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BITSAT Chemistry Flashcards

51 question-and-answer cards covering Chemistry as it is examined in BITSAT. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.

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24 sample cards from the Chemistry deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

  1. Distinguish Schottky and Frenkel point defects.

    Schottky: equal numbers of cations and anions missing; lowers density (e.g. NaCl). Frenkel: a cation dislocated to an interstitial site; density unchanged (e.g. AgCl, ZnS).

  2. Differentiate physisorption from chemisorption.

    Physisorption: weak van der Waals forces, reversible, low heat (20-40 kJ/mol), multilayer, decreases with temperature. Chemisorption: chemical bonds, irreversible, high heat (80-240 kJ/mol), monolayer, specific.

  3. What is the Tyndall effect and what distinguishes a colloid?

    The Tyndall effect is scattering of light by colloidal particles, making the beam path visible. Colloids have particle sizes 1-1000 nm, between true solutions and suspensions.

  4. State the general periodic trends for atomic radius and ionization energy across a period.

    Across a period (left to right), atomic radius decreases and ionization energy increases, due to increasing nuclear charge with same shell. Down a group, both reverse: radius increases, ionization energy decreases.

  5. Define electronegativity and name the most electronegative element.

    Electronegativity is the tendency of an atom to attract a shared pair of electrons in a bond. Fluorine is the most electronegative element (4.0 on the Pauling scale).

  6. State VSEPR-predicted shapes for 2, 3, 4, 5, and 6 electron pairs (no lone pairs).

    2: linear; 3: trigonal planar; 4: tetrahedral; 5: trigonal bipyramidal; 6: octahedral.

  7. Explain hybridization for sp, sp2, and sp3 with example geometries.

    sp: linear, 180 deg (BeCl2, acetylene). sp2: trigonal planar, 120 deg (BF3, ethene). sp3: tetrahedral, 109.5 deg (CH4, ethane).

  8. What is the bond order formula in molecular orbital theory, and is O2 paramagnetic?

    Bond order = (bonding electrons - antibonding electrons)/2. O2 has bond order 2 and is paramagnetic due to two unpaired electrons in pi-antibonding orbitals.

  9. What is diagonal relationship, with an example in the s-block?

    Diagonal relationship is the similarity between an element and the element diagonally placed in the next group/period, due to similar charge/radius ratio. Example: Li resembles Mg; Be resembles Al.

  10. Why is hydrogen placed ambiguously in the periodic table?

    Hydrogen resembles alkali metals (1 valence electron, forms H+) but also halogens (needs 1 electron to fill shell, forms H-, diatomic gas). So its placement is anomalous.

  11. What is the inert pair effect and where is it observed?

    The reluctance of the ns2 electrons to participate in bonding in heavier p-block elements, making lower oxidation states more stable (e.g. Pb2+ more stable than Pb4+, Tl+ over Tl3+).

  12. Why does nitrogen not form pentahalides while phosphorus does?

    Nitrogen has no available d-orbitals in its valence shell (only 2s and 2p), so it cannot expand its octet beyond 4 bonds; phosphorus has empty 3d orbitals and can form PCl5.

  13. What is lanthanoid contraction and one of its consequences?

    The steady decrease in atomic/ionic size across the lanthanoids due to poor shielding by 4f electrons. Consequence: Zr and Hf have nearly identical sizes, making them hard to separate.

  14. Why do transition metals show variable oxidation states and form coloured compounds?

    Variable oxidation states arise from similar energies of (n-1)d and ns electrons. Colour arises from d-d electronic transitions in partially filled d-orbitals.

  15. State Werner's coordination theory terms: ligand, coordination number, denticity.

    A ligand is an electron-pair donor bonded to the central metal. Coordination number is the number of donor atoms bonded to the metal. Denticity is the number of donor sites of one ligand (mono-, bi-, polydentate).

  16. In crystal field theory, how do octahedral and tetrahedral splittings differ?

    Octahedral: d-orbitals split into lower t2g (3) and higher eg (2). Tetrahedral: split into lower e (2) and higher t2 (3), with smaller splitting (delta-t = 4/9 delta-o), so tetrahedral complexes are usually high spin.

  17. Name a strong-field and a weak-field ligand from the spectrochemical series.

    Strong-field (large splitting, low spin): CO, CN-, NH3. Weak-field (small splitting, high spin): I-, Br-, Cl-, F-, H2O. Order: I- < Br- < Cl- < F- < H2O < NH3 < en < CN- < CO.

  18. What is the difference between roasting and calcination in metallurgy?

    Roasting: heating sulphide ore in excess air to convert it to oxide (e.g. ZnS to ZnO). Calcination: heating carbonate/hydroxide ore in limited air to give oxide and drive off CO2/water (e.g. CaCO3 to CaO).

  19. What is the role of a depressant and a collector in froth flotation?

    Froth flotation concentrates sulphide ores. A collector (e.g. pine oil/xanthate) makes ore particles water-repellent so they rise with froth; a depressant (e.g. NaCN) prevents an unwanted sulphide from floating.

  20. What causes the greenhouse effect and name two major greenhouse gases.

    The greenhouse effect is trapping of infrared radiation by atmospheric gases, warming Earth. Major greenhouse gases include CO2, CH4 (methane), water vapour, N2O, and CFCs.

  21. State Markovnikov's rule and its anti (peroxide) effect.

    Markovnikov: in addition of HX to an unsymmetrical alkene, H adds to the carbon with more hydrogens (X to the more substituted carbon). With peroxides, HBr adds anti-Markovnikov (Kharasch/peroxide effect) via free radicals.

  22. Compare SN1 and SN2 reaction mechanisms.

    SN1: two steps via carbocation, first-order rate (depends only on substrate), favoured by tertiary halides and polar protic solvents, gives racemization. SN2: one step, second-order, favoured by primary halides and polar aprotic solvents, gives inversion of configuration.

  23. Distinguish enantiomers and diastereomers; define a chiral centre.

    A chiral centre is a carbon bonded to four different groups. Enantiomers are non-superimposable mirror images (same physical properties except optical rotation). Diastereomers are stereoisomers that are not mirror images (different physical properties).

  24. Name the test reactions to distinguish aldehydes from ketones.

    Aldehydes give positive Tollens' test (silver mirror) and Fehling's/Benedict's test (red Cu2O precipitate); ketones do not. Both give 2,4-DNP (orange-yellow) precipitate confirming a carbonyl group.

What this deck covers

The Chemistry deck follows the BITSAT Chemistry syllabus — 14 chapters and 64 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 3.6 cards per chapter.

Answers are written to be recallable, not just readable — averaging about 188 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 flashcards FAQ

How many Chemistry flashcards are in this BITSAT deck?

51 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.

Are these BITSAT flashcards free?

Yes. The preview here is free to read with no signup, and the full 51-card deck is free inside the Examius app.

What do the Chemistry cards cover?

They follow the BITSAT Chemistry syllabus — 14 chapters and 64 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.