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Chapter 6 of 15
Important Questions

Surface Chemistry

Tamil Nadu Board · Class 12 · Chemistry

Most important questions from Surface Chemistry for Tamil Nadu Board Class 12 Chemistry board exam 2026. MCQs, short answer, and long answer questions with marks.

44 questions40 flashcards5 concepts

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44 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

In the Bredig's arc method for preparation of colloidal gold, an electrical arc is struck between gold electrodes in ice-cold water. Why is the water kept ice-cold during this process?

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To prevent coagulation of colloidal gold particles by keeping thermal energy of particles low

Step 1: In Bredig's arc method, gold electrodes produce an arc that vaporises gold, which then condenses in water to form colloidal particles. Step 2: The newly formed colloidal particles are extremely small (1–200 nm) and thermodynamically unstable — they tend to aggregate (coagulate) due to their high surface energy. Step 3: At higher temperatures, colloidal particles gain more kinetic energy, collide more frequently, and coagulate more readily. Ice-cold water removes this thermal energy, reducing particle motion and collision frequency. Step 4: Option A is incorrect — conductivity is not th

2multiple choice
1 marks

In the process of dialysis for purifying a colloidal solution of Fe(OH)₃, the colloidal solution is placed inside a semipermeable membrane bag suspended in flowing water. Which of the following statements CORRECTLY explains why this purification works?

Show answer

The pores of the semipermeable membrane allow small ions (like Fe³⁺ and Cl⁻ from FeCl₃ impurity) to pass through but retain larger colloidal Fe(OH)₃ particles

Step 1: Fe(OH)₃ sol is prepared by hydrolysis of FeCl₃: FeCl₃ + 3H₂O → Fe(OH)₃ (sol) + 3HCl. The impurity is excess FeCl₃ or HCl (electrolytes containing small Fe³⁺, Cl⁻ ions). Step 2: The semipermeable membrane has pores of size intermediate between colloidal particles (1–200 nm) and true solution ions (< 1 nm). Step 3: Small electrolyte ions (Fe³⁺, Cl⁻, H⁺) pass freely through the membrane pores, diffusing into the flowing water and being carried away. Colloidal Fe(OH)₃ particles (larger size) cannot pass through and are retained. Step 4: Option A is wrong — charge of colloidal particles is

3multiple choice
1 marks

Consider the following statement about chemisorption: 'When temperature is raised, chemisorption first increases and then decreases.' Which of the following CORRECTLY explains this observation?

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Initial increase occurs because chemisorption requires activation energy — higher temperature activates more molecules; decrease at high temperature is due to desorption as kinetic energy of adsorbate exceeds adsorption energy

Step 1: Chemisorption involves formation of chemical bonds between adsorbent and adsorbate, which requires activation energy (unlike physisorption). Step 2: At low temperatures, molecules do not have sufficient energy to overcome the activation energy barrier. As temperature increases, more molecules gain enough energy → chemisorption increases. This is why chemisorption is also called 'activated adsorption'. Step 3: At very high temperatures, the kinetic energy of adsorbed molecules becomes very large. These molecules overcome the adsorption forces and escape from the surface — this is called

4multiple choice
1 marks

In the Haber's process for ammonia synthesis: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), iron is used as a catalyst and molybdenum as a promoter. H₂S is a catalyst poison. Which of the following CORRECTLY explains the action of H₂S as a catalyst poison based on adsorption theory?

Show answer

H₂S molecules are preferentially and strongly adsorbed on the active centres (steps, cracks, corners) of the Fe catalyst surface, blocking N₂ and H₂ from adsorbing

Step 1: According to adsorption theory, heterogeneous catalysis occurs at 'active centres' — coordinatively unsaturated atoms at steps, cracks, and corners of the catalyst surface that have residual bonding forces. Step 2: H₂S molecules have a strong affinity for iron surfaces and get chemisorbed onto these active centres with great strength. Since chemisorption forms monolayers, these sites become permanently occupied. Step 3: With active centres blocked, N₂ and H₂ molecules cannot adsorb on the Fe surface → no activated complex formation → no catalysis. Step 4: Option A is wrong — H₂S does n

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Frequently Asked Questions

What are the important topics in Surface Chemistry for Tamil Nadu Board Class 12 Chemistry?
Key topics in Surface Chemistry include Surface Chemistry – Complete Chapter Overview, Comparison flowchart showing the key differences between adsorption and absorption processes, Comparison of physical and chemical adsorption characteristics. These are the concepts Tamil Nadu Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Surface Chemistry — Tamil Nadu Board Class 12 Chemistry?
Understand the core concepts first, then work through the 44 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many important questions are there in Surface Chemistry?
There are 44 practice questions available for Surface Chemistry. These cover multiple question types including MCQs, short answer, and long answer questions.

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