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Derivations
Session 1 Session 2 Session 3 Session 4
Reference
Periodic Table & Trends
Chemistry Derivations

Session 1 — Key Chemical Equations

Beginner-friendly step-by-step derivations for NEET. Each starts with real-life context, then builds the formula from first principles.

1. Ideal Gas Equation — PV = nRT

💡 Why this matters in real life

Why does a balloon expand when heated? Why does a tyre feel firm when you pump air into it? The ideal gas equation PV = nRT connects pressure, volume, temperature, and amount of gas. It explains how car engines work (compressing gas heats it), how weather balloons behave (expanding as they rise), and even how your lungs function. Every NEET chemistry section has at least one gas law problem.

Ideal Gas — Molecular motion & pressure Piston Pressure P T = 298 K PV = nRT    R = 8.314 J/mol·K Collisions/s 142
Fig 1.1: Gas molecules (8 types) move randomly inside a container with piston. Wall collisions (red flashes) create pressure. Piston moves to show volume change.

🧠 The Big Idea — In Plain Words

The ideal gas law combines three simpler gas laws: Boyle's law (P ∝ 1/V at constant T, n), Charles's law (V ∝ T at constant P, n), and Avogadro's law (V ∝ n at constant P, T). The unified equation is PV = nRT. R is the universal gas constant, 8.314 J/(mol·K). This equation works well for real gases at low pressure and high temperature.

🔍 Step-by-Step: Deriving PV = nRT

1
Boyle's law: At constant temperature and amount, pressure is inversely proportional to volume. P ∝ 1/V (at constant T, n). Robert Boyle discovered this in 1662: if you double the pressure, the volume halves.
2
Charles's law: At constant pressure and amount, volume is proportional to absolute temperature. V ∝ T (at constant P, n). Jacques Charles found that all gases expand by the same fraction when heated at constant pressure.
3
Avogadro's law: At constant pressure and temperature, volume is proportional to number of moles. V ∝ n (at constant P, T). Equal volumes of gases at same T and P contain equal numbers of molecules.
4
Combine all three proportionalities. Since V ∝ n, V ∝ T, and V ∝ 1/P, we get: V ∝ nT/P → V = R·nT/P. Rearranging: PV = nRT. R is the constant that makes the units work.
PV = nRT    where R = 8.314 J/(mol·K) = 0.0821 L·atm/(mol·K)

🔍 Common Mistake

Common Mistake — Units
Students forget to convert temperature to Kelvin (K = °C + 273) and to use consistent units. If P is in atm, V in L, use R = 0.0821. If P in Pa, V in m³, use R = 8.314. Also: "ideal gas" assumes no intermolecular forces — real gases deviate at high pressure and low temperature.

🎯 NEET Pattern

🎯 NEET Pattern — Ideal Gas
• "At constant T, P is halved. What happens to V?" → V doubles (Boyle's).
• "At STP, 1 mole occupies 22.4 L" comes from PV = nRT: V = (1×0.0821×273)/1 = 22.4 L.
• A common question gives P, V, T for two conditions and asks for the unknown using PV = nRT or P₁V₁/T₁ = P₂V₂/T₂.

✍️ Worked Example

Example 1: Ideal Gas
2 moles of an ideal gas at 27°C occupy 49.2 L. Find the pressure. (R = 0.0821 L·atm/mol·K)
Solution: T = 27 + 273 = 300 K. PV = nRT → P = nRT/V = (2 × 0.0821 × 300) / 49.2 = 49.26 / 49.2 = 1.0 atm. Key insight: Always convert °C to K first.

2. Nernst Equation — E = E° − (RT/nF) ln Q

💡 Why this matters in real life

Batteries and electrochemical cells produce electricity from chemical reactions. The Nernst equation tells us the voltage of a cell under non-standard conditions, which is almost always the case in real life. It explains how concentration affects battery voltage — why a dying battery has lower voltage. It's also used in pH meters, which measure the voltage of a special electrode to determine acidity.

Electrochemical cell — Nernst equation describes voltage Anode [Red] Cathode [Ox] Salt bridge e⁻ flow E = E° − (RT/nF) ln Q
Fig 2.1: Electrochemical cell with anode, cathode, and salt bridge. Electrons flow through the external circuit.

🧠 The Big Idea — In Plain Words

The Nernst equation relates cell potential (voltage) to ion concentrations. For a general cell reaction: aA + bB → cC + dD, the equation is E = E° − (RT/nF) ln Q, where Q = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ. At 298 K, using log₁₀: E = E° − (0.0591/n) log Q. E° is the standard cell potential when all concentrations are 1 M.

🔍 Step-by-Step: Deriving the Nernst Equation

1
Start with Gibbs free energy and cell potential. The maximum electrical work from a cell equals −nFE, where n = number of electrons transferred, F = Faraday constant (96485 C/mol). Gibbs free energy: ΔG = −nFE. Under standard conditions: ΔG° = −nFE°
2
Use the thermodynamic relationship for ΔG. For a reaction aA + bB → cC + dD: ΔG = ΔG° + RT·ln(Q) where Q = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇ
3
Substitute ΔG = −nFE and ΔG° = −nFE°. −nFE = −nFE° + RT·ln(Q) Multiply both sides by −1: nFE = nFE° − RT·ln(Q)
4
Divide both sides by nF to get the Nernst equation. E = E° − (RT/nF)·ln(Q). At 298 K, converting ln to log₁₀ (ln = 2.303·log₁₀): E = E° − (0.0591/n)·log(Q)
E = E° − (RT/nF)·ln Q    At 298 K: E = E° − (0.0591/n)·log Q

🔍 Common Mistake

Common Mistake — Q vs K
Students confuse Q (reaction quotient, for current conditions) with K (equilibrium constant). At equilibrium, Q = K and E = 0. Also: the Nernst equation uses activities for precise work, but NEET uses concentrations. Solids and pure liquids have activity = 1, so they don't appear in Q.

🎯 NEET Pattern

🎯 NEET Pattern — Nernst Equation
• Typically given E°, concentrations, and asked to find E, or vice versa.
• "Calculate the emf of the cell: Zn|Zn²⁺(0.1M)||Cu²⁺(0.01M)|Cu" → Apply Nernst with n = 2.
• For Daniel cell: E° = 1.1 V. If [Zn²⁺] = [Cu²⁺] = 0.1 M, E = E° because Q = 1, so E = E°.

✍️ Worked Example

Example 2: Nernst Equation
For the cell Zn|Zn²⁺(0.01M)||Cu²⁺(1M)|Cu, E° = 1.1 V. Find E at 298 K. (n = 2)
Solution: Q = [Zn²⁺]/[Cu²⁺] = 0.01/1 = 0.01. E = 1.1 − (0.0591/2)·log(0.01) = 1.1 − 0.02955 × (−2) = 1.1 + 0.0591 = 1.159 V. Key insight: Lower [Zn²⁺] makes E slightly higher than E°.

3. Integrated Rate Equations

💡 Why this matters in real life

How long does a drug remain active in your body? How long until a radioactive substance becomes safe? These questions are answered by integrated rate equations. The order of a reaction tells us how the concentration changes with time. Half-life is a key concept: for first-order reactions (like radioactive decay), the half-life is constant — every 5730 years, half the carbon-14 in a sample decays.

Arrhenius — Activation Energy & Molecular Collisions Eₐ Reactants Products ΔH Energy distribution Higher T k = A·e^{−Eₐ/RT}   ln k = ln A − Eₐ/RT
Fig 5.1: Energy barrier (top) with animated molecular collision (bottom left) and Maxwell-Boltzmann energy distribution (bottom right). Higher temperature increases the fraction of molecules with E > Eₐ.

🧠 The Big Idea — In Plain Words

The order of a reaction tells us how the rate depends on concentration. For zero order: rate is constant (rate = k). For first order: rate ∝ [A] (rate = k[A]). For second order: rate ∝ [A]² (rate = k[A]²). Integrating these rate laws gives equations for [A] as a function of time. Each order has a characteristic half-life behaviour.

🔍 Step-by-Step: Zero Order

1
Rate law: Rate = −d[A]/dt = k. The rate is constant, independent of concentration.
2
Integrate: ∫d[A] = −k∫dt → [A] = −kt + C. At t = 0, [A] = [A]₀, so C = [A]₀. Thus: [A] = [A]₀ − kt
3
Half-life: When [A] = [A]₀/2: [A]₀/2 = [A]₀ − kt_½ → t_½ = [A]₀/(2k). Half-life depends on initial concentration.

🔍 Step-by-Step: First Order

1
Rate law: Rate = −d[A]/dt = k[A]. The rate depends linearly on concentration.
2
Integrate: ∫d[A]/[A] = −k∫dt → ln[A] = −kt + C. At t = 0, [A] = [A]₀, so C = ln[A]₀. Thus: ln[A] = ln[A]₀ − kt or [A] = [A]₀e^{−kt}
3
Half-life: When [A] = [A]₀/2: ln([A]₀/2) = ln[A]₀ − kt_½ → ln(½) = −kt_½ → t_½ = ln2/k = 0.693/k. Half-life is constant — independent of concentration.

🔍 Step-by-Step: Second Order

1
Rate law: Rate = −d[A]/dt = k[A]². The rate depends on the square of concentration.
2
Integrate: ∫d[A]/[A]² = −k∫dt → −1/[A] = −kt + C. At t = 0, [A] = [A]₀, so C = −1/[A]₀. Thus: 1/[A] = 1/[A]₀ + kt
3
Half-life: When [A] = [A]₀/2: 2/[A]₀ = 1/[A]₀ + kt_½ → t_½ = 1/(k[A]₀). Half-life depends on initial concentration.
Zero: [A] = [A]₀ − kt   t_½ = [A]₀/(2k)   |   First: [A] = [A]₀e^{−kt}   t_½ = 0.693/k   |   Second: 1/[A] = 1/[A]₀ + kt   t_½ = 1/(k[A]₀)

🔍 Common Mistake

Common Mistake — Units of k
The units of rate constant k depend on the order. Zero order: mol·L⁻¹·s⁻¹. First order: s⁻¹. Second order: L·mol⁻¹·s⁻¹. Also: for a first-order reaction, you can use any units of concentration since ln([A]₀/[A]) has no units.

🎯 NEET Pattern

🎯 NEET Pattern — Rate Laws
• "A first-order reaction has t_½ = 10 min. How much remains after 30 min?" → 3 half-lives → 1/8 remains.
• "The half-life of a reaction is independent of [A]₀. What is the order?" → First order.
• "A zero-order reaction has [A]₀ = 1 M and k = 0.1 M/s. Time for 80% completion?" → [A] = 0.2 M, t = (1 − 0.2)/0.1 = 8 s.

✍️ Worked Example

Example 3: First-Order Kinetics
A first-order reaction has half-life 20 minutes. What percentage remains after 60 minutes?
Solution: 60 min = 3 half-lives. After 1st: 50% remains. After 2nd: 25%. After 3rd: 12.5%. Key insight: For 1st order, each half-life halves the remaining concentration.

4. Henderson–Hasselbalch Equation

Buffer Action — Resisting pH Change Buffer solution HA HA A⁻ A⁻ H⁺ pH 7.40 stable + HCl pH 7.38 → barely changed!
Fig 4.1: Buffer action — added H⁺ is neutralized by A⁻ (conjugate base), keeping pH nearly constant.

💡 Why this matters in real life

Your blood maintains a nearly constant pH of 7.4. This is thanks to buffer solutions — mixtures of weak acids and their conjugate bases. The Henderson-Hasselbalch equation lets us calculate the pH of a buffer. It's essential for understanding how your body resists pH changes, how medicines are formulated, and how biological systems maintain homeostasis.

🧠 The Big Idea — In Plain Words

For a weak acid HA dissociating as HA ⇌ H⁺ + A⁻, the equilibrium constant is Kₐ = [H⁺][A⁻]/[HA]. Taking negative logs: pH = pKₐ + log([A⁻]/[HA]). This is the Henderson-Hasselbalch equation. It shows that pH depends on the ratio of conjugate base to acid. When [A⁻] = [HA], pH = pKₐ. The buffer is most effective near its pKₐ.

🔍 Step-by-Step: Deriving pH = pKₐ + log([A⁻]/[HA])

1
Write the acid dissociation equilibrium. For a weak acid HA: HA ⇌ H⁺ + A⁻. The equilibrium constant: Kₐ = [H⁺][A⁻]/[HA]
2
Take the negative logarithm (base 10) of both sides. −log Kₐ = −log([H⁺][A⁻]/[HA]). By log rules: pKₐ = pH − log([A⁻]/[HA])
3
Rearrange to get the Henderson-Hasselbalch equation. pH = pKₐ + log([A⁻]/[HA]). For a basic buffer (weak base B + conjugate acid BH⁺): pOH = pK_b + log([BH⁺]/[B])
pH = pKₐ + log([conjugate base]/[acid])    Maximum buffer capacity when pH = pKₐ

🔍 Common Mistake

Common Mistake — Which Ratio?
Students often invert the ratio. The log term is log([base]/[acid]) — base on top, acid on bottom. For basic buffers, remember pOH = pK_b + log([conjugate acid]/[base]), then pH = 14 − pOH.

🎯 NEET Pattern

🎯 NEET Pattern — Buffers
• "A buffer has [CH₃COOH] = [CH₃COONa] = 0.1 M. pKₐ = 4.74. Find pH." → pH = 4.74 + log(0.1/0.1) = 4.74.
• "How does pH change when 0.01 M HCl is added to the buffer?" → Ratio changes slightly → pH changes slightly. This is the buffer's job!
• "Which buffer has greater capacity: [acid]=1M, [base]=1M or [acid]=0.1M, [base]=0.1M?" → The 1 M buffer (higher total concentration means more capacity).

✍️ Worked Example

Example 4: Buffer pH
A buffer contains 0.2 M CH₃COOH and 0.4 M CH₃COONa. pKₐ = 4.74. Find pH.
Solution: pH = 4.74 + log(0.4/0.2) = 4.74 + log(2) = 4.74 + 0.30 = 5.04. Key insight: Since [base] > [acid], pH > pKₐ.

5. Arrhenius Equation — k = A·e^{−Eₐ/RT}

💡 Why this matters in real life

Why does food spoil faster in summer? Why does a chemical reaction speed up when you heat it? The Arrhenius equation quantifies how temperature affects reaction rates. It tells us the activation energy — the minimum energy needed for a reaction to occur. Every 10°C rise roughly doubles the rate of many reactions (the rule of thumb for food storage).

Electrochemical Cell — Nernst Equation in action Anode Zn → Zn²⁺ + 2e⁻ Zn²⁺ ions Cathode Cu²⁺ + 2e⁻ → Cu Cu²⁺ → Cu Salt bridge e⁻ flow → 1.10 V E = E° − (RT/nF) ln Q
Fig 2.1: Electrochemical cell with Zn anode (oxidation) and Cu cathode (reduction). Ion migration through salt bridge, electron flow through wire, and voltmeter reading.

🧠 The Big Idea — In Plain Words

The Arrhenius equation k = A·e^{−Eₐ/RT} has three key parts: k is the rate constant, A is the frequency factor (how often molecules collide with correct orientation), and e^{−Eₐ/RT} is the fraction of molecules with enough energy to react. Taking natural logs: ln k = ln A − Eₐ/RT. This is a straight line: ln k vs 1/T gives slope = −Eₐ/R.

🔍 Step-by-Step: Deriving the Logarithmic Form

1
Start with the Arrhenius equation. k = A·e^{−Eₐ/RT}. A is the pre-exponential factor, Eₐ is activation energy, R is gas constant, T is temperature in Kelvin.
2
Take natural log of both sides. ln k = ln(A·e^{−Eₐ/RT}) = ln A + ln(e^{−Eₐ/RT}) = ln A − Eₐ/RT This is the linear form: ln k = ln A − (Eₐ/R)·(1/T)
3
For two different temperatures T₁ and T₂: Subtract the equations: ln(k₂/k₁) = −(Eₐ/R)(1/T₂ − 1/T₁) = (Eₐ/R)(1/T₁ − 1/T₂). This is used to find Eₐ from rate constants at two temperatures.
ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)

🔍 Common Mistake

Common Mistake — Temperature Units
Temperature must always be in Kelvin. Also: Eₐ and R must have consistent energy units (both in J/mol or both in kJ/mol). R = 8.314 J/(mol·K), so if Eₐ is in kJ/mol, multiply by 1000 first.

🎯 NEET Pattern

🎯 NEET Pattern — Arrhenius
• "Rate doubles when T increases from 300 to 310 K. Find Eₐ." → Use ln(2) = (Eₐ/R)(1/300 − 1/310).
• "A plot of ln k vs 1/T gives slope = −5000 K. Find Eₐ." → Eₐ = −slope × R = 5000 × 8.314 = 41570 J/mol.
• "Which has higher Eₐ: fast reaction or slow reaction?" → Slow reaction has higher Eₐ (larger energy barrier).

✍️ Worked Example

Example 5: Arrhenius
The rate constant at 300 K is 0.02 s⁻¹ and at 310 K is 0.04 s⁻¹. Find activation energy. (R = 8.314 J/mol·K)
Solution: ln(0.04/0.02) = ln(2) = 0.693. Eₐ = 0.693 × 8.314 / (1/300 − 1/310) = 0.693 × 8.314 / 0.0001075 = 53600 J/mol ≈ 53.6 kJ/mol. Key insight: A 10 K rise doubled k — that's a typical Eₐ.

6. Raoult's Law & Henry's Law

Raoult's Law & Henry's Law — Vapor Pressure Pure solvent P°ₐ High vapor pressure Solution Pₐ = P°ₐ·Xₐ Sl Sl Lower vapor pressure → ΔP = P°ₐ·X_B Henry P = K_H·X Pₐ = P°ₐ·Xₐ & ΔP = P°ₐ·X_B
Fig 6.1: Pure solvent (left) has high vapor pressure. Solute molecules (yellow) block surface → fewer solvent molecules escape → lower vapor pressure.

💡 Why this matters in real life

Raoult's law explains why adding salt to water raises its boiling point (used in cooking pasta). Henry's law explains why soda fizzes when you open it (CO₂ dissolved under pressure escapes when pressure is released). Both are essential for understanding solutions and are frequently tested in NEET.

🧠 The Big Idea — In Plain Words

Raoult's law: The partial vapor pressure of a solvent above a solution equals the vapor pressure of pure solvent times its mole fraction. Pₐ = P°ₐ·Xₐ. For a non-volatile solute, the vapor pressure decreases because the solute molecules occupy space at the surface, reducing the number of solvent molecules that can escape.

Henry's law: The partial pressure of a gas above a liquid is proportional to its mole fraction in the liquid. P = K_H·X. K_H is Henry's constant. Gases with low K_H are more soluble (like CO₂ in water).

🔍 Step-by-Step: Deriving Raoult's Law

1
Consider a solution with solvent A and non-volatile solute B. The vapor pressure above the solution comes only from solvent molecules. At equilibrium, the rate of evaporation equals the rate of condensation.
2
The rate of evaporation depends on the fraction of surface occupied by solvent molecules. This is proportional to the mole fraction of solvent: Pₐ ∝ Xₐ → Pₐ = P°ₐ·Xₐ. When Xₐ = 1 (pure solvent), Pₐ = P°ₐ.
3
The lowering of vapor pressure is ΔP = P°ₐ − Pₐ = P°ₐ(1 − Xₐ) = P°ₐ·X_B. Relative lowering: ΔP/P°ₐ = X_B. For dilute solutions, X_B ≈ n_B/nₐ, which gives colligative properties.
Raoult: Pₐ = P°ₐ·Xₐ   ΔP = P°ₐ·X_B    Henry: P = K_H·X

🔍 Common Mistake

Common Mistake — Raoult vs Henry
Raoult's law applies to the solvent in a solution. Henry's law applies to a gas dissolved in a liquid. Both are mathematically similar (P ∝ X) but the constant is different. Also: Raoult's law is for ideal solutions — real solutions deviate positively or negatively.

🎯 NEET Pattern

🎯 NEET Pattern — Solutions
• "Vapor pressure of pure water at 25°C is 23.8 mm Hg. Find vapor pressure above a solution with X_solute = 0.1." → P = 23.8 × 0.9 = 21.42 mm Hg.
• "Henry's constant for CO₂ in water is 1.67×10⁸ Pa at 25°C. Find concentration when P_CO₂ = 0.1 atm." → Convert units, apply P = K_H·X.
• "Which has higher boiling point: 0.1 M NaCl or 0.1 M glucose?" → NaCl (i = 2, more particles).

✍️ Worked Example

Example 6: Raoult's Law
Pure water has vapor pressure 23.8 mm Hg at 25°C. A solution contains 10 g glucose (M = 180 g/mol) in 100 g water (M = 18 g/mol). Find the vapor pressure.
Solution: n_glucose = 10/180 = 0.0556 mol. n_water = 100/18 = 5.556 mol. X_water = 5.556/(5.556 + 0.0556) = 0.99. P = 23.8 × 0.99 = 23.56 mm Hg. Key insight: The lowering is tiny because glucose has much higher molar mass.
Reference

Periodic Table — Complete Reference

Structure, trends, block characteristics, and mnemonics for NEET.

1. Modern Periodic Table Structure

Blocks, Groups, Periods

The modern periodic table has 118 elements arranged by increasing atomic number. 18 groups (columns) and 7 periods (rows). Elements are classified into four blocks based on the subshell receiving the last electron:

BlockGroupsSubshellElementsExamples
s-block1, 2ns¹⁻²14Li, Na, K, Be, Mg, Ca
p-block13–18np¹⁻⁶36C, N, O, Cl, Ar, Si
d-block3–12(n−1)d¹⁻¹⁰ ns¹⁻²40Fe, Cu, Zn, Cr, Mn, Ni
f-block—(n−2)f¹⁻¹⁴28 (14+14)Ce, Eu, Gd (Lanthanoids); U, Pu (Actinoids)
Period number = highest n (principal quantum number)  |  Group number = (ns + np) valence electrons for p-block, ns for s-block

Position of Hydrogen

Hydrogen (1s¹) resembles both Group 1 (forms H⁺) and Group 17 (forms H⁻). It has unique placement — ionization enthalpy 1312 kJ/mol is much higher than alkali metals. It is often placed separately at the top of the table.

Position of d-block (Transition Elements)

Elements with partially filled d-orbitals in their ground state or common oxidation state. All are metals, form coloured ions, show variable oxidation states, paramagnetism, and catalytic activity. Examples: Fe²⁺/Fe³⁺, Cu⁺/Cu²⁺, Mn²⁺/Mn⁷⁺.

Position of f-block (Inner Transition)

Two rows placed below: Lanthanoids (Ce 58 to Lu 71, 4f series) and Actinoids (Th 90 to Lr 103, 5f series). All are metals. Lanthanoids show lanthanoid contraction (steady decrease in atomic radius across the series due to poor shielding of 4f electrons).

2.1 Atomic Radius

Trend: Decreases across a period (Z_eff increases), increases down a group (new shells added).

Types: Covalent radius (bonded atoms), Metallic radius (metal lattice), van der Waals radius (non-bonded, largest).

Period 3NaMgAlSiPSClAr
Radius (pm)1861601431171101049971*
Exception
Ga (135 pm) has smaller radius than Al (143 pm) due to d-block contraction — poor shielding by 3d¹⁰ electrons increases Z_eff. Same applies to Ge vs Si.

2.2 Ionization Enthalpy (IE)

Energy required to remove the most loosely bound electron from a gaseous atom. Trend: Increases across period, decreases down group.

Period 3NaMgAlSiPSClAr
IE₁ (kJ/mol)4967385787871012100012511521
Key Exceptions for NEET
Mg (738) > Al (578): Mg has 3s² (filled subshell, stable). Al is 3s²3p¹ — removing one p-electron is easier.
P (1012) > S (1000): P has 3p³ (half-filled, extra stable). S is 3p⁴ — the 4th p-electron is paired, causing repulsion, easier to remove.
1
Z_eff = Z − S (Slater's rules). Across a period, Z increases faster than S (shielding), so Z_eff increases → stronger attraction → electrons harder to remove → IE increases.
2
Down a group: New shells are added, distance from nucleus increases, shielding effect of inner electrons increases → Z_eff decreases → IE decreases.
3
Successive IEs: IE₁ < IE₂ < IE₃ because removing an electron from a positive ion requires more energy. Sharp jump after removing all valence electrons (e.g., Na: IE₁=496, IE₂=4562 — core electron removal).

2.3 Electronegativity (EN)

Tendency of an atom to attract shared electrons in a chemical bond. Trend: Increases across period, decreases down group. F (4.0) is the most electronegative.

Pauling ScaleHLiBeBCNOF
EN2.11.01.52.02.53.03.54.0
Order for NEET
EN order: F > O > N ≈ Cl > Br > I > S ≈ C. Most electronegative elements: F, O, N, Cl. Bond polarity: ΔEN > 1.7 = ionic; 0.4–1.7 = polar covalent; < 0.4 = non-polar covalent.

2.4 Electron Affinity (EA)

Energy released when an electron is added to a neutral gaseous atom. Trend: Generally becomes more negative across a period, less negative down a group. Most negative (highest magnitude) at Cl not F.

EA Values (kJ/mol)
Cl = −349, F = −328, Br = −325, I = −295, O = −141, S = −200, N ≈ +7 (endothermic)
Why Cl > F? F's 2p orbitals are very compact → added electron experiences repulsion → less energy released. Cl has larger 3p orbitals → less repulsion → more negative EA.

3. Block Characteristics & Mnemonics

s-Block (Groups 1 & 2)

Group 1 — Alkali metals: ns¹, highly reactive, low IE, form +1 ions, never found free in nature. React violently with water (vigour increases down group).

Group 2 — Alkaline earth metals: ns², harder than alkali metals, higher melting points, form +2 ions.

Group 1: H, Li, Na, K, Rb, Cs, Fr → "Hi LiNa K Rb Cs Fr"  |  Group 2: Be, Mg, Ca, Sr, Ba, Ra → "Be Mg Ca Sr Ba Ra"

p-Block (Groups 13–18)

Includes metals, metalloids, and non-metals. Variable oxidation states. Inert-pair effect in heavier elements (Tl⁺ stable, Pb²⁺ stable, Bi³⁺ stable).

Gr 13: B, Al, Ga, In, Tl → "B Al Ga In Tl"  |  Gr 14: C, Si, Ge, Sn, Pb → "C Si Ge Sn Pb" (Carbon family)
Gr 15: N, P, As, Sb, Bi → "N P As Sb Bi" (Pnicogens)  |  Gr 16: O, S, Se, Te, Po → "O S Se Te Po" (Chalcogens)
Gr 17: F, Cl, Br, I, At → "F Cl Br I At" (Halogens — salt-formers)  |  Gr 18: He, Ne, Ar, Kr, Xe, Rn → "He Ne Ar Kr Xe Rn" (Noble gases)

d-Block (Transition Metals)

Groups 3–12, (n−1)d¹⁻¹⁰ ns¹⁻². Key properties: variable oxidation states, coloured ions, paramagnetism (unpaired e⁻), catalytic activity, formation of complexes, high melting/boiling points.

3d series: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn → "Sc Ti V Cr Mn Fe Co Ni Cu Zn"
4d series: Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Cd
5d series: La, Hf, Ta, W, Re, Os, Ir, Pt, Au, Hg

f-Block (Inner Transition)

Lanthanoids: Ce–Lu (4f¹⁻¹⁴). Common ox. state +3. Lanthanoid contraction: steady decrease in ionic radius across series due to poor shielding of 4f electrons.

Actinoids: Th–Lr (5f¹⁻¹⁴). Radioactive, show multiple oxidation states (+3 to +6).

4. Key Facts & Anomalies for NEET

PropertyTrendExceptions / Key Points
Atomic radius↓ across, ↑ downGa < Al, Ge < Si (d-block contraction)
Ionization Enthalpy↑ across, ↓ downGr 2 > Gr 13 (filled s²), Gr 15 > Gr 16 (half-filled p³)
Electronegativity↑ across, ↓ downF (4.0) highest. O > Cl? Yes! O=3.5, Cl=3.0
Electron Affinity↑ across, ↓ downCl > F (size+repulsion). N has +ve EA (stable half-filled)
Metallic character↓ across, ↑ downMost metallic: Fr. Most non-metallic: F
Oxidising power↑ across, ↓ downF₂ strongest oxidising agent. Down group: decreases
Reducing power↓ across, ↑ downLi strongest reducing agent (high hydration enthalpy)

Diagonal Relationship

Li–Mg, Be–Al, B–Si. Due to similar charge-to-size ratio. Similar properties: both form nitrides (Li₃N, Mg₃N₂), carbonates decompose on heating, hydroxides are weak bases.

Anomalous Behaviour of Second Period

Li to F show anomalous behaviour compared to rest of their groups due to: small size, high EN, no d-orbitals, maximum covalency of 4. Examples: Li resembles Mg more than Na; Be forms covalent compounds unlike Mg; B is electron-deficient (BF₃ is Lewis acid).

NEET Example: Periodic Trends
Arrange in order of increasing IE: Na, Mg, Al, Si, P, S, Cl, Ar.
Solution: Na < Al < Mg < Si < S < P < Cl < Ar. Mg (3s² filled) > Al (3s²3p¹); P (3p³ half-filled) > S (3p⁴).
NEET Example: Blocks
An element has configuration [Kr]4d⁵5s¹. Which block does it belong to? Name the element.
Solution: d-block (last electron in d-subshell, (n−1)d⁵). Element is Mo (Molybdenum, Z=42). Cr and Mo are exceptions with d⁵s¹ configuration (half-filled stability).
📌 For NEET: Know the 4 blocks, 18 groups, anomalies (Mg>Al, P>S, Cl>F EA), and diagonal relationship (Li–Mg, Be–Al, B–Si). Practice identifying block from electronic configuration.