Basic Concepts of Chemistry - JEE Main & Advanced
1. Matter and Its Classification
1.1 States of Matter
- Solid: Definite shape and volume
- Liquid: Definite volume but no fixed shape
- Gas: Neither definite shape nor volume
- Plasma: Ionized state of matter at high temperatures
- BEC (Bose-Einstein Condensate): State at extremely low temperatures
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
- Dalton's Atomic Theory: Matter made of indivisible atoms
- Thomson's Model: Plum pudding model
- Rutherford's Model: Nuclear model with empty space
- Bohr's Model: Quantized orbits for electrons
- Quantum Mechanical Model: Electron clouds/orbitals
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
- Atomic Mass Unit (amu): 1/12th mass of C-12 atom
- Mole: Amount of substance containing 6.022×10²³ particles
- Molar Mass: Mass of 1 mole of substance (g/mol)
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
- Ionic Bond: Transfer of electrons (NaCl)
- Covalent Bond: Sharing of electrons (H₂O)
- Metallic Bond: Sea of electrons (Cu, Fe)
- Hydrogen Bond: Special dipole-dipole interaction (H₂O, HF)
- van der Waals Forces: Weak intermolecular forces
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
- System: Part of universe under study
- Surroundings: Everything else
- State Functions: Properties independent of path (P, V, T, H, U, S, G)
- Extensive Properties: Depend on quantity (mass, volume)
- Intensive Properties: Independent of quantity (density, temperature)
Thermodynamic Laws:
- Zeroth Law: Defines temperature
- First Law: ΔU = q + w (Conservation of energy)
- Second Law: ΔS_univ > 0 for spontaneous process
- 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
- Understand the mole concept thoroughly - it's fundamental to all quantitative chemistry
- Memorize periodic trends and their exceptions
- Practice stoichiometry problems with different approaches
- Relate atomic structure concepts to periodic properties
- Understand the significance of thermodynamic parameters in chemical reactions
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)
- Matter composed of indivisible atoms
- Atoms of same element are identical
- Compounds form when atoms combine in fixed ratios
- Chemical reactions involve rearrangement of atoms
1.2 Thomson's Plum Pudding Model (1897)
- Discovered electrons using cathode ray tube
- Atom as positively charged sphere with embedded electrons
- Failed to explain atomic stability and spectral lines
Thomson's Model: ⊕⊕⊕⊕⊕ (positive sphere) with ••• (electrons)
1.3 Rutherford's Nuclear Model (1911)
- Gold foil experiment led to discovery of nucleus
- Most of atom is empty space
- Positive charge concentrated in small nucleus
- Electrons orbit the nucleus
- Failed to explain atomic stability (classical EM theory predicts electrons should spiral into nucleus)
Rutherford's Scattering Formula:
N(θ) ∝ 1/sin⁴(θ/2)
Where θ = scattering angle
1.4 Bohr's Model (1913)
- Postulated quantized electron orbits
- Electrons don't radiate energy in stationary orbits
- Energy emitted/absorbed when electrons jump between orbits
- Successfully explained hydrogen spectrum
- Failed for multi-electron atoms and fine spectrum
2. Quantum Mechanical Model
2.1 Wave-Particle Duality
- de Broglie hypothesis: λ = h/p = h/mv
- Davisson-Germer experiment confirmed wave nature of electrons
- Heisenberg Uncertainty Principle: Δx·Δp ≥ h/4π
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
- Electrons revolve in stationary orbits without radiating energy
- Angular momentum is quantized: mvr = nħ where ħ = h/2π
- 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
- Memorize Bohr's formulas for hydrogen-like atoms
- Understand quantum numbers and their significance
- Practice writing electronic configurations, including exceptions
- Know the shapes and orientations of different orbitals
- Be thorough with spectral series calculations
- Understand the limitations of each atomic model
Periodic Classification - JEE Main & Advanced
Periodic Classification & Periodicity - JEE Main & Advanced
1. Historical Development of Periodic Table
1.1 Dobereiner's Triads (1829)
- Groups of 3 elements with similar properties
- Atomic weight of middle element ≈ average of other two
- Example: Li (7), Na (23), K (39)
- Limited to only a few elements
1.2 Newlands' Law of Octaves (1864)
- Every 8th element had similar properties
- Worked well for lighter elements only
1.3 Mendeleev's Periodic Table (1869)
- Arranged elements in increasing atomic weights
- Left gaps for undiscovered elements (Ga, Ge, Tc)
- Predicted properties of missing elements accurately
- Anomalies: Ar/K, Co/Ni, Te/I pairs
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
- Alkali Metals: Group 1 (ns¹)
- Alkaline Earth Metals: Group 2 (ns²)
- Chalcogens: Group 16 (ns²np⁴)
- Halogens: Group 17 (ns²np⁵)
- Noble Gases: Group 18 (ns²np⁶)
- Coinage Metals: Cu, Ag, Au (Group 11)
- Volatile Metals: Zn, Cd, Hg (Group 12)
3. Periodic Trends
3.1 Atomic Radius
Type |
Definition |
Trend in Period |
Trend in Group |
Covalent Radius |
½ distance between nuclei of 2 bonded atoms |
Decreases → |
Increases ↓ |
van der Waals Radius |
½ distance between nuclei of 2 adjacent atoms |
Decreases → |
Increases ↓ |
Exceptions: Noble gases have largest vdW radius in period
Anomalous Size: Lanthanide contraction (5d series ≈ 4d series size)
3.2 Ionization Energy (IE)
Energy required to remove most loosely bound electron from gaseous atom
X(g) → X⁺(g) + e⁻; ΔH = IE₁
Order |
Trend in Period |
Trend in Group |
Exceptions |
IE₁ < IE₂ < IE₃... |
Increases → |
Decreases ↓ |
Be > B, N > O, Mg > Al, P > S |
Example Problem:
Arrange in increasing order of IE₁: N, O, F
Answer: O < N < F (N has half-filled p-subshell stability)
3.3 Electron Affinity (EA)
Energy released when electron is added to gaseous atom
X(g) + e⁻ → X⁻(g); ΔH = -EA
Trend in Period |
Trend in Group |
Exceptions |
Increases → |
Decreases ↓ |
Cl > F (small size of F causes electron repulsion) |
3.4 Electronegativity (EN)
Tendency to attract shared electron pair (Pauling scale)
Trend in Period |
Trend in Group |
Highest/Lowest |
Increases → |
Decreases ↓ |
F (4.0) > O (3.5) > Cl (3.0); Cs (0.7) lowest |
3.5 Metallic Character
Property |
Trend in Period |
Trend in Group |
Electropositivity |
Decreases → |
Increases ↓ |
Reducing Nature |
Decreases → |
Increases ↓ |
4. Other Periodic Properties
4.1 Valency
- s-block: Equal to group number
- p-block: 8 - group number (or group number for lower groups)
- d-block: Variable valency common
4.2 Oxidation States
- s-block: Fixed (+1 Group 1, +2 Group 2)
- p-block: Variable (except F always -1)
- Transition metals: Multiple oxidation states common
4.3 Melting/Boiling Points
- s-block: Decrease down group (except anomalous Li)
- p-block: Increase then decrease across period (peak at C/Si)
- d-block: Generally high, peak at middle of series
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
- Memorize all periodic trends with exceptions
- Understand why anomalies occur (half-filled/filled subshell stability)
- Practice comparative questions (which is larger/higher EN etc.)
- Know special cases like lanthanide contraction
- Remember diagonal relationships
Chemical Bonding - JEE Main & Advanced
Chemical Bonding & Molecular Structure - JEE Main & Advanced
1. Types of Chemical Bonds
1.1 Ionic (Electrovalent) Bond
- Formed by complete electron transfer (metal → non-metal)
- Governed by Fajan's Rules
- Lattice Energy (U) = k(q⁺q⁻)/r₀
- Examples: NaCl, CaO, MgCl₂
Fajan's Rules for Covalent Character in Ionic Bonds:
- Small cation size → More polarization
- Large anion size → More polarization
- High charge on ions → More polarization
- Non-polarizable cations (Noble gas config) → Less covalent
1.2 Covalent Bond
- Formed by electron sharing (Lewis theory)
- Explained by Valence Bond Theory (VBT) and Molecular Orbital Theory (MOT)
- Examples: H₂, O₂, CH₄
1.3 Coordinate Bond
- Special covalent bond where both electrons come from one atom
- Represented by "→"
- Examples: H₃O⁺, NH₄⁺, SO₄²⁻
1.4 Metallic Bond
- "Sea of electrons" model
- Responsible for metallic properties (conductivity, malleability)
- Examples: Cu, Fe, Na
2. VSEPR Theory (Valence Shell Electron Pair Repulsion)
2.1 Basic Principles
- Electron pairs (bonding & lone pairs) arrange to minimize repulsion
- Order of repulsion: LP-LP > LP-BP > BP-BP
- Bond angles affected by lone pairs
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
- Mixing of atomic orbitals to form new hybrid orbitals
- Explains molecular geometry and bond angles
- Hybridization = ½(V + M - C + A)
- Where V = valence e⁻, M = monovalent atoms, C = cation charge, A = anion charge
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
- Atomic orbitals combine to form molecular orbitals
- Bonding MO (σ, π): Lower energy than parent AOs
- Anti-bonding MO (σ*, π*): Higher energy than parent AOs
- Bond order = ½(Nb - Na)
4.2 MOT Diagrams
Homonuclear Diatomics (O₂, N₂):
σ2s ↑↓
σ*2s ↑↓
π2p ↑↓ ↑↓
σ2p ↑↓
π*2p ↑ ↑ (for O₂)
σ*2p
4.3 Magnetic Properties
- Paramagnetic: Unpaired electrons (O₂, B₂)
- Diamagnetic: All electrons paired (N₂, H₂)
Important MOT Observations:
- O₂ is paramagnetic (2 unpaired e⁻ in π*2p)
- N₂ has triple bond (σ2p + π2p pair)
- Bond order predicts stability: Higher BO → More stable
5. Hydrogen Bonding
5.1 Characteristics
- Special dipole-dipole interaction (X-H···Y)
- X, Y = F, O, N (highly electronegative)
- Strength: 5-10% of covalent bond
5.2 Types
Type |
Description |
Examples |
Intermolecular |
Between molecules |
H₂O (ice), HF (liquid) |
Intramolecular |
Within same molecule |
o-nitrophenol, salicylaldehyde |
5.3 Effects
- Increases boiling/melting points
- Increases solubility (e.g., ethanol in water)
- Affects density (ice less dense than water)
- Important in biological structures (DNA, proteins)
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
- Memorize VSEPR geometries and bond angles
- Practice hybridization calculations for complex molecules
- Understand MOT diagrams for diatomic molecules
- Know exceptions to octet rule (PCl₅, SF₆ etc.)
- Remember effects of hydrogen bonding on physical properties
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
- Postulates: Point masses, random motion, elastic collisions
- PV = ⅓ mNū² (where ū = root mean square speed)
- Kinetic energy per mole = 3/2 RT
- Average KE per molecule = 3/2 kT (k = Boltzmann constant)
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
- Deviation from ideal behavior at high P and low T
- van der Waals equation: (P + an²/V²)(V - nb) = nRT
- a = intermolecular forces correction, b = volume correction
- Boyle temperature (TB): Gas behaves ideally over widest P range
Compressibility Factor (Z = PV/nRT):
- Z = 1 for ideal gas
- Z < 1 at low P (attractive forces dominate)
- Z > 1 at high P (repulsive forces dominate)
3. Liquid State Properties
3.1 Surface Tension (γ)
- γ = F/l (N/m)
- Work done: W = γ·ΔA
- Capillary rise: h = 2γcosθ/rρg
- Effect of temperature: γ decreases with T
- Effect of additives: Surfactants decrease γ
3.2 Viscosity (η)
- F = ηA(dv/dx) (Poiseuille's law)
- SI unit: Pa·s (1 Poise = 0.1 Pa·s)
- Stokes' law: F = 6πηrv (for spherical particles)
- Effect of temperature: η decreases for liquids, increases for gases
3.3 Vapor Pressure
- Equilibrium pressure exerted by vapor at given T
- Clausius-Clapeyron equation: ln(P₂/P₁) = (ΔHvap/R)(1/T₁ - 1/T₂)
- Boiling point: When vapor pressure = atmospheric pressure
- Effect of temperature: Exponential increase with T
4. Phase Equilibria
4.1 Phase Diagrams
Water Phase Diagram:
- Triple point: 0.01°C, 4.58 mmHg (all 3 phases coexist)
- Critical point: 374°C, 218 atm (liquid-gas distinction disappears)
- Negative slope for solid-liquid line (ice less dense than water)
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
- Memorize all gas law relationships and their graphical representations
- Understand the significance of van der Waals constants a and b
- Know the three molecular speeds and their ratios
- Remember surface tension and viscosity formulas with units
- Practice phase diagram interpretation (especially water's anomalous behavior)
- Be thorough with azeotrope concepts and examples
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
- Independent of path: U (internal energy), H (enthalpy), S (entropy), G (Gibbs energy)
- Depend only on initial and final states
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:
- +q: Heat absorbed by system
- -q: Heat released by system
- +w: Work done on system (compression)
- -w: Work done by system (expansion)
2.3 Second Law
- Entropy of universe always increases: ΔSuniv > 0
- Clausius statement: Heat cannot flow from colder to hotter body without work
- Kelvin-Planck statement: No engine can convert all heat to work
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
- CP - at constant pressure
- CV - at constant volume
- CP - CV = R (for ideal gas)
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
- Memorize sign conventions for work and heat
- Understand state functions vs path functions
- Be thorough with ΔG = ΔH - TΔS applications
- Practice Hess's law problems
- Know standard enthalpy definitions
- Remember work calculations for different processes
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
- Dynamic nature - forward and reverse reactions continue
- Macroscopic properties remain constant
- Achieved in closed system at constant temperature
- Independent of path taken
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
- Understand the dynamic nature of equilibrium
- Memorize Kp-Kc relationship
- Apply Le Chatelier's principle to predict shifts
- Know all pH calculation formulas
- Remember buffer action mechanism
- Practice solubility product problems
- Recognize common ion effect applications
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
- Oxidation: Loss of electrons or increase in oxidation number.
- Reduction: Gain of electrons or decrease in oxidation number.
- Example:
Zn + CuSO₄ → ZnSO₄ + Cu
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:
- Oxidation number method
- 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
- Metallurgy (extraction of metals)
- Electrochemical cells and batteries
- Biological systems (e.g., cellular respiration)
9. Important Tips for JEE/NEET
- Practice oxidation number assignments
- Be thorough with redox titration questions
- Understand both methods of balancing
- Memorize common oxidizing and reducing agents
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
- Protium (¹H): Most common, no neutrons
- Deuterium (²H or D): One neutron, used in nuclear reactions
- Tritium (³H or T): Radioactive, used in tracer studies
4. Preparation of Dihydrogen (H₂)
Laboratory Methods:
- Action of dilute acid on metals:
Zn + 2HCl → ZnCl₂ + H₂
- Electrolysis of water:
2H₂O → 2H₂ + O₂
Commercial Methods:
- Steam reforming of methane
- Water gas shift reaction
5. Properties of Hydrogen
- Colorless, odorless, tasteless gas
- Highly combustible
- Forms covalent compounds with non-metals
- Acts as a reducing agent
6. Reactions of Dihydrogen
- With metals: Forms metal hydrides (e.g., NaH, CaH₂)
- With non-metals: Forms covalent compounds (e.g., NH₃, CH₄, HCl)
- With oxides: Reduces metal oxides (e.g.,
CuO + H₂ → Cu + H₂O
)
7. Uses of Hydrogen
- Production of ammonia (Haber process)
- Hydrogenation of vegetable oils
- As rocket fuel (liquid H₂ and O₂)
- In fuel cells to generate electricity
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
- Hydrides: Ionic (e.g., NaH), Covalent (e.g., CH₄), Metallic (e.g., TiH2)
- Water (H₂O): Essential compound with hydrogen bonding
- Hydrogen Peroxide (H₂O₂): Oxidizing agent, antiseptic
10. Tips for JEE/NEET
- Understand hydrogen's dual nature (alkali + halogen)
- Memorize isotopes and their uses
- Practice balancing redox reactions involving hydrogen
- Focus on hydrides and hydrogen bonding concepts
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
- Elements: Li, Na, K, Rb, Cs, Fr
- Valence shell configuration: ns1
- Highly reactive metals, soft, low melting points
- Stored under oil to prevent reaction with air/moisture
3. Group 2: Alkaline Earth Metals
- Elements: Be, Mg, Ca, Sr, Ba, Ra
- Valence shell configuration: ns2
- Harder and denser than alkali metals
- Less reactive than Group 1 but still reactive
4. General Trends (Both Groups)
- ↓ Atomic & Ionic size increases down the group
- ↓ Ionization enthalpy decreases
- ↓ Metallic character increases
- ↓ Reactivity increases
- ↓ Electropositive nature increases
5. Chemical Properties
- With Oxygen: Form oxides and peroxides (Na₂O₂, K₂O₂)
- With Water: Release H₂ gas (e.g.,
2Na + 2H₂O → 2NaOH + H₂
)
- With Halogens: Form ionic halides (e.g., NaCl, MgCl₂)
- With Acids: Produce hydrogen gas
6. Important Compounds
- NaOH: Strong base, used in soap industry
- Na₂CO₃: Washing soda
- CaCO₃: Limestone, chalk, marble
- Ca(OH)₂: Slaked lime, used in whitewashing
- CaSO₄·2H₂O: Plaster of Paris
7. Biological Importance
- Sodium: Regulates osmotic balance and nerve function
- Potassium: Essential for muscle contraction and nerve function
- Calcium: Bone and teeth formation, blood clotting
- Magnesium: Component of chlorophyll, enzyme activation
8. Anomalous Behavior of Lithium and Beryllium
- Due to small size and high charge density
- Show diagonal relationship with Mg and Al respectively
- Form covalent compounds (LiCl, BeCl₂)
9. Trends in Solubility
- Group 1: All salts are soluble
- Group 2: Solubility of hydroxides ↑ down the group; solubility of sulphates ↓
10. Tips for JEE/NEET
- Memorize properties and uses of important compounds
- Understand periodic trends and exceptions
- Practice reaction equations (especially with H₂O and O₂)
- Learn biological roles and anomalies of Li and Be
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
- Show variable oxidation states
- Have both metallic and non-metallic characters
- Form covalent bonds predominantly
- Inert pair effect is observed (especially in heavier elements)
3. Group 13 – Boron Family
- Elements: B, Al, Ga, In, Tl
- General electronic configuration: ns2np1
- Boron is a metalloid; others are metals
- Oxidation states: +3 (common), +1 (in Tl due to inert pair effect)
Important Compounds:
- Borax (Na₂B₄O₇·10H₂O)
- Orthoboric acid (H₃BO₃)
- Aluminium chloride (AlCl₃) – Lewis acid
4. Group 14 – Carbon Family
- Elements: C, Si, Ge, Sn, Pb
- General electronic configuration: ns2np2
- Carbon and silicon are non-metals/metalloids; Sn and Pb are metals
- Oxidation states: +4 (common), +2 (in Sn and Pb due to inert pair effect)
Important Compounds:
- CO, CO₂ – acidic oxides of carbon
- SiO₂ – strong covalent network solid
- PbCl₂ and PbCl₄ – illustrate variable oxidation states
5. Trends and Anomalies
- Electronegativity and ionization enthalpy decrease down the group
- Metallic character increases down the group
- Catenation tendency is highest in carbon
- Inert pair effect increases down the group
6. Chemical Properties
- Reactivity with acids and bases: Amphoteric behavior in elements like Al
- Reactivity with oxygen: Form oxides of the type E₂O₃ and EO₂
- Halide formation: Electron-deficient halides (e.g., BCl₃)
7. Important Points for JEE/NEET
- Understand Lewis acid behavior of BCl₃, AlCl₃
- Remember all group trends and exceptions (e.g., catenation, inert pair effect)
- Practice naming and reactions of borates, silicates, and other compounds
- Study the structure and acidic nature of boric acid
Class 11 Organic Chemistry Notes
Class 11 Chemistry Notes: Organic Chemistry
1. Introduction to Organic Chemistry
- Study of carbon compounds (except CO, CO₂, carbonates, etc.)
- Due to catenation and tetravalency, carbon forms large number of compounds
2. Classification of Organic Compounds
- Acyclic (Aliphatic): Open chain compounds (e.g., ethane)
- Cyclic: Closed chain or ring compounds
- Aromatic (e.g., benzene)
- Alicyclic (e.g., cyclohexane)
3. Functional Groups
- Alcohols (–OH), Aldehydes (–CHO), Ketones (–CO–), Acids (–COOH)
- Determine chemical properties of organic compounds
4. Homologous Series
- Series of compounds with same functional group and general formula
- Each member differs by –CH₂– group
5. Nomenclature (IUPAC)
- Steps: Longest carbon chain → Numbering → Prefix/suffix → Functional group
CH₃CH₂CH₂OH
→ Propanol
CH₃COOH
→ Ethanoic acid
6. Isomerism
- Structural Isomerism: Different connectivity
- Chain isomerism
- Position isomerism
- Functional isomerism
- Stereoisomerism: Same structure, different spatial arrangement
- Geometrical (cis-trans)
- Optical isomerism
7. Reaction Mechanism Basics
- Types of reagents: Nucleophiles (e⁻ rich), Electrophiles (e⁻ deficient)
- Types of reactions: Substitution, Addition, Elimination, Rearrangement
- Types of fission: Homolytic (→ free radicals), Heterolytic (→ ions)
8. Electron Effects in Organic Molecules
- Inductive effect (–I, +I): Due to electronegativity difference
- Resonance: Delocalization of electrons (e.g., benzene)
- Hyperconjugation: No-bond resonance (alkyl stabilization)
- Electromeric effect: Temporary shift of electrons in double/triple bonds
9. Purification Techniques
- Crystallization, Distillation, Sublimation
- Chromatography – separates based on polarity/adsorption
10. Qualitative & Quantitative Analysis
- Detect elements: Lassaigne’s test for N, S, halogens
- Determine % composition: Dumas method, Kjeldahl’s method for nitrogen
11. Tips for JEE/NEET
- Master functional groups and IUPAC naming rules
- Understand electron effects and reaction mechanisms clearly
- Practice identifying isomers and drawing them
- Remember qualitative tests for elements
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:
- Alkanes (saturated)
- Alkenes (unsaturated, double bond)
- Alkynes (unsaturated, triple bond)
- Aromatic hydrocarbons (e.g., benzene)
2. Alkanes (Paraffins)
- General formula: CnH2n+2
- Single bonds only (sp³ hybridized C atoms)
- Examples: Methane, Ethane, Propane
Preparation:
- Hydrogenation of alkenes/alkynes
- Wurtz reaction:
2R–X + 2Na → R–R + 2NaX
Reactions:
- Halogenation:
CH₄ + Cl₂ → CH₃Cl + HCl
(in presence of sunlight)
- Combustion:
CH₄ + 2O₂ → CO₂ + 2H₂O
3. Alkenes (Olefins)
- General formula: CnH2n
- One double bond (sp² hybridized C atoms)
- Examples: Ethene, Propene
Preparation:
- Dehydration of alcohols:
CH₃CH₂OH → CH₂=CH₂ + H₂O
- Dehydrohalogenation of alkyl halides
Reactions:
- Addition of H₂ (hydrogenation)
- Addition of halogens (Br₂, Cl₂)
- Markovnikov's Rule:
CH₃CH=CH₂ + HBr → CH₃CHBrCH₃
4. Alkynes
- General formula: CnH2n–2
- One triple bond (sp hybridized C atoms)
- Examples: Ethyne (acetylene), Propyne
Preparation:
- Dehalogenation of vicinal dihalides
- From calcium carbide:
CaC₂ + H₂O → HC≡CH + Ca(OH)₂
Reactions:
- Addition of H₂, halogens, and HCl
- Oxidation with KMnO₄ → CO₂ + H₂O
5. Aromatic Hydrocarbons
- Contain benzene ring (delocalized π-electrons)
- Examples: Benzene, Toluene, Xylene
Reactions of Benzene:
- Electrophilic substitution:
- Nitration:
C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
- Halogenation:
C₆H₆ + Cl₂ → C₆H₅Cl + HCl
(with FeCl₃)
- Friedel-Crafts alkylation/acylation
6. Physical Properties
- Hydrocarbons are generally non-polar
- Insoluble in water, soluble in organic solvents
- Boiling points increase with molecular weight
7. Combustion of Hydrocarbons
CH₄ + 2O₂ → CO₂ + 2H₂O
(complete combustion)
- Incomplete combustion → CO or C (soot)
8. Important Tips for JEE/NEET
- Learn reaction mechanisms (free radical, electrophilic addition)
- Understand Markovnikov's and anti-Markovnikov’s rule
- Practice naming and drawing hydrocarbons
- Memorize key reagents (HBr, KMnO₄, AlCl₃, etc.)
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
- Pollutants: Substances that cause pollution
- Types: Air, Water, Soil, Noise, Thermal, Radioactive
3. Air Pollution
Main Pollutants:
- Carbon monoxide (CO): Binds with hemoglobin, causes suffocation
- Carbon dioxide (CO₂): Contributes to global warming
- Oxides of nitrogen (NO, NO₂): Cause acid rain and respiratory issues
- Oxides of sulfur (SO₂, SO₃): Acid rain
- Particulate matter (PM2.5, PM10): Respiratory problems
Control: Catalytic converters, switching to clean fuels
4. Greenhouse Effect & Global Warming
- CO₂, CH₄, N₂O trap heat → Increase Earth's temperature
- Causes melting of glaciers, rise in sea levels
- Prevention: Reduce fossil fuel use, afforestation
5. Ozone Layer Depletion
- Ozone (O₃) absorbs harmful UV rays
- CFCs (chlorofluorocarbons) break down ozone → Ozone hole
- Montreal Protocol: International agreement to phase out CFCs
6. Water Pollution
Causes:
- Industrial waste (heavy metals, acids)
- Domestic sewage
- Thermal pollution (hot water from industries)
Indicators:
- BOD (Biological Oxygen Demand)
- COD (Chemical Oxygen Demand)
Control: Wastewater treatment, reducing chemical discharge
7. Soil Pollution
- Causes: Pesticides (DDT), industrial waste, plastics
- Effects: Reduces soil fertility, contaminates food
- Prevention: Use of biodegradable products, organic farming
8. Industrial Waste Management
- Segregation of waste at source
- Neutralization of acids/bases before disposal
- Recycling and reuse of materials
9. Green Chemistry
- Design of chemical products and processes to reduce or eliminate hazardous substances
- Examples:
- Use of H₂O₂ instead of Cl₂ for bleaching
- Use of enzymes in place of traditional reagents
- Principles: Atom economy, less hazardous synthesis, energy efficiency
10. Key Terms for JEE/NEET
- BOD: Oxygen required by microbes to decompose organic matter
- COD: Oxygen required to chemically oxidize pollutants
- Acid rain: Rainwater with pH < 5.6 due to SO₂ and NO₂
- CFCs: Cause depletion of ozone