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AFMC / NEET-linked AFMC Admission Chemistry (NEET-UG Academic Component) Flashcards
55 question-and-answer cards covering Chemistry (NEET-UG Academic Component) as it is examined in AFMC / NEET-linked AFMC Admission. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Chemistry (NEET-UG Academic Component) deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
Distinguish between sigma and pi bonds.
A sigma bond forms by head-on (axial) overlap of orbitals and is stronger; a pi bond forms by sidewise (lateral) overlap of p-orbitals and is weaker. Single bonds are sigma; double/triple bonds contain pi bonds.
Why does the s-block contain alkali and alkaline earth metals, and what is their general reactivity?
The s-block has groups 1 (alkali metals, ns1) and 2 (alkaline earth metals, ns2). They are highly reactive, electropositive metals; reactivity increases down each group.
What is the diagonal relationship, with an example from the s-block?
Certain elements in period 2 resemble the element diagonally placed in period 3, group adjacent. Example: lithium resembles magnesium, and beryllium resembles aluminium.
Why is the +1 oxidation state increasingly stable down group 13 and 14 (inert pair effect)?
The inert pair effect: the ns2 electrons become reluctant to participate in bonding down the group due to poor shielding by d and f electrons, stabilizing a lower oxidation state (e.g., Tl+1, Pb+2).
Define transition elements and explain why they show variable oxidation states.
Transition (d-block) elements have partially filled d-orbitals in atoms or common ions. They show variable oxidation states because the energies of (n-1)d and ns electrons are similar, so both can be lost.
Why are most transition metal compounds coloured?
Because of d-d electronic transitions: when d-orbitals split in a ligand field, electrons absorb visible light to jump between d levels, and the complementary colour is transmitted.
What are lanthanoid contraction and its consequence?
Lanthanoid contraction is the steady decrease in atomic/ionic size across the lanthanoid series due to poor shielding by 4f electrons. It causes second- and third-row transition elements (e.g., Zr and Hf) to have similar sizes.
State Werner's coordination theory in brief.
Metals show two valencies: primary (ionizable, satisfied by anions) and secondary (non-ionizable, satisfied by ligands, equal to coordination number). Secondary valencies are directional and fix the geometry.
State the rule for naming the coordination sphere and give the formula meaning of [Co(NH3)6]Cl3.
Ligands are named alphabetically before the metal, with the metal's oxidation state in Roman numerals. [Co(NH3)6]Cl3 is hexaamminecobalt(III) chloride: Co3+ with six ammine ligands, coordination number 6.
What is crystal field splitting in an octahedral complex?
In an octahedral field the five degenerate d-orbitals split into a lower t2g set (dxy, dyz, dxz) and a higher eg set (dx2-y2, dz2), separated by the crystal field splitting energy (delta_o).
State Markovnikov's rule for addition of HX to an unsymmetrical alkene.
The negative part (X) of the reagent adds to the carbon bearing fewer hydrogen atoms, while H adds to the carbon with more hydrogens, giving the more stable carbocation intermediate.
Define and order the stability of carbocations.
A carbocation is a positively charged carbon species. Stability order: tertiary > secondary > primary > methyl, due to increasing hyperconjugation and the +I (inductive) effect of alkyl groups.
Distinguish between SN1 and SN2 reaction mechanisms.
SN1 is two-step via a carbocation, first-order kinetics, favoured by tertiary substrates, and gives racemization. SN2 is one-step concerted, second-order kinetics, favoured by primary substrates, and proceeds with inversion of configuration.
What is Markovnikov's anti rule (peroxide/Kharasch effect)?
In the presence of peroxides, HBr adds to an unsymmetrical alkene against Markovnikov's rule (anti-Markovnikov) via a free-radical mechanism, placing Br on the carbon with more hydrogens.
Name the reactions for: (a) aldehyde/ketone + HCN, and (b) two molecules of aldehyde under dilute base.
(a) Nucleophilic addition forming a cyanohydrin. (b) Aldol condensation, giving a beta-hydroxy aldehyde/ketone that can dehydrate to an alpha,beta-unsaturated carbonyl.
What is the difference between an aldehyde and a ketone in terms of structure and oxidation?
An aldehyde has the carbonyl carbon bonded to at least one H (R-CHO) and is easily oxidized to a carboxylic acid. A ketone has the carbonyl carbon bonded to two carbons (R-CO-R') and resists mild oxidation.
Describe the Hofmann bromamide degradation reaction.
A primary amide is treated with Br2 and aqueous (or alcoholic) KOH/NaOH to give a primary amine with one fewer carbon atom: R-CONH2 to R-NH2.
Differentiate addition polymers from condensation polymers with one example each.
Addition polymers form by repeated addition of monomers with no by-product (e.g., polythene from ethene). Condensation polymers form by combination of monomers with loss of a small molecule like water (e.g., nylon-6,6).
Name the four major classes of biomolecules and their building blocks.
Carbohydrates (monosaccharides), proteins (amino acids), nucleic acids (nucleotides), and lipids (fatty acids and glycerol).
What are essential and non-essential amino acids?
Essential amino acids cannot be synthesized by the body and must be obtained from the diet; non-essential amino acids can be synthesized within the body.
Distinguish between BOD and COD as water-quality indicators.
BOD (Biochemical Oxygen Demand) is the oxygen used by microorganisms to decompose organic matter biologically. COD (Chemical Oxygen Demand) is the oxygen required to chemically oxidize all organic/inorganic matter; COD is usually higher than BOD.
Define green chemistry and one of its core goals.
Green chemistry is the design of chemical products and processes that reduce or eliminate the use and generation of hazardous substances; a core goal is preventing waste rather than treating it after formation.
How does Lassaigne's test detect nitrogen in an organic compound?
The compound is fused with sodium to convert nitrogen to sodium cyanide (NaCN); adding ferrous sulphate and ferric ions produces Prussian blue (ferric ferrocyanide), confirming nitrogen.
On what principle does chromatography separate components of a mixture?
Differential distribution (adsorption or partition) of components between a stationary phase and a moving mobile phase, so components travel at different rates and separate.
What this deck covers
The Chemistry (NEET-UG Academic Component) deck follows the AFMC / NEET-linked AFMC Admission Chemistry (NEET-UG Academic Component) syllabus — 4 chapters and 19 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 13.8 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 191 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Chemistry (NEET-UG Academic Component) flashcards FAQ
How many Chemistry (NEET-UG Academic Component) flashcards are in this AFMC / NEET-linked AFMC Admission deck?
55 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these AFMC / NEET-linked AFMC Admission flashcards free?
Yes. The preview here is free to read with no signup, and the full 55-card deck is free inside the Examius app.
What do the Chemistry (NEET-UG Academic Component) cards cover?
They follow the AFMC / NEET-linked AFMC Admission Chemistry (NEET-UG Academic Component) syllabus — 4 chapters and 19 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.