Comprehensive chemistry lessons with theory, reactions, and solved examples for JEE, NEET, and board exams.
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Atomic structure describes the composition of atoms, including subatomic particles, electron configurations, and quantum mechanical models that explain atomic behavior.
Atomic structure covers the discovery of subatomic particles, atomic models, quantum numbers, electronic configuration, and periodic properties of elements.
| Particle | Symbol | Charge | Mass | Location | Discoverer |
|---|---|---|---|---|---|
| Proton | p⁺ | +1.6 × 10⁻¹⁹ C | 1.672 × 10⁻²⁷ kg | Nucleus | Rutherford (1919) |
| Neutron | n⁰ | 0 | 1.675 × 10⁻²⁷ kg | Nucleus | Chadwick (1932) |
| Electron | e⁻ | -1.6 × 10⁻¹⁹ C | 9.109 × 10⁻³¹ kg | Orbitals | Thomson (1897) |
Number of protons in nucleus
\( Z = \text{Number of protons} \)
Total number of protons and neutrons
\( A = Z + N \)
Where N = number of neutrons
Radius of nth orbit
\( r_n = \frac{n^2h^2}{4\pi^2mZe^2} \)
For hydrogen: \( r_n = 0.529 × n^2 \) Å
Energy of nth orbit
\( E_n = -\frac{2\pi^2me^4Z^2}{n^2h^2} \)
For hydrogen: \( E_n = -\frac{13.6}{n^2} \) eV
Velocity of electron
\( v_n = \frac{2\pi Ze^2}{nh} \)
For hydrogen: \( v_n = \frac{2.188 × 10^6}{n} \) m/s
Wavelength of emitted radiation
\( \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \)
R = Rydberg constant = 1.097 × 10⁷ m⁻¹
| Quantum Number | Symbol | Values | Significance |
|---|---|---|---|
| Principal | n | 1, 2, 3, ... | Energy level, size of orbital |
| Azimuthal | l | 0 to (n-1) | Shape of orbital, sublevel |
| Magnetic | ml | -l to +l | Orientation of orbital |
| Spin | ms | +½, -½ | Electron spin direction |
l = 0
Spherical shape
1 orientation
Max 2 electrons
l = 1
Dumbbell shape
3 orientations
Max 6 electrons
l = 2
Cloverleaf shape
5 orientations
Max 10 electrons
l = 3
Complex shape
7 orientations
Max 14 electrons
Electrons occupy orbitals in order of increasing energy
1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p
No two electrons in an atom can have the same set of four quantum numbers
An orbital can hold maximum 2 electrons with opposite spins
Electrons occupy degenerate orbitals singly before pairing
Maximum multiplicity rule - electrons prefer parallel spins
Problem: Write all possible sets of quantum numbers for electrons in 3d orbital.
Solution:
For 3d: n=3, l=2
ml = -2, -1, 0, +1, +2
For each ml, ms = +½, -½
Total 10 possible sets:
(3,2,-2,±½), (3,2,-1,±½), (3,2,0,±½), (3,2,1,±½), (3,2,2,±½)
Problem: Write electronic configuration for Chromium (Z=24).
Solution:
Expected: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴
Actual (with half-filled stability): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
Chromium shows exception due to extra stability of half-filled d-subshell
Problem: Calculate radius and energy of first orbit in hydrogen atom.
Solution:
Radius: r₁ = 0.529 Å = 5.29 × 10⁻¹¹ m
Energy: E₁ = -13.6 eV = -2.18 × 10⁻¹⁸ J
1. Calculate wavelength of radiation emitted when electron in hydrogen atom jumps from n=4 to n=2.
[Answer: λ = 486 nm]
2. Write electronic configuration and quantum numbers for valence electrons of Oxygen (Z=8).
[Answer: 1s² 2s² 2p⁴, (2,1,-1,±½), (2,1,0,±½), (2,1,1,±½)]
3. An element has atomic number 26. Identify the element and write its electronic configuration.
[Answer: Iron, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶]
States of matter describe the distinct forms that different phases of matter take on. The four fundamental states are solid, liquid, gas, and plasma.
States of matter are characterized by differences in molecular arrangement, energy, and intermolecular forces. Transitions between states involve energy changes and follow specific thermodynamic principles.
Characteristics: Definite shape and volume
Molecular Arrangement: Regular, closely packed
Motion: Vibrational only
Examples: Ice, iron, wood
Characteristics: Definite volume, indefinite shape
Molecular Arrangement: Random, close together
Motion: Vibrational + rotational + limited translational
Examples: Water, oil, mercury
Characteristics: Indefinite shape and volume
Molecular Arrangement: Random, far apart
Motion: Complete freedom of movement
Examples: Air, oxygen, steam
Characteristics: Ionized gas, conducts electricity
Molecular Arrangement: Charged particles
Motion: Highly energetic charged particles
Examples: Stars, neon signs, lightning
| Property | Solid | Liquid | Gas |
|---|---|---|---|
| Shape | Fixed | Takes container shape | Takes container shape |
| Volume | Fixed | Fixed | Volume of container |
| Density | High | Medium | Low |
| Compressibility | Negligible | Very low | High |
| Intermolecular Forces | Very strong | Strong | Weak |
Melting (Fusion)
Solid → Liquid
Energy absorbed: Latent heat of fusion
Freezing
Liquid → Solid
Energy released: Latent heat of fusion
Vaporization
Liquid → Gas
Energy absorbed: Latent heat of vaporization
Condensation
Gas → Liquid
Energy released: Latent heat of vaporization
Sublimation
Solid → Gas
Energy absorbed: Latent heat of sublimation
Deposition
Gas → Solid
Energy released: Latent heat of sublimation
Heat for Melting/Freezing
\( Q = mL_f \)
L_f = Latent heat of fusion
Heat for Vaporization/Condensation
\( Q = mL_v \)
L_v = Latent heat of vaporization
Heat for Temperature Change
\( Q = mc\Delta T \)
c = Specific heat capacity
Boyle's Law
\( P \propto \frac{1}{V} \) (constant T)
\( P_1V_1 = P_2V_2 \)
Charles's Law
\( V \propto T \) (constant P)
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
Gay-Lussac's Law
\( P \propto T \) (constant V)
\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
Avogadro's Law
\( V \propto n \) (constant P,T)
\( \frac{V_1}{n_1} = \frac{V_2}{n_2} \)
\( PV = nRT \)
Where:
Combined Gas Law
\( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)
Dalton's Law of Partial Pressures
\( P_{\text{total}} = P_1 + P_2 + P_3 + \cdots \)
Pressure from Kinetic Theory
\( P = \frac{1}{3}\frac{N}{V}m\overline{v^2} \)
N = number of molecules, m = molecular mass
Root Mean Square Speed
\( v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \)
M = molar mass (kg/mol)
Average Kinetic Energy
\( \overline{KE} = \frac{3}{2}k_BT \)
k_B = Boltzmann constant
Most Probable Speed
\( v_{\text{mp}} = \sqrt{\frac{2RT}{M}} \)
Speed of maximum molecules
Correction to ideal gas law accounting for molecular volume and intermolecular forces:
\( \left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT \)
Correction for Intermolecular Forces
\( \frac{an^2}{V^2} \)
a = measure of attractive forces
Correction for Molecular Volume
\( (V - nb) \)
b = volume occupied by molecules
Compressibility Factor
\( Z = \frac{PV}{nRT} \)
Problem: How much heat is required to convert 2kg of ice at -10°C to steam at 120°C?
Given: c_ice = 2100 J/kg·K, c_water = 4186 J/kg·K, c_steam = 2010 J/kg·K, L_f = 3.34×10⁵ J/kg, L_v = 2.26×10⁶ J/kg
Solution:
1. Heat ice from -10°C to 0°C: Q₁ = mc_iceΔT = 2×2100×10 = 42,000 J
2. Melt ice at 0°C: Q₂ = mL_f = 2×3.34×10⁵ = 668,000 J
3. Heat water from 0°C to 100°C: Q₃ = mc_waterΔT = 2×4186×100 = 837,200 J
4. Vaporize water at 100°C: Q₄ = mL_v = 2×2.26×10⁶ = 4,520,000 J
5. Heat steam from 100°C to 120°C: Q₅ = mc_steamΔT = 2×2010×20 = 80,400 J
Total Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 6,147,600 J
Problem: A gas occupies 2L at 300K and 1 atm. Find volume at 400K and 2 atm.
Solution:
Using combined gas law: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)
\( \frac{1 \times 2}{300} = \frac{2 \times V_2}{400} \)
\( V_2 = \frac{1 \times 2 \times 400}{300 \times 2} = \frac{800}{600} = 1.33 \) L
Problem: Calculate RMS speed of oxygen molecules at 27°C (M = 0.032 kg/mol)
Solution:
\( v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \)
\( v_{\text{rms}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} \)
\( v_{\text{rms}} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233,831} = 483.6 \) m/s
1. Calculate the heat required to convert 500g of water at 25°C to steam at 100°C.
[Given: c_water = 4.186 J/g°C, L_v = 2260 J/g]
[Answer: 1,305,650 J]
2. A gas occupies 5L at 2 atm pressure. What volume will it occupy at 4 atm at constant temperature?
[Answer: 2.5 L]
3. Find the RMS speed of nitrogen molecules at 0°C (M = 0.028 kg/mol).
[Answer: 493 m/s]
Chemical thermodynamics deals with the relationship between heat, work, and chemical reactions or physical changes of state.
Gibbs Free Energy: ΔG = ΔH - TΔS
Enthalpy: ΔH = ΔU + PΔV
Entropy: ΔS = ∫(dQrev/T)
Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant.
For a general reaction: aA + bB ⇌ cC + dD
\( K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)
If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
Problem: For the reaction N₂ + 3H₂ ⇌ 2NH₃, Kc = 4.0 at 400°C. If [N₂] = 0.5 M, [H₂] = 0.8 M, and [NH₃] = 1.2 M, is the system at equilibrium?
Solution: Calculate Q = [NH₃]²/([N₂][H₂]³) = (1.2)²/((0.5)(0.8)³) = 5.625
Since Q > K, the reaction will shift left to reach equilibrium.
Chemical thermodynamics deals with the study of energy changes during chemical reactions and the relationship between heat, work, and chemical energy.
Chemical thermodynamics involves concepts like internal energy, enthalpy, entropy, Gibbs free energy, and their applications in predicting reaction spontaneity and equilibrium.
System: The part of universe under study
Surroundings: Everything else in the universe
Boundary: The surface separating system and surroundings
Open: Can exchange both energy and matter
Closed: Can exchange only energy
Isolated: Cannot exchange energy or matter
If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
Basis for temperature measurement
Energy can neither be created nor destroyed, only converted from one form to another.
Mathematical Form: \( \Delta U = q + w \)
Where ΔU = change in internal energy, q = heat, w = work
The entropy of an isolated system always increases for spontaneous processes.
\( \Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0 \)
For spontaneous processes
The entropy of a perfect crystal at absolute zero temperature is zero.
\( S \rightarrow 0 \) as \( T \rightarrow 0 \) K
Provides absolute entropy values
Total energy of a system
State function, depends only on initial and final states
\( \Delta U = q + w \)
Heat content at constant pressure
\( H = U + PV \)
\( \Delta H = \Delta U + P\Delta V \)
Measure of disorder or randomness
\( \Delta S = \frac{q_{\text{rev}}}{T} \)
Units: J/mol·K
Predicts spontaneity at constant T and P
\( G = H - TS \)
\( \Delta G = \Delta H - T\Delta S \)
Constant temperature: ΔT = 0
ΔU = 0 (for ideal gas)
q = -w
No heat exchange: q = 0
ΔU = w
Temperature changes
Constant volume: ΔV = 0
w = 0
ΔU = qv
Constant pressure: ΔP = 0
w = -PΔV
ΔH = qp
\( C_v = \left(\frac{\partial U}{\partial T}\right)_v \)
\( q_v = nC_v\Delta T \)
ΔU = nCvΔT
\( C_p = \left(\frac{\partial H}{\partial T}\right)_p \)
\( q_p = nC_p\Delta T \)
ΔH = nCpΔT
\( C_p - C_v = R \) (for ideal gases)
\( \gamma = \frac{C_p}{C_v} \) (adiabatic index)
ΔG < 0
Spontaneous process
ΔG = 0
Equilibrium state
ΔG > 0
Non-spontaneous process
Temperature Dependence:
\( \Delta G = \Delta H - T\Delta S \)
ΔH < 0, ΔS > 0
Always spontaneous
ΔH > 0, ΔS < 0
Never spontaneous
ΔH < 0, ΔS < 0
Spontaneous at low T
ΔH > 0, ΔS > 0
Spontaneous at high T
Problem: A system absorbs 500 J of heat and does 300 J of work on surroundings. Calculate ΔU.
Solution:
Using first law: \( \Delta U = q + w \)
Heat absorbed: q = +500 J
Work done by system: w = -300 J
\( \Delta U = 500 + (-300) = 200 \) J
Problem: For a reaction at 298K, ΔH = -50 kJ and ΔS = -100 J/K. Is the reaction spontaneous?
Solution:
\( \Delta G = \Delta H - T\Delta S \)
\( \Delta G = -50000 - (298 \times -100) \)
\( \Delta G = -50000 + 29800 = -20200 \) J
Since ΔG < 0, reaction is spontaneous
Problem: Calculate entropy change when 1 mole of ice melts at 0°C (ΔHfus = 6.01 kJ/mol)
Solution:
\( \Delta S = \frac{\Delta H_{fus}}{T} \)
\( \Delta S = \frac{6010}{273} = 22.0 \) J/mol·K
Entropy increases during melting
1. Calculate work done when 2 moles of ideal gas expand isothermally at 300K from 1L to 10L.
[Answer: w = -11488 J]
2. For a reaction with ΔH = 80 kJ and ΔS = 200 J/K, find the temperature above which it becomes spontaneous.
[Answer: T > 400 K]
3. Calculate ΔG° at 298K for a reaction with K = 105.
[Answer: ΔG° = -28.5 kJ/mol]
Physical equilibrium occurs when physical processes (not chemical reactions) reach a state where the forward and reverse processes occur at the same rate, resulting in no net change in the system.
Physical equilibrium involves phase changes, dissolution processes, and physical transformations where the chemical composition remains unchanged.
Dynamic Equilibrium: Forward and reverse processes occur at equal rates
Macroscopic Properties: Constant (no observable change)
Rate Equality: Forward rate = Reverse rate
Closed System: Required for equilibrium
Equilibrium between solid and liquid phases at melting point
Equilibrium between liquid and vapor phases
Equilibrium between solid and vapor phases (sublimation)
Equilibrium between dissolved and undissolved solute
\( F = C - P + 2 \)
Where:
F = Degrees of freedom
C = Number of components
P = Number of phases
Pressure exerted by vapor in equilibrium with liquid
Increases with temperature
Independent of surface area
Problem: At 25°C, the vapor pressure of water is 23.8 mmHg. Explain what happens when volume is decreased at constant temperature.
Solution:
Initially, some water vapor condenses to maintain equilibrium vapor pressure
Once equilibrium is reestablished, vapor pressure returns to 23.8 mmHg
Mass of liquid increases, mass of vapor decreases
Problem: A saturated solution of NaCl contains undissolved salt. What happens when more water is added?
Solution:
Adding water makes the solution unsaturated
More salt dissolves until saturation is reached again
Equilibrium: NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq)
1. Explain why vapor pressure of liquid is constant at constant temperature.
[Answer: Rate of evaporation equals rate of condensation]
2. What happens to solid-liquid equilibrium when pressure is increased?
[Answer: Depends on density difference; for water, melting point decreases]
3. Why does equilibrium require a closed system?
[Answer: To prevent mass transfer with surroundings]
Chemical equilibrium occurs in reversible chemical reactions when the rates of forward and reverse reactions become equal, and the concentrations of reactants and products remain constant.
Chemical equilibrium involves the equilibrium constant, reaction quotient, Le Chatelier's principle, and factors affecting equilibrium position.
Forward Reaction: Reactants → Products
Reverse Reaction: Products → Reactants
Dynamic Equilibrium: Both reactions occur simultaneously
Constant Concentrations: No net change
Macroscopic Properties: Remain constant
Microscopic Activity: Continuous molecular activity
For a general reaction: \( aA + bB \rightleftharpoons cC + dD \)
Equilibrium Constant (Kc)
\( K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)
Concentration-based
Equilibrium Constant (Kp)
\( K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \)
Pressure-based for gases
\( K_p = K_c (RT)^{\Delta n} \)
Where:
Δn = (c + d) - (a + b) = moles of gaseous products - moles of gaseous reactants
R = Gas constant (0.0821 L·atm·mol⁻¹·K⁻¹)
T = Temperature in Kelvin
For reaction: \( aA + bB \rightleftharpoons cC + dD \)
\( Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \) (at any time)
Q > K
Reaction proceeds backward
Too much product
Q = K
At equilibrium
No net change
Q < K
Reaction proceeds forward
Too much reactant
"If a system at equilibrium is subjected to a change, the equilibrium shifts to counteract the change."
All reactants and products in same phase
Reactants and products in different phases
Problem: For reaction \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \), at equilibrium [SO₂] = 0.4M, [O₂] = 0.2M, [SO₃] = 0.8M. Find Kc.
Solution:
\( K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} = \frac{(0.8)^2}{(0.4)^2 (0.2)} \)
\( K_c = \frac{0.64}{0.16 \times 0.2} = \frac{0.64}{0.032} = 20 \)
Problem: For \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) ΔH = -92 kJ, predict effect of increasing pressure.
Solution:
Reactants: 4 moles gas, Products: 2 moles gas
Increasing pressure favors side with fewer gas moles
Equilibrium shifts toward products (NH₃)
Problem: For reaction with Kc = 50, if Qc = 25, which way will reaction proceed?
Solution:
Since Qc (25) < Kc (50)
Reaction will proceed in forward direction
More products will form until Qc = Kc
1. For \( 2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g) \), Kc = 100 at 25°C. If [NO] = 0.1M, [O₂] = 0.2M, [NO₂] = 0.5M, is the system at equilibrium?
[Answer: Qc = 125 > Kc, so reverse reaction favored]
2. For endothermic reaction, what is the effect of increasing temperature?
[Answer: Forward reaction favored, K increases]
3. Write Kc expression for \( CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \)
[Answer: Kc = [CO₂]]
Surface chemistry deals with phenomena that occur at the interfaces of heterogeneous systems, including adsorption, catalysis, colloids, and emulsions.
Surface chemistry studies processes at phase boundaries, focusing on adsorption, catalysis, colloid formation, and their applications in industry and daily life.
Interface: Boundary between two phases
Surface: Boundary where one phase is solid/gas or solid/liquid
Definition: Excess energy at surface compared to bulk
Unit: J/m² or N/m
Adsorption
Absorption
Physical Adsorption (Physisorption)
Chemical Adsorption (Chemisorption)
Nature of Adsorbent
Nature of Adsorbate
Temperature
Pressure
Empirical relationship between amount adsorbed and pressure
Equation: \( \frac{x}{m} = kP^{1/n} \)
Where x/m = amount adsorbed per unit mass, P = pressure, k and n are constants
Logarithmic Form:
\( \log\frac{x}{m} = \log k + \frac{1}{n}\log P \)
Plot of log(x/m) vs log P gives straight line
Theoretical model based on monolayer adsorption
Assumptions:
Equation: \( \frac{x}{m} = \frac{aP}{1 + bP} \)
Where a and b are Langmuir constants
Linear Form:
\( \frac{P}{x/m} = \frac{1}{a} + \frac{b}{a}P \)
Plot of P/(x/m) vs P gives straight line
Homogeneous Catalysis
Heterogeneous Catalysis
Enzyme Catalysis
| Dispersed Phase | Dispersion Medium | Type | Examples |
|---|---|---|---|
| Solid | Liquid | Sol | Paint, ink |
| Liquid | Liquid | Emulsion | Milk, cream |
| Gas | Liquid | Foam | Whipped cream |
| Solid | Gas | Aerosol | Smoke, dust |
Optical Properties
Kinetic Properties
Electrical Properties
Stability
Types of Emulsions
Emulsifying Agents
Definition
Semi-solid systems in which liquid is dispersed in solid medium
Examples
Industrial Applications
Daily Life Applications
Problem: For Freundlich isotherm, log k = 0.6020 and 1/n = 0.5. Find amount adsorbed when pressure is 4 atm.
Solution:
Using Freundlich equation: \( \frac{x}{m} = kP^{1/n} \)
\( \frac{x}{m} = (10^{0.6020}) \times (4^{0.5}) \)
\( \frac{x}{m} = 4 \times 2 = 8 \) g/g adsorbent
Problem: A reaction takes 50 minutes to complete without catalyst and 10 minutes with catalyst. Calculate rate enhancement.
Solution:
Rate enhancement = \( \frac{\text{Time without catalyst}}{\text{Time with catalyst}} \)
Rate enhancement = \( \frac{50}{10} = 5 \) times
Problem: Describe Bredig's Arc method for preparing colloidal solution of gold.
Solution:
• Gold electrodes immersed in water containing NaOH
• Electric arc struck between electrodes
• Gold vaporizes and condenses to form colloidal particles
• NaOH stabilizes the colloidal particles
1. Differentiate between physisorption and chemisorption based on enthalpy of adsorption, specificity, and temperature dependence.
[Answer: Compare enthalpy, specificity, and temperature effects]
2. A Langmuir isotherm gives intercept = 0.2 and slope = 0.5 when P/(x/m) is plotted against P. Find Langmuir constants a and b.
[Answer: a = 5, b = 2.5]
3. Explain the role of emulsifying agents in stabilizing emulsions with examples.
[Answer: Reduce interfacial tension, form protective layer]
Organic chemistry is the study of carbon-containing compounds and their properties, structure, composition, reactions, and preparation.
Organic chemistry covers hydrocarbons, functional groups, isomerism, nomenclature, and fundamental reaction mechanisms involving carbon compounds.
Sigma Bonds (σ)
Pi Bonds (π)
Hybridization
sp³
Tetrahedral
109.5°
Alkanes
sp²
Trigonal planar
120°
Alkenes
sp
Linear
180°
Alkynes
sp³d
Trigonal bipyramidal
90°, 120°
P compounds
Saturated
Unsaturated
Benzene Structure
Aromaticity Rules
| Functional Group | Formula | Suffix/Prefix | Example |
|---|---|---|---|
| Alkane | -C-C- | -ane | CH₃-CH₃ (Ethane) |
| Alkene | C=C | -ene | CH₂=CH₂ (Ethene) |
| Alkyne | C≡C | -yne | CH≡CH (Ethyne) |
| Alcohol | -OH | -ol | CH₃OH (Methanol) |
| Aldehyde | -CHO | -al | HCHO (Methanal) |
| Ketone | -CO- | -one | CH₃COCH₃ (Propanone) |
| Carboxylic Acid | -COOH | -oic acid | CH₃COOH (Ethanoic acid) |
Different carbon skeletons
Example: n-butane vs isobutane
Different positions of functional group
Example: 1-propanol vs 2-propanol
Different functional groups
Example: Ethanol vs Dimethyl ether
Cis-trans or E-Z isomers
Example: Cis-2-butene vs Trans-2-butene
Chiral molecules, enantiomers
Example: Lactic acid enantiomers
Different spatial arrangements
Example: Staggered vs eclipsed ethane
Step 1: Select Longest Chain
Identify the parent hydrocarbon chain
Step 2: Number the Chain
Give lowest numbers to substituents
Step 3: Identify Substituents
List alkyl groups in alphabetical order
CH₃-CH₂-CH₂-CH₃
Butane
CH₃-CH(CH₃)-CH₃
2-Methylpropane
CH₃-CH₂-OH
Ethanol
CH₃-CH₂-COOH
Propanoic acid
S_N1 and S_N2 mechanisms
Example: CH₃Cl + OH⁻ → CH₃OH + Cl⁻
Aromatic compounds
Example: Nitration of benzene
Alkenes and alkynes
Example: CH₂=CH₂ + HBr → CH₃-CH₂Br
Carbonyl compounds
Example: Aldehydes and ketones
Formation of alkenes
Example: Dehydrohalogenation
Migration of atoms/groups
Example: Wagner-Meerwein rearrangement
Problem: Give IUPAC name for: CH₃-CH₂-CH(CH₃)-CH₂-CH₃
Solution:
1. Longest chain: 5 carbons (pentane)
2. Numbering gives methyl at position 3
3. IUPAC name: 3-Methylpentane
Problem: Draw all structural isomers of C₄H₁₀O
Solution:
Alcohols: Butan-1-ol, Butan-2-ol, 2-Methylpropan-1-ol, 2-Methylpropan-2-ol
Ethers: Ethoxyethane, Methoxypropane
Total: 7 structural isomers
Problem: Determine hybridization in C₂H₂, C₂H₄, C₂H₆
Solution:
C₂H₂ (Ethyne): Triple bond → sp hybridization
C₂H₄ (Ethene): Double bond → sp² hybridization
C₂H₆ (Ethane): Single bonds → sp³ hybridization
1. Write IUPAC names for: CH₃-CH₂-CH₂-CH₂-OH and CH₃-CH(OH)-CH₃
[Answer: Butan-1-ol and Propan-2-ol]
2. Identify type of isomerism between CH₃CH₂OH and CH₃OCH₃
[Answer: Functional group isomerism]
3. Determine hybridization of carbon in H₂C=O and HC≡N
[Answer: sp² in H₂C=O, sp in HC≡N]
Hydrocarbons are organic compounds consisting entirely of hydrogen and carbon atoms. They are the fundamental building blocks of organic chemistry and are primarily obtained from petroleum and natural gas.
Hydrocarbons are classified based on their structure and bonding into aliphatic (alkanes, alkenes, alkynes) and aromatic compounds. They form the basis for understanding organic reactions and functional groups.
General Formula
CnH2n+2
For non-cyclic alkanes
Hybridization
sp3 hybridization
Tetrahedral geometry
First 10 Straight-Chain Alkanes:
| Name | Formula | Boiling Point (°C) |
|---|---|---|
| Methane | CH4 | -162 |
| Ethane | C2H6 | -89 |
| Propane | C3H8 | -42 |
| Butane | C4H10 | -1 |
| Pentane | C5H12 | 36 |
Structural Isomerism
Example: Butane Isomers
n-Butane: CH3-CH2-CH2-CH3
Isobutane: (CH3)2CH-CH3
General Formula
CnH2n
For non-cyclic alkenes
Hybridization
sp2 hybridization
Trigonal planar geometry
Common Alkenes:
Due to restricted rotation around double bonds
Cis-Isomer
Similar groups on same side
Higher dipole moment
Trans-Isomer
Similar groups on opposite sides
Lower dipole moment
General Formula
CnH2n-2
For non-cyclic alkynes
Hybridization
sp hybridization
Linear geometry
Important Alkynes:
Molecular Formula
C6H6
Planar hexagonal structure
Resonance
Two equivalent Kekulé structures
Delocalized π-electrons
Aromaticity Criteria (Hückel's Rule):
Toluene
C6H5CH3
Xylene
C6H4(CH3)2
Naphthalene
C10H8
Halogenation
CH4 + Cl2 → CH3Cl + HCl (in sunlight)
Free radical substitution
Combustion
CH4 + 2O2 → CO2 + 2H2O + Heat
Complete combustion
Addition Reactions
CH2=CH2 + H2 → CH3-CH3 (Hydrogenation)
CH2=CH2 + Br2 → CH2Br-CH2Br (Bromination)
Markovnikov's Rule
"The rich get richer" - H adds to carbon with more H atoms
Addition Reactions
HC≡CH + 2H2 → CH3-CH3
HC≡CH + 2Br2 → CHBr2-CHBr2
Acidic Nature
Terminal alkynes react with NaNH2
HC≡CH + NaNH2 → HC≡C⁻Na⁺ + NH3
CH3-CH2-CH(CH3)-CH3
2-Methylbutane
CH3-CH=CH-CH3
2-Butene
Problem: Draw all structural isomers of pentane (C5H12).
Solution:
1. n-Pentane: CH3-CH2-CH2-CH2-CH3
2. Isopentane: (CH3)2CH-CH2-CH3
3. Neopentane: (CH3)4C
Problem: Name the compound: CH3-CH2-C(CH3)=CH-CH3
Solution:
Longest chain: 4 carbons (butene)
Numbering from right: CH3-CH2-C(CH3)=CH-CH3
Double bond at position 2, methyl at position 3
3-Methyl-2-pentene
Problem: Predict the product when propene reacts with HBr.
Solution:
Following Markovnikov's rule:
CH3-CH=CH2 + HBr → CH3-CHBr-CH3
2-Bromopropane
1. Write the structural formula for 2,3-dimethylbutane.
[Answer: (CH3)2CH-CH(CH3)2]
2. How many structural isomers are possible for C4H8? Draw them.
[Answer: 6 isomers including chain and position isomers]
3. Explain why benzene undergoes substitution rather than addition reactions.
[Answer: Due to resonance stabilization energy]
Nitrogen-containing organic compounds are a major class of organic molecules that include amines, amides, nitriles, nitro compounds, and heterocyclic compounds with nitrogen atoms.
This section covers the structure, nomenclature, properties, preparation methods, and reactions of major nitrogen-containing organic compounds, with emphasis on amines and their derivatives.
Primary Amines (1°)
R-NH₂
One alkyl/aryl group attached to N
Example: CH₃NH₂ (Methylamine)
Secondary Amines (2°)
R₂NH
Two alkyl/aryl groups attached to N
Example: (CH₃)₂NH (Dimethylamine)
Tertiary Amines (3°)
R₃N
Three alkyl/aryl groups attached to N
Example: (CH₃)₃N (Trimethylamine)
Common Names
IUPAC Names
Examples:
CH₃CH₂CH₂NH₂ → 1-Propanamine
CH₃NHCH₂CH₃ → N-Methylethanamine
C₆H₅NH₂ → Benzenamine (Aniline)
Boiling Points
Solubility
Ammonolysis
R-X + NH₃ → R-NH₂ + HX
Gives mixture of 1°, 2°, 3° amines and quaternary salts
Gabriel Phthalimide
Gives pure 1° amines only
No contamination with 2° or 3° amines
From Nitro Compounds
R-NO₂ + 3H₂ → R-NH₂ + 2H₂O
Using Sn/HCl, Fe/HCl, or catalytic hydrogenation
From Nitriles
R-CN + 4[H] → R-CH₂-NH₂
Using LiAlH₄ or catalytic hydrogenation
Hofmann Degradation
RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Amide to amine with one less carbon
Reductive Amination
RCHO + R'NH₂ → RCH₂NHR'
Aldehyde/ketone with amine under reduction
Reaction with Acids
R-NH₂ + HCl → R-NH₃⁺Cl⁻
Forms ammonium salts
Basicity Order
Aliphatic 2° > Aliphatic 1° > Aliphatic 3° > NH₃ > Aromatic
Due to +I effect and resonance
1° and 2° amines react with acid chlorides/anhydrides
R-NH₂ + R'-COCl → R-NH-COR' + HCl
Forms amides, used as protecting group
Only for 1° amines (test for 1° amines)
R-NH₂ + CHCl₃ + 3KOH → R-NC + 3KCl + 3H₂O
Produces foul-smelling isocyanides
Distinguishes between 1°, 2°, and 3° amines
1° Amine
+ Benzenesulfonyl chloride → Soluble in alkali
2° Amine
+ Benzenesulfonyl chloride → Insoluble precipitate
3° Amine
+ Benzenesulfonyl chloride → No reaction
Important reaction of aromatic 1° amines
Ar-NH₂ + NaNO₂ + 2HCl → Ar-N₂⁺Cl⁻ + NaCl + 2H₂O (0-5°C)
Diazonium salts used in synthesis
Replacement Reactions
Coupling Reactions
Preparation
Reactions
Problem: Arrange the following in increasing order of basicity: NH₃, CH₃NH₂, (CH₃)₂NH, C₆H₅NH₂
Solution:
C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH
Aromatic amine is weakest due to resonance. Among aliphatic, 2° > 1° due to +I effect and solvation factors.
Problem: How will you convert aniline to iodobenzene?
Solution:
Step 1: Diazotization
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ (0-5°C)
Step 2: Reaction with KI
C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂ + KCl
Problem: How will you distinguish between 1°, 2°, and 3° amines?
Solution:
Hinsberg Test:
• 1° amine: Forms sulfonamide soluble in alkali
• 2° amine: Forms sulfonamide insoluble in alkali
• 3° amine: No reaction with benzenesulfonyl chloride
1. Write the IUPAC names of: (CH₃)₂CHNH₂, C₆H₅NHCH₃, CH₃CH₂NHCH₃
[Answer: 2-Propanamine, N-Methylbenzenamine, N-Methylethanamine]
2. How will you prepare ethylamine from acetonitrile?
[Answer: CH₃CN + 4[H] → CH₃CH₂NH₂ (using LiAlH₄ or H₂/Ni)]
3. Give the chemical reaction to distinguish between aniline and N-methylaniline.
[Answer: Carbylamine test - aniline gives foul smell, N-methylaniline does not react]
Biomolecules are organic compounds that are essential for life processes. They include carbohydrates, proteins, lipids, nucleic acids, and vitamins.
Biomolecules are the building blocks of living organisms and perform various biological functions. They can be classified based on their structure, composition, and function.
Micromolecules: Small molecular weight (100-1000 Da)
Macromolecules: Large molecular weight (>1000 Da)
Structural: Provide framework
Functional: Perform specific tasks
Polyhydroxy aldehydes or ketones or compounds that produce them on hydrolysis
Simple sugars (Glucose, Fructose)
Cannot be hydrolyzed
2-10 units (Sucrose, Lactose)
Yield 2-10 monosaccharides
Many units (Starch, Cellulose)
Yield many monosaccharides
Amino Acid Structure
\( \text{H}_2\text{N}-\text{CHR}-\text{COOH} \)
Amphoteric nature
Peptide Bond
\( -\text{CO}-\text{NH}- \)
Formed by condensation
Amino acid sequence
Covalent bonds
α-helix, β-pleated sheets
H-bonding
3D folding
Various interactions
Multiple subunits
e.g., Hemoglobin
Fatty Acids
Saturated: No double bonds (Stearic acid)
Unsaturated: Double bonds present (Oleic acid)
Saponification
\( \text{Fat} + \text{NaOH} \rightarrow \text{Soap} + \text{Glycerol} \)
Hydrolysis of esters
Nucleotide Components
Base Pairing Rules
Adenine ⇌ Thymine (DNA)
Adenine ⇌ Uracil (RNA)
Guanine ⇌ Cytosine
Biological catalysts that speed up biochemical reactions
Characteristics
Mechanism
Problem: Glucose and fructose both have formula C₆H₁₂O₆ but different properties. Explain.
Solution:
Glucose is an aldose (aldehyde group) while fructose is a ketose (ketone group)
Different functional groups lead to different chemical properties
Both are structural isomers with different functional groups
Problem: What type of structure is represented by α-helix in proteins?
Solution:
α-helix represents the secondary structure of proteins
It is stabilized by hydrogen bonds between -NH and -CO groups of peptide bonds
Right-handed helix with 3.6 amino acids per turn
Problem: If a DNA strand has 30% adenine, what are the percentages of other bases?
Solution:
According to Chargaff's rules: A = T and G = C
If A = 30%, then T = 30%
Total A + T = 60%, so G + C = 40%
Therefore G = 20% and C = 20%
1. Differentiate between reducing and non-reducing sugars with examples.
[Reducing: Glucose, Maltose; Non-reducing: Sucrose]
2. What are essential and non-essential amino acids? Give two examples of each.
[Essential: Cannot be synthesized; Non-essential: Can be synthesized]
3. Explain the double helix structure of DNA with a neat diagram.
[Two antiparallel strands, base pairing, hydrogen bonding]
4. What are enzymes? Explain the lock and key mechanism of enzyme action.
[Biological catalysts, specificity, active site]
Polymers are large molecules composed of repeating structural units called monomers, connected by covalent bonds. They play a crucial role in both natural and synthetic materials.
Polymers encompass natural and synthetic macromolecules, their classification, polymerization mechanisms, properties, and applications in daily life and industry.
Monomer: Small repeating unit that forms polymers
Polymer: Large molecule made of many monomers
Degree of Polymerization: Number of monomer units
Homopolymer: Single type of monomer
Copolymer: Multiple types of monomers
Oligomer: Short polymer chain
Natural Polymers
Semi-synthetic
Synthetic
Long straight chains
e.g., HDPE, Nylon
Chains with branches
e.g., LDPE, Amylopectin
Chains connected by bonds
e.g., Bakelite, Vulcanized rubber
Monomers add together without loss of small molecules
Free Radical Mechanism
Examples
Monomers join with elimination of small molecules (H₂O, HCl, etc.)
Mechanism
\( nHOOC-R-COOH + nH_2N-R'-NH_2 \)
\( → [-OC-R-CONH-R'-NH-]_n + 2nH_2O \)
Examples
| Property | Addition | Condensation |
|---|---|---|
| By-products | None | Small molecules eliminated |
| Monomers | Unsaturated compounds | Bifunctional monomers |
| Growth | Fast, chain reaction | Step-wise |
| Examples | PE, PP, PVC | Nylon, Polyester |
Polyethylene (PE)
Polyvinyl Chloride (PVC)
Polystyrene (PS)
Polypropylene (PP)
Nylon
Polyester
Natural Rubber
Synthetic Rubbers
Molecular Weight
Chain Structure
Intermolecular Forces
Crystallinity
Problem: Identify the polymer formed from ethylene glycol and terephthalic acid. Write the reaction.
Solution:
Polymer: Polyethylene terephthalate (PET)
Reaction:
\( nHO-CH_2-CH_2-OH + nHOOC-C_6H_4-COOH \)
\( → [-O-CH_2-CH_2-O-OC-C_6H_4-CO-]_n + 2nH_2O \)
Problem: Calculate degree of polymerization for PVC with molecular weight 62,500. (Monomer MW = 62.5)
Solution:
Degree of polymerization = \( \frac{\text{Polymer MW}}{\text{Monomer MW}} \)
\( = \frac{62500}{62.5} = 1000 \)
The polymer contains 1000 monomer units.
Problem: What is vulcanization of rubber? How does it improve properties?
Solution:
Vulcanization is the process of heating natural rubber with sulfur.
Effects:
1. Write the structure of Nylon-6,6 and name its monomers.
[Answer: Hexamethylenediamine + Adipic acid]
2. Differentiate between thermoplastics and thermosetting polymers with examples.
[Answer: Thermoplastics soften on heating (PE, PVC), Thermosetting harden permanently (Bakelite)]
3. What is the monomer of natural rubber? Write its structure.
[Answer: Isoprene (2-methyl-1,3-butadiene)]
4. Calculate molecular weight of polymer with DP=500 and monomer MW=100.
[Answer: 50,000 g/mol]
Organic chemistry plays a crucial role in our daily lives through medicines, food additives, cosmetics, cleaning agents, and many other products we use regularly.
This chapter explores how organic chemistry principles apply to pharmaceuticals, soaps, detergents, food additives, and other consumer products that impact our daily lives.
By Therapeutic Action
By Molecular Action
Aspirin
Acetylsalicylic acid
Analgesic, antipyretic, anti-inflammatory
Paracetamol
N-acetyl-p-aminophenol
Analgesic and antipyretic
Penicillin
β-lactam antibiotic
Bacterial infection treatment
Saponification Reaction:
Fat/Oil + NaOH → Soap + Glycerol
Soaps
Detergents
Prevent spoilage
Examples:
Sodium benzoate
Potassium metabisulfite
Prevent oxidation
Examples:
BHA, BHT
Vitamin C, Vitamin E
Sugar substitutes
Examples:
Aspartame
Saccharin
Soaps
Sodium/potassium salts of fatty acids
Synthetic Detergents
Alkyl benzene sulfonates
Shampoos
Surfactants + Conditioners
Micelle Formation:
Examples:
Cellulose (cotton, paper)
Starch (food)
Proteins (silk, wool)
DNA/RNA (genetic material)
Examples:
Polythene (bags)
PVC (pipes)
Nylon (clothing)
Teflon (non-stick)
Examples:
Conductive polymers
Biodegradable plastics
Superabsorbent polymers
Shape-memory polymers
Problem: Write the chemical reaction for preparation of soap from glyceryl tripalmitate.
Solution:
(C₁₅H₃₁COO)₃C₃H₅ + 3NaOH → 3C₁₅H₃₁COONa + C₃H₅(OH)₃
Glyceryl tripalmitate + Sodium hydroxide → Sodium palmitate (soap) + Glycerol
Problem: Classify the following drugs: Aspirin, Chloramphenicol, Luminal, Seldane
Solution:
Problem: Why is sodium benzoate used in food products? Write its chemical structure.
Solution:
Sodium benzoate is used as a preservative in food products to prevent microbial growth.
Chemical structure: C₆H₅COONa
It works by inhibiting the growth of bacteria, yeast, and molds.
1. What is the difference between antiseptics and disinfectants? Give one example of each.
[Antiseptics: Safe for living tissues (Dettol); Disinfectants: For non-living surfaces (Bleach)]
2. Why do soaps not work in hard water? How do detergents solve this problem?
[Soaps form insoluble scum with Ca²⁺/Mg²⁺ ions; Detergents don't form precipitates]
3. Name the sweetening agent used in preparing sweets for diabetics. Why is it preferred?
[Saccharin/Aspartame; They don't metabolize to provide calories]
4. What are biodegradable and non-biodegradable detergents? Give one example of each.
[Biodegradable: Straight chain alkyl group (Sodium lauryl sulfate); Non-biodegradable: Branched chain (ABS)]
The periodic table is a systematic arrangement of elements that helps in understanding and predicting their properties based on atomic structure and electronic configuration.
The periodic classification organizes elements based on atomic number and electronic configuration, revealing periodic trends in physical and chemical properties.
Concept: Groups of three elements with similar properties
Example: Li, Na, K
Limitation: Limited to few elements
Concept: Every 8th element has similar properties
Limitation: Failed beyond calcium
Analogy: Musical octaves
Basis: Atomic mass
Merits: Predicted new elements
Demerits: Position anomalies
Basis: Atomic number (Z)
Law: "Properties of elements are periodic functions of their atomic numbers"
Significance: Resolved position anomalies
1st Period
2 elements
H to He
2nd & 3rd Period
8 elements each
Li to Ne, Na to Ar
4th & 5th Period
18 elements each
K to Kr, Rb to Xe
6th & 7th Period
32 elements each
Cs to Rn, Fr to Og
Groups 1-2
ns¹⁻²
Reactive metals
Groups 13-18
ns²np¹⁻⁶
Metals to noble gases
Groups 3-12
(n-1)d¹⁻¹⁰ns⁰⁻²
Transition metals
Lanthanides & Actinides
(n-2)f¹⁻¹⁴
Inner transition
Across a Period (→)
Decreases
Due to increasing nuclear charge
Down a Group (↓)
Increases
Due to increasing shells
Across a Period (→)
Increases
Due to decreasing atomic size
Down a Group (↓)
Decreases
Due to increasing atomic size
Across a Period (→)
Increases
More negative values
Down a Group (↓)
Decreases
Less negative values
Across a Period (→)
Increases
Pauling scale: F = 4.0
Down a Group (↓)
Decreases
Except transition elements
First Element of Each Group
Examples
Similarities between elements of period 2 and period 3 elements of next group
Li ⇄ Mg
Be ⇄ Al
B ⇄ Si
Other pairs
Problem: Arrange in increasing order of atomic size: Na, Mg, Al, Si
Solution:
All elements belong to period 3
Atomic size decreases across a period due to increasing nuclear charge
Order: Si < Al < Mg < Na
Problem: An element has atomic number 20. Identify its period, group, and block.
Solution:
Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Valence shell: n = 4 → 4th period
Valence electrons: 2 → Group 2
Last electron in s-orbital → s-block
Answer: Element is Calcium (4th period, Group 2, s-block)
Problem: Why does Be have higher first ionization energy than B?
Solution:
Be: 1s² 2s² (fully filled s-subshell, stable)
B: 1s² 2s² 2p¹ (easier to remove electron from p-orbital)
Reason: Fully filled subshell configuration of Be provides extra stability
1. Arrange in order of increasing ionization energy: K, Na, Li, Rb
[Answer: Rb < K < Na < Li]
2. Which has higher electron affinity: F or Cl? Explain.
[Answer: Cl > F due to small size and electron repulsion in F]
3. An element has configuration [Xe]4f¹⁴5d¹⁰6s²6p⁴. Identify period, group, block.
[Answer: 6th period, Group 16, p-block - Element Polonium]
4. Why is the size of Zn²⁺ ion less than Zn atom?
[Answer: Due to increased effective nuclear charge per electron]
Coordination compounds are compounds in which a central metal atom or ion is bonded to a group of ions or molecules.
1. Cation named before anion
2. Ligands named in alphabetical order
3. Prefixes indicate number of ligands
4. Metal oxidation state in Roman numerals
Example: [Co(NH₃)₆]Cl₃ - Hexaamminecobalt(III) chloride
General principles cover the fundamental concepts that govern the behavior and properties of inorganic compounds, including atomic structure, periodic properties, chemical bonding, and reaction mechanisms.
This chapter explores atomic structure, periodic trends, chemical bonding, acid-base theories, coordination chemistry, and reaction mechanisms that form the foundation of inorganic chemistry.
Proton: Positive charge, mass ~1 amu
Neutron: Neutral, mass ~1 amu
Electron: Negative charge, mass ~1/1836 amu
n: Principal quantum number (energy level)
l: Azimuthal quantum number (subshell)
m: Magnetic quantum number (orbital)
s: Spin quantum number (±½)
Aufbau Principle
Electrons fill orbitals from lowest to highest energy
1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p...
Hund's Rule
Orbitals are filled with parallel spins before pairing
Maximum multiplicity
Pauli Exclusion Principle
No two electrons can have same set of four quantum numbers
Atomic Radius
Decreases across period, increases down group
Due to effective nuclear charge
Ionization Energy
Increases across period, decreases down group
Energy to remove electron
Electron Affinity
Increases across period, decreases down group
Energy released adding electron
Electronegativity
Increases across period, decreases down group
Pauling scale: F = 4.0
s-block
Groups 1-2
Alkali & Alkaline Earth Metals
p-block
Groups 13-18
Representative Elements
d-block
Groups 3-12
Transition Metals
f-block
Lanthanides & Actinides
Inner Transition Metals
Formation
Electron transfer from metal to non-metal
Na → Na⁺ + e⁻, Cl + e⁻ → Cl⁻
Properties
Bond Formation
Electron sharing between atoms
H + H → H:H or H-H
Bond Parameters
sp³
Tetrahedral, 109.5°
CH₄, NH₃
sp²
Trigonal planar, 120°
BF₃, C₂H₄
sp
Linear, 180°
BeCl₂, C₂H₂
dsp²
Square planar, 90°
Ni(CO)₄
Acid: Produces H⁺ ions in water
Base: Produces OH⁻ ions in water
Limited to aqueous solutions
Acid: Proton donor
Base: Proton acceptor
Conjugate acid-base pairs
Acid: Electron pair acceptor
Base: Electron pair donor
Most general definition
Coordination Entity
Central metal atom/ion surrounded by ligands
[Fe(CN)₆]⁴⁻
Ligands
Molecules/ions bonded to central metal
NH₃, H₂O, CN⁻, Cl⁻
Coordination Number
Number of ligand donor atoms bonded to metal
Common: 4, 6
Oxidation State
Charge on central metal atom
Fe in [Fe(CN)₆]⁴⁻: +2
Structural Isomerism
Stereoisomerism
Problem: Write electronic configuration of Cr (Z=24) and explain exception.
Solution:
Expected: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴
Actual: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
Exception due to extra stability of half-filled d-subshell
Problem: Arrange in increasing atomic radius: Na, Mg, Al, Si
Solution:
All elements in period 3
Atomic radius decreases across period due to increasing nuclear charge
Order: Si < Al < Mg < Na
Increasing radius: Si → Al → Mg → Na
Problem: Determine hybridization in SF₆ and predict geometry.
Solution:
S atom: 6 bond pairs, 0 lone pairs
Steric number = 6
Hybridization = sp³d²
Geometry: Octahedral
Bond angle: 90°
1. Calculate effective nuclear charge for 3p electron in chlorine atom.
[Answer: Z_eff = 7.5]
2. Predict hybridization and geometry for XeF₄ molecule.
[Answer: sp³d², Square planar]
3. Identify type of isomerism in [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br.
[Answer: Ionization isomerism]
Environmental chemistry is the study of chemical and biochemical phenomena that occur in natural environments, focusing on the sources, reactions, transport, effects, and fates of chemical species.
Environmental chemistry deals with atmospheric chemistry, water chemistry, soil chemistry, pollution, and environmental protection measures.
Composition of Dry Air:
N₂: 78.08%, O₂: 20.95%, Ar: 0.93%, CO₂: 0.04%
Emitted directly
Formed in atmosphere
Cause global warming
Formation:
SO₂ + O₂ → SO₃
SO₃ + H₂O → H₂SO₄
NO + O₂ → NO₂
2NO₂ + H₂O → HNO₃ + HNO₂
Effects:
Mechanism:
CFCl₃ → CFCl₂ + Cl•
Cl• + O₃ → ClO• + O₂
ClO• + O → Cl• + O₂
One Cl atom can destroy 100,000 O₃ molecules
Effects:
Greenhouse Effect:
Global Warming Potential:
CO₂: 1 (reference)
CH₄: 25
N₂O: 298
CFC-12: 10,900
pH
6.5-8.5 for drinking water
DO
Dissolved Oxygen > 4 mg/L
BOD
Biological Oxygen Demand
COD
Chemical Oxygen Demand
Mineral Matter
45%
Water & Air
25% each
Organic Matter
5%
Major Contaminants:
Effects:
Reduce
Minimize waste generation
Reuse
Use items multiple times
Recycle
Process materials for new use
Problem: A water sample has initial DO 8.2 mg/L and after 5 days DO 4.8 mg/L. Calculate BOD₅.
Solution:
BOD₅ = Initial DO - Final DO
BOD₅ = 8.2 - 4.8 = 3.4 mg/L
This indicates moderate organic pollution
Problem: Calculate pH of rainwater in equilibrium with atmospheric CO₂ (PCO₂ = 10⁻³.⁵ atm, KH = 10⁻¹.⁵, Ka1 = 10⁻⁶.³).
Solution:
[H₂CO₃] = KH × PCO₂ = 10⁻¹.⁵ × 10⁻³.⁵ = 10⁻⁵ M
[H⁺] = √(Ka1 × [H₂CO₃]) = √(10⁻⁶.³ × 10⁻⁵) = √(10⁻¹¹.³) = 10⁻⁵.⁶⁵
pH = -log[H⁺] = 5.65
Problem: If Earth's temperature increases by 2°C, by what percentage would sea levels rise due to thermal expansion? (Assume coefficient of thermal expansion for seawater is 2.1 × 10⁻⁴ °C⁻¹)
Solution:
Percentage rise = α × ΔT × 100
= (2.1 × 10⁻⁴) × 2 × 100
= 0.042% rise in sea level
1. Calculate the pH of acid rain formed when SO₂ dissolves in water to form 10⁻⁴ M H₂SO₃ (Ka1 = 1.7 × 10⁻²).
[Answer: pH = 3.38]
2. A lake has 200 mg/L of Ca²⁺ and 60 mg/L of Mg²⁺. Calculate total hardness in ppm of CaCO₃.
[Answer: 686 ppm]
3. If the ozone layer thickness decreases from 300 DU to 200 DU, by what percentage has UV radiation increased? (Assume Beer-Lambert law applies)
[Answer: 50% increase]
Titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution.
\( M_1V_1 = M_2V_2 \)
Where M₁ and V₁ are molarity and volume of known solution,
M₂ and V₂ are molarity and volume of unknown solution
Carbohydrates are biological molecules consisting of carbon, hydrogen, and oxygen atoms, usually with a hydrogen:oxygen atom ratio of 2:1.
Glucose: C₆H₁₂O₆ - Primary energy source
Sucrose: C₁₂H₂₂O₁₁ - Table sugar
Starch: (C₆H₁₀O₅)ₙ - Energy storage in plants
Cellulose: (C₆H₁₀O₅)ₙ - Structural component of plants
Lipids are a diverse group of hydrophobic or amphipathic molecules that include fats, oils, waxes, phospholipids, and steroids. They serve crucial roles in energy storage, membrane structure, and signaling.
Lipids are characterized by their insolubility in water and solubility in nonpolar solvents. They include simple lipids (fats, waxes), complex lipids (phospholipids, glycolipids), and derived lipids (steroids, fat-soluble vitamins).
Esters of fatty acids with alcohols
Contain additional groups
Derived from simple or complex lipids
General Formula
R-COOH where R is hydrocarbon chain
Chain length: 4-36 carbons
Nomenclature
Δx system: Double bond position from carboxyl end
ω system: Double bond position from methyl end
Glycerol + 3 fatty acids
Ester bonds formed by dehydration
CH₂OCOR₁ | CHOCOR₂ | CH₂OCOR₃
Oils
Liquid at room temperature
High unsaturated fatty acids
Fats
Solid at room temperature
High saturated fatty acids
Mixed Triglycerides
Different fatty acids
Most natural fats
Hydrophilic Head
Hydrophobic Tail
Proposed by Singer and Nicolson (1972)
Integral Proteins
Span the membrane
Transmembrane
Peripheral Proteins
Associated with surface
Easily removed
Lipid-Anchored
Covalently attached to lipids
Glycosylphosphatidylinositol
Sterol
Cholesterol
Membrane component
Steroid Hormones
Cortisol, Testosterone
Signaling molecules
Bile Acids
Cholic acid
Digestive emulsifiers
Problem: Compare melting points of stearic acid (18:0) and oleic acid (18:1Δ9). Explain the difference.
Solution:
Stearic acid (saturated): Higher melting point (69°C)
Oleic acid (unsaturated): Lower melting point (13°C)
Explanation: Saturated fatty acids pack tightly due to linear structure, requiring more energy to melt. Unsaturated fatty acids have kinks from double bonds, preventing tight packing and lowering melting point.
Problem: How would increasing unsaturated fatty acids in membrane phospholipids affect membrane fluidity?
Solution:
Increasing unsaturated fatty acids increases membrane fluidity.
Reason: Unsaturated fatty acids have cis double bonds that create kinks in the hydrocarbon chains. These kinks prevent tight packing of phospholipids, maintaining fluidity even at lower temperatures.
Problem: Calculate the ATP yield from complete oxidation of one molecule of palmitic acid (16:0).
Solution:
Palmitic acid (16:0) undergoes 7 cycles of β-oxidation:
• 7 NADH × 2.5 ATP = 17.5 ATP
• 7 FADH₂ × 1.5 ATP = 10.5 ATP
• 8 Acetyl-CoA × 10 ATP = 80 ATP
• Activation cost: -2 ATP
Total: 17.5 + 10.5 + 80 - 2 = 106 ATP
1. Explain why lipids aggregate in aqueous solutions but carbohydrates dissolve readily.
[Answer: Lipids are hydrophobic/nonpolar while carbohydrates have multiple hydroxyl groups making them hydrophilic]
2. Compare the structure and function of triacylglycerols and phospholipids.
[Answer: TAGs have 3 fatty acids for energy storage; phospholipids have 2 fatty acids + phosphate group for membranes]
3. Describe how cholesterol affects membrane fluidity at different temperatures.
[Answer: At high temps, cholesterol decreases fluidity; at low temps, it prevents solidification by disrupting packing]
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