Updated NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry + Important Questions 2026

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Class 12 Chemistry Chapter 5

Surface Chemistry Complete NCERT Resource Guide

Hey Lucky! Are you looking to score full marks in your board exams? This ultimate guide provides Updated NCERT Solutions for Class 12 Chemistry Chapter 5, Surface Chemistry. This chapter is highly theoretical, scoring, and critical for your CBSE boards and competitive exams like NEET or JEE. Let's break down every concept with simple explanations and clear answers to boost your preparation!

Chapter NameSurface Chemistry
SubjectChemistry
ClassClass 12
BoardCBSE & State Boards
Important TopicsPhysisorption vs Chemisorption, Freundlich Isotherm, Enzyme Catalysis, Colloids Classification, Tyndall Effect, Hardy-Schulze Rule
Difficulty LevelEasy to Moderate (Theory Based)
Exam Weightage4-5 Marks
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Learning Objectives

After completing this chapter, students will be able to:

Key Concepts, Definitions & Formulas

Let's quickly revise the core terms and equations before moving to the solutions:

Full NCERT Solutions (Step-by-Step)

Here are the complete, detailed NCERT Solutions for Class 12 Chemistry Chapter 5 following the exact CBSE evaluation standards.

Question 5.1: Distinguish between adsorption and absorption with examples.

Step 1: Comparison Table. The primary differences between adsorption and absorption are as follows:

Feature Adsorption Absorption
Definition Surface phenomenon where particles accumulate only at the surface. Bulk phenomenon where particles penetrate uniformly throughout the substance.
Concentration Higher at the surface than in the bulk. Uniform throughout the bulk of the system.
Rate Rapid in the beginning and slows down gradually. Occurs at a uniform rate throughout the process.
Example Water vapor adsorbed on silica gel. Water vapor absorbed by anhydrous calcium chloride.

Question 5.2: What is the difference between physisorption and chemisorption?

Step 1: Mechanistic Differences. These two types of adsorption differ based on the nature of forces between the adsorbate and adsorbent:

Feature Physisorption (Physical Adsorption) Chemisorption (Chemical Adsorption)
Forces Weak van der Waals forces. Strong chemical bonds (covalent or ionic).
Specificity Non-specific; any gas can be adsorbed. Highly specific; occurs only if bond formation is possible.
Reversibility Reversible ($\text{Gas} + \text{Solid} \rightleftharpoons \text{Gas/Solid} + \text{Heat}$). Irreversible.
Enthalpy Low enthalpy ($20\text{--}40 \text{ kJ mol}^{-1}$). High enthalpy ($80\text{--}240 \text{ kJ mol}^{-1}$).
Layers Forms multimolecular layers. Forms a monomolecular layer.
Activation Energy No appreciable activation energy required. Requires measurable activation energy.

Question 5.3: Give reasons why a finely divided substance is more effective as an adsorbent.

Step 1: Surface Area Analysis. Adsorption is a surface phenomenon. The extent of adsorption depends directly on the available surface area of the adsorbent.
Step 2: Particle Sub-division. When a solid substance is finely divided or powdered, its effective surface area increases significantly.
Step 3: Site Multiplication. This provides a much larger number of active sites for the adsorbate particles to bind, making it a highly effective adsorbent.

Question 5.4: What are the factors which influence the adsorption of a gas on a solid?

Step 1: Core Influencing Parameters. Several factors influence the adsorption of gases on solid surfaces:
  • 1. Nature of the Adsorbate Gas: Easily liquefiable gases (with higher critical temperatures like $\text{CO}_2$, $\text{NH}_3$, $\text{SO}_2$) are adsorbed more readily because their intermolecular van der Waals forces are stronger.
  • 2. Surface Area of the Adsorbent: Greater surface area translates directly to higher adsorption capacity.
  • 3. Effect of Temperature: Physisorption decreases with an increase in temperature, whereas chemisorption first increases (due to activation energy requirements) and then decreases.
  • 4. Effect of Pressure: At constant temperature, the adsorption of a gas increases with an increase in pressure.

Question 5.5: What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

Step 1: Functional Definition. An adsorption isotherm is a graph that plots the variation in the mass of gas adsorbed per unit mass of solid adsorbent ($\frac{x}{m}$) against the equilibrium pressure ($p$) at a constant temperature.
Step 2: Empirical Framework. Freundlich gave an empirical equation to describe the behavior of adsorption over a limited range of pressure: $$\frac{x}{m} = k \cdot p^{1/n}$$ Where $x$ is the mass of gas adsorbed, $m$ is the mass of adsorbent, $p$ is the equilibrium pressure, and $k$ and $n$ are constants depending on the nature of the gas and adsorbent at a particular temperature ($n > 1$).
Step 3: Linear Validation. Taking log on both sides gives a linear relationship used to verify the isotherm: $$\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log p$$

Question 5.6: What do you understand by activation of adsorbent? How is it achieved?

Step 1: Definition of Activation. Activation of an adsorbent means increasing its adsorbing power or capacity by expanding its effective surface area.
Step 2: Implementation Methods. It can be achieved through the following methods:
  • Mechanical Powdering: Breaking the adsorbent into fine particles to increase surface area.
  • Chemical or Thermal Treatment: Heating charcoal in superheated steam or vacuum to remove previously adsorbed gases and unpack blocked pores (creating activated charcoal).

Question 5.7: What role does adsorption play in heterogeneous catalysis?

Step 1: Mechanistic Steps. Heterogeneous catalysis primarily operates based on the Adsorption Theory, which involves five main steps:
  1. Diffusion of reactant molecules to the surface of the catalyst.
  2. Adsorption of reactant molecules onto the active sites of the catalyst surface.
  3. Chemical reaction on the catalyst surface to form an intermediate complex.
  4. Desorption of reaction products away from the catalyst surface to make sites available again.
  5. Diffusion of product molecules away from the catalyst.

Question 5.8: Why is adsorption always exothermic?

Step 1: Surface Energy Resolution. Adsorption involves the accumulation of particles on a surface, which satisfies residual attractive unbalanced forces of the surface. Consequently, there is a decrease in surface energy, which is released as heat. Therefore, the enthalpy change ($\Delta H$) is always negative (exothermic).
Step 2: Entropy Considerations. Furthermore, gas molecules lose their freedom of movement upon adsorption, meaning the entropy change ($\Delta S$) is also negative.
Step 3: Gibbs Free Energy Equilibrium. For a spontaneous process, $\Delta G = \Delta H - T\Delta S$ must be negative, which is only possible when $\Delta H$ is sufficiently negative to overcome the positive $-T\Delta S$ term.

Question 5.9: How are colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?

Step 1: Complete Layout Grid. Based on the physical states of the dispersed phase and dispersion medium, there are 8 types of colloidal systems:

Dispersed Phase Dispersion Medium Type of Colloid Common Examples
Solid Solid Solid Sol Colored gemstones, milky glass
Solid Liquid Sol Paints, cell fluids, ink
Solid Gas Aerosol Smoke, dust storm
Liquid Solid Gel Cheese, butter, jellies
Liquid Liquid Emulsion Milk, hair cream, vanishing cream
Liquid Gas Aerosol Fog, mist, cloud, insecticide sprays
Gas Solid Solid Foam Pumice stone, foam rubber
Gas Liquid Foam Froth, whipped cream, soap lather

Note: A gas-in-gas system always forms a homogeneous mixture/solution, not a heterogeneous colloidal system.

Question 5.10: Discuss the effect of pressure and temperature on the adsorption of gases on solids.

Step 1: Effect of Pressure. Adsorption increases with increasing pressure. According to the Freundlich isotherm, at low pressure $\frac{x}{m} \propto p^1$, at intermediate pressure $\frac{x}{m} \propto p^{1/n}$, and at high pressure, the surface becomes completely saturated, making adsorption independent of pressure ($\frac{x}{m} \propto p^0$).
Step 2: Effect of Temperature. Adsorption is an exothermic process. According to Le Chatelier's principle, increasing the temperature shifts the equilibrium backward, decreasing physisorption. Chemisorption initially increases with temperature due to the requirement of activation energy, but decreases at higher temperatures as bonds begin to break.

Question 5.11: What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?

Step 1: Lyophilic Sols (Liquid-loving). Colloidal sols formed directly by mixing substances (like gum, gelatine, starch) with a suitable liquid dispersion medium. They are highly stable and reversible.
Step 2: Lyophobic Sols (Liquid-hating). Sols prepared only by special methods because the substances (like metals, metal sulfides) have no affinity for the dispersion medium. They are unstable and irreversible.
Step 3: Coagulation Vulnerability. Hydrophobic particles are stabilized solely by the presence of identical electrical charges on their surface which cause them to repel one another. When an electrolyte is added, these stabilizing surface charges are neutralized, causing the particles to aggregate and settle out (coagulate) easily.

Question 5.12: What is the difference between multimolecular, macromolecular and associated colloids? Give one example of each.

Step 1: Particle Architecture. The classification is based on the nature of the particles of the dispersed phase:
  • Multimolecular Colloids: Formed when a large number of atoms or smaller molecules (diameter $< 1 \text{ nm}$) aggregate together to form species of colloidal dimensions. Example: Gold sol, Sulfur sol ($S_8$).
  • Macromolecular Colloids: Formed when large molecules (macromolecules like polymers) having high molecular masses are dissolved in a suitable solvent such that their individual size falls in the colloidal range ($1\text{--}1000 \text{ nm}$). Example: Starch, Cellulose, Nylon.
  • Associated Colloids (Micelles): Substances that behave as normal strong electrolytes at low concentrations, but aggregate to exhibit colloidal behavior at higher concentrations above a specific temperature called the Kraft temperature ($T_k$) and concentration called the Critical Micelle Concentration (CMC). Example: Soap solution, synthetic detergents.

Question 5.13: What are enzymes? Write a short note on their mechanism.

Step 1: Structural Profile. Enzymes are biological catalysts. They are complex nitrogenous organic compounds (globular proteins) produced by living plants and animals that catalyze biochemical reactions with high specificity and efficiency.
Step 2: Lock and Key Model. The surface of an enzyme molecule contains active sites with specific geometric shapes and functional groups (like $-\text{NH}_2, -\text{COOH}, -\text{OH}$). The reactant molecules (substrates), matching these shapes like a key into a lock, bind to form an unstable intermediate complex, which then decomposes into products.

[Image diagram showing Lock and Key Mechanism of Enzyme Catalysis]

Step 3: Step-by-Step Kinetics:
  • Step 1: Binding of enzyme ($E$) to substrate ($S$) to form complex ($ES$): $E + S \rightarrow ES$
  • Step 2: Decomposition of the complex to yield products ($P$) and release the regenerated enzyme ($E$): $ES \rightarrow E + P$

Question 5.14: How are colloids classified on the basis of the nature of interactions between dispersed phase and dispersion medium?

Step 1: Affinity-Based Categories. Based on the nature of attractive forces and interactions between the phases, colloids are divided into two main categories:
  1. Lyophilic Colloids: Strong interaction/affinity between the dispersed phase and dispersion medium. They are self-stabilizing, reversible, and do not easily precipitate.
  2. Lyophobic Colloids: Minimal or no interaction between the dispersed phase and dispersion medium. They require stabilizing agents, are irreversible, and easily coagulate upon adding minimal electrolyte.

Question 5.15: Explain what is observed when:
(i) A beam of light is passed through a colloidal sol.
(ii) An electrolyte, $\text{NaCl}$, is added to hydrated ferric oxide sol.
(iii) An electric current is passed through a colloidal sol.

Step 1: Observation (i) - Tyndall Effect. The path of the beam becomes visible as a bright cone of light because colloidal particles are large enough to scatter light rays in all directions.
Step 2: Observation (ii) - Coagulation. Hydrated ferric oxide sol is positively charged. When $\text{NaCl}$ is added, the negatively charged chloride ions ($\text{Cl}^-$) neutralize the positive charges on the colloidal particles, causing them to clump together and precipitate out.
Step 3: Observation (iii) - Electrophoresis. The colloidal particles migrate toward the oppositely charged electrode due to their surface electrical charge. For instance, positive particles move to the cathode, where they discharge and precipitate.

Question 5.16: What are emulsions? What are their different types? Give one example of each type.

Step 1: Core Definition. An emulsion is a colloidal system in which both the dispersed phase and the dispersion medium are immiscible or partially miscible liquids.
Step 2: Sub-Type Breakdown. There are two main types of emulsions:
  1. Oil dispersed in Water (O/W type): Oil acts as the dispersed phase, and water acts as the continuous dispersion medium. Example: Milk, vanishing cream.
  2. Water dispersed in Oil (W/O type): Water acts as the dispersed phase, and oil acts as the continuous dispersion medium. Example: Butter, cod liver oil, cold cream.

Question 5.17: What is demulsification? Name two demulsifiers.

Step 1: Process Definition. Demulsification is the process of breaking an emulsion down into its constituent liquid layers.
Step 2: Implementation Channels. It can be achieved by heating, freezing, centrifuging, or adding chemical agents that destroy the emulsifying agent.
Step 3: Demulsifier Selection. Dehydrating alcohols, chemical surfactants/strong acids, or physical methods like high-velocity centrifuges serve as standard solutions.

Question 5.18: Action of soap is due to emulsification and micelle formation. Comment.

Step 1: Amphiphilic Structure. Soap molecules ($\text{RCOO}^-\text{Na}^+$) consist of a long hydrophobic hydrocarbon tail ($\text{R}-$) and a hydrophilic polar ionic head ($-\text{COO}^-$).
Step 2: Micellar Sequestration. When applied to greasy cloth in water, the hydrophobic tails dissolve in the oil/grease droplet, while the hydrophilic heads stick out into the surrounding water. Above the critical micelle concentration (CMC), these molecules form a spherical aggregate called a micelle.
Step 3: Emulsified Cleansing. The grease droplet is pulled away from the fabric surface into the water, forming a stable oil-in-water emulsion that is easily rinsed away during washing.

Question 5.19: Give four examples of heterogeneous catalysis.

Step 1: Industrial Implementations. Four industrial examples where the catalyst and reactants are in different physical states:
  1. Haber’s Process for Ammonia synthesis (Gas reactants, Solid catalyst): $$\text{N}_2(g) + 3\text{H}_2(g) \xrightarrow{\text{Fe}(s)} 2\text{NH}_3(g)}$$
  2. Contact Process for Manufacturing $\text{SO}_3$ (Gas reactants, Solid catalyst): $$2\text{SO}_2(g) + \text{O}_2(g) \xrightarrow{\text{V}_2\text{O}_5(s)} 2\text{SO}_3(g)$$
  3. Ostwald’s Process for Oxidation of Ammonia (Gas reactants, Solid catalyst): $$4\text{NH}_3(g) + 5\text{O}_2(g) \xrightarrow{\text{Pt}(s)} 4\text{NO}(g) + 6\text{H}_2\text{O}(g)$$
  4. Hydrogenation of Vegetable Oils into vanaspati ghee (Liquid/Gas reactants, Solid catalyst): $$\text{Vegetable Oil}(l) + \text{H}_2(g) \xrightarrow{\text{Ni}(s)} \text{Vegetable Ghee}(s)$$

Question 5.20: What do you understand by shape-selective catalysis? Why are zeolites good shape-selective catalysts?

Step 1: Shape Selectivity. Shape-selective catalysis is a catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules.
Step 2: Zeolite Micro-architecture. Zeolites are highly effective shape-selective catalysts due to their honey-comb like porous framework. They are microporous aluminosilicates with a three-dimensional network of silicates in which some silicon atoms are replaced by aluminum atoms, forming an $\text{Al-O-Si}$ framework.

Question 5.21: Explain the following purification and optical terms related to colloids:
(i) Dialysis
(ii) Tyndall Effect

Step 1: Dialysis Mechanics. Dialysis is a purification technique used to separate colloidal particles from dissolved crystalloids and impurities by diffusion through a suitable semi-permeable membrane (dialysis bag).
Step 2: Tyndall Effect Optical Profile. The scattering of light by colloidal particles when a strong beam of light is passed through a colloidal solution, making the path of light visible in a dark room.

Question 5.22: Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sols are more stable than lyophobic sols.
(iii) Sky appears blue in color.

Step 1: Reason (i) - Tanning Dynamics. Animal hides are colloidal in nature and carry positive electrical charges. Tanning materials contain negatively charged colloidal particles. When the hide is soaked in tannin, mutual coagulation occurs, resulting in the hardening of leather.
Step 2: Reason (ii) - Sol Stability Profiles. Lyophilic sols are highly stable because their particles are extensively solvated (hydrated) by a protective layer of the dispersion medium, in addition to carrying charge barriers.
Step 3: Reason (iii) - Light Scattering Mechanics. Dust particles and water molecules suspended in the atmosphere form a colloidal dispersion that selectively scatters shorter wavelengths of light (blue light) more effectively than longer wavelengths, causing the sky to appear blue.

Extra Important Questions (Board Style)

Target verification tasks matching high-yield board testing frameworks engineered to lock down full marks in your 2026 exams.

Section A: Multiple Choice Questions (MCQs)

Q1. What is the value of $1/n$ in the Freundlich adsorption isotherm under normal conditions?

A) 0 and 1
B) 2 and 4
C) 1 and 10
D) Always equal to 1

Correct Answer Choice: A
Step 1: Parameter Verification. The value of $1/n$ typically ranges from 0 to 1 under normal conditions, reflecting the mathematical behavior of adsorption across intermediate pressure ranges.

Section B: Assertion-Reason Questions

Directions: Choose option (A) if both statements are true and the reason provides the exact explanation. Choose (B) if both are true but independent. Choose (C) if the assertion is true but the reason is false. Choose (D) if assertion is false.

Q2. Assertion (A): Micelles are formed only above a specific concentration called Critical Micelle Concentration (CMC).
Reason (R): At higher concentrations, soap molecules behave as normal strong electrolytes.

Correct Answer Choice: C
Step 1: Structural Validation. The assertion is true. However, the reason is false because soap molecules behave as normal electrolytes at low concentrations and aggregate to form micelles at higher concentrations.

Section C: Short & Long Answer Questions

Q3. What is the "Gold Number"? Explain its significance.

Step 1: Definition. The gold number is the minimum weight in milligrams of a protective lyophilic colloid required to prevent the coagulation of $10 \text{ mL}$ of a standard gold sol upon the addition of $1 \text{ mL}$ of a $10\%$ $\text{NaCl}$ solution.
Step 2: Protective Index. It serves as a quantitative measure of the protective efficiency of a colloid; a smaller gold number signifies a higher protective power.

Q4. Explain how a delta is formed at the meeting place of river water and sea water.

Step 1: Coagulation Dynamics. River water is a colloidal solution of clay, sand, and organic matter carrying negative charges. Sea water contains a high concentration of dissolved electrolytes ($\text{Na}^+, \text{Mg}^{2+}, \text{Cl}^-$).
Step 2: Deposition. When river water meets sea water, the electrolytes present in sea water neutralize the electrical charges on the clay particles, causing them to coagulate and settle as vast sand deposits, forming a delta.

Common Mistakes Students Make

Exam Preparation Tips

Frequently Asked Questions (FAQs)

1. Is Class 12 Surface Chemistry a numerical-heavy chapter?
No, Surface Chemistry is almost entirely theoretical and conceptual. The only minor mathematical calculations are related to the Freundlich Adsorption Isotherm or calculation of the coagulating value.
2. What happens to physisorption when the temperature increases?
Physisorption decreases continuously with an increase in temperature because it is an exothermic process, and higher thermal kinetic energy disrupts the weak van der Waals forces.
3. Why are lyophilic sols called reversible sols?
They are called reversible because if the dispersion medium is separated from the dispersed phase (by evaporation, for example), the sol can be easily reconstituted simply by remixing them with the medium.
4. Where can I find more board questions for CBSE 2026?
You can practice previous year question papers (PYQs) alongside these Updated NCERT Solutions to get a complete command over all formats of questions.

Conclusion: Surface Chemistry is one of the easiest chapters to score full marks if you understand the conceptual definitions. Make sure to revise your notes consistently, practice the textbook questions in this guide, and solve your mock papers confidently. Keep working hard, Lucky!

More Class 12 Chemistry Chapters

Ch 1: The Solid State Ch 2: Solutions Ch 3: Electrochemistry Ch 4: Chemical Kinetics Ch 5: Surface Chemistry Ch 6: General Principles Ch 7: The p-Block Elements Ch 8: The d- & f-Block Elements Ch 9: Coordination Compounds Ch 10: Haloalkanes Ch 11: Alcohols & Phenols Ch 12: Aldehydes & Ketones Ch 13: Amines Ch 14: Biomolecules All Subjects Dashboard