers, mixing leads to the formation of mechanical blends.

10. The blend is said to be additive for the mechanical property if the term (I) in the empirical formula is ____________
a) Imaginary
b) Positive
c) Negative
d) Zero
Answer: d
Clarification: The blend is said to be additive for the mechanical property if the term (I) in the empirical formula is zero. For amorphous and semi-crystalline polymers, mixing leads to the formation of mechanical blends.

Engineering Chemistry Multiple Choice Questions on “Biodegradable Polymers”.

1. Which of the following is not required for the biodegradation process?
a) Micro-organism
b) Environment conditions
c) Adhesives
d) Substrate
Answer: c
Clarification: There are three essential conditions for the biodegradable process, micro-organism, environment and substrate. Micro-organisims biodegrade the substrate in the presence of a suitable environment.

2. Greater the hydrophilicity of the polymers ____________ is the rate of biodegradation.
a) Larger
b) Smaller
Answer: a
Clarification: Greater the hydrophilicity of the polymers, larger is the rate of biodegradation. Biodegradation increases with more hydrophilicity. Hydrophilicity means having an affinity for water; readily absorbing or dissolving in water.

3. Biodegradation will be more for ____________
a) More molecular weights and high crystallinity
b) Low molecular weights and high crystallinity
c) More molecular weights and less crystallinity
d) Low molecular weights and less crystallinity
Answer: d
Clarification: Biodegradation will be more for low molecular weights and less crystallinity. Low molecular weight compound can easily be broken into pieces by micro-organisms and hence improve the rate of biodegradation.

4. Which of the following is not an example of a natural biodegradable polymer?
a) Collagen
b) Polyvinyl alcohol
c) Lignin
d) Natural rubber
Answer: b
Clarification: Collagen, natural rubber, lignin etc are some of the examples of natural biodegradable polymer. Polyvinyl alcohol is not an example of a natural biodegradable polymer.

5. Which of the following is not an example of synthetic biodegradable polymer?
a) Polyvinyl alcohol
b) Poly gamma-glutamic acid
c) Polyanhydrides
d) PHBV
Answer: b
Clarification: Polyvinyl alcohol, polyanhydrides and PBV are examples of the synthetic biodegradable polymer. Poly gamma-glutamic acid is not an example of a synthetic biodegradable polymer.

6. Biodegradable polymers do not need to be land-filled, they will re-enter normal geo-chemical cycles over time.
a) True
b) False
Answer: a
Clarification: Biodegradable polymers do not need to be land-filled, they will re-enter normal geo-chemical cycles over time. They are easily decomposed in the natural environment and does not require any external agent.

7. PHB is used in ____________
a) Agricultural applications
b) Medical applications
c) Manufacture of shampoo bottles
d) Adhesives
Answer: c
Clarification: PHB is used in manufacture of shampoo bottles. Polyhydroxybutyrate (PHB) is a polyhydroxyalkanoate (PHA), a polymer belonging to the polyesters class that are of interest as biodegradable plastics.

8. Select the incorrect statement from the following option.
a) Biodegradable polymers are not suitable candidates in the recycling of commingled plastics
b) Biodegradable polymers are very expensive
c) Biodegradable polymers are an attractive option for addressing the solid waste and marine pollution
d) Biodegradable polymers are easily available
Answer: d
Clarification: Biodegradable polymers are not easily available. All the other options are correct. Biodegradable polymers are not suitable candidates in the recycling of commingled plastics. They are very expensive but an attractive option for addressing solid waste and marine pollution.