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.

300 fundamentals

ChemistryChemical Kinetics· Class 12

Collision theory

“Think of two pool balls: they must hit each other hard enough and with the right spin to pocket a ball, just like reactants need sufficient energy and proper orientation.”

Collision theory states a reaction occurs only when reactant particles collide with energy ≥ Ea and with the correct orientation (steric factor). Both factors determine the frequency of effective collisions, which sets the rate constant. Trap: ignoring orientation can make a seemingly fast reaction appear slow on paper.

Memory trick: Energy ≥ Ea + right angle = effective collision.

ChemistryChemical Kinetics· Class 12

Pseudo first-order reaction

“When you dissolve a drop of dye in a huge tank of water, the dye’s disappearance looks like a first-order decay because water’s concentration stays essentially constant.”

If one reactant is in large excess, its concentration change is negligible, so the rate law k[A]^m[B]ⁿ reduces to k' [A]^m where k' = k[B]ⁿ (constant). The reaction then follows apparent first-order kinetics. Trap: verify that the excess reactant truly remains constant throughout the experiment.

Memory trick: Excess ≈ constant → absorb into k'.

ChemistryChemical Kinetics· Class 12

Rate-determining step

“In a relay race, the slowest runner sets the team’s finish time; similarly, the slowest elementary step controls the overall reaction rate.”

The step with the highest activation energy (or smallest rate constant) is the rate-determining step; the overall rate law is derived from this step alone, with intermediates expressed via steady-state or pre-equilibrium approximations. Trap: writing the overall rate from a fast step gives an incorrect expression.

Memory trick: RDS = slowest link → overall rate.

ChemistryStates of Matter· Class 11

Critical temperature

“Carbon dioxide can’t be turned into liquid by squeezing a soda can at room temperature because its critical temperature is 31 °C.”

Critical temperature is the highest temperature at which a gas can be liquefied by pressure alone. Above this temperature, no amount of pressure will condense the gas. Each substance has its own Tc, a key property for liquefaction.

Memory trick: Critical T = ‘Can’t’ condense above it

ChemistryChemical Bonding & Molecular Structure· Class 11

Formal charge

“A school ledger balances total marks; each subject’s credit adjusts the final score just as formal charge balances electrons on an atom.”

Formal charge = valence electrons – (non-bonding electrons + ½ bonding electrons). The best Lewis structure has the smallest absolute FC on each atom and puts negative charge on the more electronegative atom. In NO₃⁻ only one O carries –1 in any resonance form; the overall –1 charge is delocalised over the three oxygens, not –1 on each.

Memory trick: FC = V – (L + B/2)

ChemistryChemical Bonding & Molecular Structure· Class 11

Fajans' rules

“Lithium iodide melts at a lower temperature than sodium chloride because Li⁺ small size pulls electron density from I⁻, giving partial covalent character.”

Fajans' rules state that a small, highly charged cation and a large, polarizable anion increase covalent character of an ionic bond. This explains LiI’s lower melting point compared with NaCl. High charge on either ion (e.g., Al³⁺) also pushes covalency.

Memory trick: Small⁺ + Large⁻ + high charge → covalent

ChemistryChemical Bonding & Molecular Structure· Class 11

Molecular orbital theory: O₂ is paramagnetic

“Liquid oxygen sticks to a magnet because two electrons occupy the degenerate π* antibonding orbitals singly, giving O₂ two unpaired spins.”

MOT places O₂’s 12 valence electrons in σ2s, σ*2s, σ2p, π2p, π*2p; the last two electrons occupy separate π* orbitals, resulting in two unpaired electrons and paramagnetism. Mistake to avoid: using the wrong orbital order, which would predict diamagnetism.

Memory trick: O₂: π* each gets 1 e⁻ → paramag

ChemistryOrganic Chemistry — Basic Concepts· Class 11

Hyperconjugation

“Like a group of friends sharing a heavy bag, a σ-bond donates electron density to a neighboring empty p-orbital, easing the carbocation.”

Hyperconjugation is delocalisation of σ-C–H or σ-C–C electrons into an adjacent empty or partially filled orbital, stabilising carbocations, alkenes and radicals. The overlap lowers the system’s energy. Do not mix it up with resonance; only σ bonds participate.

Memory trick: Hyper = over, conj = join → σ-C–H joins empty p

ChemistryOrganic Chemistry — Basic Concepts· Class 12

Chirality

“Like a left-handed glove fits only the left hand, a molecule that cannot be overlaid on its mirror image is chiral.”

A molecule is chiral when it has no improper symmetry element (plane, centre, Sn). A tetrahedral carbon with four different groups often gives chirality, but axial, planar or helical arrangements can also be chiral. Conversely, a molecule with stereogenic centres can be achiral (meso) if an internal symmetry plane exists.

Memory trick: Four different groups → chiral; internal mirror → meso.

ChemistryAldehydes, Ketones & Carboxylic Acids· Class 12

Tautomerism

“Keto-enol tautomerism is like a dancer swapping a hat (hydrogen) and a cape (double bond) between carbon and oxygen.”

Tautomerism is the rapid interconversion of isomers that differ by the position of a proton and a double bond, the classic example being keto-enol equilibrium. The shift is facilitated by acid or base catalysis and is significant when the enol form is stabilised by conjugation or intramolecular H-bond. Do not confuse it with constitutional isomerism; only certain functional groups show tautomerism.

Memory trick: TAUT = Transfer A Ubiquitous Hydrogen.

ChemistryThe p-Block Elements· Class 12

Haber process

“Ammonia fertilizer is made in a Haber reactor where N₂ and H₂ are forced together like a high-pressure soda can.”

The Haber reaction N₂ + 3H₂ ⇌ 2NH₃ is exothermic. According to Le Chatelier, high pressure (≈150–200 atm) shifts equilibrium toward NH₃, while moderate temperature (400–500 °C) balances rate and yield. An iron catalyst with promoters lowers activation energy. Trap: raising temperature improves speed but cuts yield; optimal compromise is essential.

N₂+3H₂⇌2NH₃

Memory trick: Haber: high P, moderate T, Fe catalyst.

ChemistryThe p-Block Elements· Class 12

Contact process

“The towering contact converters at a fertilizer plant turn SO₂ into SO₃ over V₂O₅, the heart of sulfuric acid production.”

In the Contact process SO₂ is oxidised to SO₃ using O₂ over a V₂O₅ catalyst at 450–550 °C. The catalyst provides a lower-energy pathway, increasing rate without being consumed. The reaction is exothermic, so too high a temperature shifts equilibrium left; plants use a compromise temperature and recycle gases. A frequent mistake is writing Pt as the catalyst or ignoring the temperature-equilibrium trade-off.

Memory trick: V-2-O-5 = V for Victory over SO₂ → SO₃.

ChemistryThe p-Block Elements· Class 12

Ostwald process

“A fertilizer plant’s bright platinum-rhodium gauze ignites ammonia, turning it stepwise into nitric acid – the heart of the Ostwald process.”

NH₃ is first oxidised to NO over a Pt–Rh gauze at ~900 °C, then NO is oxidised to NO₂ and absorbed in water to give HNO₃. The high temperature and catalyst make the otherwise slow oxidation feasible. Trap: NO forms first, not HNO₃ directly.

Memory trick: Pt-Rh spark → NH₃ → NO → HNO₃

ChemistryHydrogen· Class 11

Hydrogen peroxide

“When you clean a kitchen countertop, the fizz you see is H₂O₂ breaking into water and oxygen.”

Hydrogen peroxide is a pale-blue liquid and a strong oxidising agent used as bleach and antiseptic. It readily decomposes: 2 H₂O₂ → 2 H₂O + O₂, releasing oxygen. In redox questions remember H₂O₂ can act as oxidiser or reducer; overlooking this flips the half-reaction.

H₂O₂

Memory trick: H2O2 = H2O + extra O → oxidiser

ChemistryHydrogen· Class 11

Hard vs soft water

“Rinsing a glass bottle and seeing a cloudy ring shows Ca²⁺/Mg²⁺ deposits – the same ions that make tap water hard.”

Hard water contains high Ca²⁺ and Mg²⁺; they react with soap to form insoluble salts, so lather drops. Hardness is expressed as mg CaCO₃ per litre: 1 mmol Ca²⁺ (or Mg²⁺) = 100 mg CaCO₃, and 1 meq L⁻¹ = 50 mg CaCO₃ L⁻¹. Remember CaCO₃’s equivalent weight (50 g eq⁻¹) – confusing mmol with mg or ignoring the valence factor gives wrong hardness values.

Memory trick: Hardness → CaCO₃: 1 mmol =100 mg, 1 meq =50 mg.

ChemistryThe p-Block Elements· Class 12

Interhalogen compounds

“ICl behaves like a mixed-flavor ice-cream, two different halogens blend to give a polar molecule that reacts more readily than pure iodine.”

Interhalogen molecules contain two different halogens (e.g., ICl, BrF₅). The bond is polar because of the electronegativity difference, making them often more reactive than the heavier parent halogen. Their reactivity depends on the specific pair; they readily hydrolyze, giving HX and halogen oxides. A common trap: assuming all interhalogens are stronger oxidizers than Cl₂ – check the actual reaction conditions.

Memory trick: Inter-halogen = INter-different → polarity ↑ reactivity ↑

ChemistryCoordination Compounds· Class 12

Crystal field splitting

“The vivid blue of a copper(II) sulfate solution comes from d-orbital splitting of Cu²⁺ by water ligands.”

In an octahedral field ligands split the five d-orbitals into lower-energy t₂g and higher-energy e_g sets; tetrahedral complexes reverse the order (e lower, t₂ higher) and the splitting Δ_t is only 4/9 of Δ₀. Remember that Δ and pairing energy together decide high- or low-spin – a common exam slip is to ignore Δ_t’s smaller size.

Memory trick: Octa: t₂g<e_g; Tet: e<t₂; Δ_t=4/9Δ₀

ChemistryCoordination Compounds· Class 12

Spectrochemical series

“Just as spices are ranked from mild to hot, ligands are ordered from weak to strong field in the spectrochemical series.”

The spectrochemical series orders ligands by the crystal-field strength they generate, affecting Δ₀. Weak-field ligands (I⁻, Br⁻) give small Δ, strong-field ligands (CN⁻, CO) give large Δ. Exam trap: remember the series follows observed Δ values, not alphabetical order.

Memory trick: I⁻<Br⁻<Cl⁻<F⁻<H₂O<NH₃<NO₂⁻<CN⁻

ChemistryCoordination Compounds· Class 12

Isomerism in complexes

“Identical sauce bottles on a rack, one holding ketchup, another mustard, illustrate how complexes with the same formula can have different ligands attached.”

Complexes exhibit structural isomerism (ionization, coordination, solvate/hydrate, linkage) where connectivity changes, and stereoisomerism (geometrical, optical) where spatial arrangement changes. Optical activity demands true chirality – no mirror plane, inversion centre or improper rotation axis. A frequent trap is to mention only geometrical, optical and linkage isomers, omitting the other structural types.

Memory trick: I C S L G O – Ionization, Coordination, Solvate, Linkage, Geometrical, Optical

ChemistryAldehydes, Ketones & Carboxylic Acids· Class 12

Grignard reagent

“Like a Lego brick snapping onto a carbonyl’s ‘sticky’ carbon, a Grignard adds its R group, later giving an alcohol after acid work-up.”

Grignard reagents (R–MgX) attack the electrophilic carbon of aldehydes and ketones, giving after H⁺ work-up a secondary or tertiary alcohol. They do NOT react with carboxylic acids (protonate MgX) and with esters they add twice, yielding a tertiary alcohol. The exam trap is to assume every carbonyl gives a single-addition product.

Memory trick: Grignard + COOH = NO; + Ester = 2-additions; + Ald/Ket = 1-addition

ChemistryHydrocarbons· Class 11

Ozonolysis

“When ozone passes over a double-bonded rubber band, it snaps the bond, leaving two separate ends that become carbonyl groups.”

Ozonolysis adds O₃ across a C=C bond, forming an unstable ozonide that, after reductive or oxidative work-up, splits into carbonyl compounds (aldehydes or ketones) at each former double-bond carbon. It reveals the position of unsaturation.

Memory trick: Cuts at the C=C.

ChemistryAldehydes, Ketones & Carboxylic Acids· Class 12

Tollens' test

“When you dip a clean silver spoon into an aldehyde solution, the surface shines back like a mirror.”

In Tollens’ test Ag(NH₃)₂⁺ is reduced to metallic Ag, depositing a silver mirror. Aldehydes do this readily; α-hydroxy ketones, some sugars and enediols also reduce the reagent, while most ketones do not. The common mistake is to mark any silver mirror as proof of an aldehyde only.

Memory trick: Ag⁺↓ with aldehyde or α-OH carbonyl → mirror

ChemistryAldehydes, Ketones & Carboxylic Acids· Class 12

Iodoform test

“Adding iodine to a bottle of rubbing alcohol makes a yellow, fluffy precipitate – the classic iodoform cloud.”

The iodoform test is positive for compounds containing a CH₃-CO- group. Ethanol and secondary alcohols of type CH₃-CH(OH)-R are first oxidised to a methyl ketone, then three iodinations and base-induced cleavage give yellow CHI₃. Assuming only methyl ketones work misses ethanol and oxidisable secondary alcohols.

Memory trick: CH₃CO- or CH₃CH(OH)- → CHI₃ yellow

ChemistryAldehydes, Ketones & Carboxylic Acids· Class 12

Esterification is reversible

“Esterification is like a seesaw: acid + alcohol on one side, ester + water on the other; removing water tips it toward ester.”

Acid-catalyzed esterification protonates the carbonyl, alcohol attacks, and water is eliminated, giving ester. The reaction is an equilibrium; Le Chatelier’s principle says removing water or using excess alcohol drives it forward. Forgetting to shift the balance yields low ester yield.

Memory trick: Ester ⇌ acid+alcohol; pull by removing H₂O

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