Notes and important questions/answers for Class 12 Chemistry for the upcoming MP Board half-yearly/annual examinations (2026-27) are provided here. In addition to Hindi, this website also offers notes and important questions/answers for subjects including English, Sanskrit, Physics, Chemistry (Science/Mathematics stream), Arts stream (History, Political Science, etc.), Commerce stream (Accountancy, Business Studies), and the Agriculture stream. Please ensure you check the website at least once a day leading up to the board examinations, as it will be updated regularly.
Do not wait for the links to be shared in your school groups; make a note of the website name/link yourself and access the study material directly on your mobile, tablet, or computer.
For Hindi medium: https://mpboardstudy.com/
For English medium: Visit www.mpboardenglishmedium.com
For Vocational Education: Visit www.vocationaleducation.in
कक्षा 12 Chemistry Notes/ Imp Q&Answers (MP Board English Medium) Half Yearly/Annual Exam 2026-27
Section A: 1 Mark Objective Questions (Total: 50 Questions)
1. Multiple Choice Questions (MCQs)
2. Fill in the Blanks
3. True or False
4. Match the Following
Q16 to Q20. Match the chemical reaction phenomena in Column A with their accurate parameters/names in Column B:
| Column A | Column B (Correct Matching Value) |
|---|---|
| 16. Lucas Reagent Composition | Anhydrous ZnCl2 + Concentrated HCl |
| 17. Rosenmund Reduction Catalyst | Pd on BaSO4 poisoned with Quinoline |
| 18. Hinsberg Reagent | Benzenesulfonyl chloride (C6H5SO2Cl) |
| 19. Kohlrausch’s Law Application | Determination of Λ0m for weak electrolytes |
| 20. Lanthanoid Contraction Cause | Imperfect shielding effect of 4f electrons |
5. One Word / One Sentence Answers
6. Short Conceptual & Core Lookups (Variants 26 to 50)
Q26. What is an azeotropic mixture?
Answer: A binary liquid mixture having a constant boiling point and distilling over completely without any change in composition.
Q27. Write the unit of cell constant for a conductivity cell.
Answer: cm-1 (or m-1).
Q28. What is the order of a reaction whose rate law is Rate = k[A]1[B]1?
Answer: Second Order (1 + 1 = 2).
Q29. Why do transition elements form a large number of interstitial compounds?
Answer: Because small non-metal atoms (H, C, N) fit into the vacant spaces inside their crystal lattices.
Q30. State the oxidation state of Nickel in [Ni(CO)4].
Answer: Zero (0).
Q31. Name the product of the reaction of an alkyl halide with aqueous KOH.
Answer: Alcohol.
Q32. Write the structural formula of Picric acid.
Answer: 2,4;6-Trinitrophenol.
Q33. Which carbonyl compound shows a highly positive response to the Fehling’s solution test?
Answer: Aliphatic Aldehydes.
Q34. What is the product of the reduction of a nitrile with LiAlH4?
Answer: Primary Amine (1° Amine).
Q35. Define ebullioscopic constant.
Answer: The elevation in boiling point produced when 1 mole of non-volatile solute is dissolved in 1 kg of solvent.
Q36. What is a secondary cell?
Answer: An electrochemical cell that can be recharged by passing an electric current through it in the reverse direction.
Q37. State the effect of a catalyst on the Gibbs free energy change (ΔG) of a reaction.
Answer: A catalyst has no effect on ΔG.
Q38. Why is E° value of Mn3+/Mn2+ highly positive compared to Cr3+/Ref2+?
Answer: Because Mn2+ has a highly stable, half-filled d5 electronic configuration.
Q39. What is a homoleptic complex?
Answer: A complex in which the central metal atom or ion is bound to only one kind of donor ligand configuration.
Q40. Write the IUPAC name of DDT.
Answer: 2,2-bis(4-chlorophenyl)-1,1,1-trichloroethane.
Q41. Why does o-nitrophenol have a lower boiling point than p-nitrophenol?
Answer: Due to the presence of intramolecular hydrogen bonding in o-nitrophenol.
Q42. Name the named organic process used to prepare benzaldehyde from toluene using chromyl chloride.
Answer: Étard Reaction.
Q43. Which configuration is more basic: Metylamine or Aniline?
Answer: Methylamine (Aniline’s lone pair is delocalized into the benzene ring via resonance).
Q44. State Raoult’s law for volatile solute components.
Answer: The partial vapor pressure of each component in a solution is directly proportional to its mole fraction.
Q45. Write the Nernst equation for a generic single electrode reaction: Mn+ + ne– → M.
Answer: E = E° – (RT / nF) ln(1 / [Mn+]).
Q46. Define pseudo-first-order reaction.
Answer: A reaction which is bimolecular but behaves kinetically as a first-order profile because one reactant is present in huge excess.
Q47. What are the common oxidation states shown by Lanthanoids?
Answer: +3 is principal, occasionally +2 and +4 are shown.
Q48. Define the term ‘Chelate Ligand’.
Answer: A di- or polydentate ligand that coordinates using multiple donor sites simultaneously to form a cyclic ring complex matrix.
Q49. Why do haloarenes show lower reactivity towards nucleophilic substitution compared to haloalkanes?
Answer: Due to the partial double-bond character of the C-X bond generated via resonance.
Q50. What is the test used to distinguish primary, secondary, and tertiary amines using nitrous acid?
Answer: Nitrous acid test / Hinsberg test profile.
Section B: 2 Mark Short Answer Questions (Total: 30 Questions)
Q1. Define the terms Molarity and Molality. Why is molality preferred over molarity for temperature-varying experiments?
Answer: Molarity is the number of moles of solute dissolved per liter of solution. Molality is the number of moles of solute per kilogram of solvent. Molality is preferred because it is based on mass, which does not change with temperature, unlike volume-dependent molarity.
Q2. State the difference between an ideal and a non-ideal solution framework based on ΔH and ΔV parameters.
Answer: For an ideal solution, the enthalpy of mixing (ΔHmix) and volume change of mixing (ΔVmix) are equal to zero. For a non-ideal solution, ΔHmix ≠ 0 and ΔVmix ≠ 0.
Q3. What are colligative properties? List any two examples.
Answer: Properties of dilute solutions that depend entirely on the number of solute particles present, irrespective of their chemical nature. Examples: Relative lowering of vapor pressure, Osmotic pressure.
Q4. State Faraday’s First Law of Electrolysis and write its mathematical expression.
Answer: It states that the mass (w) of a substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity (Q) passed. Formula: w = z · Q = z · I · t, where z is the electrochemical equivalent.
Q5. Define specific conductance (conductivity) and molar conductivity. State their units.
Answer: Conductivity (κ) is the reciprocal of resistivity. Unit: S cm-1. Molar conductivity (Λm) is the conducting power of all ions produced by dissolving 1 mole of electrolyte. Unit: S cm2 mol-1.
Q6. State Kohlrausch’s Law of Independent Migration of Ions.
Answer: It states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of its constituent anions and cations at infinite dilution.
Q7. Differentiate between Molecularity and Order of a chemical reaction.
Answer: Order is an experimental value representing the sum of powers of concentration terms in the rate law. It can be a fraction or zero. Molecularity is a theoretical value representing the number of reacting species taking part in an elementary reaction step. It is always a whole number.
Q8. Define half-life period of a reaction. Write the mathematical relationship linking half-life to the rate constant for a first-order reaction profile.
Answer: The time required for the concentration of a reactant to decrease to exactly half of its initial value. For a first-order reaction: t1/2 = 0.693 / k.
Q9. What is activation energy? How does a catalyst change the activation energy path?
Answer: The minimum extra energy required by reactant molecules to cross the energy barrier and form the activated complex. A catalyst lowers this activation energy barrier by providing an alternative reaction pathway.
Q10. Why do transition elements show catalytic properties? Give two reasons.
Answer: 1) They possess vacant d-orbitals to form unstable intermediate compounds with reactants. 2) They can exhibit multiple oxidation states to alter chemical pathways.
Q11. Explain why transition elements show a high tendency to form coordination complexes.
Answer: Due to their relatively small ionic sizes, high nuclear charge, and the availability of vacant d-orbitals to accept lone pairs of electrons from donor ligands.
Q12. What is Lanthanoid Contraction? State one consequence of this phenomenon.
Answer: The steady, gradual decrease in the atomic and ionic radii of Lanthanoid elements with an increase in atomic number. Consequence: The atomic sizes of the 4d and 5d transition series become nearly identical (e.g., Zr and Hf).
Q13. Differentiate between a Double Salt and a Complex Compound structure.
Answer: A double salt dissociates completely into its simple constituent ions when dissolved in water and loses its identity (e.g., Mohr’s salt). A complex compound retains its identity in solution and does not dissociate completely into simple ions (e.g., K4[Fe(CN)6]).
Q14. Write the IUPAC names of: a) [Co(NH3)5Cl]Cl2 b) K3[Fe(C2O4)3].
Answer: a) Pentaamminechloridocobalt(III) chloride b) Potassium trioxalatoferrate(III).
Q15. Explain Werner’s Coordination Theory postulates regarding Primary and Secondary valencies.
Answer: Primary valency is ionizable, non-directional, and corresponds to the oxidation state of the metal ion. Secondary valency is non-ionizable, directional, and corresponds to the coordination number of the metal ion.
Q16. What is an asymmetric/chiral carbon atom? What is its role in optical activity?
Answer: A carbon atom that is bonded to four different atoms or groups of atoms. The presence of a chiral carbon atom makes the molecule asymmetric, leading to non-superimposable mirror images, which causes optical activity.
Q17. Explain the terms: Enantiomers and Racemic Mixture.
Answer: Enantiomers are stereoisomers that are non-superimposable mirror images of each other and rotate plane-polarized light in opposite directions. A racemic mixture is an equimolar mixture of d- and l-enantiomers, making it optically inactive due to external compensation.
Q18. Write the chemical equation for the Finkelstein reaction profile.
Answer: R-Cl + NaI → R-I + NaCl (in the presence of dry acetone solvent matrix).
Q19. Write the chemical mechanism steps or equations for Wurtz-Fittig and Fittig reactions.
Answer:
Wurtz-Fittig: C6H5-X + 2Na + R-X → C6H5-R + 2NaX (Alkylbenzene formation).
Fittig: 2C6H5-X + 2Na → C6H5-C6H5 + 2NaX (Diphenyl formation under dry ether conditions).
Q20. Why are alcohols highly soluble in water compared to their corresponding hydrocarbon configurations?
Answer: Because alcohol molecules possess an -OH group which enables them to form strong intermolecular hydrogen bonds with water molecules. Hydrocarbons cannot form these bonds.
Q21. Explain the mechanism of the acid-catalyzed dehydration of ethanol to form ethene.
Answer: It involves three key steps: 1) Protonation of ethanol to form an oxonium ion. 2) Formation of a carbocation intermediate via the loss of a water molecule (slow step). 3) Elimination of a proton to generate the ethene double bond.
Q22. Write the chemical equation for Reimer-Tiemann reaction.
Answer: C6H5OH + CHCl3 + 3NaOH → o-Hydroxybenzaldehyde (Salicylaldehyde) + 3NaCl + 2H2O.
Q23. Write the chemical steps involved in the Williamson Synthesis of ethers. What limitation does it possess?
Answer: It involves the nucleophilic attack of an alkoxide ion on a primary alkyl halide: R-O– + R’-X → R-O-R’ + X–. Limitation: If a tertiary alkyl halide (3°) is used, elimination takes precedence over substitution, yielding an alkene instead of an ether.
Q24. Write the chemical equation for the Kolbe’s Reaction of phenol.
Answer: Phenol is treated with NaOH to form sodium phenoxide, which then reacts with CO2 followed by acidification to produce 2-Hydroxybenzoic acid (Salicylic acid).
Q25. Distinguish between Aldehydes and Ketones using Tollens’ Reagent test.
Answer: Aldehydes reduce Tollens’ reagent to form a shining silver mirror on the inner walls of the test tube. Ketones do not reduce Tollens’ reagent and show no silver mirror formation.
Q26. Write the chemical equation for the Clemmensen Reduction.
Answer: >C=O + 4[H] → >CH2 + H2O (in the presence of Zinc Amalgam / Zn-Hg and concentrated HCl catalyst).
Q27. Write the chemical steps for the Wolff-Kishner Reduction.
Answer: Carbonyl compounds are treated with Hydrazine (NH2NH2) to form a hydrazone, which is then heated with KOH in ethylene glycol solvent to yield a hydrocarbon (>CH2).
Q28. Why are carboxylic acids stronger acids compared to phenols?
Answer: The carboxylate ion obtained by losing a proton is stabilized by two equivalent resonance structures where the negative charge resides on electronegative oxygen atoms. The phenoxide ion has non-equivalent resonance structures where the negative charge resides on less electronegative carbon atoms.
Q29. Write the chemical equation for the Hell-Volhard-Zelinsky (HVZ) Reaction.
Answer: R-CH2-COOH + X2 → R-CH(X)-COOH + HX (in the presence of red phosphorus and X = Cl, Br, creating α-halo carboxylic acid configurations).
Q30. Explain why aromatic amines (like aniline) are weaker bases than aliphatic amines.
Answer: In aromatic amines, the lone pair of electrons on the nitrogen atom is delocalized into the π-system of the benzene ring via resonance, making it less available for protonation compared to aliphatic amines, where alkyl groups increase electron density via the +I effect.
Section C: 3 Mark Long Answer Questions (Total: 15 Questions)
Q1. State Raoult’s law for non-volatile solutes. Derive the mathematical expression linking the relative lowering of vapor pressure to the molecular mass of the non-volatile solute.
Answer: Raoult’s law states that the relative lowering of vapor pressure of a dilute solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution phase.
Derivation: (P° – P) / P° = x2 = n2 / (n1 + n2).
For a highly dilute solution, n2 << n1, so the denominator simplifies to n1.
Therefore, (P° – P) / P° = n2 / n1 = (w2 · M1) / (M2 · w1).
Rearranging for solute molecular mass (M2) gives: M2 = (w2 · M1 · P°) / [w1 · (P° – P)].
Q2. Define Osmotic Pressure. Show that it is a colligative property. How can the molecular mass of a polymer biomolecule be determined using osmotic pressure measurements?
Answer: The excess hydrostatic pressure that must be applied to the solution side to prevent the flow of pure solvent into it through a semi-permeable membrane. According to the Van ‘t Hoff equation: π = CRT = (n2/V)RT. Since π depends directly on the number of moles of solute particles per unit volume (concentration) and not on their chemical identity, it is a colligative property.
Rearranging for molecular mass: π = (w2 · R · T) / (M2 · V) ⇒ M2 = (w2 · R · T) / (π · V). This method is highly preferred for biomolecules because measurements are taken at room temperature and do not alter delicate macromolecular configurations.
Q3. State the Nernst Equation for a complete electrochemical cell setup. Calculate the cell potential (E) at 298 K for the following system: Mg(s) | Mg2+(0.001 M) || Cu2+(0.0001 M) | Cu(s). [Given: E°Mg2+/Mg = -2.37 V, E°Cu2+/Cu = +0.34 V].
Answer: The complete Nernst Equation is: Ecell = E°cell – (0.0591 / n) · log([Products] / [Reactants]).
Step 1: Calculate E°cell = E°cathode – E°anode = 0.34 – (-2.37) = +2.71 V.
Step 2: Here, n = 2 electrons transferred. The overall cell reaction is Mg + Cu2+ → Mg2+ + Cu.
Step 3: Substitute concentrations into Nernst expression:
Ecell = 2.71 – (0.0591 / 2) · log([Mg2+] / [Cu2+])
Ecell = 2.71 – 0.0295 · log(10-3 / 10-4) = 2.71 – 0.0295 · log(101)
Ecell = 2.71 – 0.0295 · 1 = 2.6805 V.
Q4. Derive the integrated rate equation for a first-order chemical reaction profile: A → Products. Write its graphical presentation variables.
Answer: For a first-order reaction, the rate is proportional to the concentration of the reactant: -d[A]/dt = k[A].
Rearranging variables gives: d[A]/[A] = -k · dt.
Integrating both sides yields: ln[A] = -kt + I, where I is the integration constant.
At initial time t = 0, [A] = [A]0, so I = ln[A]0.
Substituting I back into the equation: ln[A] = -kt + ln[A]0 ⇒ kt = ln([A]0 / [A]).
Converting to base 10 log gives the final first-order expression: k = (2.303 / t) · log([A]0 / [A]).
Graphical presentation: A plot of log[A] versus time (t) gives a straight line with a negative slope equal to -k/2.303.
Q5. State the rate constant equation according to the Arrhenius Theory. The rate constant of a reaction triples when the temperature is raised from 20°C to 50°C. Calculate the activation energy (Ea) for this reaction. [R = 8.314 J mol-1 K-1].
Answer: The Arrhenius equation is k = A · e-Ea/RT. In log form for two temperatures: log(k2 / k1) = (Ea / 2.303R) · [(T2 – T1) / (T1 · T2)].
Given: T1 = 20 + 273 = 293 K; T2 = 50 + 273 = 323 K; k2/k1 = 3.
log(3) = (Ea / (2.303 × 8.314)) · [(323 – 293) / (293 × 323)]
0.4771 = (Ea / 19.147) · [30 / 94639]
0.4771 = Ea × (1.656 × 10-5)
Ea = 0.4771 / (1.656 × 10-5) = 28810.38 J mol-1 = 28.81 kJ mol-1.
Q6. Give reasons for the following transition metal behaviors:
a) Transition metals form colored ions.
b) Transition metals and their compounds show paramagnetic behaviors.
c) Highest oxidation state of a transition metal is exhibited in its oxide or fluoride compounds.
Answer:
a) Due to the presence of vacant or unpaired d-orbitals. When visible light falls on these ions, d-d electronic transitions take place by absorbing specific wavelengths, and the transmitted complementary light imparts color.
b) Due to the presence of one or more unpaired electrons in their (n-1)d subshell configurations. The magnetic moment increases with the number of unpaired electrons.
c) Oxygen and fluorine have small atomic sizes and exceptionally high electronegativities, which allow them to oxidize transition metals to their highest possible oxidation states.
Q7. Explain Crystal Field Theory (CFT) splitting parameters. Draw a neatly labeled energy diagram illustrating d-orbital splitting in an octahedral coordinate field configuration.
Answer: Crystal Field Theory assumes the bonding between ligands and metal ions to be purely electrostatic. In an isolated gaseous atom, all five d-orbitals have equal energy (degenerate). When ligands approach in an octahedral field, the d-orbitals split into two sets due to asymmetric repulsion: a lower energy triplet set called t2g (dxy, dyz, dzx) and a higher energy doublet set called eg (dx2-y2, dz2). The energy gap between these two sets is called the Octahedral Crystal Field Splitting Energy, denoted by Δo. The eg set is raised by +3/5 Δo, and the t2g set is lowered by -2/5 Δo relative to the barycenter.
Section D: 4 Marks Long Answer Questions (Remaining & Detailed Answers)
1. Two Types of Valencies: Central metal atoms in coordination complexes exhibit two distinct types of valencies: Primary Valency and Secondary Valency.
2. Primary Valency:
• This corresponds to the oxidation state of the central metal atom.
• It is ionizable in nature and is always satisfied by negative ions.
• It is non-directional and is represented by dotted lines (. . . . .) in structural formulas.
3. Secondary Valency:
• This corresponds to the coordination number (number of donor atoms linked) of the central metal atom.
• It is non-ionizable and can be satisfied by negative ions, neutral molecules, or occasionally positive ions.
• It is strictly directional in space, which gives a definite geometry (like octahedral or tetrahedral) to the coordination complex. It is represented by solid lines (—————).
4. Example Explanation [Co(NH₃)₆]Cl₃:
• When silver nitrate (AgNO₃) is added to a solution containing 1 mole of this complex, 3 moles of AgCl precipitate immediately. This proves that the 3 Cl− ions satisfy the Primary Valency (outside the coordination sphere) and are ionizable.
• The 6 NH₃ molecules are directly bound to the Co³+ ion inside the square bracket, satisfying its Secondary Valency (coordination number = 6). They are non-ionizable and project towards the corners of a regular octahedron, giving the complex an octahedral geometry.
| Feature | SN1 Mechanism | SN2 Mechanism |
|---|---|---|
| Full Name | Substitution Nucleophilic Unimolecular | Substitution Nucleophilic Bimolecular |
| Reaction Steps | Two steps (Formation of carbocation intermediate, followed by nucleophilic attack). | Single concerted step (Simultaneous bond-breaking and bond-making via a transition state). |
| Kinetics & Order | First-order kinetics. Rate depends only on substrate concentration: Rate = k[R-X]. | Second-order kinetics. Rate depends on both reactants: Rate = k[R-X][Nu−]. |
| Reactivity Order | 3° > 2° > 1° > CH₃X (governed by carbocation stability). | CH₃X > 1° > 2° > 3° (governed by minimal steric hindrance). |
| Stereochemistry | Racemization occurs due to planar carbocation structure attacked from both sides. | Complete Walden Inversion occurs due to strict backside attack of the nucleophile. |
(a) Alcohols and Phenols (b) Aldehydes and Ketones.
(a) Distinguishing Tests for Alcohols and Phenols:
1. Ferric Chloride Test: Add a few drops of neutral FeCl₃ solution to the compound. Phenol gives a characteristic deep violet/purple coloration due to complex formation. Alcohols do not produce any such color change.
2. Br₂ Water Test: Phenol reacts rapidly with bromine water to give a white precipitate of 2,4,6-tribromophenol. Alcohols do not react with bromine water under identical conditions.
3. Lucas Test (For Alcohols): Used to distinguish 1°, 2°, and 3° alcohols. Tertiary alcohols produce immediate turbidity with Lucas reagent (conc. HCl + anhyd. ZnCl₂), secondary alcohols produce turbidity within 5 minutes, while primary alcohols do not produce turbidity at room temperature.
(b) Distinguishing Tests for Aldehydes and Ketones:
1. Tollen’s Test (Silver Mirror Test): Warm the organic compound with freshly prepared ammoniacal silver nitrate solution (Tollen’s Reagent). Aldehydes readily oxidize and form a brilliant silver mirror coating on the inner walls of the test tube. Ketones show no reaction.
2. Fehling’s Test: Heat the compound with an equal mixture of Fehling’s solution A and B. Aliphatic aldehydes reduce it to form a distinct brick-red precipitate of cuprous oxide (Cu₂O). Ketones do not reduce Fehling’s solution.
Mathematical Expression: For an electrolyte AxBy:
Λ°m = xλ°+ + yλ°−
Where Λ°m is the limiting molar conductivity of the electrolyte, and λ°+ and λ°− are the limiting molar conductivities of the cation and anion respectively.
Applications:
1. Calculation of Limiting Molar Conductivity of Weak Electrolytes: Weak electrolytes (like CH₃COOH) do not dissociate completely even at high dilutions, so their Λ°m cannot be determined graphically. Kohlrausch’s law allows us to calculate it using the values of strong electrolytes:
Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl).
2. Determination of Degree of Dissociation (α): The degree of dissociation of a weak electrolyte at a given concentration C can be calculated using the ratio of its molar conductivity (Λm) to its limiting molar conductivity (Λ°m):
α = Λm / Λ°m.
Example: Acid-catalyzed hydrolysis of ethyl acetate:
CH₃COOC₂H₅ + H₂O (∞) —[H+]→ CH₃COOH + C₂H₅OH
Since water is present in vast excess, its concentration practically remains constant during the reaction. Therefore, the rate depends only on the concentration of ethyl acetate: Rate = k'[CH₃COOC₂H₅].
Derivation of First-Order Integrated Rate Equation:
For a reaction R → P, the differential rate law is: −d[R]/dt = k[R]
Rearranging the equation: d[R]/[R] = −k · dt
Integrating both sides: ln[R] = −kt + I —- (Equation 1)
At t = 0, [R] = [R]₀ (initial concentration). Substituting these values in Eq 1: ln[R]₀ = I
Substitute the value of I back into Eq 1: ln[R] = −kt + ln[R]₀
kt = ln[R]₀ − ln[R] ⇒ kt = ln([R]₀ / [R])
Converting natural log (ln) to base 10 log (log₁₀):
k = (2.303 / t) · log([R]0 / [R])
Fundamental Cause: As the atomic number increases within the lanthanoid series, one nuclear proton and one 4f electron are added at each step. Because 4f orbitals have a highly diffused shape, they offer very poor shielding to outer electrons from the growing nuclear charge. Consequently, the effective nuclear charge pulling the valence shell inward increases steadily, shrinking the atom’s size.
Consequences:
1. Identical Size of 4d and 5d Elements: Due to the intervention of the lanthanoid series, elements of the 5d series (e.g., Hf, Ta, W) have atomic radii almost identical to the elements of the 4d series lying directly above them (e.g., Zr, Nb, Mo).
2. Similarity in Properties: Because pairs like Zr-Hf have matching sizes and valence configurations, their chemical and physical properties are nearly identical, making their separation from natural minerals extremely difficult.
3. Trend in Basic Strength of Hydroxides: Due to Lanthanoid Contraction, the size of Ln³+ ions decreases from La³+ to Lu³+. According to Fajan’s rules, smaller sizes increase covalent character. Hence, basic strength decreases from La(OH)₃ (most basic) to Lu(OH)₃ (least basic).
(a) Williamson’s Ether Synthesis (b) Kolbe’s Reaction.
• Equation: R−O−Na+ + R’−X → R−O−R’ + NaX
• Example: Sodium ethoxide reacting with bromoethane yields diethyl ether:
C₂H₅ONa + C₂H₅Br → C₂H₅−O−C₂H₅ + NaBr
• Note: The alkyl halide must be primary (1°). If a tertiary (3°) alkyl halide is used, elimination dominates over substitution, yielding an alkene instead of an ether.
(b) Kolbe’s Reaction: Phenol undergoes electrophilic aromatic substitution when treated with sodium hydroxide followed by carbon dioxide under pressure (4-7 atm) at 400K to produce salicylic acid.
• Step 1: Phenol reacts with NaOH to form the highly reactive phenoxide ion.
• Step 2: Phenoxide ion undergoes electrophilic attack by the weak electrophile CO₂ at the ortho-position.
• Equation: C₆H₅OH + NaOH → C₆H₅ONa —[1. CO₂, 2. H+]→ o-HOC₆H₄COOH (Salicylic acid / 2-Hydroxybenzoic acid).
1. Phthalimide is treated with ethanolic KOH to form potassium phthalimide.
2. Potassium phthalimide is heated with an alkyl halide (R-X). It undergoes a nucleophilic substitution (SN2) reaction to form N-alkylphthalimide.
3. N-alkylphthalimide is hydrolyzed with aqueous sodium hydroxide (NaOH) to yield a pure primary aliphatic amine (R-NH₂) and sodium phthalate.
Why Aromatic Amines (Aniline) Cannot Be Prepared:
Preparing primary aromatic amines like aniline requires reacting potassium phthalimide with an aryl halide (e.g., chlorobenzene). However, aryl halides do not undergo nucleophilic substitution (SN2) reactions easily. This resistance is due to the partial double-bond character of the C−X bond caused by resonance, as well as steric hindrance from the aromatic ring blocking backside nucleophilic attack. Therefore, N-arylphthalimide cannot form, and aniline cannot be synthesized using this method.
Section E: Advanced Numerical Drills & Core Mechanisms (4 Marks Each)
• Mass of solute (w₂) = 18 g
• Molar mass of solute (M₂) = 180 g mol−¹
• Mass of solvent (w₁) = 1 kg = 1000 g
• Molal boiling point elevation constant (Kb) = 0.52 K kg mol−¹
• Boiling point of pure solvent (T°b) = 373.15 K
2. Calculate the molality (m) of the solution:
Number of moles of glucose = w₂ / M₂ = 18 / 180 = 0.1 mol
Molality (m) = Moles of solute / Mass of solvent in kg = 0.1 mol / 1 kg = 0.1 mol kg−¹
3. Calculate the elevation in boiling point (ΔTb):
ΔTb = Kb × m = 0.52 K kg mol−¹ × 0.1 mol kg−¹ = 0.052 K
4. Determine the boiling point of the solution (Tb):
Tb = T°b + ΔTb
Tb = 373.15 K + 0.052 K = 373.202 K
Therefore, the glucose solution will boil at 373.202 K (or 100.052 °C).
Zn(s) + Cu²+(aq) → Zn²+(aq) + Cu(s)
(Given: E°Zn²+/Zn = −0.76 V, E°Cu²+/Cu = +0.34 V, F = 96487 C mol−¹, R = 8.314 J K−¹ mol−¹).
E°cell = E°cathode − E°anode = E°Cu²+/Cu − E°Zn²+/Zn
E°cell = 0.34 V − (−0.76 V) = +1.10 V
Here, number of electrons transferred (n) = 2.
2. Calculate Standard Gibbs Free Energy Change (ΔG°) for maximum work:
ΔG° = −nFE°cell
ΔG° = −2 × 96487 C mol−¹ × 1.10 V = −212271.4 J mol−¹ = −212.27 kJ mol−¹
(The maximum work obtainable is 212.27 kJ mol−¹).
3. Calculate the Equilibrium Constant (Kc):
Using the Nernst formula at 298 K: E°cell = (0.0591 / n) × log Kc
1.10 = (0.0591 / 2) × log Kc
1.10 = 0.02955 × log Kc
log Kc = 1.10 / 0.02955 = 37.225
Kc = Antilog(37.225) = 1.68 × 10³⁷
The equilibrium constant Kc is 1.68 × 10³⁷.
For a first-order reaction: t1/2 = 0.693 / k
Given k = 5.5 × 10−¹⁴ s−¹
t1/2 = 0.693 / (5.5 × 10−¹⁴ s−¹) = 1.26 × 10¹³ seconds.
Part 2: Calculate Rate Constant (k) for 30% decomposition:
Let initial concentration [R]₀ = 100.
Since 30% decomposes, remaining concentration [R] = 100 − 30 = 70.
Time (t) = 40 minutes = 40 × 60 = 2400 seconds.
Using the integrated rate expression: k = (2.303 / t) × log([R]₀ / [R])
k = (2.303 / 2400 s) × log(100 / 70)
k = (2.303 / 2400) × log(1.428)
Since log(1.428) ≈ 0.1548:
k = (2.303 / 2400) × 0.1548 = 0.0009595 × 0.1548 = 1.48 × 10−⁴ s−¹ (or 8.91 × 10−³ min−¹).
• Carbonyl carbon is sp² hybridized and planar. Due to the high electronegativity of oxygen, the C&dbnd;O bond is highly polarized, leaving a partial positive charge on the carbon.
• Step 1: The nucleophile attacks the electrophilic carbonyl carbon perpendicularly to the planar orbital. The hybridization of carbon changes from sp² to sp³, forming a tetrahedral alkoxide intermediate.
• Step 2: The alkoxide intermediate captures a proton (H+) from the reaction medium to give the final neutral addition product.
Why Aldehydes are more reactive than Ketones:
1. Steric Reasons: Aldehydes have only one bulky alkyl group attached to the carbonyl group, whereas ketones have two. The presence of two bulky groups in ketones hinders the approach of the incoming nucleophile to the carbon atom, reducing reactivity.
2. Electronic Reasons: Alkyl groups are electron-donating in nature due to the inductive effect (+I effect). Ketones have two alkyl groups that push electron density toward the carbonyl carbon, reducing its partial positive charge and making it less electrophilic. Aldehydes, having only one alkyl group, retain a stronger positive charge on the carbon, making them easier targets for nucleophiles.
(a) Aldol Condensation (b) Cannizzaro Reaction.
• Equation: 2 CH₃−CHO —[Dilute NaOH]→ CH₃−CH(OH)−CH₂−CHO (3-Hydroxybutanal / Aldol)
• When heated, the aldol loses a water molecule to form an α,β-unsaturated carbonyl compound:
CH₃−CH(OH)−CH₂−CHO —[Δ / −H₂O]→ CH₃−CH&dbnd;CH−CHO (But-2-enal).
(b) Cannizzaro Reaction: Aldehydes that do not contain any α-hydrogen atom (such as formaldehyde or benzaldehyde) undergo self-oxidation and reduction (disproportionation) when heated with concentrated alkali (50% NaOH). One molecule is reduced to an alcohol, while the other is oxidized to a carboxylic acid salt.
• Equation: 2 HCHO (Formaldehyde) + Conc. NaOH → CH₃OH (Methanol) + HCOONa (Sodium formate)
• Example with Benzaldehyde: 2 C₆H₅CHO + Conc. NaOH → C₆H₅CH₂OH (Benzyl alcohol) + C₆H₅COONa (Sodium benzoate).