Ch 1Metallurgy
5-Mark Questions
What are the differences between minerals and ores?
Mineral: naturally occurring chemical compound or element; may or may not be of economic value. Ore: a mineral (or mixture) from which a metal can be profitably extracted.
What are the various steps involved in the extraction of pure metals from their ores?
Concentration → Calcination/Roasting → Reduction/Smelting → Refining.
2-Mark Questions
What is the role of quick lime in the extraction of Iron from its oxide Fe2O3?
Quicklime (CaO) acts as a flux to form a removable slag with acidic impurities (silica).
Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.
Sulfide (and other hydrophobic) ores, e.g. galena (PbS), sphalerite (ZnS), chalcopyrite (CuFeS2).
1-Mark Questions (MCQ)
Write a short note on electrochemical principles of metallurgy.
Electrochemical metallurgy uses electrode potentials and cell EMFs to predict and carry out extraction/refining of metals by electrolysis or electrowinning.
Ch 2p-Block Elements-I
5-Mark Questions
Write a short note on anomalous properties of the first element of p-block.
Anomalous properties of boron (first p-block element)
Describe briefly allotropism in p- block elements with specific reference to carbon.
Allotropy in p-block elements — example of carbon
2-Mark Questions
Give one example for each of the following (i) icosagens (ii) tetragens (iii) pnictogens (iv) chalcogens
(i) Icosagen: Boron (B) (ii) Tetragen: Carbon (C) (iii) Pnictogen: Nitrogen (N) (iv) Chalcogen: Oxygen (O)
1-Mark Questions (MCQ)
Match items in column - I with the items of column - II and assign the correct code. Column-I: A Borazole B(OH)3 B Boric acid B3N3H6 C Quartz Na2[B4O5(OH)4]·8H2O D Borax SiO2
A–B3N3H6, B–B(OH)3, C–SiO2, D–Na2[B4O5(OH)4]·8H2O
Ch 3p-Block Elements-II
5-Mark Questions
What is inert pair effect?
Inert pair effect is the tendency of the two electrons in the outermost ns2 subshell of heavy p‑block elements to remain non‑ionized or non‑shared, causing lower than expected oxidation states.
Chalcogens belongs to p-block. Give reason.
Chalcogens (group 16 elements) belong to the p‑block because their valence electrons occupy the ns2 np4 configuration; the outermost electrons are in p orbitals.
2-Mark Questions
Give the uses of helium.
Common uses of helium include: Filling balloons and airships (non-flammable, lighter-than-air gas). Breathing gas mixtures for deep-sea diving to avoid nitrogen narcosis. Liquid helium for cryogenics and cooling superconducting magnets (e.g., MRI machines). Inert shielding gas in welding and in semiconductor production. Leak detection and as an inert atmosphere for manufacture of reactive metals.
Give the uses of argon.
Uses of argon include: Inert shielding gas in arc welding (prevents oxidation of the weld). Filling incandescent and fluorescent lamps to prevent filament oxidation and prolong life. Providing an inert atmosphere in the production of reactive metals and in semiconductor manufacturing. Used in gas-discharge tubes and as an inert blanket in laboratory and industrial processes.
Write the valence shell electronic configuration of group-15 elements.
ns2 np3 (e.g. N: 2s2 2p3; P: 3s2 3p3; As: 4s2 4p3; Sb: 5s2 5p3; Bi: 6s2 6p3).
1-Mark Questions (MCQ)
Explain why fluorine always exhibit an oxidation state of -1?
-1
Ch 4Transition and Inner Transition Elements
5-Mark Questions
What are transition metals? Give four examples.
Transition metals are d-block elements whose atoms or common ions have incompletely filled d-subshells. Examples: Fe, Co, Ni, Cr.
Explain the oxidation states of 4d series elements.
4d-series elements show a range of oxidation states from +1 to +6 (rarely up to +7/+8 for some species), with lower oxidation states more common for the heavier members.
2-Mark Questions
Write the electronic configuration of Ce4+ and Co2+.
Ce4+: [Xe]. Co2+: [Ar] 3d7.
Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
Copper has electronic configuration 3d 10 4s¹. It can easily lose 4s¹ electron to give the stable 3d 10 configuration. Hence, it shows +1 oxidation state.
1-Mark Questions (MCQ)
Why do transition metals show high melting points?
Because of strong metallic bonding from delocalised d and s electrons and many unpaired d electrons leading to high cohesive energy.
Ch 5Coordination Chemistry
5-Mark Questions
Write the IUPAC names for the following complexes. i) Na[Ni(EDTA)] ii) [Ag(CN)2]- iii) [Co(en)2(SO4)?] (?) iv) [Co(ONO)(NH3)5]2+ v) [Pt(NH3)Cl(NO2)] (forms shown in source OCR)
i) sodium nickel(II) ethylenediaminetetraacetate (Na[Ni(EDTA)]) ii) dicyanidoargentate(I) ([Ag(CN)2]−) iii) sulfato-bis(ethylenediamine)cobalt(III) (sulfato-bis(ethylenediamine)cobalt(III) complex) iv) pentaamminenitrito-Ocobalt(III) ion ([Co(NH3)5(ONO)]2+) v) amminechloridonitroplatinum(II) (e.g. [Pt(NH3)Cl(NO2)])
Write the formula for the following coordination compounds. a) Potassiumhexacyanidoferrate(II) b) Pentacarbonyliron(0) c) Pentaamminenitrito-κNcobalt(III)ion d) Hexaamminecobalt(III)sulphate e) Sodiumtetrafluoridodihydroxidochromate(III)
a) K4[Fe(CN)6] b) Fe(CO)5 c) [Co(NH3)5(NO2)]2+ d) [Co(NH3)6]2(SO4)3 e) Na3[CrF4(OH)2]
2-Mark Questions
Give an example of coordination compound used in medicine and two examples of biologically important coordination compounds.
Medicine: cisplatin, Pt(NH3)2Cl2. Biologically important: heme (iron porphyrin in haemoglobin) and chlorophyll (magnesium porphyrin); also vitamin B12 (cobalamin).
In an octahedral crystal field, show the splitting of the five d orbitals.
In an octahedral field the five d orbitals split into two sets: the lower‑energy t2g set (dxy, dxz, dyz) and the higher‑energy eg set (dz2, dx2−y2). The t2g orbitals are stabilized by −0.4Δo each and the eg orbitals are destabilized by +0.6Δo each; the energy gap between the sets is Δo.
What is the coordination entity formed when excess of liquid ammonia is added to an aqueous solution of copper sulphate?
[Cu(NH3)4(H2O)2]^{2+} (tetraamminecopper(II) complex, deep blue).
1-Mark Questions (MCQ)
What is linkage isomerism? Explain with an example.
Linkage isomerism arises when an ambidentate ligand can bind through two different donor atoms. Example: nitrite ion NO2− binds via N (nitro) or via O (nitrito): [Co(NH3)5(NO2)]2+ (nitro, N‑bound) vs [Co(NH3)5(ONO)]2+ (nitrito, O‑bound).
Ch 6Solid State
5-Mark Questions
Explain briefly seven types of unit cell.
The seven crystal systems (types of unit cells): cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, trigonal (rhombohedral).
Distinguish between hexagonal close packing and cubic close packing.
hcp stacking sequence = ABAB...; ccp (fcc) stacking sequence = ABCABC.... Both have packing efficiency 74% and coordination number 12, but unit cells differ: hcp has 6 atoms per conventional cell, ccp (fcc) has 4 atoms per unit cell.
2-Mark Questions
Define unit cell.
A unit cell is the smallest repeating structural unit of a crystal lattice that, by translation in three dimensions, reproduces the entire crystal. It is defined by the edge lengths a, b, c and interaxial angles α, β, γ and contains the arrangement of atoms for the crystal.
Give any three characteristics of ionic crystals.
1. High melting and boiling points due to strong electrostatic forces. 2. Hard but brittle (cleave along planes). 3. Conduct electricity only in molten state or in solution (ions mobile); poor conductors as solids.
Differentiate crystalline solids and amorphous solids.
Crystalline solids: long-range order, definite geometric shape, sharp melting points, anisotropic. Amorphous solids: no long-range order, no definite shape, show glass transition rather than sharp melting, isotropic.
1-Mark Questions (MCQ)
Write a note on Frenkel defect.
Frenkel defect: cation vacancy–interstitial pair; stoichiometry unchanged.
Ch 7Chemical Kinetics
5-Mark Questions
Define average rate and instantaneous rate.
Average rate: change in concentration of a reactant or product divided by time interval Δt: average rate = −Δ[A]/Δt (for reactant) or +Δ[P]/Δt (for product). Instantaneous rate: rate at a specific time t; mathematically the derivative: instantaneous rate = −d[A]/dt (or d[P]/dt) evaluated at that time.
Define rate law and rate constant.
Rate law: an expression relating the reaction rate to the concentrations of reactants (and sometimes products or catalysts), e.g. rate = k [A]^m [B]^n, where m,n are orders determined experimentally. Rate constant (k): proportionality constant in the rate law; depends on temperature and catalyst but not on reactant concentrations; units depend on overall order.
2-Mark Questions
Describe the graphical representation of first order reaction.
For first order: [A] = [A]0 e^{-kt}. Plots: (i) ln[A] vs t is a straight line with slope −k and intercept ln[A]0. (ii) log10[A] vs t is straight with slope −k/2.303. (iii) [A] vs t is an exponential decay curve. From ln[A] vs t one obtains k from slope.
Write the rate law for the following reactions. (a) A reaction that is 3/2 order in x and zero order in y. (b) A reaction that is second order in NO and first order in Br2.
(a) rate = k [x]^{3/2} (since zero order in y means [y]^0 = 1). (b) rate = k [NO]^2 [Br2]^1 = k [NO]^2 [Br2].
The rate of formation of a dimer in a second order reaction is 7.5 × 10^{-5} mol L^{-1} s^{-1} at 0.05 mol L^{-1} monomer concentration. Calculate the rate constant.
k = 0.03 L mol^{-1} s^{-1}
1-Mark Questions (MCQ)
The rate law for a reaction of A, B and L has been found to be rate = k [A][B][L]. How would the rate of reaction change when (i) Concentration of [L] is quadrupled (ii) Concentration of both [A] and [B] are doubled (iii) Concentration of [A] is halved (iv) Concentration of [A] is reduced to 1/4 and concentration of [L] is quadrupled.
(i) 4 times (ii) 4 times (iii) 1/2 times (iv) unchanged (1 times).
Ch 8Ionic Equilibrium
5-Mark Questions
Discuss the Lowry–Bronsted concept of acids and bases.
Bronsted–Lowry acid: proton (H+) donor. Bronsted–Lowry base: proton acceptor. Reactions occur as acid + base ⇌ conjugate base + conjugate acid.
Identify the conjugate acid–base pairs for the following reactions in aqueous solution: (i) HS−(aq) + HF ⇌ F−(aq) + H2S(aq) (ii) HPO4^2− + SO4^2− ⇌ PO4^3− + HSO4− (iii) NH3 + HCO3− ⇌ NH4+ + CO3^2−
(i) HF/F− and H2S/HS− (ii) HPO4^2−/PO4^3− and HSO4−/SO4^2− (iii) NH4+/NH3 and HCO3−/CO3^2−
2-Mark Questions
What are Lewis acids and bases? Give two examples for each.
Lewis acid: electron-pair acceptor. Examples: BF3, AlCl3. Lewis base: electron-pair donor. Examples: NH3, OH−.
When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetramminecopper(II) complex, [Cu(H2O)4]2+ + 4NH3 ⇌ [Cu(NH3)4]2+ + 4H2O. Between H2O and NH3 which is the stronger Lewis base?
NH3 is the stronger Lewis base.
The concentration of hydroxide ion in a water sample is found to be 2.5 × 10^-6 M. Identify the nature of the solution.
Basic (pH ≈ 8.40).
1-Mark Questions (MCQ)
Ksp of Al(OH)3 is 1.0×10^-33. At what pH does 1.0×10^-3 M Al3+ precipitate on the addition of buffer of NH4Cl and NH4OH solution?
pH = 4.00 (precipitation occurs when pH > 4.00)
Ch 9Electro Chemistry
5-Mark Questions
Define anode and cathode
Anode: electrode where oxidation occurs (electrons are produced). Cathode: electrode where reduction occurs (electrons are consumed). In a galvanic cell the anode is negative and the cathode is positive; in an electrolytic cell polarities are reversed.
Why does conductivity of a solution decrease on dilution of the solution?
Specific conductance κ depends on ionic concentration and mobility. On dilution the number of charge carriers per unit volume decreases, so κ decreases (even though molar conductivity increases due to reduced inter-ionic interactions).
2-Mark Questions
A copper electrode is immersed in 0.10 M CuSO4 at 25°C. Calculate the electrode potential of the copper electrode. (E°(Cu2+/Cu) = +0.34 V.)
E = 0.310 V (approximately).
1-Mark Questions (MCQ)
Ionic conductances at infinite dilution of Al3+ and SO42− are 189 and 160 mho·cm2·equiv−1 respectively. Calculate the equivalent and molar conductance at infinite dilution of the electrolyte Al2(SO4)3.
Equivalent conductance Λ°eq = 349 mho·cm2·equiv−1. Molar conductance Λ°m = 2094 mho·cm2·mol−1.
Ch 10Surface Chemistry
5-Mark Questions
Give two important characteristics of physisorption
(i) It involves weak van der Waals forces and is usually reversible. (ii) Heat of adsorption is low and multilayer adsorption is possible.
Differentiate physisorption and chemisorption
Key differences: physisorption — weak van der Waals forces, low ΔH, reversible, multilayer, favoured at low T. Chemisorption — chemical bond formation, high ΔH, often irreversible, monolayer, may require activation energy and favoured at moderate T.
2-Mark Questions
Which will be adsorbed more readily on the surface of charcoal and why? NH3 or O2 ?
NH3 will be adsorbed more readily.
What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?
Mutual coagulation (precipitation) occurs due to neutralization of opposite surface charges.
Give three uses of emulsions.
Uses: food products (milk, mayonnaise), pharmaceuticals/cosmetics (creams, lotions), and paints/coatings or insecticide formulations.
1-Mark Questions (MCQ)
Describe adsorption theory of catalysis.
Adsorption theory: catalysis occurs because reactant molecules are adsorbed on the catalyst surface, which brings them together, weakens bonds, provides active centres and proper orientation, thereby lowering activation energy and increasing reaction rate.
Ch 11Hydroxy Compounds and Ethers
5-Mark Questions
Identify the product (s) is / are formed when 1 - methoxy propane is heated with excess HI. Name the mechanism involved in the reaction
Methyl iodide and 1-propanol (and with excess HI and heat, 1-iodopropane). Mechanism: protonation of ether followed by SN2 (and further SN2 on the alcohol with excess HI).
Predict the major product, when 2-methyl but -2-ene is converted into an alcohol in each of the following methods. (i.) Acid catalysed hydration (ii.) Hydroboration (iii.) Hydroxylation using Baeyer's reagent
(i) 2‑Methylbutan‑2‑ol (Markovnikov addition, OH at C‑2). (ii) 2‑Methylbutan‑3‑ol (anti‑Markovnikov addition, OH at the less substituted carbon C‑3). (iii) 2‑Methylbutane‑2,3‑diol (vicinal diol at C‑2 and C‑3).
2-Mark Questions
Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent of HI
Propanal (CH3CH2CHO) and ethyl iodide (C2H5I) — major carbonyl product: propanal.
Suggest a suitable reagent to prepare secondary alcohol with identical group using Grignard reagent.
React a Grignard reagent R–MgX with an aldehyde of the same R group, R–CHO, followed by acid workup: R–MgX + R–CHO → R–CH(OMgX)–R → (H^+/H2O) → R–CH(OH)–R.
What is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis.
Tertiary alcohol: phenyl(ethyl)2-carbinol i.e. Ph–C(OH)(Et)2 (a tertiary alcohol obtained by double addition).
1-Mark Questions (MCQ)
3,3‑Dimethylbutan‑2‑ol on treatment with conc. H2SO4 gives tetramethyl ethylene as a major product. Suggest a suitable mechanism.
Mechanism: acid‑catalyzed dehydration (E1). Protonation of OH → loss of H2O to give a secondary carbocation; a methyl shift (1,2‑shift) from the gem‑dimethyl center produces a more stable tertiary carbocation; deprotonation of the tertiary carbocation gives the highly substituted alkene (tetramethyl ethylene, i.e. (CH3)2C=C(CH3)2).
Ch 12Carbonyl Compounds and Carboxylic Acids
5-Mark Questions
Identify X and Y. \(CH_3COCH_2CH_2COOC_2H_5 \xrightarrow{CH_3MgBr} X \xrightarrow{H_3O^+} Y\)
Starting compound is a β‑keto ester (ethyl 4‑oxobutanoate, shown as CH3COCH2CH2COOEt). Addition of methyl Grignard (1 equiv.) attacks the ketone carbonyl to give the magnesium alkoxide intermediate X; acidic workup gives the corresponding tertiary/secondary alcohol Y.
Identify A, B, C and D: ethanoic acid --A--> ? --Pd/BaSO4--> B --NaOH--> C --SOCl2--> D (identify intermediates/products)
One convenient sequence (common textbook transformations): (1) D = CH3COCl (acetyl chloride) obtained from ethanoic acid by SOCl2: \(CH3COOH + SOCl2 \to CH3COCl + SO2 + HCl\). (2) Acetyl chloride (D) on Rosenmund reduction (H2/Pd–BaSO4) gives acetaldehyde (A): \(CH3COCl \xrightarrow{H2, Pd/BaSO4} CH3CHO\). (3) Acetaldehyde (A) on treatment with NaOH undergoes aldol condensation to give the aldol product (B) = 3‑hydroxybutanal which on heating dehydrates to crotonaldehyde (C). …
2-Mark Questions
Identify A, B and C: benzoic acid --PCl5--> A; Benzene + Anhydrous AlCl3 + A --> B; H+ + C2H5OH on B gives C; C6H5MgBr ...
benzoic acid --(PCl5)--> A = benzoyl chloride (C6H5COCl). Benzene + benzoyl chloride/AlCl3 --> B = benzophenone (C6H5COC6H5) (Friedel–Crafts acylation). B + H^+ / C2H5OH (acidic ethanol) converts the ketone to its acetal (diethyl acetal): C = benzophenone diethyl acetal, (C6H5)2C(OC2H5)2. (Alternatively, protonation/acetalization of the ketone with ethanol produces the corresponding ketal/acetal.)
What is the action of HCN on (i) propanone (ii) 2,4-dichlorobenzaldehyde (iii) ethanal
HCN adds to carbonyl compounds to give cyanohydrins. (i) Propanone (acetone): \((CH3)2CO + HCN \to (CH3)2C(OH)CN\) (acetone cyanohydrin, 2‑hydroxy‑2‑methylpropanenitrile). (ii) 2,4‑Dichlorobenzaldehyde: \(Cl2C6H3CHO + HCN \to Cl2C6H3CH(OH)CN\) (the corresponding aromatic cyanohydrin). (iii) Ethanal (acetaldehyde): \(CH3CHO + HCN \to CH3CH(OH)CN\) (acetaldehyde cyanohydrin, 2‑hydroxypropanenitrile). Cyanohydrins can be further hydrolysed to α‑hydroxy acids on acidic hydrolysis.
A carbonyl compound A having molecular formula C5H10O forms crystalline precipitate with sodium bisulphite and gives positive iodoform test. A does not reduce Fehling solution. Identify A.
Positive iodoform and no Fehling reduction identifies a methyl ketone (CH3CO–R). A C5H10O methyl ketone is pentan‑2‑one (CH3COCH2CH2CH3) or pentan‑2‑one (also written 2‑pentanone). 2‑pentanone forms a bisulphite adduct (bisulphite soluble) and gives iodoform. Therefore A = pentan‑2‑one (2‑pentanone).
1-Mark Questions (MCQ)
How is propanoic acid is prepared starting from (a) an alcohol (b) an alkylhalide (c) an alkene
See solution
Ch 13Organic Nitrogen Compounds
5-Mark Questions
Write down the possible isomers of the C3H7NO2 and give their IUPAC names.
Isomers of C3H7NO2 (common constitutional isomers — amino acids): 1) 2‑Aminopropanoic acid (alanine) — CH3–CH(NH2)–COOH. 2) 3‑Aminopropanoic acid (β‑alanine) — NH2–CH2–CH2–COOH.
There are two isomers with the formula C2H5NO2. How will you distinguish between them?
Two constitutional isomers of C2H5NO2 are nitroethane (CH3CH2NO2) and aminoacetic acid (glycine, NH2CH2COOH). Distinguishing tests: (1) Ninhydrin test: glycine (an amino acid) gives a purple colour with ninhydrin, nitroethane does not. (2) Reaction with NaNO2/HCl (nitrous acid): the primary amine in glycine reacts (deamination) whereas nitroethane (a nitro compound) is inert to nitrous acid under these conditions. (3) Acid–base behaviour: glycine shows zwitterionic behaviour and forms salts with base/acid; nitroethane does not behave as an amino acid.
2-Mark Questions
How will you convert diethylamine into i) N,N‑diethylacetamide ii) N‑nitrosodiethylamine?
i) Diethylamine (Et2NH) + acetyl chloride (CH3COCl) or acetic anhydride → N,N‑diethylacetamide (Et2N–COCH3) via acylation (Schotten–Baumann conditions or pyridine). ii) N‑nitrosation: Diethylamine treated with NaNO2/HCl at 0–5 °C (or with nitrosyl chloride) gives N‑nitrosodiethylamine (Et2N–NO).
Identify A in: aniline + benzaldehyde → A → ? (Schiff base formation).
Aniline (PhNH2) reacts with benzaldehyde (PhCHO) to give an imine (Schiff base): N‑benzylideneaniline (Ph–CH=N–Ph) (A).
1-Mark Questions (MCQ)
Account for the following: i. Aniline does not undergo Friedel–Crafts reaction. ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. iii. pKb of aniline is more than that of methylamine. iv. Gabriel phthalimide synthesis is preferred for synthesising primary amines. v. Ethylamine is soluble in water whereas aniline is not. vi. Amines are more basic than amides. vii. Although –NH2 is o‑ and p‑ directing, nitration of aniline gives a substantial amount of m‑nitroaniline.
Short accounts: i) Aniline is strongly deactivated toward Friedel–Crafts because the –NH2 coordinates to Lewis acids (AlCl3), forming anilinium salt or complex; this removes the lone pair from resonance donation, preventing the electrophilic substitution and often destroying the catalyst. ii) Aromatic diazonium salts (Ar–N2+) are stabilized by resonance with the aromatic ring; aliphatic diazonium ions lack such resonance and readily decompose, so they are unstable. …
Ch 14Biomolecules
5-Mark Questions
Give the differences between primary and secondary structure of proteins.
Primary: amino acid sequence linked by peptide bonds. Secondary: local folding (α‑helix, β‑sheet) stabilized by H‑bonds between backbone C=O and N‑H.
Write a short note on peptide bond
Peptide bond is an amide linkage formed between the carboxyl of one amino acid and the amino group of another with elimination of water; it has partial C–N double‑bond character and is planar.
2-Mark Questions
What type of linkages hold together monomers of DNA?
Phosphodiester linkages (and N‑glycosidic bonds between sugar and base).
Name the Vitamins whose deficiency cause i) rickets ii) scurvy
i) Vitamin D ii) Vitamin C (ascorbic acid)
Write the Zwitter ion structure of alanine
Zwitterion: H3N+–CH(CH3)–COO−
1-Mark Questions (MCQ)
Is the following sugar, D - sugar or L - sugar? CHO H OH H OH H OH CH2 - OH
D-sugar
Ch 15Chemistry in Everyday Life
5-Mark Questions
Write a note on synthetic detergents
Synthetic detergents are surfactants (anionic, cationic, nonionic or amphoteric) produced from petrochemical or oleochemical feedstocks; they lower surface tension and emulsify oils and dirt.
How do antiseptics differ from disinfectants?
Antiseptics are safe for use on living tissues to prevent infection; disinfectants are stronger agents used on inanimate surfaces and usually harmful to living tissues.
2-Mark Questions
What are antibiotics?
Antibiotics are substances (natural or synthetic) that kill or inhibit the growth of microorganisms, especially bacteria, at low concentrations. They can be bactericidal or bacteriostatic; examples include penicillin and streptomycin.
Name one substance which can act as both analgesic and antipyretic
Aspirin (acetylsalicylic acid) or paracetamol (acetaminophen).
1-Mark Questions (MCQ)
Which one of the following structures represents nylon 6,6 polymer? (intended answer: repeating unit of nylon-6,6)
d
Frequently asked questions
- What are the differences between minerals and ores?
- Mineral: naturally occurring chemical compound or element; may or may not be of economic value. Ore: a mineral (or mixture) from which a metal can be profitably extracted.
- What are the various steps involved in the extraction of pure metals from their ores?
- Concentration → Calcination/Roasting → Reduction/Smelting → Refining.
- What is the role of quick lime in the extraction of Iron from its oxide Fe2O3?
- Quicklime (CaO) acts as a flux to form a removable slag with acidic impurities (silica).
- Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.
- Sulfide (and other hydrophobic) ores, e.g. galena (PbS), sphalerite (ZnS), chalcopyrite (CuFeS2).
These important questions are selected from the Samacheer Kalvi Class 12 Chemistry textbook book-back exercises to help you revise the most useful questions. Mark weightage (5/2/1) follows the usual exam pattern and may vary by exam — always check your latest syllabus and question pattern. Open each chapter for the complete set of questions and answers.