Chemistry Grade 11 — CBSE (2025-26)

Welcome to this comprehensive Grade 11 Chemistry textbook, aligned with the Central Board of Secondary Education (CBSE) 2025-26 syllabus.

About This Book

This book is designed to help students:

• Master core chemistry concepts across all 9 units of the CBSE Class XI syllabus
• Develop competency-based problem-solving skills through previous years’ CBSE questions
• Visualize abstract concepts through detailed SVG diagrams and illustrations
• Practice extensively with questions drawn from the last 5 years of CBSE board exams and official sample papers

Syllabus Overview

How to Use This Book

Each chapter follows a structured format:

1. Theory — Key concepts explained with examples and illustrations
2. Worked Examples — Step-by-step solutions to important problem types
3. SVG Diagrams — Vector graphics for molecular structures, graphs, and experimental setups
4. Practice Questions — Competency-based questions organized as:
   • Multiple Choice Questions (MCQs)
   • Short Answer Questions (2-3 marks)
   • Long Answer Questions (5 marks)
   • Numerical Problems
   • Assertion-Reason Questions

Note: Mathematical equations are rendered using LaTeX. Inline math appears as \( E = mc^2 \) and display math appears as:

\[ PV = nRT \]

Based on the CBSE Senior Secondary Chemistry Syllabus 2025-26. Questions sourced from CBSE Board Examinations and Official Sample Papers (2020-2025).

Copyright © 2026 by Abhishek Kumar (support@abhishek-kumar.co.in). All rights reserved.

Chapter 1: Some Basic Concepts of Chemistry

1.1 Importance and Scope of Chemistry

Chemistry is the branch of science that deals with the composition, structure, properties, and transformations of matter. It is often called the central science because it connects physics with other natural sciences such as biology, geology, and environmental science.

Applications of Chemistry

Chemistry plays a vital role in nearly every aspect of our daily lives:

• Medicine & Health: Development of drugs, anaesthetics, and antibiotics
• Agriculture: Fertilizers, pesticides, and herbicides
• Materials: Polymers, alloys, ceramics, and nanomaterials
• Energy: Fuels, batteries, and solar cells
• Environment: Water purification, pollution control, and green chemistry
• Food: Preservatives, artificial sweeteners, and food processing

1.2 Nature of Matter

Matter is anything that has mass and occupies space. Matter can be classified in two ways:

Classification Based on Physical State

Classification Based on Chemical Composition

1.3 Properties of Matter and Their Measurement

Physical Properties

Physical properties can be measured or observed without changing the identity of the substance. Examples include colour, odour, melting point, boiling point, and density.

SI Units (International System of Units)

Important Derived Quantities

Temperature Conversions

\[ T(\text{K}) = T(°\text{C}) + 273.15 \]

\[ T(°\text{F}) = \frac{9}{5},T(°\text{C}) + 32 \]

Significant Figures

Rules for determining significant figures:

1. All non-zero digits are significant. (e.g., 285 has 3 significant figures)
2. Zeros between non-zero digits are significant. (e.g., 2.005 has 4)
3. Leading zeros are not significant. (e.g., 0.0025 has 2)
4. Trailing zeros in a number with a decimal point are significant. (e.g., 2.500 has 4)
5. Trailing zeros in a number without a decimal point may or may not be significant.

1.4 Laws of Chemical Combination

1. Law of Conservation of Mass (Lavoisier, 1789)

“In all physical and chemical changes, the total mass of the reactants is equal to the total mass of the products.”

\[ \text{Mass of reactants} = \text{Mass of products} \]

2. Law of Definite Proportions (Proust, 1799)

“A given compound always contains exactly the same proportion of elements by weight, regardless of the source or method of preparation.”

Example: Water (H₂O) always contains hydrogen and oxygen in the mass ratio 1:8.

3. Law of Multiple Proportions (Dalton, 1803)

“If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.”

Example: Carbon forms two oxides with oxygen:

Ratio of oxygen = 16 : 32 = 1 : 2 (simple whole numbers)

4. Gay-Lussac’s Law of Gaseous Volumes (1808)

“When gases react together, they do so in volumes which bear a simple whole number ratio to each other and to the volumes of the products, if gaseous — all volumes measured at same temperature and pressure.”

\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2,\text{HCl}(g) \] \[ 1;\text{vol} : 1;\text{vol} : 2;\text{vol} \]

5. Avogadro’s Law (1811)

“Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.”

1.5 Dalton’s Atomic Theory

John Dalton proposed the atomic theory in 1808:

1. Matter consists of indivisible atoms.
2. Atoms of a given element are identical in mass and properties.
3. Atoms of different elements differ in mass and properties.
4. Compounds are formed when atoms of different elements combine in fixed ratios.
5. Chemical reactions involve reorganisation of atoms — atoms are neither created nor destroyed.

Limitations:

• Atoms can be further divided into subatomic particles (electrons, protons, neutrons).
• Atoms of the same element can have different masses (isotopes).
• Atoms of different elements can have the same mass (isobars).

1.6 Atomic and Molecular Masses

Atomic Mass Unit (amu or u)

\[ 1;\text{amu} = \frac{1}{12} \times \text{mass of one } {}^{12}\text{C atom} = 1.66054 \times 10^{-24};\text{g} \]

Average Atomic Mass

For elements with isotopes, the average atomic mass is calculated as:

\[ \text{Average atomic mass} = \sum_{i} f_i \times m_i \]

where \( f_i \) is the fractional abundance and \( m_i \) is the mass of isotope \( i \).

Example: Chlorine has two isotopes: \({}^{35}\text{Cl}\) (75.77%) and \({}^{37}\text{Cl}\) (24.23%)

\[ \text{Average atomic mass} = (0.7577 \times 35) + (0.2423 \times 37) = 35.48;\text{u} \]

Molecular Mass

The molecular mass is the sum of atomic masses of all atoms in a molecule.

Example: Molecular mass of H₂SO₄:

\[ = 2(1.008) + 32.06 + 4(16.00) = 98.08;\text{u} \]

Formula Mass

Used for ionic compounds which do not exist as discrete molecules.

Example: Formula mass of NaCl = 23.0 + 35.5 = 58.5 u

1.7 Mole Concept and Molar Mass

The Mole

One mole of any substance contains exactly \( 6.022 \times 10^{23} \) elementary entities (atoms, molecules, ions, etc.). This number is called Avogadro’s number (\( N_A \)).

\[ 1;\text{mol} = 6.022 \times 10^{23};\text{entities} \]

Molar Mass

The molar mass of a substance is the mass of one mole, expressed in g/mol. It is numerically equal to the atomic/molecular mass in u.

Relationship Between Moles, Mass, and Number

Key Formulas:

\[ n = \frac{\text{Given mass (w)}}{\text{Molar mass (M)}} \]

\[ n = \frac{N}{N_A} \]

\[ n = \frac{V(\text{at STP})}{22.4;\text{L}} \quad \text{(for gases)} \]

1.8 Percentage Composition

The mass percentage of an element in a compound:

\[ \text{Mass %} = \frac{\text{Mass of element in 1 mol of compound}}{\text{Molar mass of compound}} \times 100 \]

Example: Percentage composition of H₂O:

\[ \text{Mass % of H} = \frac{2 \times 1.008}{18.02} \times 100 = 11.19% \]

\[ \text{Mass % of O} = \frac{16.00}{18.02} \times 100 = 88.81% \]

1.9 Empirical and Molecular Formula

\[ \text{Molecular formula} = n \times \text{Empirical formula} \]

where \( n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} \)

Example: Glucose has empirical formula CH₂O (empirical formula mass = 30 u) and molar mass = 180 u.

\[ n = \frac{180}{30} = 6 \]

Therefore, molecular formula = \( \text{C}6\text{H}{12}\text{O}_6 \)

Steps to Determine Empirical Formula

1. Convert percentage composition to grams (assume 100 g of compound).
2. Convert mass to moles by dividing by atomic mass.
3. Divide each mole value by the smallest mole value.
4. If ratios are not whole numbers, multiply by a suitable integer.

1.10 Chemical Reactions and Stoichiometry

Balanced Chemical Equations

A balanced chemical equation has equal numbers of atoms of each element on both sides.

Example:

\[ 4,\text{Fe}(s) + 3,\text{O}_2(g) \rightarrow 2,\text{Fe}_2\text{O}_3(s) \]

This equation tells us:

• 4 atoms of Fe react with 3 molecules of O₂
• 4 mol Fe reacts with 3 mol O₂ to form 2 mol Fe₂O₃
• \( 4 \times 55.85 = 223.4 \) g Fe reacts with \( 3 \times 32 = 96 \) g O₂

Limiting Reagent

The limiting reagent is the reactant that is completely consumed in a reaction, determining the maximum amount of product formed.

Worked Example:

Calculate the mass of water formed when 4 g of H₂ reacts with 16 g of O₂.

\[ 2,\text{H}_2 + \text{O}_2 \rightarrow 2,\text{H}_2\text{O} \]

Moles of H₂ = \( \frac{4}{2} = 2 \) mol

Moles of O₂ = \( \frac{16}{32} = 0.5 \) mol

From stoichiometry: 2 mol H₂ requires 1 mol O₂.

Available ratio: \( \frac{2}{2} = 1 \) for H₂ and \( \frac{0.5}{1} = 0.5 \) for O₂.

Since 0.5 < 1, O₂ is the limiting reagent.

Moles of H₂O formed = \( 2 \times 0.5 = 1 \) mol

Mass of H₂O = \( 1 \times 18 = 18 \) g

Reactions in Solutions — Concentration Terms

Molarity (M):

\[ M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}} \]

Molality (m):

\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \]

Worked Examples

Example 1: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its molar mass is 180 g/mol. Find the molecular formula.

Solution:

Empirical formula = CH₂O (empirical formula mass = 30 u)

\[ n = \frac{180}{30} = 6 \]

Molecular formula = C₆H₁₂O₆ (Glucose)

Example 2: How many molecules of methane are present in 0.80 g of CH₄?

Molar mass of CH₄ = 16 g/mol

\[ n = \frac{0.80}{16} = 0.05;\text{mol} \]

\[ N = n \times N_A = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22};\text{molecules} \]

Practice Questions

Multiple Choice Questions (MCQs)

1. The number of atoms in 4.25 g of NH₃ is approximately:

 (a) \( 6.022 \times 10^{23} \)

 (b) \( 1.505 \times 10^{23} \)

 (c) \( 4 \times 6.022 \times 10^{23} \)

 (d) \( 6.022 \times 10^{22} \)

2. The empirical formula of a compound with 40% sulphur and 60% oxygen is:

 (a) SO₂

 (b) SO₃

 (c) SO

 (d) S₂O₃

3. The number of moles of CO₂ containing 8.0 g of oxygen is:

 (a) 0.25

 (b) 0.50

 (c) 1.0

 (d) 2.0

4. If 1.0 g of a metal oxide contains 0.5 g of oxygen, the equivalent weight of the metal is:

 (a) 4

 (b) 8

 (c) 16

 (d) 32

5. Which has the maximum number of molecules?

 (a) 1 g of CO₂

 (b) 1 g of N₂

 (c) 1 g of H₂

 (d) 1 g of CH₄

Short Answer Questions (2–3 Marks)

6. Define the law of conservation of mass. Give an example.

7. Calculate the molarity of a solution prepared by dissolving 5.85 g of NaCl in water to make 500 mL of solution.

8. What is the difference between empirical formula and molecular formula? Explain with an example.

9. A compound has the following composition: Na = 29.11%, S = 40.51%, O = 30.38%. Determine its empirical formula.

10. Calculate the mass percentage of each element in calcium carbonate (CaCO₃).

Long Answer Questions (5 Marks)

11. (a) State and explain the law of multiple proportions with a suitable example.

 (b) Calculate the amount of carbon dioxide that could be produced when:

  (i) 1 mole of carbon is burnt in air

  (ii) 1 mole of carbon is burnt in 16 g of dioxygen

12. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).

 (a) Express this in percent by mass.

 (b) Determine the molality of chloroform in the water sample.

13. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g CO₂ and 0.690 g H₂O and no other products. A volume of 10.0 L (measured at STP) of this welding gas weighs 11.6 g.

 (a) Find the empirical formula.

 (b) Find the molar mass and molecular formula.

Numerical Problems

14. Calculate the number of atoms of hydrogen present in 52 g of methane (CH₄). What would be the mass of an individual atom of hydrogen?

15. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

\[ \text{CuSO}_4 + \text{Fe} \rightarrow \text{FeSO}_4 + \text{Cu} \]

16. Determine the molecular formula of an oxide of iron in which the mass percentage of iron and oxygen are 69.9% and 30.1% respectively. (Given: molar mass ≈ 160 g/mol)

Assertion-Reason Questions

In each of the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is NOT the correct explanation of A.

(c) A is true but R is false.

(d) A is false but R is true.

17. Assertion (A): One mole of different gases at STP have different volumes.

Reason (R): At STP, the molar volume of any ideal gas is 22.4 L.

18. Assertion (A): The empirical formula and molecular formula of glucose are the same.

Reason (R): The empirical formula gives the simplest whole number ratio of atoms.

Answer Key

Chapter 2: Structure of Atom

2.1 Discovery of Subatomic Particles

Discovery of Electron — Cathode Ray Experiment (J.J. Thomson, 1897)

When a high voltage is applied across a cathode ray discharge tube at very low pressure (~0.001 mm Hg), a stream of particles moves from the cathode to the anode. These are called cathode rays.

Properties of Cathode Rays:

1. Travel in straight lines
2. Consist of negatively charged particles (electrons)
3. Independent of the material of the cathode or gas in the tube
4. Possess kinetic energy and can rotate a paddle wheel

Charge-to-mass ratio of electron (J.J. Thomson):

\[ \frac{e}{m_e} = 1.758820 \times 10^{11};\text{C/kg} \]

Charge of electron (R.A. Millikan — Oil Drop Experiment):

\[ e = 1.6022 \times 10^{-19};\text{C} \]

Mass of electron:

\[ m_e = 9.1094 \times 10^{-31};\text{kg} \]

Discovery of Proton — Anode Rays (Canal Rays)

When a perforated cathode is used in a discharge tube, a stream of positively charged particles moves towards the cathode. These are anode rays or canal rays.

Properties:

• The charge-to-mass ratio depends on the gas in the tube
• The lightest particles are obtained with hydrogen → protons

\[ m_p = 1.6726 \times 10^{-27};\text{kg}, \quad e_p = +1.6022 \times 10^{-19};\text{C} \]

Discovery of Neutron (James Chadwick, 1932)

Chadwick bombarded beryllium with α-particles and observed the emission of electrically neutral particles with a mass slightly greater than that of protons — neutrons.

\[ {}^{9}_{4}\text{Be} + {}^{4}_{2}\text{He} \rightarrow {}^{12}_{6}\text{C} + {}^{1}_{0}\text{n} \]

\[ m_n = 1.6750 \times 10^{-27};\text{kg} \]

2.2 Atomic Number, Mass Number, Isotopes, and Isobars

Atomic number (Z) = Number of protons = Number of electrons (in a neutral atom)

Mass number (A) = Number of protons + Number of neutrons

\[ A = Z + \text{number of neutrons} \]

Notation: \( {}^{A}_{Z}\text{X} \), e.g., \( {}^{12}_{6}\text{C} \), \( {}^{23}_{11}\text{Na} \)

2.3 Thomson’s Model of Atom (1904)

J.J. Thomson proposed the “plum pudding” model: the atom is a sphere of positive charge in which electrons are embedded, like plums in a pudding.

Limitation: Could not explain the results of Rutherford’s scattering experiment.

2.4 Rutherford’s Nuclear Model (1911)

The α-Particle Scattering Experiment

Rutherford bombarded a thin gold foil (0.0004 cm thick) with α-particles from a radioactive source.

Observations:

1. Most α-particles passed through undeflected → atom is mostly empty space
2. A small fraction was deflected by small angles → positive charge is concentrated
3. Very few (~1 in 20,000) bounced back → the positive charge occupies a very small volume (nucleus)

Rutherford’s Conclusions:

1. The atom has a tiny, dense, positively charged centre called the nucleus (radius ~ \(10^{-15}\) m)
2. Nearly all mass is concentrated in the nucleus
3. Electrons revolve around the nucleus in circular orbits
4. The atom is mostly empty space (radius ~ \(10^{-10}\) m)

Limitations:

• Could not explain the stability of atoms (accelerating charged particles should radiate energy and spiral into the nucleus)
• Could not explain line spectra of atoms

2.5 Bohr’s Model of the Hydrogen Atom (1913)

Niels Bohr proposed a model for hydrogen-like atoms based on quantum ideas:

Postulates

1. Electrons revolve in fixed circular orbits (called stationary states or shells) without radiating energy.

2. Quantized angular momentum: The angular momentum of an electron in a stationary state is an integral multiple of \(\frac{h}{2\pi}\):

\[ m_e v r = n \frac{h}{2\pi}, \quad n = 1, 2, 3, \ldots \]

3. Energy transitions: When an electron jumps from a higher orbit (\(n_2\)) to a lower orbit (\(n_1\)), energy is emitted as a photon:

\[ \Delta E = E_{n_2} - E_{n_1} = h\nu \]

Key Results for Hydrogen-like Species

Radius of \(n\)-th orbit:

\[ r_n = \frac{n^2 a_0}{Z} \]

where \( a_0 = 52.9;\text{pm} \) (Bohr radius), \(Z\) = atomic number.

Energy of \(n\)-th orbit:

\[ E_n = -\frac{13.6,Z^2}{n^2};\text{eV} = -\frac{2.18 \times 10^{-18},Z^2}{n^2};\text{J} \]

Velocity of electron:

\[ v_n = \frac{2.18 \times 10^6 , Z}{n};\text{m/s} \]

Hydrogen Spectrum

When excited hydrogen atoms return to lower energy levels, they emit photons of specific wavelengths, producing line spectra.

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( R_H = 1.097 \times 10^7;\text{m}^{-1} \) (Rydberg constant)

Limitations of Bohr’s Model:

1. Works only for hydrogen-like (one-electron) species
2. Could not explain the fine structure of spectral lines
3. Could not explain the Zeeman effect or Stark effect
4. Does not account for electron-electron repulsion in multi-electron atoms

2.6 Dual Nature of Matter and Radiation

Wave-Particle Duality

Light exhibits both wave and particle nature:

• Wave nature: Diffraction, interference
• Particle nature: Photoelectric effect, black-body radiation

Planck’s quantum theory:

\[ E = h\nu = \frac{hc}{\lambda} \]

where \( h = 6.626 \times 10^{-34};\text{J·s} \) (Planck’s constant)

Photoelectric effect (Einstein):

\[ h\nu = h\nu_0 + \frac{1}{2}m_e v^2 \]

where \( \nu_0 \) = threshold frequency

de Broglie Relation

Louis de Broglie (1924) proposed that matter also has a dual nature:

\[ \lambda = \frac{h}{mv} = \frac{h}{p} \]

where \( \lambda \) = wavelength, \( m \) = mass, \( v \) = velocity, \( p \) = momentum.

Note: The wave nature is significant only for microscopic particles (electrons, protons) and negligible for macroscopic objects.

2.7 Heisenberg’s Uncertainty Principle

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

or equivalently:

\[ \Delta x \cdot m\Delta v \geq \frac{h}{4\pi} \]

It is impossible to simultaneously determine the exact position and exact momentum of an electron. This principle makes the concept of fixed orbits (Bohr model) meaningless.

2.8 Quantum Mechanical Model of the Atom

Schrödinger Wave Equation

\[ \hat{H}\psi = E\psi \]

The full time-independent equation:

\[ \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} + \frac{8\pi^2 m}{h^2}(E - V)\psi = 0 \]

• \(\psi\) = wave function
• \(|\psi|^2\) = probability density of finding the electron
• \(E\) = total energy
• \(V\) = potential energy

Quantum Numbers

Each electron in an atom is described by a set of four quantum numbers:

Subshell notation:

2.9 Shapes of Orbitals

s Orbitals (\(l = 0\)) — Spherical

• Spherically symmetric
• Node at nucleus for \(n \geq 2\) (radial node)
• Number of radial nodes = \(n - l - 1\)

p Orbitals (\(l = 1\)) — Dumbbell shaped

• Each p orbital has two lobes with a nodal plane at the nucleus
• Three p orbitals are oriented along x, y, and z axes (mutually perpendicular)

d Orbitals (\(l = 2\)) — Cloverleaf shapes

There are five d orbitals: \(d_{xy}\), \(d_{yz}\), \(d_{xz}\), \(d_{x^2-y^2}\), \(d_{z^2}\).

• \(d_{xy}\), \(d_{yz}\), \(d_{xz}\): four lobes between the axes
• \(d_{x^2-y^2}\): four lobes along x and y axes
• \(d_{z^2}\): two lobes along z-axis with a doughnut-shaped ring in the xy plane

2.10 Rules for Filling Electrons in Orbitals

1. Aufbau Principle

Electrons fill orbitals in order of increasing energy (\(n + l\) rule). If \(n + l\) values are equal, the orbital with the lower \(n\) fills first.

Energy order:

\[ 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f \ldots \]

2. Pauli’s Exclusion Principle

No two electrons in an atom can have the same set of four quantum numbers. Consequently, each orbital can hold a maximum of 2 electrons with opposite spins.

3. Hund’s Rule of Maximum Multiplicity

Electrons are distributed among orbitals of a subshell in such a way as to give the maximum number of unpaired electrons with parallel spins. Pairing occurs only after all degenerate orbitals are singly occupied.

2.11 Electronic Configuration of Atoms

The electronic configuration is written as: \(nl^x\), where \(n\) = principal quantum number, \(l\) = subshell, \(x\) = number of electrons.

★ Anomalous configurations: Cr and Cu achieve extra stability from half-filled (\(d^5\)) and completely filled (\(d^{10}\)) d-subshells respectively.

Stability of Half-filled and Completely Filled Subshells

Two factors contribute:

1. Symmetrical distribution of electrons → lower energy
2. Exchange energy — electrons with parallel spins in degenerate orbitals can exchange positions, releasing energy. More exchanges → greater stability.

Practice Questions

Multiple Choice Questions (MCQs)

1. The number of angular nodes for a 4d orbital is:

 (a) 1

 (b) 2

 (c) 3

 (d) 4

2. Which of the following sets of quantum numbers is NOT possible?

 (a) \(n = 2,\ l = 1,\ m_l = 0,\ m_s = +\frac{1}{2}\)

 (b) \(n = 3,\ l = 2,\ m_l = -2,\ m_s = -\frac{1}{2}\)

 (c) \(n = 2,\ l = 2,\ m_l = 0,\ m_s = +\frac{1}{2}\)

 (d) \(n = 4,\ l = 0,\ m_l = 0,\ m_s = -\frac{1}{2}\)

3. If uncertainty in position of an electron is zero, the uncertainty in its momentum would be:

 (a) zero

 (b) \(\geq \frac{h}{4\pi}\)

 (c) \(\lt \frac{h}{4\pi}\)

 (d) infinite

4. The wavelength of a ball of mass 100 g moving with a velocity of 100 m/s is (\(h = 6.6 \times 10^{-34}\) J·s):

 (a) \(6.6 \times 10^{-35}\) m

 (b) \(6.6 \times 10^{-34}\) m

 (c) \(6.6 \times 10^{-33}\) m

 (d) \(6.6 \times 10^{-32}\) m

5. The electronic configuration of Cu (Z = 29) is:

 (a) \([Ar],3d^9,4s^2\)

 (b) \([Ar],3d^{10},4s^1\)

 (c) \([Ar],3d^{10},4s^2\)

 (d) \([Ar],3d^8,4s^2,4p^1\)

Short Answer Questions (2–3 Marks)

6. Calculate the wavelength of an electron moving with a velocity of \(2.05 \times 10^7\) m/s.

7. Write the electronic configuration of Fe²⁺ and Fe³⁺ ions. Which one is more stable and why?

8. What is the maximum number of electrons that can have the quantum numbers \(n = 3\), \(l = 2\)?

9. State Heisenberg’s uncertainty principle. Why is it significant for microscopic particles but not for macroscopic objects?

10. Write the four quantum numbers for the last electron of sodium (Z = 11).

Long Answer Questions (5 Marks)

11. (a) Explain the Bohr model of hydrogen atom. Derive the expression for the radius and energy of the \(n\)-th orbit.

 (b) Calculate the wavelength of the first line in the Balmer series of hydrogen spectrum.

12. (a) Draw the shapes of the five d orbitals. How do \(d_{z^2}\) and \(d_{x^2-y^2}\) differ from the other three d orbitals?

 (b) Explain why Cr has the electronic configuration \([Ar],3d^5,4s^1\) and not \([Ar],3d^4,4s^2\).

13. (a) State and explain the photoelectric effect. How did it provide evidence for the particle nature of light?

 (b) A photon of wavelength 4 × 10⁻⁷ m strikes a metal surface, the work function of the metal being 2.13 eV. Calculate the kinetic energy and velocity of the emitted photoelectron.

Assertion-Reason Questions

14. Assertion (A): The energy of 2s orbital is less than that of 2p orbital in multi-electron atoms.

Reason (R): 2s electrons have greater penetration power than 2p electrons.

15. Assertion (A): The total number of nodes for 3p orbital is 2.

Reason (R): Total nodes = \(n - 1\), angular nodes = \(l\), radial nodes = \(n - l - 1\).

Answer Key

Chapter 3: Classification of Elements and Periodicity in Properties

3.1 Why Do We Need to Classify Elements?

At present, 118 elements are known. Studying each element individually would be a daunting task. Classification helps us:

• Organize elements systematically
• Predict properties of unknown elements
• Correlate properties of similar elements
• Simplify the study of chemistry

3.2 Brief History of the Development of the Periodic Table

Döbereiner’s Triads (1829)

Döbereiner observed that elements could be grouped in triads where the atomic mass of the middle element was approximately the arithmetic mean of the other two.

Limitation: Could not classify all elements known at the time.

Newlands’ Law of Octaves (1866)

When elements are arranged in order of increasing atomic mass, every eighth element has properties similar to the first — like the octave in music.

Limitation: Valid only up to calcium; failed for heavier elements.

Mendeleev’s Periodic Table (1869)

“The physical and chemical properties of elements are a periodic function of their atomic masses.”

Key Features:

• Arranged 63 elements in order of increasing atomic mass
• Left gaps for undiscovered elements (Eka-aluminium → Gallium, Eka-silicon → Germanium)
• Correctly predicted properties of undiscovered elements

Limitations:

• Position of hydrogen was ambiguous
• Isotopes could not be accommodated
• Some elements with higher atomic mass were placed before those with lower atomic mass (e.g., Co before Ni)

3.3 Modern Periodic Law and the Present Form of the Periodic Table

Modern Periodic Law (Moseley, 1913)

“The physical and chemical properties of elements are a periodic function of their atomic numbers.”

The Long Form of the Periodic Table

The modern periodic table has:

• 7 periods (horizontal rows) — corresponding to the principal quantum number \(n\)
• 18 groups (vertical columns)
• 4 blocks: s, p, d, and f

Nomenclature of Elements with Z > 100

Elements with atomic number greater than 100 are named using the IUPAC nomenclature based on their atomic number:

Example: Element 109 → Un-nil-ennium → Unnilennium (Une) (now named Meitnerium, Mt)

3.4 Periodic Trends in Properties

1. Atomic Radius

Atomic radius is the distance from the centre of the nucleus to the outermost shell of electrons.

Types:

• Covalent radius: Half the internuclear distance between two bonded atoms of the same element
• Van der Waals radius: Half the distance between adjacent atoms of the same element in non-bonded state
• Metallic radius: Half the internuclear distance between adjacent atoms in a metallic crystal

Trends:

2. Ionic Radius

• Cations are smaller than parent atoms (loss of electrons)
• Anions are larger than parent atoms (gain of electrons)
• In an isoelectronic series (same number of electrons), the ion with greater nuclear charge is smaller

Example: Isoelectronic species (all have 10 electrons):

\[ \text{O}^{2-} > \text{F}^- > \text{Ne} > \text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+} \]

3. Ionization Enthalpy (IE)

The energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state:

\[ \text{X}(g) \rightarrow \text{X}^+(g) + e^- \quad (\Delta_i H) \]

Trends:

• Across a period: Generally increases (due to increasing nuclear charge)
• Down a group: Decreases (outermost electron is farther from nucleus)

Successive ionization enthalpies: \( IE_1 < IE_2 < IE_3 < \ldots \) (each successive electron is harder to remove)

Anomalies:

• \(IE_1\) of B < \(IE_1\) of Be (electron from 2p is easier to remove than from 2s)
• \(IE_1\) of O < \(IE_1\) of N (paired electron in O’s 2p is easier to remove)

4. Electron Gain Enthalpy (\(\Delta_{eg}H\))

The energy change when an electron is added to an isolated gaseous atom:

\[ \text{X}(g) + e^- \rightarrow \text{X}^-(g) \quad (\Delta_{eg}H) \]

• Usually negative (energy is released) for most elements
• Halogens have the most negative values (high tendency to gain electrons)
• Noble gases have positive values (stable electron configuration resists addition)

Trends:

• Across a period: Becomes more negative (more exothermic)
• Down a group: Becomes less negative (exception: Cl has more negative \(\Delta_{eg}H\) than F due to F’s small size and electron-electron repulsion)

5. Electronegativity

The tendency of an atom to attract shared electrons towards itself in a chemical bond.

Pauling Scale: F is the most electronegative element (EN = 4.0)

Trends:

• Across a period: Increases
• Down a group: Decreases

6. Valency

Valency is the combining capacity of an element.

Valency increases from 1 to 4 in Groups 1-14, then decreases from 4 to 0 in Groups 14-18.

Practice Questions

Multiple Choice Questions (MCQs)

1. The element with atomic number 35 belongs to which block of the periodic table?

 (a) s-block

 (b) p-block

 (c) d-block

 (d) f-block

2. Which of the following is the correct order of ionization enthalpy?

 (a) \(\text{B} \gt \text{Be} \gt \text{N} \gt \text{C}\)

 (b) \(\text{Be} \gt \text{B} \gt \text{C} \gt \text{N}\)

 (c) \(\text{N} \gt \text{C} \gt \text{Be} \gt \text{B}\)

 (d) \(\text{N} \gt \text{O} \gt \text{F} \gt \text{Ne}\)

3. In the isoelectronic series \(\text{Na}^+\), \(\text{Mg}^{2+}\), \(\text{Al}^{3+}\), the correct order of ionic radii is:

 (a) \(\text{Na}^+ \gt \text{Mg}^{2+} \gt \text{Al}^{3+}\)

 (b) \(\text{Al}^{3+} \gt \text{Mg}^{2+} \gt \text{Na}^+\)

 (c) \(\text{Mg}^{2+} \gt \text{Na}^+ \gt \text{Al}^{3+}\)

 (d) All have same ionic radii

4. Which of the following has the most negative electron gain enthalpy?

 (a) F

 (b) Cl

 (c) Br

 (d) I

5. The IUPAC name for element with atomic number 112 is:

 (a) Ununbium

 (b) Copernicium

 (c) Unbibium

 (d) Unnilbium

Short Answer Questions (2–3 Marks)

6. What is the modern periodic law? How does it differ from Mendeleev’s periodic law?

7. Explain why the ionization enthalpy of N is more than that of O.

8. Arrange the following in order of increasing atomic radius: C, N, Si, P.

9. What are isoelectronic species? Write four isoelectronic species with 18 electrons each.

10. How would you use the periodic table to predict the formula of a stable compound formed between potassium and sulphur?

Long Answer Questions (5 Marks)

11. (a) Discuss the trends in ionization enthalpy across a period and down a group. Explain any exceptions in Period 2.

 (b) What is electronegativity? Why does fluorine have the highest electronegativity?

12. (a) Explain the process of naming elements with atomic number greater than 100 according to IUPAC nomenclature. Give two examples.

 (b) Which of the following elements would have the highest electron gain enthalpy and why: N, O, F, S?

13. (a) Compare the first ionization enthalpies of Be, B, N, and O. Explain the observed trend with reasons.

 (b) Define ionic radius. How does it vary in an isoelectronic species?

Assertion-Reason Questions

14. Assertion (A): The first ionization enthalpy of Be is greater than that of B.

Reason (R): 2s electrons are more tightly held than 2p electrons.

15. Assertion (A): Noble gases have positive electron gain enthalpy.

Reason (R): Noble gases have stable, completely filled electronic configuration.

Answer Key

Chapter 4: Chemical Bonding and Molecular Structure

4.1 Introduction

Atoms combine to achieve a stable noble gas electronic configuration. The attractive force that holds atoms together in a molecule or crystal is called a chemical bond.

Lewis Symbols (Electron Dot Structures): The valence electrons of an atom are represented as dots around its symbol.

4.2 Types of Chemical Bonds

Ionic Bond (Electrovalent Bond)

Formed by the transfer of electrons from a metal to a non-metal.

Example: Formation of NaCl

\[ \text{Na} \rightarrow \text{Na}^+ + e^- \quad (\text{loses 1 electron}) \] \[ \text{Cl} + e^- \rightarrow \text{Cl}^- \quad (\text{gains 1 electron}) \]

Factors favouring ionic bond formation:

• Low ionization enthalpy of the metal
• High electron gain enthalpy of the non-metal
• High lattice enthalpy

Lattice Enthalpy: The energy required to completely separate one mole of an ionic solid into gaseous ions.

\[ \text{NaCl}(s) \rightarrow \text{Na}^+(g) + \text{Cl}^-(g) \quad \Delta_{\text{lattice}}H^\circ = +786;\text{kJ/mol} \]

Covalent Bond

Formed by the sharing of electrons between two atoms (usually non-metals).

Lewis Structures represent bonded atoms showing shared and lone pairs.

Formal Charge

\[ \text{Formal charge} = (\text{Valence electrons}) - (\text{Lone pair electrons}) - \frac{1}{2}(\text{Bonded electrons}) \]

Bond Parameters

\[ \text{Bond order} = \frac{1}{2}(N_b - N_a) \]

where \(N_b\) = number of bonding electrons, \(N_a\) = number of antibonding electrons (in MO theory).

4.3 The Octet Rule and Its Limitations

Octet Rule: Atoms tend to achieve 8 electrons in their valence shell (or 2 for H and He) — a duet.

Limitations:

1. Incomplete octet: BF₃ (B has 6 electrons), BeCl₂ (Be has 4)
2. Expanded octet: PCl₅ (P has 10 electrons), SF₆ (S has 12), possible for elements in Period 3+ due to available d-orbitals
3. Odd-electron molecules: NO, NO₂ have an odd number of electrons

4.4 Polar and Nonpolar Covalent Bonds

When bonded atoms have different electronegativities, the shared pair is displaced towards the more electronegative atom, creating a polar covalent bond.

Dipole moment (\(\mu\)):

\[ \mu = q \times d \]

where \(q\) = magnitude of charge, \(d\) = distance of separation. Unit: Debye (D), \(1;\text{D} = 3.336 \times 10^{-30};\text{C·m}\)

• \(\mu = 0\) for symmetrical molecules (e.g., CO₂, CCl₄, BF₃)
• \(\mu \neq 0\) for asymmetrical molecules (e.g., H₂O, NH₃, CHCl₃)

4.5 Fajan’s Rules — Covalent Character of Ionic Bonds

Ionic bonds develop covalent character when:

1. Small cation with high charge (high polarising power)
2. Large anion with high charge (high polarisability)

4.6 VSEPR Theory (Valence Shell Electron Pair Repulsion)

The shape of a molecule is determined by the repulsion between electron pairs (bonding and lone pairs) in the valence shell of the central atom.

Order of repulsion: lp–lp > lp–bp > bp–bp

4.7 Valence Bond Theory (VBT)

• A covalent bond is formed when half-filled orbitals of two atoms overlap.
• The greater the overlap → stronger the bond.
• Types of overlap:
  • Sigma (σ) bond: Head-on overlap (s-s, s-p, p-p along bond axis)
  • Pi (π) bond: Lateral/sideways overlap (p-p perpendicular to bond axis)

A single bond consists of one σ bond. A double bond = 1σ + 1π. A triple bond = 1σ + 2π.

4.8 Hybridization

Hybridization is the intermixing of atomic orbitals of similar energies to form new, equivalent hybrid orbitals.

Important: Lone pairs also occupy hybrid orbitals:

• NH₃: sp³ hybridized, 3 bp + 1 lp → trigonal pyramidal
• H₂O: sp³ hybridized, 2 bp + 2 lp → bent

4.9 Molecular Orbital Theory (MOT)

Key Concepts

1. Atomic orbitals of comparable energy combine to form molecular orbitals (MOs)
2. Number of MOs formed = Number of atomic orbitals combined
3. Bonding MOs (lower energy) and Antibonding MOs (higher energy)
4. Filling follows Aufbau, Pauli, and Hund’s rules

MO Energy Level Diagram for Homonuclear Diatomic Molecules

For O₂ and F₂ (\(Z \geq 8\)):

\[ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z \]

For Li₂ to N₂ (\(Z < 8\)):

\[ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z \]

Bond Order:

\[ \text{Bond order} = \frac{N_b - N_a}{2} \]

4.10 Resonance

When a single Lewis structure cannot explain all properties of a molecule, we write resonance structures (canonical forms). The actual structure is a resonance hybrid.

Example: Carbonate ion \(\text{CO}_3^{2-}\) has three equivalent resonance structures. All C–O bond lengths are equal (127 pm), intermediate between C=O (122 pm) and C–O (143 pm).

4.11 Hydrogen Bond

A type of electrostatic attraction between a hydrogen atom bonded to a highly electronegative atom (F, O, N) and a lone pair on another electronegative atom.

Types:

• Intermolecular: Between different molecules (e.g., H₂O, HF)
• Intramolecular: Within the same molecule (e.g., o-nitrophenol)

Consequences:

• High boiling point of H₂O, NH₃, HF compared to hydrides of their group members
• Ice is less dense than liquid water (hydrogen bonding creates an open structure)

Practice Questions

Multiple Choice Questions (MCQs)

1. The correct order of bond angles is:

 (a) \(\text{H}_2\text{O} \gt \text{NH}_3 \gt \text{CH}_4\)

 (b) \(\text{CH}_4 \gt \text{NH}_3 \gt \text{H}_2\text{O}\)

 (c) \(\text{NH}_3 \gt \text{H}_2\text{O} \gt \text{CH}_4\)

 (d) \(\text{CH}_4 \gt \text{H}_2\text{O} \gt \text{NH}_3\)

2. The hybridization of Xe in XeF₄ is:

 (a) sp³

 (b) sp³d

 (c) sp³d²

 (d) dsp²

3. O₂ is paramagnetic because it has:

 (a) unpaired electrons in bonding MOs

 (b) unpaired electrons in antibonding MOs

 (c) no unpaired electrons

 (d) more antibonding electrons than bonding electrons

4. The bond order in NO⁺ is:

 (a) 2

 (b) 2.5

 (c) 3

 (d) 1.5

5. Which molecule has the shortest bond length?

 (a) O₂

 (b) O₂⁻

 (c) O₂²⁻

 (d) O₂⁺

Short Answer Questions (2–3 Marks)

6. Using VSEPR theory, predict the shapes of BrF₅ and XeOF₂.

7. Draw the resonance structures of ozone (O₃). What is the O–O bond order in ozone?

8. Explain why the bond angle in H₂O (104.5°) is less than in NH₃ (107°).

9. Define hydrogen bonding. Why is the boiling point of HF much higher than that of HCl?

10. Use MO theory to explain why He₂ does not exist but He₂⁺ does.

Long Answer Questions (5 Marks)

11. (a) What is hybridization? Explain sp, sp², and sp³ hybridization with examples.

 (b) What is the hybridization of each carbon atom in CH₂=C=CH₂?

12. (a) Draw the MO energy level diagram for N₂. Calculate its bond order and predict its magnetic behaviour.

 (b) Compare the bond order and stability of N₂ and N₂⁺.

13. (a) Predict the shape and hybridization of the following: SF₆, ClF₃, ICl₂⁻.

 (b) Explain why PCl₅ exists but NCl₅ does not.

Assertion-Reason Questions

14. Assertion (A): The bond angle in PH₃ is less than in NH₃.

Reason (R): P is less electronegative than N.

15. Assertion (A): CO₂ has zero dipole moment though it has two polar C=O bonds.

Reason (R): CO₂ is a linear molecule and the bond dipoles cancel each other.

Answer Key

Chapter 5: Chemical Thermodynamics

5.1 Thermodynamic Terms

System and Surroundings

• System: The part of the universe under study
• Surroundings: Everything else in the universe

\[ \text{Universe} = \text{System} + \text{Surroundings} \]

Types of Systems

State Functions and Path Functions

• State functions: Depend only on the state, not the path (e.g., \(U\), \(H\), \(S\), \(G\), \(P\), \(V\), \(T\))
• Path functions: Depend on the path (e.g., work \(w\), heat \(q\))

Extensive and Intensive Properties

Extensive properties depend on the amount of matter. Intensive properties do not.

5.2 Internal Energy (U)

The internal energy of a system is the total energy (kinetic + potential) of all particles.

• \(U\) is a state function — we can measure only the change \(\Delta U\), not absolute \(U\).

\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = q + w \]

where \(q\) = heat absorbed by the system, \(w\) = work done on the system.

Work

For gases, the work of expansion/compression against a constant external pressure:

\[ w = -P_{\text{ext}} \Delta V \]

For reversible processes:

\[ w = -nRT \ln\frac{V_f}{V_i} \]

Sign convention (IUPAC):

• Work done on the system → \(w > 0\)
• Work done by the system → \(w < 0\)
• Heat absorbed by the system → \(q > 0\)
• Heat released by the system → \(q < 0\)

5.3 First Law of Thermodynamics

“Energy can neither be created nor destroyed; it can only be transformed from one form to another.”

\[ \Delta U = q + w \]

Special Cases

1. At constant volume (\(\Delta V = 0\)): \(w = 0\), so \(\Delta U = q_V\)
2. At constant pressure: \(\Delta H = q_P\)

5.4 Enthalpy (H)

\[ H = U + PV \]

For a process at constant pressure:

\[ \Delta H = \Delta U + P\Delta V \]

For reactions involving gases:

\[ \Delta H = \Delta U + \Delta n_g RT \]

where \(\Delta n_g\) = (moles of gaseous products) − (moles of gaseous reactants)

Heat Capacity

• Heat capacity at constant volume: \(C_V = \frac{q_V}{\Delta T} = \frac{\Delta U}{\Delta T}\)
• Heat capacity at constant pressure: \(C_P = \frac{q_P}{\Delta T} = \frac{\Delta H}{\Delta T}\)

For an ideal gas:

\[ C_P - C_V = R = 8.314;\text{J mol}^{-1}\text{K}^{-1} \]

5.5 Measurement of ΔU and ΔH — Calorimetry

Bomb Calorimeter (Constant Volume)

Used to measure \(\Delta U\) for combustion reactions.

5.6 Enthalpy Changes for Various Types of Reactions

Standard Enthalpy of Formation (\(\Delta_f H^\circ\))

Enthalpy change when one mole of a compound is formed from its elements in their standard states at 1 bar and specified temperature (usually 298 K).

\[ \Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) \]

By convention: \(\Delta_f H^\circ\) of elements in their standard state = 0

Standard Enthalpy of Combustion (\(\Delta_c H^\circ\))

Enthalpy change when one mole of a substance undergoes complete combustion in O₂.

\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta_c H^\circ = -890.3;\text{kJ/mol} \]

Other Important Enthalpy Changes

5.7 Hess’s Law of Constant Heat Summation

“The total enthalpy change for a reaction is the same whether the reaction takes place in one step or in a series of steps.”

This is because \(H\) is a state function.

Application — Born-Haber Cycle:

Used to calculate lattice enthalpy of ionic compounds indirectly.

5.8 Bond Enthalpy

The enthalpy of a reaction can be estimated from bond enthalpies:

\[ \Delta_r H^\circ \approx \sum (\text{B.E. of bonds broken}) - \sum (\text{B.E. of bonds formed}) \]

Bonds broken → endothermic (\(+\)); Bonds formed → exothermic (\(-\))

5.9 Second Law of Thermodynamics

The second law introduces the concept of entropy (\(S\)) — a measure of the disorder or randomness of a system.

“In any spontaneous process, the total entropy of the universe always increases.”

\[ \Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0 \]

For a reversible process at constant temperature:

\[ \Delta S = \frac{q_{\text{rev}}}{T} \]

Entropy Changes

• \(\Delta S > 0\): Increase in disorder (e.g., melting, vaporization, dissolution, gas expansion)
• \(\Delta S < 0\): Increase in order (e.g., freezing, condensation)

5.10 Gibbs Energy (G)

\[ G = H - TS \]

At constant T and P:

\[ \Delta G = \Delta H - T\Delta S \]

Criteria for Spontaneity

Equilibrium Condition

\[ \Delta G = 0 \quad \text{(at equilibrium)} \]

\[ \Delta G^\circ = -RT \ln K \]

where \(K\) is the equilibrium constant.

5.11 Third Law of Thermodynamics (Brief Introduction)

“The entropy of a perfectly crystalline substance at absolute zero (0 K) is zero.”

\[ S_{0,\text{K}} = 0 \quad \text{(for a perfect crystal)} \]

This law provides a reference point for calculating absolute entropy values.

Practice Questions

Multiple Choice Questions (MCQs)

1. For an adiabatic process, the correct expression is:

 (a) \(\Delta U = q\)

 (b) \(\Delta U = w\)

 (c) \(\Delta U = q + w\)

 (d) \(q = 0, \Delta U = w\)

2. The enthalpy of combustion of carbon to CO₂ is −393.5 kJ/mol. The heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas is:

 (a) −393.5 kJ

 (b) −314.8 kJ

 (c) +314.8 kJ

 (d) −787.0 kJ

3. A reaction has \(\Delta H = -40;\text{kJ}\) and \(\Delta S = -120;\text{J/K}\). The reaction is spontaneous at:

 (a) all temperatures

 (b) temperatures below 333 K

 (c) temperatures above 333 K

 (d) no temperature

4. The relationship between \(\Delta H\) and \(\Delta U\) for the reaction \(2\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)\) is:

 (a) \(\Delta H = \Delta U + 2RT\)

 (b) \(\Delta H = \Delta U - 2RT\)

 (c) \(\Delta H = \Delta U\)

 (d) \(\Delta H = \Delta U + RT\)

5. For which of the following reactions is \(\Delta H = \Delta U\)?

 (a) \(\text{PCl}_5(g) \rightarrow \text{PCl}_3(g) + \text{Cl}_2(g)\)

 (b) \(\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)\)

 (c) \(\text{H}_2(g) + \text{I}_2(g) \rightarrow 2\text{HI}(g)\)

 (d) \(2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)\)

Short Answer Questions (2–3 Marks)

6. State Hess’s law. Why is it a consequence of the first law of thermodynamics?

7. Calculate the standard enthalpy of formation of CH₃OH(l) using the following data:

\[ \text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta_c H^\circ = -726;\text{kJ/mol} \]

\[ \Delta_f H^\circ[\text{CO}_2(g)] = -393.5;\text{kJ/mol}, \quad \Delta_f H^\circ[\text{H}_2\text{O}(l)] = -285.8;\text{kJ/mol} \]

8. What is the sign of entropy change (\(\Delta S\)) for the following processes?

 (i) Freezing of water

 (ii) Evaporation of ethanol

 (iii) Dissolving NaCl in water

9. Explain why \(C_P > C_V\) for gases.

10. Define: (i) Standard enthalpy of atomization (ii) Lattice enthalpy

Long Answer Questions (5 Marks)

11. (a) State and explain the first law of thermodynamics. Derive the relationship between \(\Delta H\) and \(\Delta U\).

 (b) Calculate the \(\Delta U\) for the reaction: \(2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)\), given \(\Delta H = -198;\text{kJ}\) at 298 K.

12. (a) Explain the Gibbs energy equation and how it determines the spontaneity of a reaction.

 (b) A reaction has \(\Delta H = 52;\text{kJ}\) and \(\Delta S = 165;\text{J/K}\). At what temperature will it become spontaneous?

Assertion-Reason Questions

13. Assertion (A): An endothermic reaction can be spontaneous.

Reason (R): Increase in entropy can drive a reaction forward.

14. Assertion (A): State functions are path-independent.

Reason (R): The value of a state function depends only on the present state, not on how that state was reached.

Answer Key

Chapter 6: Equilibrium

6.1 Equilibrium in Physical Processes

Physical equilibrium is the equilibrium established in physical processes such as:

• Solid ⇌ Liquid (melting/freezing at melting point)
• Liquid ⇌ Gas (evaporation/condensation at boiling point)
• Solute in Solid ⇌ Solute in Solution (dissolution/crystallization at saturation)
• Gas ⇌ Gas dissolved in liquid (Henry’s law)

At equilibrium, the rate of the forward process equals the rate of the reverse process.

6.2 Equilibrium in Chemical Processes — Dynamic Equilibrium

In a reversible reaction at equilibrium:

\[ \text{Rate of forward reaction} = \text{Rate of backward reaction} \]

Characteristics of chemical equilibrium:

1. Equilibrium is dynamic (both forward and reverse reactions continue)
2. Equilibrium can be reached from either direction
3. A catalyst does not change the equilibrium position, only speeds up attainment
4. At equilibrium, the concentrations remain constant

6.3 Law of Mass Action and Equilibrium Constant

For a general reversible reaction:

\[ a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \]

The equilibrium constant expression (in terms of concentrations):

\[ K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \]

In terms of partial pressures (for gaseous reactions):

\[ K_p = \frac{(P_\text{C})^c(P_\text{D})^d}{(P_\text{A})^a(P_\text{B})^b} \]

Relationship between Kp and Kc

\[ K_p = K_c(RT)^{\Delta n_g} \]

where \(\Delta n_g\) = (moles of gaseous products) − (moles of gaseous reactants)

Important Relationships

6.4 Factors Affecting Equilibrium — Le Chatelier’s Principle

“If a system at equilibrium is subjected to a change, the equilibrium will shift in the direction that tends to counteract the change.”

Example: Haber’s process:

\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92.4;\text{kJ/mol} \]

Favourable conditions: High pressure (forward has fewer moles of gas), Low temperature (exothermic), use of Fe catalyst (for faster rate).

6.5 Ionic Equilibrium in Solution

Electrolytes

• Strong electrolytes: Completely ionize in solution (e.g., NaCl, HCl, NaOH, KNO₃)
• Weak electrolytes: Partially ionize in solution (e.g., CH₃COOH, NH₃, H₂CO₃)

Degree of Ionization (\(\alpha\))

\[ \alpha = \frac{\text{Number of moles ionized}}{\text{Total moles of electrolyte}} \]

6.6 Ionization of Acids and Bases

Arrhenius Concept

• Acid: Produces H⁺ in water
• Base: Produces OH⁻ in water

Brønsted-Lowry Concept

• Acid: Proton (H⁺) donor
• Base: Proton (H⁺) acceptor

Conjugate acid-base pairs:

\[ \text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+ \]

Here, CH₃COOH/CH₃COO⁻ is a conjugate acid-base pair.

Lewis Concept

• Acid: Electron pair acceptor (e.g., BF₃, AlCl₃)
• Base: Electron pair donor (e.g., NH₃, H₂O)

6.7 Ionization Constant of Weak Acids (Ka)

For a weak acid HA:

\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]

If \(c\) = initial concentration and \(\alpha\) = degree of ionization:

\[ K_a = \frac{c\alpha^2}{1 - \alpha} \approx c\alpha^2 \quad (\text{if } \alpha \ll 1) \]

\[ \alpha = \sqrt{\frac{K_a}{c}} \]

Polybasic Acids

For acids with more than one ionizable hydrogen (e.g., H₃PO₄):

\[ K_{a_1} \gg K_{a_2} \gg K_{a_3} \]

Each successive proton is harder to remove.

6.8 Ionization Constant of Weak Bases (Kb)

\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]

Relationship between Ka and Kb for a conjugate pair:

\[ K_a \times K_b = K_w = 1.0 \times 10^{-14} \quad (\text{at 25°C}) \]

6.9 Concept of pH

Ionic Product of Water (Kw)

\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]

\[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \quad (\text{at 25°C}) \]

pH Scale

\[ \text{pH} = -\log[\text{H}^+] \]

\[ \text{pOH} = -\log[\text{OH}^-] \]

\[ \text{pH} + \text{pOH} = 14 \quad (\text{at 25°C}) \]

6.10 Hydrolysis of Salts

Hydrolysis is the reaction of a salt with water, producing acidic or basic solutions.

For a salt of weak acid + strong base (e.g., CH₃COONa):

\[ \text{pH} = 7 + \frac{1}{2}\text{p}K_a + \frac{1}{2}\log c \]

For a salt of strong acid + weak base (e.g., NH₄Cl):

\[ \text{pH} = 7 - \frac{1}{2}\text{p}K_b - \frac{1}{2}\log c \]

6.11 Buffer Solutions

A buffer solution resists changes in pH on the addition of small amounts of acid or base.

Types:

1. Acidic buffer: Weak acid + its conjugate base (salt), e.g., CH₃COOH + CH₃COONa
2. Basic buffer: Weak base + its conjugate acid (salt), e.g., NH₃ + NH₄Cl

Henderson-Hasselbalch Equation

For acidic buffer:

\[ \text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} \]

For basic buffer:

\[ \text{pOH} = \text{p}K_b + \log\frac{[\text{Salt}]}{[\text{Base}]} \]

6.12 Solubility Equilibria — Solubility Product (Ksp)

For a sparingly soluble salt \(\text{A}_x\text{B}_y\):

\[ \text{A}_x\text{B}_y(s) \rightleftharpoons x\text{A}^{y+}(aq) + y\text{B}^{x-}(aq) \]

\[ K_{sp} = [\text{A}^{y+}]^x[\text{B}^{x-}]^y \]

Relationship between \(K_{sp}\) and solubility (\(s\)):

For AB type (e.g., AgCl): \(K_{sp} = s^2\)

For AB₂ type (e.g., PbCl₂): \(K_{sp} = 4s^3\)

Common Ion Effect

The solubility of a sparingly soluble salt decreases in the presence of a common ion.

Example: Solubility of AgCl decreases in NaCl solution (Cl⁻ is the common ion).

Practice Questions

Multiple Choice Questions (MCQs)

1. The equilibrium constant for the reaction \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\) is \(K\). The equilibrium constant for \(\text{NH}_3 \rightleftharpoons \frac{1}{2}\text{N}_2 + \frac{3}{2}\text{H}_2\) is:

 (a) \(\frac{1}{K}\)

 (b) \(K^{1/2}\)

 (c) \(\frac{1}{K^{1/2}}\)

 (d) \(K^2\)

2. The pH of a 0.001 M HCl solution is:

 (a) 1

 (b) 2

 (c) 3

 (d) 4

3. A buffer solution can be prepared by mixing:

 (a) NaCl + HCl

 (b) CH₃COOH + CH₃COONa

 (c) NaOH + NaCl

 (d) HCl + NaOH

4. The solubility product of AgCl is \(1.6 \times 10^{-10}\). Its solubility in mol/L is:

 (a) \(1.26 \times 10^{-5}\)

 (b) \(1.6 \times 10^{-5}\)

 (c) \(4.0 \times 10^{-5}\)

 (d) \(1.6 \times 10^{-10}\)

5. For which reaction does \(K_p = K_c\)?

 (a) \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\)

 (b) \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\)

 (c) \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)\)

 (d) \(2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)\)

Short Answer Questions (2–3 Marks)

6. State Le Chatelier’s principle. Predict the direction of shift for the reaction \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 + \text{heat}\) when (i) pressure is increased, (ii) temperature is increased.

7. Calculate the pH of a 0.01 M Ba(OH)₂ solution.

8. The \(K_a\) for HF is \(6.8 \times 10^{-4}\). Calculate the degree of ionization of HF in its 0.1 M solution.

9. Write the Henderson-Hasselbalch equation. Calculate the pH of a buffer made by mixing 0.1 M CH₃COOH and 0.1 M CH₃COONa. (\(K_a = 1.8 \times 10^{-5}\))

10. What is the common ion effect? How does it affect the solubility of AgCl in NaCl solution?

Long Answer Questions (5 Marks)

11. (a) Derive the relationship between \(K_p\) and \(K_c\).

 (b) For the reaction \(\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)\), the value of \(K_c = 0.36\) at 100°C. If the initial concentration of N₂O₄ is 0.1 M, calculate the concentrations at equilibrium.

12. (a) Define ionic product of water. What is its value at 25°C? How does it vary with temperature?

 (b) The pH of gastric juice is 1.2. Calculate the concentration of HCl in gastric juice.

 (c) What is the pH of a 0.002 M NaOH solution?

13. (a) Explain hydrolysis of salts with examples. Why does a solution of sodium acetate have a pH > 7?

 (b) Calculate the pH of 0.1 M NH₄Cl solution. (\(K_b\) for NH₃ = \(1.8 \times 10^{-5}\))

Assertion-Reason Questions

14. Assertion (A): Addition of an inert gas at constant volume does not affect the equilibrium.

Reason (R): The concentrations (partial pressures) of the reactants and products remain unchanged.

15. Assertion (A): The pH of 10⁻⁸ M HCl is not 8.

Reason (R): For very dilute solutions of acids, the contribution of H⁺ from water must be considered.

Answer Key

Chapter 7: Redox Reactions

7.1 Classical Concept of Oxidation and Reduction

Example:

\[ \text{CuO}(s) + \text{H}_2(g) \rightarrow \text{Cu}(s) + \text{H}_2\text{O}(l) \]

• CuO is reduced (loss of oxygen)
• H₂ is oxidised (gain of oxygen)

7.2 Redox Reactions in Terms of Electron Transfer

Remember: OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons)

7.3 Oxidation Number (Oxidation State)

The oxidation number is the charge an atom would have if all bonds were ionic.

Rules for Assigning Oxidation Numbers

1. Free elements: Oxidation number = 0 (e.g., O₂, Na, Fe)
2. Monoatomic ions: Oxidation number = charge on ion (e.g., Na⁺ = +1, Cl⁻ = −1)
3. Hydrogen: Usually +1, except in metal hydrides (NaH, CaH₂) where it is −1
4. Oxygen: Usually −2, except in:
   • Peroxides (H₂O₂, Na₂O₂): −1
   • Superoxides (KO₂): −1/2
   • OF₂: +2
5. Fluorine: Always −1
6. Sum of oxidation numbers in a neutral molecule = 0; in an ion = charge on ion

Examples:

In \(\text{KMnO}_4\): Let Mn = \(x\)

\[ (+1) + x + 4(-2) = 0 \implies x = +7 \]

In \(\text{Cr}_2\text{O}_7^{2-}\): Let Cr = \(x\)

\[ 2x + 7(-2) = -2 \implies x = +6 \]

Redox Reaction Identification Using Oxidation Numbers

• Oxidation: Increase in oxidation number
• Reduction: Decrease in oxidation number
• Disproportionation: Same element is simultaneously oxidised and reduced

Example of disproportionation:

\[ 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \]

Oxygen in H₂O₂ is −1. In H₂O it becomes −2 (reduced), in O₂ it becomes 0 (oxidised).

7.4 Balancing Redox Reactions

Method 1: Oxidation Number Method

1. Write the skeleton equation
2. Assign oxidation numbers to all atoms
3. Identify atoms whose oxidation number changes
4. Equalize the increase and decrease in oxidation number
5. Balance the remaining atoms (O using H₂O, H using H⁺ in acidic medium or OH⁻ in basic)
6. Balance charge using electrons

Method 2: Half-Reaction (Ion-Electron) Method

Steps (in acidic medium):

1. Separate into oxidation and reduction half-reactions
2. Balance atoms other than O and H
3. Balance O by adding H₂O
4. Balance H by adding H⁺
5. Balance charge by adding electrons
6. Equalize electrons in both half-reactions
7. Add the half-reactions

Steps (in basic medium): Follow steps 1–7 for acidic medium, then add OH⁻ to both sides to neutralize H⁺.

Worked Example: Balance \(\text{Fe}^{2+} + \text{MnO}_4^- \rightarrow \text{Fe}^{3+} + \text{Mn}^{2+}\) in acidic medium.

Oxidation half-reaction:

\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

Reduction half-reaction:

\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

Multiply oxidation by 5:

\[ 5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^- \]

Balanced equation:

\[ \text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \]

7.5 Types of Redox Reactions

7.6 Applications of Redox Reactions

1. Electrochemical cells (batteries, fuel cells)
2. Corrosion of metals (rusting of iron)
3. Extraction of metals from ores
4. Biological processes (photosynthesis, respiration)
5. Quantitative analysis (titrations with KMnO₄, K₂Cr₂O₇)

Practice Questions

Multiple Choice Questions (MCQs)

1. The oxidation number of Cr in \(\text{Cr}_2\text{O}_7^{2-}\) is:

 (a) +3

 (b) +4

 (c) +6

 (d) +7

2. In which of the following, hydrogen has an oxidation number of −1?

 (a) H₂O

 (b) HF

 (c) NaH

 (d) H₂O₂

3. Which of the following is a disproportionation reaction?

 (a) \(2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2\)

 (b) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}\)

 (c) \(\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2\)

 (d) \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)

4. The number of electrons involved in the reduction of \(\text{MnO}_4^-\) to \(\text{Mn}^{2+}\) is:

 (a) 1

 (b) 3

 (c) 5

 (d) 7

5. The oxidation state of S in \(\text{Na}_2\text{S}_2\text{O}_3\) is:

 (a) +2

 (b) −2

 (c) +4

 (d) +6

Short Answer Questions (2–3 Marks)

6. Assign oxidation numbers to all atoms in: (i) H₂SO₄ (ii) KMnO₄ (iii) Na₂Cr₂O₇

7. Balance the following reaction in acidic medium using the ion-electron method:

\[ \text{Cr}_2\text{O}_7^{2-} + \text{I}^- \rightarrow \text{Cr}^{3+} + \text{I}_2 \]

8. Identify the oxidising and reducing agents in: \(2\text{FeCl}_3 + \text{H}_2\text{S} \rightarrow 2\text{FeCl}_2 + \text{S} + 2\text{HCl}\)

9. What is a disproportionation reaction? Give an example.

10. The oxidation number of phosphorus in \(\text{Ba}(\text{H}_2\text{PO}_4)_2\) is?

Long Answer Questions (5 Marks)

11. (a) Explain the concept of oxidation number with examples.

 (b) Balance the following reaction in basic medium:

\[ \text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} \rightarrow \text{MnO}_2 + \text{CO}_3^{2-} \]

12. (a) Classify the following reactions as combination, decomposition, displacement, or disproportionation:

 (i) \(2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}\)

 (ii) \(\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2\)

 (iii) \(2\text{Cu}_2\text{O} \rightarrow 4\text{Cu} + \text{O}_2\) (at high temperature)

 (b) Give two applications of redox reactions in daily life.

Assertion-Reason Questions

13. Assertion (A): Fluorine always shows an oxidation state of −1.

Reason (R): Fluorine is the most electronegative element.

14. Assertion (A): In H₂O₂, the oxidation state of oxygen is −1.

Reason (R): H₂O₂ has an O–O bond.

Answer Key

Chapter 8: Organic Chemistry – Some Basic Principles and Techniques

8.1 General Introduction

Organic chemistry is the study of carbon compounds (excluding simple substances like CO, CO₂, carbonates, and cyanides).

Vital Force Theory (Berzelius): Organic compounds could only be made by living organisms.

Disproved by Wöhler (1828): Synthesised urea from ammonium cyanate — an inorganic compound.

\[ \text{NH}_4\text{CNO} \xrightarrow{\text{heat}} \text{NH}_2\text{CONH}_2 \]

Unique Properties of Carbon

• Tetravalency: Carbon forms 4 bonds
• Catenation: Carbon atoms bond with each other to form long chains, branches, and rings
• Multiple bonding (single, double, triple bonds)
• Small size → strong bonds

8.2 Methods of Purification

Chromatography

Principle: Components of a mixture are distributed between a stationary phase and a mobile phase based on their differential adsorption.

Types:

• Column chromatography: Stationary phase = column packed with alumina/silica
• Thin Layer Chromatography (TLC): Stationary phase = thin layer of silica on glass plate
• Paper chromatography: Stationary phase = water trapped in paper

\[ R_f = \frac{\text{Distance moved by component}}{\text{Distance moved by solvent front}} \]

8.3 Qualitative Analysis of Organic Compounds

Detection of Elements

Quantitative Analysis

Carbon and Hydrogen (Liebig’s method):

\[ \text{% C} = \frac{12 \times \text{Mass of CO}_2}{44 \times \text{Mass of compound}} \times 100 \]

\[ \text{% H} = \frac{2 \times \text{Mass of H}_2\text{O}}{18 \times \text{Mass of compound}} \times 100 \]

Nitrogen:

• Dumas’ method: N₂ gas collected over KOH
• Kjeldahl’s method: \(\text{% N} = \frac{1.4 \times M \times V}{\text{Mass of compound}}\)

where \(M\) = molarity of acid, \(V\) = volume of acid used.

8.4 Classification and IUPAC Nomenclature

Classification of Organic Compounds

IUPAC Nomenclature

General format: Prefix + Root word + Suffix

Functional Groups

8.5 Isomerism

8.6 Electronic Displacement Effects in Covalent Bonds

1. Inductive Effect (permanent)

The shifting of σ-electrons along a chain of atoms due to difference in electronegativity.

• −I effect (electron-withdrawing): −NO₂, −CN, −COOH, −F, −Cl, −Br, −I, −OH, −NH₂
• +I effect (electron-donating): alkyl groups (−CH₃, −C₂H₅, etc.)

Order of +I effect: \(-\text{C(CH}_3)_3 > -\text{CH(CH}_3)_2 > -\text{C}_2\text{H}_5 > -\text{CH}_3 > -\text{H}\)

2. Resonance (Mesomeric) Effect

Delocalisation of π-electrons or lone pairs in conjugated systems.

• +M effect (electron-donating to the ring): −OH, −NH₂, −OCH₃, −OR
• −M effect (electron-withdrawing from the ring): −NO₂, −CN, −CHO, −COOH

3. Electromeric Effect (temporary)

Temporary transfer of π-electrons to one atom on demand of an attacking reagent.

4. Hyperconjugation (σ–π conjugation)

The delocalisation of σ-electrons of C–H bonds adjacent to a multiple bond or positive charge.

• Stability of carbocations: \(\text{3°} > \text{2°} > \text{1°} > \text{CH}_3^+\) (more hyperconjugation → more stable)

8.7 Homolytic and Heterolytic Fission

Homolytic Fission (free radicals)

\[ \text{A:B} \xrightarrow{\text{UV/heat}} \text{A}· + \text{B}· \]

Produces free radicals (species with unpaired electrons). Common in non-polar solvents and gas phase.

Heterolytic Fission (ions)

\[ \text{A:B} \rightarrow \text{A}^+ + \text{B}^- \quad \text{or} \quad \text{A}^- + \text{B}^+ \]

Produces carbocations (\(\text{C}^+\)) and carbanions (\(\text{C}^-\)).

Stability of carbocations: \(\text{3°} > \text{2°} > \text{1°} > \text{CH}_3^+\) (due to hyperconjugation and +I effect)

Stability of carbanions: \(\text{CH}_3^- > \text{1°} > \text{2°} > \text{3°}\) (opposite order)

8.8 Types of Organic Reactions

Electrophiles: Electron-loving species (Lewis acids): H⁺, NO₂⁺, Cl⁺, BF₃, AlCl₃

Nucleophiles: Nucleus-loving species (Lewis bases): OH⁻, CN⁻, NH₃, H₂O, R−O⁻

Practice Questions

Multiple Choice Questions (MCQs)

1. The IUPAC name of (CH₃)₃C–CH₂–CH(CH₃)₂ is:

 (a) 2,2,4-Trimethylpentane

 (b) 2,4,4-Trimethylpentane

 (c) 2,2,4-Trimethylhexane

 (d) 2-Tert-butyl-3-methylbutane

2. Which of the following shows +I effect?

 (a) −NO₂

 (b) −Cl

 (c) −CH₃

 (d) −COOH

3. Which is the most stable carbocation?

 (a) \(\text{CH}_3^+\)

 (b) \(\text{(CH}_3)_2\text{CH}^+\)

 (c) \(\text{(CH}_3)_3\text{C}^+\)

 (d) \(\text{C}_2\text{H}_5^+\)

4. Lassaigne’s test is used to detect:

 (a) C and H only

 (b) N, S, and halogens

 (c) Molecular formula

 (d) Functional groups

5. The number of structural isomers of C₄H₁₀ is:

 (a) 1

 (b) 2

 (c) 3

 (d) 4

Short Answer Questions (2–3 Marks)

6. What is the inductive effect? Arrange the following in order of increasing +I effect: −H, −CH₃, −C₂H₅, −(CH₃)₃C.

7. Draw all the structural isomers of C₅H₁₂ and give their IUPAC names.

8. Distinguish between electrophiles and nucleophiles with examples.

9. Explain hyperconjugation with a suitable example.

10. Write the IUPAC names of: (i) CH₃CH(OH)CH₂CHO (ii) CH₂=CHCH₂Br

Long Answer Questions (5 Marks)

11. (a) Explain resonance with the example of benzene. What are the conditions for resonance?

 (b) Compare the stability of the following carbocations and explain:

  \(\text{CH}_3^+,;\text{C}_2\text{H}_5^+,;\text{(CH}_3)_2\text{CH}^+,;\text{(CH}_3)_3\text{C}^+\)

12. (a) What are the different types of isomerism in organic chemistry? Give one example of each.

 (b) Write the IUPAC name of:

  \(\text{CH}_3\text{CH}_2\text{C(CH}_3)_2\text{CH}_2\text{CH(CH}_3)\text{CH}_2\text{OH}\)

13. (a) Describe the following purification methods with diagrams: (i) Distillation (ii) Crystallisation.

 (b) How is nitrogen estimated by Kjeldahl’s method?

Assertion-Reason Questions

14. Assertion (A): Tertiary carbocations are more stable than primary carbocations.

Reason (R): Hyperconjugation and +I effect of alkyl groups stabilise carbocations.

15. Assertion (A): −NO₂ group shows −I and −M effects.

Reason (R): −NO₂ is an electron-withdrawing group.

Answer Key

Chapter 9: Hydrocarbons

9.1 Introduction

Hydrocarbons are organic compounds containing only carbon and hydrogen.

PART A: ALIPHATIC HYDROCARBONS

9.2 Alkanes (Paraffins)

Nomenclature (IUPAC)

Isomerism in Alkanes (Chain Isomerism)

C₄H₁₀ has 2 isomers: n-butane, isobutane (2-methylpropane)

C₅H₁₂ has 3 isomers: pentane, 2-methylbutane, 2,2-dimethylpropane (neopentane)

Conformation of Ethane

Conformations are different spatial arrangements obtained by rotation about a C–C single bond.

Energy difference: Eclipsed form is ~12 kJ/mol higher in energy than staggered form due to torsional strain.

Physical Properties of Alkanes

• First four members (C₁–C₄) are gases; C₅–C₁₇ are liquids; C₁₈+ are solids
• Boiling point increases with molecular mass
• Branching decreases boiling point
• Insoluble in water (non-polar), soluble in organic solvents

Chemical Reactions of Alkanes

1. Halogenation (Free Radical Substitution)

\[ \text{CH}_4 + \text{Cl}_2 \xrightarrow{h\nu} \text{CH}_3\text{Cl} + \text{HCl} \]

Mechanism (Free Radical Chain Reaction):

Step 1: Initiation

\[ \text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}· \]

Step 2: Propagation

\[ \text{CH}_4 + \text{Cl}· \rightarrow \text{CH}_3· + \text{HCl} \]

\[ \text{CH}_3· + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}· \]

Step 3: Termination

\[ \text{CH}_3· + \text{Cl}· \rightarrow \text{CH}_3\text{Cl} \]

Reactivity of H atoms: 3° > 2° > 1°

2. Combustion

\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \Delta_c H = -890;\text{kJ/mol} \]

General: \(\text{C}n\text{H}{2n+2} + \frac{3n+1}{2}\text{O}_2 \rightarrow n\text{CO}_2 + (n+1)\text{H}_2\text{O}\)

3. Pyrolysis (Cracking)

\[ \text{C}{10}\text{H}{22} \xrightarrow{500°\text{C}} \text{C}5\text{H}{12} + \text{C}5\text{H}{10} \]

Higher alkanes → smaller alkanes + alkenes (thermal decomposition)

9.3 Alkenes (Olefins)

Structure of Double Bond (Ethene)

• Each carbon is sp² hybridised
• Bond angle ≈ 120°, planar geometry
• One σ bond (head-on overlap of sp² orbitals) + One π bond (lateral overlap of unhybridised p orbitals)

Geometrical (cis-trans) Isomerism

Alkenes with two different groups on each doubly-bonded carbon show geometrical isomerism due to restricted rotation about the C=C bond.

Methods of Preparation of Alkenes

1. Dehydrohalogenation of alkyl halides (Saytzeff’s rule):

\[ \text{CH}_3\text{CHBrCH}_3 \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH=CH}_2 + \text{HBr} \]

2. Dehydration of alcohols:

\[ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{H}_2\text{SO}_4, 443\text{K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \]

Chemical Reactions of Alkenes

Addition Reactions

(a) Addition of hydrogen (hydrogenation):

\[ \text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni/Pt/Pd}} \text{CH}_3\text{CH}_3 \]

(b) Addition of halogens (halogenation):

\[ \text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br} \]

This is used as a test for unsaturation — decolorization of brown Br₂ water.

(c) Addition of HX (Markovnikov’s rule):

“The negative part of the addendum (X) adds to the carbon bearing fewer hydrogen atoms.”

\[ \text{CH}_3\text{CH=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CHBrCH}_3 \quad \text{(Markovnikov product)} \]

(d) Anti-Markovnikov addition (Peroxide effect / Kharash effect):

In the presence of organic peroxides (e.g., benzoyl peroxide), HBr adds in anti-Markovnikov fashion:

\[ \text{CH}_3\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \]

This works only with HBr, not with HCl or HI.

(e) Addition of water (hydration):

\[ \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{OH} \]

Oxidation

• With cold, dilute KMnO₄ (Baeyer’s test): Forms glycol

\[ \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} + [\text{O}] \xrightarrow{\text{KMnO}_4} \text{HOCH}_2\text{CH}_2\text{OH} \]

(Decolorization of pink KMnO₄ → test for unsaturation)

• Ozonolysis: Alkene + O₃ → ozonide → cleaved by Zn/H₂O → carbonyl compounds

\[ \text{R}_1\text{R}_2\text{C=CR}_3\text{R}_4 \xrightarrow{1.,\text{O}_3,\ 2.,\text{Zn/H}_2\text{O}} \text{R}_1\text{R}_2\text{C=O} + \text{R}_3\text{R}_4\text{C=O} \]

9.4 Alkynes

Structure of Triple Bond (Ethyne)

• Each carbon is sp hybridised
• Linear geometry, bond angle = 180°
• One σ bond + Two π bonds

Acidic Character of Alkynes

Terminal alkynes (with C≡C-H) show acidic character because the sp-hybridised carbon has more s-character (50%) → holds electrons more tightly → makes H more acidic.

\[ \text{CH≡CH} + \text{Na} \rightarrow \text{CH≡CNa} + \frac{1}{2}\text{H}_2 \]

Acidity order: HC≡CH > H₂C=CH₂ > H₃C−CH₃ (sp > sp² > sp³)

Chemical Reactions of Alkynes

(a) Addition of H₂:

\[ \text{CH≡CH} \xrightarrow{\text{H}_2/\text{Pd-BaSO}_4} \text{CH}_2=\text{CH}_2 \xrightarrow{\text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_3 \]

Lindlar’s catalyst (Pd/BaSO₄ + quinoline) → gives cis-alkene (partial reduction)

(b) Addition of HX: Two molecules add (Markovnikov’s rule applies)

\[ \text{CH≡CH} \xrightarrow{\text{HCl}} \text{CH}_2=\text{CHCl} \xrightarrow{\text{HCl}} \text{CH}_3\text{CHCl}_2 \]

(c) Addition of water (hydration):

\[ \text{CH≡CH} + \text{H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4/\text{HgSO}_4} \text{CH}_3\text{CHO} \quad \text{(acetaldehyde)} \]

9.5 Aromatic Hydrocarbons

Benzene: Structure and Aromaticity

Benzene (C₆H₆) was proposed by Kekulé (1865) as a cyclic structure with alternating single and double bonds.

Actual structure: All C–C bonds in benzene are equal (139 pm, intermediate between C–C 154 pm and C=C 134 pm) due to resonance / delocalised π system.

Hückel’s Rule for Aromaticity

A cyclic planar compound is aromatic if it has \((4n + 2)\) π electrons, where \(n = 0, 1, 2, 3, \ldots\)

Electrophilic Aromatic Substitution (EAS)

The characteristic reaction of benzene is electrophilic substitution (not addition), because substitution preserves aromaticity.

General Mechanism:

\[ \text{ArH} + \text{E}^+ \rightarrow [\text{Arenium ion}] \rightarrow \text{ArE} + \text{H}^+ \]

Directive Influence of Substituents

Existing substituents on benzene direct incoming groups to specific positions:

Carcinogenicity and Toxicity

• Benzene is a known carcinogen (causes leukaemia)
• Polynuclear aromatic hydrocarbons (PAHs) like benzo[a]pyrene (found in coal tar and tobacco smoke) are potent carcinogens
• Toluene poisoning can cause damage to the central nervous system

Practice Questions

Multiple Choice Questions (MCQs)

1. The product formed when propene reacts with HBr in the presence of peroxide is:

 (a) 2-Bromopropane

 (b) 1-Bromopropane

 (c) 2-Bromobutane

 (d) Propan-2-ol

2. Which of the following is NOT aromatic?

 (a) Benzene

 (b) Naphthalene

 (c) Cyclooctatetraene

 (d) Pyridine

3. The electrophile in Friedel-Crafts acylation is:

 (a) R⁺

 (b) RCO⁺

 (c) AlCl₃

 (d) Cl⁺

4. Lindlar’s catalyst converts alkynes to:

 (a) Alkanes

 (b) trans-Alkenes

 (c) cis-Alkenes

 (d) Alkynes remain unchanged

5. The correct order of decreasing acidity is:

 (a) \(\text{HC≡CH} > \text{H}_2\text{C=CH}_2 > \text{H}_3\text{C-CH}_3\)

 (b) \(\text{H}_3\text{C-CH}_3 > \text{H}_2\text{C=CH}_2 > \text{HC≡CH}\)

 (c) \(\text{H}_2\text{C=CH}_2 > \text{HC≡CH} > \text{H}_3\text{C-CH}_3\)

 (d) \(\text{HC≡CH} > \text{H}_3\text{C-CH}_3 > \text{H}_2\text{C=CH}_2\)

6. Which substituent is meta-directing?

 (a) −OH

 (b) −CH₃

 (c) −NO₂

 (d) −NH₂

Short Answer Questions (2–3 Marks)

7. State Markovnikov’s rule. What is the exception (anti-Markovnikov addition)?

8. Give the mechanism of free radical halogenation of methane.

9. What are conformational isomers? Draw the eclipsed and staggered conformations of ethane.

10. Write the reactions of ethyne with: (i) HCl (ii) water

11. Explain why benzene undergoes electrophilic substitution rather than addition.

Long Answer Questions (5 Marks)

12. (a) Explain Hückel’s rule for aromaticity. Using this rule, predict whether the following are aromatic: (i) Cyclopentadienyl anion (ii) Cycloheptatrienyl cation (iii) Cyclooctatetraene.

 (b) What are polynuclear aromatic hydrocarbons? Why are they considered carcinogenic?

13. (a) Compare the reactivity of alkanes, alkenes, and alkynes towards addition reactions.

 (b) Write the products formed by ozonolysis of: (i) But-1-ene (ii) But-2-ene

 (c) An alkene on ozonolysis gives methanal and ethanal. Identify the alkene.

14. (a) What is Friedel-Crafts reaction? Explain both alkylation and acylation with mechanisms.

 (b) Classify the following substituents as ortho/para-directing or meta-directing: −Cl, −OH, −NO₂, −CH₃, −COOH.

Assertion-Reason Questions

15. Assertion (A): Terminal alkynes are weakly acidic.

Reason (R): sp-hybridised carbon has more s-character, making the C–H bond more polar.

16. Assertion (A): Toluene is ortho/para-directing in electrophilic substitution.

Reason (R): The methyl group has +I effect and activates the ring.

Answer Key