Basic Concepts of Chemistry - JEE Main & Advanced

1. Matter and Its Classification

1.1 States of Matter

1.2 Classification of Matter

Pure Substances Mixtures
  • Elements (Metals, Non-metals, Metalloids)
  • Compounds (Organic, Inorganic)
  • Homogeneous (Solutions)
  • Heterogeneous (Colloids, Suspensions)

2. Atomic Structure

2.1 Fundamental Particles

Particle Charge Mass (amu) Discoverer
Electron (e⁻) -1.6×10⁻¹⁹ C 0.0005486 J.J. Thomson
Proton (p⁺) +1.6×10⁻¹⁹ C 1.007276 Goldstein
Neutron (n⁰) 0 1.008665 Chadwick

2.2 Atomic Models

Important Formulas:

Bohr's radius: rₙ = 0.529 × n²/Z Å

Energy of electron in nth orbit: Eₙ = -13.6 × Z²/n² eV

de Broglie wavelength: λ = h/mv

3. Periodic Table and Periodicity

3.1 Modern Periodic Law

"The physical and chemical properties of elements are periodic functions of their atomic numbers."

3.2 Periodic Trends

Property Trend Across Period Trend Down Group
Atomic Radius Decreases Increases
Ionization Energy Increases Decreases
Electron Affinity Increases Decreases
Electronegativity Increases Decreases

4. Stoichiometry and Mole Concept

4.1 Basic Definitions

Key Formulas:

Number of moles (n) = Given mass (w)/Molar mass (M)

Number of particles = n × Nₐ (Avogadro's number = 6.022×10²³)

At STP: 1 mole gas = 22.4 L

Example Problem:

Calculate the number of moles in 5.6 g of nitrogen gas (N₂).

Solution: Molar mass of N₂ = 28 g/mol
Number of moles = 5.6/28 = 0.2 moles

5. Chemical Bonding

5.1 Types of Bonds

5.2 Hybridization

Hybridization Geometry Bond Angle Examples
sp Linear 180° C₂H₂, BeCl₂
sp² Trigonal planar 120° BF₃, C₂H₄
sp³ Tetrahedral 109.5° CH₄, NH₄⁺

6. Basic Thermodynamics

6.1 Important Terms

Thermodynamic Laws:

  1. Zeroth Law: Defines temperature
  2. First Law: ΔU = q + w (Conservation of energy)
  3. Second Law: ΔS_univ > 0 for spontaneous process
  4. Third Law: S → 0 as T → 0 K

ΔG = ΔH - TΔS (Gibbs free energy)

ΔG° = -RT ln K (Relation with equilibrium constant)

Practice Questions (JEE Level)

Question 1:

The ratio of radii of second orbit of He⁺ to the first orbit of H atom is:

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

Answer: (b) 2:1

Question 2:

Which of the following has the highest electron affinity?

(a) F (b) Cl (c) Br (d) I

Answer: (b) Cl (Fluorine has lower EA than Cl due to small size and electron repulsion)

Question 3:

Calculate the number of atoms in 4.25 g of NH₃.

Solution:
Moles of NH₃ = 4.25/17 = 0.25 moles
Molecules = 0.25 × 6.022×10²³ = 1.5055×10²³
Each NH₃ has 4 atoms (1N + 3H)
Total atoms = 4 × 1.5055×10²³ = 6.022×10²³ atoms

Key Takeaways

2. Atomic Structure - JEE Main & Advanced

2. Atomic Structure - JEE Main & Advanced

1. Historical Development of Atomic Models

1.1 Dalton's Atomic Theory (1808)

1.2 Thomson's Plum Pudding Model (1897)

Thomson's Model: ⊕⊕⊕⊕⊕ (positive sphere) with ••• (electrons)

1.3 Rutherford's Nuclear Model (1911)

Rutherford's Scattering Formula:

N(θ) ∝ 1/sin⁴(θ/2)

Where θ = scattering angle

1.4 Bohr's Model (1913)

2. Quantum Mechanical Model

2.1 Wave-Particle Duality

Example Calculation:

Calculate de Broglie wavelength of electron moving at 1% speed of light.

Solution:
v = 0.01 × 3×10⁸ = 3×10⁶ m/s
λ = h/mv = (6.626×10⁻³⁴)/(9.1×10⁻³¹ × 3×10⁶) = 2.43×10⁻¹⁰ m = 2.43 Å

2.2 Schrödinger Wave Equation

Ĥψ = Eψ

Where:
Ĥ = Hamiltonian operator
ψ = Wave function
E = Energy of system

2.3 Quantum Numbers

Quantum Number Symbol Values Significance
Principal n 1, 2, 3,... Energy level/shell (K, L, M,...)
Azimuthal l 0 to n-1 Subshell shape (s=0, p=1, d=2, f=3)
Magnetic m -l to +l Orbital orientation
Spin s +½ or -½ Electron spin direction

2.4 Orbitals and Their Shapes

Orbital l value Number of Orbitals Shape
s 0 1 Spherical
p 1 3 (pₓ, pᵧ, p_z) Dumbbell
d 2 5 Double dumbbell/cloverleaf
f 3 7 Complex

3. Bohr's Model in Detail

3.1 Postulates

  1. Electrons revolve in stationary orbits without radiating energy
  2. Angular momentum is quantized: mvr = nħ where ħ = h/2π
  3. Energy is emitted/absorbed when electrons jump between orbits: ΔE = E₂ - E₁ = hν

Bohr's Formulas for Hydrogen-like Atoms:

Radius of nth orbit: rₙ = 0.529 × n²/Z Å

Velocity of electron: vₙ = 2.18×10⁶ × Z/n m/s

Energy of nth orbit: Eₙ = -13.6 × Z²/n² eV

Wave number of spectral lines: 1/λ = RZ²(1/n₁² - 1/n₂²)

Where R = Rydberg constant = 1.097×10⁷ m⁻¹

3.2 Spectral Series

Series Transition Region Discovery Year
Lyman n→1 UV 1906-1914
Balmer n→2 Visible 1885
Paschen n→3 IR 1908
Brackett n→4 Far IR 1922
Pfund n→5 Far IR 1924

Example Problem:

Calculate the wavelength of light emitted when electron in H atom falls from n=4 to n=2.

Solution:
1/λ = R(1/n₁² - 1/n₂²) = 1.097×10⁷(1/4 - 1/16) = 1.097×10⁷ × 3/16 = 2.056×10⁶ m⁻¹
λ = 1/(2.056×10⁶) = 4.86×10⁻⁷ m = 486 nm (Balmer series, blue-green color)

4. Electronic Configuration

4.1 Aufbau Principle

Electrons fill orbitals in order of increasing energy.

Order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s

4.2 Pauli Exclusion Principle

No two electrons can have all four quantum numbers identical.

Maximum 2 electrons per orbital with opposite spins.

4.3 Hund's Rule

For degenerate orbitals, electrons occupy separate orbitals with parallel spins before pairing.

Example Configurations:

Carbon (6 electrons): 1s² 2s² 2p² (not 1s² 2s² 2p¹ₓ 2p¹ᵧ)

Iron (Fe, 26 electrons): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

Copper (Cu, 29 electrons): [Ar] 4s¹ 3d¹⁰ (exception due to stability)

Practice Questions (JEE Level)

Question 1:

The orbital angular momentum of an electron in 3d orbital is:

(a) √6(h/2π) (b) √2(h/2π) (c) √3(h/2π) (d) 0

Answer: (a) √6(h/2π) [L = √l(l+1)ħ = √2(2+1)ħ = √6ħ]

Question 2:

Which of the following sets of quantum numbers is not possible?

(a) n=3, l=2, m=-2, s=+½

(b) n=3, l=3, m=0, s=-½

(c) n=2, l=1, m=1, s=+½

(d) n=4, l=0, m=0, s=-½

Answer: (b) [l cannot be equal to n]

Question 3:

The ratio of radii of first orbits of H, He⁺ and Li²⁺ is:

(a) 1:2:3 (b) 1:½:⅓ (c) 1:4:9 (d) 1:1:1

Answer: (b) 1:½:⅓ [r ∝ 1/Z]

Key Takeaways

Periodic Classification - JEE Main & Advanced

Periodic Classification & Periodicity - JEE Main & Advanced

1. Historical Development of Periodic Table

1.1 Dobereiner's Triads (1829)

1.2 Newlands' Law of Octaves (1864)

1.3 Mendeleev's Periodic Table (1869)

1.4 Modern Periodic Law (Moseley, 1913)

"The physical and chemical properties of elements are periodic functions of their atomic numbers."

2. Modern Periodic Table

2.1 Structure

Block Groups Orbitals Filled Elements
s-block 1, 2 ns¹⁻² Alkali & Alkaline earth metals
p-block 13-18 np¹⁻⁶ Metalloids, Non-metals, Halogens, Noble gases
d-block 3-12 (n-1)d¹⁻¹⁰ns⁰⁻² Transition metals
f-block - (n-2)f¹⁻¹⁴(n-1)d⁰⁻¹ns² Lanthanides & Actinides

2.2 Special Groups

4. Other Periodic Properties

4.1 Valency

4.2 Oxidation States

4.3 Melting/Boiling Points

4.4 Diagonal Relationship

Similarities between 2nd period element and 3rd period element of next group:

Pair Similar Properties
Li-Mg Form nitrides, carbonates decompose on heating
Be-Al Amphoteric oxides, form polymeric hydrides
B-Si Form acidic oxides, covalent compounds

Practice Questions (JEE Level)

Question 1:

Which has highest first ionization energy?

(a) N (b) O (c) F (d) Ne

Answer: (d) Ne (noble gas configuration)

Question 2:

Correct order of atomic radii is:

(a) N < O < F (b) F < O < N (c) O < F < N (d) F < N < O

Answer: (b) F < O < N (size decreases across period)

Question 3:

Element with electronic configuration [Xe]4f¹⁴5d¹6s² belongs to:

(a) d-block (b) f-block (c) p-block (d) s-block

Answer: (a) d-block (last electron enters d-orbital)

Key Takeaways

Chemical Bonding - JEE Main & Advanced

Chemical Bonding & Molecular Structure - JEE Main & Advanced

1. Types of Chemical Bonds

1.1 Ionic (Electrovalent) Bond

Fajan's Rules for Covalent Character in Ionic Bonds:

  1. Small cation size → More polarization
  2. Large anion size → More polarization
  3. High charge on ions → More polarization
  4. Non-polarizable cations (Noble gas config) → Less covalent

1.2 Covalent Bond

1.3 Coordinate Bond

1.4 Metallic Bond

2. VSEPR Theory (Valence Shell Electron Pair Repulsion)

2.1 Basic Principles

2.2 Molecular Geometries

Steric Number Lone Pairs Shape Bond Angle Examples
2 0 Linear 180° BeCl₂, CO₂
3 0 Trigonal planar 120° BF₃, SO₃
3 1 Bent/V-shape <120° SO₂, O₃
4 0 Tetrahedral 109.5° CH₄, NH₄⁺
4 1 Trigonal pyramidal 107° NH₃, PCl₃
4 2 Bent/V-shape 104.5° H₂O, H₂S
5 0 Trigonal bipyramidal 90°, 120° PCl₅, PF₅
6 0 Octahedral 90° SF₆, [AlF₆]³⁻

3. Hybridization

3.1 Concept

3.2 Types of Hybridization

Hybridization Geometry Bond Angle Examples
sp Linear 180° C₂H₂, BeCl₂
sp² Trigonal planar 120° C₂H₄, BF₃
sp³ Tetrahedral 109.5° CH₄, NH₄⁺
sp³d Trigonal bipyramidal 90°, 120° PCl₅, SF₄
sp³d² Octahedral 90° SF₆, [Fe(CN)₆]³⁻
dsp² Square planar 90° [Ni(CN)₄]²⁻, XeF₄

Example Problem:

Determine hybridization in XeOF₄

Solution:
Steric number = 6 (1 Xe-O σ + 4 Xe-F σ + 1 lone pair)
Hybridization = sp³d² (octahedral geometry with 1 lone pair)

4. Molecular Orbital Theory (MOT)

4.1 Key Concepts

4.2 MOT Diagrams

Homonuclear Diatomics (O₂, N₂):

σ2s      ↑↓
σ*2s     ↑↓
π2p      ↑↓ ↑↓
σ2p      ↑↓
π*2p     ↑ ↑   (for O₂)
σ*2p     

4.3 Magnetic Properties

Important MOT Observations:

5. Hydrogen Bonding

5.1 Characteristics

5.2 Types

Type Description Examples
Intermolecular Between molecules H₂O (ice), HF (liquid)
Intramolecular Within same molecule o-nitrophenol, salicylaldehyde

5.3 Effects

Practice Questions (JEE Level)

Question 1:

Which has maximum bond angle?

(a) NH₃ (b) NF₃ (c) NCl₃ (d) All equal

Answer: (a) NH₃ (F is more electronegative than H, pulls e⁻ density reducing bond angle)

Question 2:

Correct order of bond length in NO species is:

(a) NO⁻ < NO < NO⁺ (b) NO⁺ < NO < NO⁻ (c) NO < NO⁺ < NO⁻ (d) NO⁻ < NO⁺ < NO

Answer: (b) NO⁺ < NO < NO⁻ (Bond order: NO⁺=3, NO=2.5, NO⁻=2)

Question 3:

Hybridization of central atom in XeF₄ is:

(a) sp³ (b) sp³d (c) sp³d² (d) sp²d²

Answer: (c) sp³d² (Steric number = 6, square planar geometry)

Key Takeaways

States of Matter: Gases & Liquids - JEE Main & Advanced

States of Matter: Gases & Liquids - JEE Main & Advanced

1. Gas Laws and Kinetic Theory

1.1 Fundamental Gas Laws

Law Equation Conditions Graph
Boyle's Law P ∝ 1/V (T constant) Isothermal Hyperbola in P-V
Charles' Law V ∝ T (P constant) Isobaric Straight line in V-T
Gay-Lussac's Law P ∝ T (V constant) Isochoric Straight line in P-T
Avogadro's Law V ∝ n (P,T constant) - Straight line in V-n

Ideal Gas Equation:

PV = nRT

Where:
P = Pressure (atm), V = Volume (L), n = moles,
R = Universal gas constant = 0.0821 L·atm·K⁻¹·mol⁻¹,
T = Temperature (K)

1.2 Kinetic Theory of Gases

2. Gas Properties and Deviations

2.1 Molecular Speeds

Speed Type Formula Ratio
Average speed (vav) √(8RT/πM) 1 : 1.128 : 1.224
Root mean square speed (urms) √(3RT/M) (vav : urms : vmp)
Most probable speed (vmp) √(2RT/M)

Maxwell-Boltzmann Distribution:

Peak (vmp) < vav < urms

Curve shifts right and flattens with temperature increase

2.2 Real Gases and Deviations

Compressibility Factor (Z = PV/nRT):

3. Liquid State Properties

3.1 Surface Tension (γ)

3.2 Viscosity (η)

3.3 Vapor Pressure

4. Phase Equilibria

4.1 Phase Diagrams

Water Phase Diagram:

4.2 Azeotropes

Type Behavior Examples
Minimum boiling Boils at lower T than components 95.6% ethanol + 4.4% water (78.1°C)
Maximum boiling Boils at higher T than components 68% HNO₃ + 32% water (122°C)

Practice Questions (JEE Level)

Question 1:

The rms speed of O₂ at temperature T is v. At what temperature will the rms speed of SO₂ be v?

(a) T (b) 2T (c) T/2 (d) 4T

Answer: (b) 2T (v ∝ √(T/M), M(SO₂)=64, M(O₂)=32 ⇒ T₂ = 2T₁)

Question 2:

For a real gas at 25°C and high pressure, the compression factor is:

(a) =1 (b) <1 (c) >1 (d) Can't predict

Answer: (c) >1 (At high P, repulsive forces dominate causing Z > 1)

Question 3:

The surface tension of liquid at its boiling point:

(a) Becomes zero (b) Becomes infinite (c) Is same as at RT (d) None

Answer: (a) Becomes zero (At critical temperature, surface tension vanishes)

Key Takeaways

Thermodynamics - JEE Main & Advanced

Thermodynamics - JEE Main & Advanced

1. Basic Concepts

1.1 System and Surroundings

System Type Mass Transfer Energy Transfer Example
Isolated No No Thermos flask
Closed No Yes Piston with gas
Open Yes Yes Open beaker

1.2 State Functions

1.3 Thermodynamic Processes

Process Condition Relation
Isothermal ΔT = 0 ΔU = 0
Adiabatic q = 0 PVγ = constant
Isobaric ΔP = 0 W = -PΔV
Isochoric ΔV = 0 W = 0
Cyclic Returns to initial state ΔU = 0, ΔH = 0

2. Laws of Thermodynamics

2.1 Zeroth Law

Defines temperature - if A is in thermal equilibrium with B, and B with C, then A is in equilibrium with C.

2.2 First Law

ΔU = q + w

Where:
ΔU = change in internal energy
q = heat absorbed by system
w = work done on system

Sign Convention:

2.3 Second Law

2.4 Third Law

Entropy of perfect crystal approaches zero as temperature approaches absolute zero.

3. Thermodynamic Potentials

3.1 Enthalpy (H)

H = U + PV

ΔH = ΔU + PΔV (at constant pressure)

3.2 Entropy (S)

Measure of disorder/multiplicity

ΔS = qrev/T

ΔSsurroundings = -ΔHsystem/T

3.3 Gibbs Free Energy (G)

G = H - TS

ΔG = ΔH - TΔS

Spontaneity criteria:
ΔG < 0: Spontaneous
ΔG > 0: Non-spontaneous
ΔG = 0: Equilibrium

3.4 Important Relations

Relation Equation
Gibbs-Helmholtz ΔG = ΔH + T(∂(ΔG)/∂T)P
Maxwell Relations (∂T/∂V)S = -(∂P/∂S)V
van't Hoff Equation d(lnK)/dT = ΔH°/RT²

4. Thermochemistry

4.1 Heat Capacity

C = q/ΔT

4.2 Hess's Law

Enthalpy change is same regardless of path taken.

Example:

Given:
C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5 kJ/mol
CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = -283.0 kJ/mol
Find ΔH for C(s) + ½O₂(g) → CO(g)

Solution: ΔH = ΔH₁ - ΔH₂ = -110.5 kJ/mol

4.3 Bond Enthalpy

ΔH°rxn = Σ(bond energies)reactants - Σ(bond energies)products

Average energy required to break 1 mole of bonds in gaseous state.

4.4 Standard Enthalpies

Type Definition Example
Formation (ΔHf) Formation of 1 mole from elements C(s) + 2H₂(g) → CH₄(g)
Combustion (ΔHc) Complete burning in O₂ CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Neutralization Acid + base → salt + water HCl + NaOH → NaCl + H₂O
Sublimation Solid → gas I₂(s) → I₂(g)

Practice Questions (JEE Level)

Question 1:

For an ideal gas, the work done in isothermal reversible expansion is:

(a) Zero (b) -nRTln(V₂/V₁) (c) -PΔV (d) ΔH

Answer: (b) -nRTln(V₂/V₁)

Question 2:

For a spontaneous process at constant T and P:

(a) ΔG > 0 (b) ΔG < 0 (c) ΔS < 0 (d) ΔH > 0

Answer: (b) ΔG < 0

Question 3:

The enthalpy of combustion of glucose is -2800 kJ/mol. 180g glucose gives energy:

(a) 2800 kJ (b) 5600 kJ (c) 28000 kJ (d) 1680 kJ

Answer: (c) 28000 kJ (180g = 1 mole glucose, 10 moles C-H bonds)

Key Takeaways

Equilibrium - JEE Main & Advanced

Equilibrium - JEE Main & Advanced

1. Chemical Equilibrium

1.1 Law of Mass Action

For a reaction: aA + bB ⇌ cC + dD

Kc = [C]c[D]d/[A]a[B]b

Kp = (PC)c(PD)d/(PA)a(PB)b

Relation between Kp and Kc:

Kp = Kc(RT)Δng

Where Δng = (c+d) - (a+b) = moles gaseous products - moles gaseous reactants

1.2 Characteristics of Equilibrium

1.3 Le Chatelier's Principle

"When a system at equilibrium is disturbed, it shifts to counteract the change."

Stress Effect on Equilibrium
Concentration increase (reactant) Shifts toward products
Pressure increase (gaseous) Shifts toward side with fewer moles
Temperature increase (endothermic) Shifts toward products
Catalyst added No shift (speeds both forward and reverse)

2. Acid-Base Equilibrium

2.1 Theories

Theory Acid Definition Base Definition
Arrhenius H⁺ producer OH⁻ producer
Brønsted-Lowry H⁺ donor H⁺ acceptor
Lewis e⁻ pair acceptor e⁻ pair donor

2.2 Ionization Constants

Ka = [H⁺][A⁻]/[HA] (for acid HA ⇌ H⁺ + A⁻)
Kb = [OH⁻][BH⁺]/[B] (for base B + H₂O ⇌ BH⁺ + OH⁻)

Ka × Kb = Kw = 10-14 (at 25°C)

pH = -log[H⁺], pOH = -log[OH⁻], pH + pOH = 14

2.3 pH Calculations

Solution Type pH Formula
Strong acid pH = -log[H⁺]
Strong base pOH = -log[OH⁻], then pH = 14 - pOH
Weak acid pH = ½(pKa - log C)
Weak base pOH = ½(pKb - log C), then pH = 14 - pOH
Buffer (acidic) pH = pKa + log([salt]/[acid])
Buffer (basic) pOH = pKb + log([salt]/[base])

3. Solubility Equilibrium

3.1 Solubility Product (Ksp)

For salt AxBy(s) ⇌ xAy+(aq) + yBx-(aq)

Ksp = [Ay+]x[Bx-]y

3.2 Common Ion Effect

Solubility decreases when common ion is added

Example: Solubility of AgCl decreases in NaCl solution due to common Cl⁻ ion

3.3 Precipitation Condition

Ionic Product (IP) > Ksp → Precipitation occurs

IP = Ksp → Saturated solution

IP < Ksp → Unsaturated, no precipitation

3.4 Simultaneous Solubility

For salts with common ion (e.g., AgCl and AgBr):

[Ag⁺] = √(Ksp(AgCl) + Ksp(AgBr))

Practice Questions (JEE Level)

Question 1:

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ. Maximum yield of NH₃ is obtained at:

(a) High P, low T (b) Low P, high T (c) High P, high T (d) Low P, low T

Answer: (a) High P (fewer moles product side), low T (exothermic)

Question 2:

pH of 0.01M CH₃COOH (Ka = 1.8×10-5) is:

(a) 2.37 (b) 3.37 (c) 4.37 (d) 5.37

Answer: (b) 3.37 (pH = ½(pKa - log C) = ½(4.74 - log0.01))

Question 3:

Ksp of AgCl is 1.8×10-10. Its solubility in 0.1M NaCl is:

(a) 1.8×10-9 M (b) 1.8×10-10 M (c) 1.8×10-11 M (d) 1.8×10-12 M

Answer: (a) 1.8×10-9 M (Ksp = [Ag⁺][Cl⁻] ⇒ 1.8×10-10 = s × 0.1)

Key Takeaways

Redox Reactions - Class 11 Chemistry Notes

Class 11 Chemistry Notes: Redox Reactions

1. Introduction

Redox reactions are chemical reactions that involve the transfer of electrons between two species. They consist of two parts: Oxidation (loss of electrons) and Reduction (gain of electrons).

2. Oxidation and Reduction

3. Oxidizing and Reducing Agents

- Oxidizing agent: Accepts electrons and gets reduced.
- Reducing agent: Donates electrons and gets oxidized.

4. Oxidation Number

The apparent charge assigned to an atom based on electron accounting rules.
Rules: Free elements = 0, H = +1 (usually), O = –2 (usually), etc.

Example: In H2O, H = +1, O = -2

5. Balancing Redox Reactions

Two common methods:

  1. Oxidation number method
  2. Half-reaction (ion-electron) method

Steps include assigning oxidation numbers, balancing atoms and charges, and adding electrons.

6. Disproportionation Reactions

A reaction in which the same element is simultaneously oxidized and reduced.
Example: 2H₂O₂ → 2H₂O + O₂

7. Redox Titrations

Involve titrating an oxidizing agent with a reducing agent.
Common examples: KMnO₄ vs Fe²⁺, I₂ vs S₂O₃²⁻.

8. Applications of Redox Reactions

9. Important Tips for JEE/NEET

Hydrogen - Class 11 Chemistry Notes

Class 11 Chemistry Notes: Hydrogen

1. Introduction

Hydrogen is the first element in the periodic table and the most abundant element in the universe. It is a non-metal and exists as a diatomic molecule (H2).

2. Position in the Periodic Table

- Resembles alkali metals (Group 1) due to its electronic configuration (1s1)
- Also resembles halogens (Group 17) as it can gain one electron to form H-

3. Isotopes of Hydrogen

4. Preparation of Dihydrogen (H₂)

Laboratory Methods:

Commercial Methods:

5. Properties of Hydrogen

6. Reactions of Dihydrogen

7. Uses of Hydrogen

8. Hydrogen Economy

A proposed system where hydrogen is used as a clean fuel. It can store and transport energy efficiently and reduce carbon emissions.

9. Important Compounds of Hydrogen

10. Tips for JEE/NEET

S-Block Elements - Class 11 Chemistry Notes

Class 11 Chemistry Notes: S-Block Elements

1. Introduction

S-block elements include Group 1 (Alkali metals) and Group 2 (Alkaline earth metals). They have their valence electrons in the 's' orbital.

2. Group 1: Alkali Metals

3. Group 2: Alkaline Earth Metals

4. General Trends (Both Groups)

5. Chemical Properties

6. Important Compounds

7. Biological Importance

8. Anomalous Behavior of Lithium and Beryllium

9. Trends in Solubility

10. Tips for JEE/NEET

P-Block Elements - Class 11 Chemistry Notes

Class 11 Chemistry Notes: P-Block Elements

1. Introduction

P-block elements are those in which the last electron enters a p-orbital. Class 11 covers groups 13 (Boron family) and 14 (Carbon family).

2. General Characteristics

3. Group 13 – Boron Family

Important Compounds:

4. Group 14 – Carbon Family

Important Compounds:

5. Trends and Anomalies

6. Chemical Properties

7. Important Points for JEE/NEET

Class 11 Organic Chemistry Notes

Class 11 Chemistry Notes: Organic Chemistry

1. Introduction to Organic Chemistry

2. Classification of Organic Compounds

3. Functional Groups

4. Homologous Series

5. Nomenclature (IUPAC)

6. Isomerism

7. Reaction Mechanism Basics

8. Electron Effects in Organic Molecules

9. Purification Techniques

10. Qualitative & Quantitative Analysis

11. Tips for JEE/NEET

Hydrocarbons - Class 11 Chemistry Notes

Class 11 Chemistry Notes: Hydrocarbons

1. Introduction

Hydrocarbons are organic compounds composed entirely of carbon and hydrogen atoms. They are classified into:

2. Alkanes (Paraffins)

Preparation:

Reactions:

3. Alkenes (Olefins)

Preparation:

Reactions:

4. Alkynes

Preparation:

Reactions:

5. Aromatic Hydrocarbons

Reactions of Benzene:

6. Physical Properties

7. Combustion of Hydrocarbons

8. Important Tips for JEE/NEET

Environmental Chemistry - Class 11 Notes

Class 11 Chemistry Notes: Environmental Chemistry

1. Introduction

Environmental Chemistry is the branch of chemistry that deals with the study of chemical changes in the environment and their impact on living organisms.

2. Environmental Pollution

3. Air Pollution

Main Pollutants:

Control: Catalytic converters, switching to clean fuels

4. Greenhouse Effect & Global Warming

5. Ozone Layer Depletion

6. Water Pollution

Causes:

Indicators:

Control: Wastewater treatment, reducing chemical discharge

7. Soil Pollution

8. Industrial Waste Management

9. Green Chemistry

10. Key Terms for JEE/NEET