Fundamentals

1,200 must-knows for NEET & JEE

The core facts every aspirant should own — each a titled nugget with a real-world story, the concept in plain words, and a memory trick. Works even when the internet doesn't.

1,200 fundamentals

ChemistryThe s-Block Elements· Class 11

Alkaline earth metals

“When a magnesium ribbon is ignited it burns a brilliant white flame, illustrating the harder, less reactive nature of alkaline-earth metals.”

Group-2 elements have a filled ns² shell, are harder and have higher melting points than alkali metals. They lose two electrons to form M²⁺ ions. Reactivity rises down the group but is still lower than Group-1; only Ca, Sr, Ba react noticeably with water, giving hydroxides, while oxides form on burning in oxygen. Do not claim all react with hot water.

Memory trick: 2 valence e⁻ → +2 ion; reactivity ↑ down group

ChemistryThe p-Block Elements· Class 12

Inert pair effect

“A lead-acid battery’s Pb²⁺ plates stay in the +2 state because the 6s electrons refuse to join the bond.”

In heavy p-block elements (Sn, Pb, Bi) the ns² pair is reluctant to participate in bonding, so the lower oxidation state (+2 for group-14, +1 for group-13) becomes more stable than the higher one. The effect grows down the group as relativistic contraction lowers the s-orbital energy. Exam trap: don’t assume the trend for lighter elements; inert-pair effect is negligible for C, Si, Al.

Memory trick: Inert pair = ‘stay-in-place’; heavy → lower oxidation state favoured.

ChemistryThe p-Block Elements· Class 12

Nitrogen's inertness

“Air is 78 % nitrogen, yet a spark won’t ignite it because the N≡N triple bond is like a super-strong steel chain.”

N₂ has a bond dissociation energy ≈945 kJ mol⁻¹, arising from one σ and two π bonds, making the molecule kinetically inert. Although thermodynamically stable, breaking the triple bond requires high temperature or a catalyst (e.g., Fe in Haber). Trap: confusing stability with reactivity; N₂ does not support combustion without activation.

N≡N

Memory trick: N₂ triple bond ≈945 kJ mol⁻¹ → need catalyst.

ChemistryThe p-Block Elements· Class 12

Oxides: acidic, basic, amphoteric

“Aluminium foil dissolves in both vinegar and NaOH, showing its oxide Al₂O₃ is amphoteric.”

Oxides of non-metals (SO₂, CO₂) react with water to give acids → acidic. Oxides of metals (Na₂O, CaO) form bases with water → basic. Some oxides like Al₂O₃, ZnO react with acids and bases, hence amphoteric. Trap: assuming every metal oxide is basic; transition-metal oxides often deviate.

Memory trick: Non-metal oxide = acid, Metal oxide = base, Some = amphoteric.

ChemistryThe p-Block Elements· Class 12

Halogens

“Like a magnet pulling a metal nail, the smaller fluorine atom grabs an electron more tightly than larger iodine.”

Halogens gain one electron to reach a noble-gas configuration. Reactivity decreases down the group because atomic radius and shielding increase, reducing the effective nuclear attraction on the added electron. Exam trap: do not link reactivity to X–X bond energy; F₂ has an anomalously weak bond, yet fluorine is the most reactive.

Memory trick: F → Cl → Br → I: Small → big, grab ↓

ChemistryThe d- and f-Block Elements· Class 12

Transition metals

“The deep violet of MnO₄⁻ (potassium permanganate) shows a d⁰ ion can still be intensely coloured via charge-transfer.”

d-block elements have (n)s and (n-1)d orbitals close in energy, so electrons may be removed from either, giving many oxidation states. Their ions are coloured by d-d transitions when d electrons exist, or by charge-transfer transitions (e.g., MnO₄⁻). Unpaired d electrons cause paramagnetism, while other magnetic orders arise from collective interactions.

Memory trick: V-O-C-M: Variable oxidation, Colour, Catalysis, Magnetism

ChemistryThe d- and f-Block Elements· Class 12

Lanthanide contraction

“Like a spring that tightens with each added coil, the atomic radius shrinks steadily from lanthanum to lutetium because 4f electrons shield poorly.”

Across the lanthanide series the 4f electrons shield the nuclear charge poorly, so effective nuclear charge rises and atomic/ionic radii decrease steadily (lanthanide contraction). This makes later lanthanides almost as small as many 3d metals, influencing bond lengths and ionization energies. Trap: the contraction continues to Lu; it does not stop at Ce.

Memory trick: 4f poor shield → radius contracts.

ChemistryCoordination Compounds· Class 12

Coordination number

“Think of a round dining table: each chair (donor atom) placed around the host (metal) counts as one seat, regardless of chair material.”

Coordination number (CN) is the total number of donor atoms directly attached to the central metal ion, irrespective of element or ligand type. It is not the number of ligands or the oxidation state. A common trap is to count ligands instead of donor atoms (e.g., bidentate ligands count twice).

Memory trick: CN = count of all donor atoms, not ligands.

ChemistryCoordination Compounds· Class 12

Ligand

“A ligand is like a friend handing over a spare key (electron pair) to unlock the metal’s empty orbital.”

A ligand is any ion or molecule possessing at least one lone pair that can be donated to a metal, forming a coordinate bond. It may be neutral (NH₃) or anionic (Cl⁻). Exam trap: the whole ion donates one pair; do not split a monodentate ligand into multiple donors.

Memory trick: L = Lone-pair donor

ChemistryCoordination Compounds· Class 12

Chelate

“A chelate ligand is like a claw that grabs a metal with two fingers, making a tighter grip than a single finger.”

A chelating ligand binds through two or more donor atoms to the same metal, creating a ring (chelate) and giving extra stability (chelate effect). Exam trap: for coordination number count each donor atom, not the ligand as a single unit.

Memory trick: Chela = claw = multi-grip

ChemistryThe Solid State· Class 12

Types of solids

“A quartz crystal in a watch keeps time because its atoms repeat in a regular pattern, unlike window glass which lacks long-range order.”

Solids are crystalline when atoms form a periodic lattice, giving a sharp melting point. Amorphous solids have no long-range order and soften over a temperature range. In exams, only crystalline solids show a definite melting point; amorphous ones do not.

Memory trick: Crystal = Clear point (sharp melt); Glass = Gradual melt.

ChemistryThe Solid State· Class 12

Unit cell

“A LEGO brick is the smallest piece that repeats to build the whole LEGO wall.”

A unit cell is the smallest repeatable volume of a crystal lattice; translating it along three axes reproduces the whole crystal. Its edge lengths (a,b,c) and angles (α,β,γ) define the lattice type. Beware mixing it up with the primitive cell, which may contain fewer lattice points.

Memory trick: Unit = one, cell = repeat block.

ChemistryThe Solid State· Class 12

Packing efficiency

“Stacking oranges in a grocery crate shows the 74% close-packing of fcc/hcp.”

Packing efficiency is the fraction of space filled by identical hard spheres in a lattice. fcc and hcp give π/√18 ≈ 0.74, bcc ≈ 0.68, simple cubic ≈ 0.52. Remember the value applies only to equal-sized spheres; ionic radii differences alter the effective packing.

Memory trick: 74% = fcc/hcp, 68% = bcc, 52% = sc.

ChemistryThe Solid State· Class 12

Crystal defects

“A few phosphorus atoms replace silicon in a crystal, supplying extra electrons that boost conductivity, like adding extra runners to speed a relay.”

Point defects include vacancies, interstitials, and substitutional atoms. Doping silicon with pentavalent phosphorus (donor) adds free electrons, while trivalent boron (acceptor) creates holes. Beware of swapping donor and acceptor roles or assuming any defect always reduces conductivity.

Memory trick: Donor adds e⁻, Acceptor adds holes.

ChemistrySolutions· Class 12

Raoult's law

“Adding table salt to water makes the water’s vapor pressure fall in direct proportion to the salt’s mole fraction.”

Raoult’s law: for an ideal solution with a non-volatile solute, Psolvent = Xsolvent·P°solvent. Since Xsolvent = 1-Xsolute, the pressure drop is proportional to the solute’s mole fraction. Trap: it applies only to ideal solutions and non-volatile solutes; volatile solutes need separate terms.

Memory trick: P = (1-Xsolute)·P° – ‘1-X’ is the solvent’s share.

ChemistrySolutions· Class 12

Colligative properties

“Sprinkling salt on ice makes it melt quicker, because the freezing point falls with more dissolved particles.”

Colligative properties (ΔTf, ΔTb, ΔPv, π) depend solely on the total number of solute particles, not their nature. ΔTf = i·Kf·m and ΔTb = i·Kb·m. Forgetting the van’t Hoff factor i for electrolytes leads to wrong numerical answers.

Memory trick: ΔT = i·K·m (count particles)

ChemistrySolutions· Class 12

Henry's law

“A sealed soda bottle stays fizzy because the high CO₂ pressure forces the gas to stay dissolved.”

Henry's law: C = k_H·P, where C is the concentration of a gas in a liquid and P its partial pressure above the liquid. Valid for low pressures and constant temperature; k_H varies strongly with temperature, a common source of error.

Memory trick: C ∝ P (k_H constant)

ChemistrySolutions· Class 12

Molality

“Dissolving 58 g NaCl in exactly 1 kg water gives a molality of 1 m, no matter the temperature.”

Molality (m) = moles of solute per kilogram of solvent. Because it uses mass, m is independent of temperature, unlike molarity which uses volume. Using solution volume instead of solvent mass is a frequent mistake.

m = mol/kg

Memory trick: m = n(solute)/kg solvent

ChemistrySurface Chemistry· Class 12

Adsorption vs absorption

“Dust clings to a car windshield, while water soaks into a sponge – surface vs bulk.”

Adsorption is the accumulation of molecules only on the outer surface of a solid; absorption involves the same species entering the interior of the material. Surface forces (van-der-Waals, chemisorption) limit adsorption, while absorption requires diffusion. In exams, never treat them as synonyms; the question will specify “surface” or “throughout”.

Memory trick: Adsorb = A (At surface); Absorb = B (Bulk).

ChemistrySurface Chemistry· Class 12

Colloids and the Tyndall effect

“Sunlight becomes a visible beam when it passes through a glass of milk, a classic colloid.”

Colloids contain particles 1–1000 nm, large enough to scatter visible light, producing the Tyndall effect. True solutions have ≤1 nm particles and stay clear. In exams, a milky appearance or a visible light beam signals a colloid, not a true solution.

Memory trick: Colloid 1-1000 nm → Tyndall; ≤1 nm → clear.

ChemistryGeneral Principles of Metallurgy· Class 12

Roasting and calcination

“Baking limestone in a kiln turns it white – that’s calcination, while heating zinc sulfide in a furnace with oxygen “toasts” it – that’s roasting.”

Roasting: heat a sulphide ore in the presence of air/oxygen, producing metal oxide and SO₂. Calcination: heat a carbonate ore without a reducing/oxidising agent, releasing CO₂ and forming metal oxide. Exam trap – don’t mix the gas environment: air → roasting, no air → calcination.

Memory trick: Roast = O₂ present; Calcine = CO₂ released, O₂ absent.

ChemistryChemical Bonding & Molecular Structure· Class 11

Bond order

“A double-lane road lets cars travel side-by-side, just like a double bond pulls atoms closer than a single lane.”

Bond order = (Nb – Na)/2, where Nb and Na are the numbers of electrons in bonding and antibonding molecular orbitals. It predicts bond strength and length: higher BO → stronger, shorter bond. BO = 0 means no bond; half-integer values are allowed.

Memory trick: BO = (Nb-Na)/2

ChemistryChemical Bonding & Molecular Structure· Class 11

Dipole moment

“A water droplet clings to a charged glass rod because its bent shape gives a permanent dipole.”

Dipole moment μ is a vector equal to charge separation (μ = δ·r). Molecules like H₂O have a non-zero μ, while symmetric CO₂ has μ = 0 because bond dipoles cancel. Even with μ = 0 a molecule can be attracted to a charge via induced dipole, a common trap.

Memory trick: μ = charge × distance; symmetry → zero.

ChemistryChemical Thermodynamics· Class 11

Bond enthalpy

“Snapping a wooden stick needs energy – breaking each C–C bond costs its bond enthalpy.”

Bond enthalpy is the energy required to break one mole of a specific bond in the gas phase. Reaction ΔH ≈ Σ(bonds broken) – Σ(bonds formed). The trap: using average values for molecules with resonance or forgetting that breaking bonds is endothermic (+) and forming bonds is exothermic (−).

Memory trick: Break = +, Form = – (bond enthalpy)

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