Surface Chemistry JEE Advanced Previous Year Questions with Solutions are provided here with simple step-by-step explanations. It is important for JEE aspirants to practice all problems and going through the problems and solutions listed here will help them to gain better clarity and marks in the main examination. The problems listed here are all important questions from past JEE Advanced papers that will further help them revise and clear concepts found in solving Physical Chemistry numerical. Students will also get a fair idea of the question types, difficulty level and pattern of the paper. They can further make a self-assessment by solving the JEE Advanced Surface Chemistry Questions with Solutions.
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JEE Advanced Previous Year Questions on Surface Chemistry
Question 1. Methylene blue from its aqueous solution is adsorbed on activated charcoal at 25oC. For this process, the correct statement is:
A. The adsorption requires activation at 25oC
B. The adsorption is accompanied by a decrease in enthalpy
C. Athe adsorption increases with increases in temperature
D. The adsorption is irreversible
Solution: (B)
The adsorption is accompanied by a decrease in enthalpy. Adsorption of methylene blue on activated charcoal is physical adsorption hence it is characterized by a decrease in enthalpy.
Hence (b) is correct.
Question 2. When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The TRUE statement(s) regarding this adsorption is(are)
A. O2 is physisorbed
B. Heat is released
C. Occupancy of π2p of O2 is increased
D. Bond length of O2 is increased
Solution: (B, C and D)
When O2 is adsorbed on a metallic surface, there is a transfer of electrons from the metal to O2.
Now if we look at the MO configuration of O2 it is as follows;
KK(σ2s)2(σ∗2s)2(σ2px)2(π2py)2(π2pz)2(π∗2py)1(π∗2pz)1
On the other hand, the MO configuration of O2– is;
KK(σ2s)2(σ∗2s)2(σ2px)2(π2py)2(π2pz)2(π∗2py)1(π∗2pz)1
(A) Oxygen is chemisorbed. So, statement (A) is false.
(B) Chemisorption is an exothermic process. Heat is released. Statement (B) is true.
(C) The occupancy of π2p* of O2 is increased. The added electron enters π2p* MO. Hence, statement (C) is also true.
(D) The bond order is decreased (as the electron enters antibonding MO) and the bond length of O2 is increased. Thus, the statement (D) is true.
Question 3. Match the catalysts to the correct processes.
| Catalyst | Process | ||
|---|---|---|---|
| (A) | TiCl3 | (i) | Wacker process |
| (B) | PdCl2 | (ii) | Ziegler- Natta polymerisation |
| (C) | CuCl2 | (iii) | Contact process |
| (D) | V2O5 | (iv) | Deacon’s process |
A. (A) – (iii), (B) – (ii), (C) – (iv), (D) – (i)
B. (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii)
C. (A) – (ii), (B) – (iii), (C) – (iv), (D) – (i)
D. (A) – (iii), (B) – (i), (C) – (ii), (D) – (iv)
Solution: (B)
A Ziegler–Natta catalyst, named after Karl Ziegler and Giulio Natta, is a catalyst used in the synthesis of polymers of 1-alkenes (alpha-olefins). The catalyst TiCl3 is used for Ziegler – Natta polymerization.
The Wacker Oxidation is an industrial process, which allows the synthesis of ethanal from ethene by palladium-catalyzed oxidation with oxygen. The catalyst PdCl2 is used for the Wacker process.
The Deacon process is a process used during the manufacture of alkalis (the initial end product was sodium carbonate) by the Leblanc process. The catalyst CuCl2 is used for Deacon’s process.
The contact process is the current method of producing sulfuric acid in the high concentrations needed for industrial processes. The catalyst V2O5 is used for the Contact process.
Hence, option B is correct.
Question 4. Rate of physisorption increases with:
A. Decrease in temperature
B. Increase in temperature
C. Decrease in pressure
D. Decrease in surface area
Solution: (A)
We have seen that physisorption usually increases with an increase in pressure. And like all adsorption processes, physisorption would increase with an increase of surface area.
Since adsorption in general or physisorption is an exothermic process, we would see a decrease in physisorption with an increase in temperature. So the reverse would be true.
Therefore, physisorption would increase with a decrease in temperature.
Question 5. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is:
A. CH3(CH2)15N+(CH3)3Br–
B. CH3(CH2)11OSO–3Na+
C. CH3(CH2)6COO–Na+
D. CH3(CH2)11N+(CH3)3Br–
Solution: (A)
Among the given options, the surfactant that will form micelles in an aqueous solution at the lowest molar concentration at ambient conditions is CH3(CH2)15N+(CH3)3Br–. So, larger the hydrophobic fragment of surfactant easier will be the micellisation and smaller the critical micelle concentration.
Question 6. The qualitative sketches I, II and III given below show the variation of surface tension with a molar concentration of three different aqueous solutions of KCl, CH3OH and CH3(CH2)111OSO3–Na+ at room temperature.
The correct assignment of the sketches is:
A. I:KCl; II:CH3OH; III:CH3(CH2)11OSO3– Na+
B. I:CH3(CH2)11OSO3– Na+: II:CH3OH; III:KCl
C. I:KCl; II:CH3(CH2)11OSO3– Na+; III:CH3OH
D. I:CH3OH; II:KCl; III:CH3(CH2)11OSO3– Na+
Solution: (D)
A solution of CH3OH and water shows positive deviation from Raoult’s law, it means by adding
CH3OH intermolecular force of attraction decreases and hence surface tension decreases.
By adding KCl in water, the intermolecular force of attraction increases slightly, so surface tension increases by a small value.
By adding surfactants like CH3(CH2)11OSO3– Na+ surface tension decreases rapidly and after forming micelle it slightly increases.
Question 7. The correct statement(s) about surface properties is(are):
A. Adsorption is accompanied by a decrease in enthalpy and a decrease in entropy of the system
B. The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on the same amount of activated charcoal at a given temperature
C. Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium
D. Brownian motion of colloidal particles does not depend on the size of the particles but depends on the viscosity of the solution
Solution: (A and B)
A. Since adsorption is spontaneous, ΔG for the process is – ve. Adsorption is accompanied by a decrease in randomness. Therefore ΔS and TΔS for the process is also negative. As ΔS for the process is –ve and the process is spontaneous, ΔH for the process has to be –ve i.e, enthalpy of the system decreases.
B. Under a given set of conditions of temperature and pressure the easily liquefiable gases e.g. C2H6, NH3 and HCl are adsorbed more than the gases like N2, H2 and CO. The ease with which a gas can be liquefied is determined by its critical temperature.
The critical temperature is the minimum temperature above which a gas can be liquified. This implies that gases with high critical temperature values can be easily liquified as compared to gases with low critical temperature values.
Question 8. Choose the correct reason(s) for the stability of the lyophobic colloidal particles.
A. Preferential adsorption of ions of their surface from the solution
B. Preferential adsorption of solvent on their surface from the solution
C. Attraction between different particles having opposite charges on their surface
D. Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles
Solution: (A and D)
Lyophobic sol gets stabilised by preferential adsorption of ions on their surface, thus developing a potential difference between the fixed layer and the diffused layer. Thus, options (A) and (D) are correct.
Question 9. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is (are);
A. Adsorption is always exothermic
B. Physisorption may transform into chemisorption at high temperatures
C. Physisorption increases with increasing temperature but chemisorption decreases with increasing temperature
D. Chemisorption is more exothermic than physisorption; however, it is very slow due to the higher energy of activation
Solution: (A, B and D)
A. Adsorption is always exothermic (factual statement).
B. It is also a factual statement.
C. The statement is wrong. Physisorption decreases with an increase in temperature. But chemisorption first increases with an increase in temperature and then decreases. The effect is called activated adsorption.
D. It is a factual statement.
Question 10. The given graph/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct?
A. I is physisorption and II is chemisorption
B. I is physisorption and III is chemisorption
C. IV is chemisorption and II is chemisorption
D. IV is chemisorption and III is chemisorption
Solution: (A and C)
Graph I highlights physisorption wherein absorbents are bonded to adsorbate through weak van der Waals’ force. Increasing temperature increases the kinetic energy of adsorbed particles increasing the rate of desorption, hence adsorption decreases.
Graph II shows chemisorption because in chemisorption, the amount of adsorption increase with the increase in temperature.
Graph – IV represents chemisorption as the graph is a characteristic of bonding forces. Attractive forces dominate until bond formation. When we try to reduce the distance after this, repulsive forces dominate.
Chemical Kinetics and Surface Chemistry – JEE Advanced Questions
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