HBSE Class 12 Chemistry Pre-Board Sample Paper 2026 Answer Key
SECTION – A (1 Mark)
1. Which one is a colligative property?
(A) Boiling Point
(B) Freezing Point
(C) Vapour Pressure
(D) Osmotic Pressure
Answer – (D) Osmotic Pressure
2. Rate law for the reaction A + 2B → C is found to be Rate = K[A][B]. Concentration of reactant B is doubled keeping the concentration of A constant, the value of rate constant will be :
(A) the same
(B) doubled
(C) quadrupled
(D) halved
Answer – (A) the same
Rate constant is independent of concentrations, it depends only on temperature.
3. How much charge is required for the reduction of 1 mol of Al3+ to Al (s) :
(A) 96500 C
(B) 3 × 96500 C
(C) 3 F
(D) both (B) and (C)
Answer – (D) both (B) and (C)
4. Which of the following species will impart colour to an aqueous solution?
(A) Ti4+
(B) Cu+
(C) Zn2+
(D) Cr3+
Answer – (D) Cr3+
5. The stability of ferric ion is due to :
(A) Half-filled d-orbitals
(B) completely filled f-orbitals
(C) Half-filled f-orbitals
(D) completely filled d-orbitals
Answer – (A) Half-filled d-orbitals
6. Consider the following sequence of reactions :
Compound A + NaCN → B + Partial Hydrolysis → CH3CONH2
Identify the compound A in the above reaction?
(A) CH3CH2CI
(B) CH3CI
(C) CH3CN
(D) CH3NO2
Answer – (B) CH3CI
7. Diethyl ether on heating with conc. HI will give :
(A) One molo ethyl alcohol and one mole of ethyl iodide
(B) Two moles of ethyl alcohol
(C) Two moles of ethyl iodide
(D) Two moles of methyl iodide
Answer – (C) Two moles of ethyl iodide
8. A primary alcohol can be prepared by the reaction of grignard reagent with :
(A) Acetaldehyde
(B) Acetone
(C) Formaldehyde
(D) Chloroethane
Answer – (C) Formaldehyde
9. The presence of five alcoholic groups in glucose is established by its reaction with ……………
(A) HI
(B) Bromine water
(C) Acetic Anhydride
(D) conc. HNO3
Answer – (C) Acetic Anhydride
10. 40% aqueous solution of formaldehyde is known as ……………
Answer – Formalin
11. Carbolic acid is the common name of ……………
Answer – Phenol (C6H5OH)
12. The structure which determine the sequence of various amino acids in a polypeptide chain is ……………
Answer – primary structure
13. Why are lower alcohols soluble in water?
Answer – Because they form hydrogen bonds with water molecules.
14. The van’t hoff factor (i) for a dilute solution of K3[Fe(CN)6] is ………..
Answer : 4
15. Which chemical test is used to distinguish between Aniline and Ethyl amine?
Answer – Azo dye test
16. Assertion (A) : When NaCl is added to water, a depression in freezing point is observed.
Reason (R) : The lowering of vapour pressure of a solution causes depression in the freezing point.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.
Answer – (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
17. Assertion (A) : Half-life of a first order reaction is variable.
Reason (R) : For first order reaction, half-life is independent of initial concentration of reactant.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.
Answer – (D) (A) is false, but (R) is true.
18. Assertion (A) : The boiling point of alkyl halides decreases in the order RI > RBr > RCl > RF for the same alkyl group.
Reason (R) : The boiling point of alkyl halide having chloride, bromide and iodides are higher than that of the hydrocarbon of comparable molecular mass.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.
Answer – (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
SECTION – B (2 Marks)
19. What are primary cells? Give the construction and uses of Mercury cell.
Answer – Primary cells are electrochemical cells in which the chemical reactions are irreversible, therefore the cells cannot be recharged. A mercury cell consists of zinc-mercury amalgam as the anode, mercuric-oxide as the cathode and potassium hydroxide as the electrolyte. It is commonly used in watches, calculators and hearing aids due to its constant voltage output.
20. Explain how a catalyst will increase the rate of a chemical reaction?
Answer – A catalyst increases the rate of a chemical reaction by providing an alternative reaction pathway with lower activation energy. Because of the lowered activation energy, a greater number of reactant molecules can undergo effective collisions per unit time, resulting in an increased rate of reaction.
21. Define Kohlrausch law.
Answer – Kohlrausch’s law, also known as the law of independent migration of ions, states that at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the individual contributions of its constituent cation and anion, as each ion migrates independently and contributes separately to the total molar conductivity.
22. The time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, then calculate the rate constant of the reaction.
Answer – For a first-order reaction :
t½ = 0.693 / k
k = 0.693 / t½ = 0.693 / 60 = 0.01155 min–1 = 1.16 × 10–2 min–1
Rate constant (k) = 1.16 × 10–2 min–1
23. Compare the basic strength of Ammonia and Aniline.
Answer – Ammonia is more basic than aniline. In aniline, the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring, which reduces its availability for protonation. In ammonia, the lone pair is freely available on nitrogen, making it more basic than aniline.
24. Write the IUPAC name of the following coordination compounds :
(i) [Co(NH3)5CO3]Cl
Answer – Pentaamminecarbonatocobalt(III) chloride
(ii) K3[Fe(C2O4)3
Answer – Potassium tris(oxalato)ferrate(III)
25. Write short notes on the following :
(i) Carbyl Amine reaction
Answer – Primary amines when heated with chloroform and alcoholic potassium hydroxide give foul-smelling isocyanides. This reaction is used as a test for primary amines.
(ii) Diazotisation reaction
Answer – Primary aromatic amines react with nitrous acid at 0-5 °C to form diazonium salts. This reaction is known as diazotisation and is important in the preparation of dyes.
SECTION – C (3 Marks)
26. Write down the differences between rate of reaction and specific reaction rate.
Answer – Rate of reaction is defined as the change in concentration of a reactant or product per unit time. It shows how fast a reaction is proceeding and depends on the concentration of reactants at a given time.
• Specific reaction rate (rate constant, k) is the proportionality constant in the rate law that relates the rate of reaction to the concentrations of reactants. For a given reaction, it has a fixed value at a particular temperature and does not depend on the concentrations of reactants.
27. Convert :
(i) Propene into Propan-2-ol
Answer – Propene on hydration with dilute sulphuric acid follows Markovnikov’s rule, in which the –OH group attaches to the more substituted carbon atom, forming propan-2-ol.
CH3–CH=CH2 + H2O/H2SO4 → CH3–CHOH–CH3
(ii) Benzyl Chloride into Benzyl Alcohol
Answer – Benzyl chloride undergoes nucleophilic substitution with aqueous sodium hydroxide to give benzyl alcohol.
C6H5–CH2Cl + NaOH (aq) → C6H5–CH2OH + NaCl
28. Explain why actinoid contraction is greater than that of lanthanoid contraction.
Answer – Actinoid contraction is greater than lanthanoid contraction because in the actinoid series the 5f orbitals are being filled, whereas in the lanthanoid series the 4f orbitals are filled. The 5f orbitals have a poorer shielding effect than the 4f orbitals. As a result, the effective nuclear charge experienced by the outer electrons increases more rapidly across the actinoid series. This stronger attraction pulls the electron cloud closer to the nucleus, leading to a greater decrease in atomic and ionic sizes in actinoids compared to lanthanoids.
29. (i) Write down the differences between DNA and RNA.
Answer – DNA contains deoxyribose sugar, whereas RNA contains ribose sugar. DNA has thymine as a nitrogenous base, while RNA has uracil instead of thymine. Structurally, DNA is generally double-stranded and forms a double helix, whereas RNA is usually single-stranded. Functionally, DNA is responsible for the storage and transmission of genetic information, while RNA plays an important role in protein synthesis.
(ii) Define reducing sugar with an example.
Answer – A reducing sugar is a carbohydrate that contains a free aldehyde or ketone group and can reduce Tollens’ or Fehling’s reagent. Example: Glucose.
30. (i) Write the IUPAC name of the amide that gives propanamine by Hoffmann’s bromamide reaction.
Answer – Butanamide (CH3CH2CH2CONH2)
(ii) Arrange the following in the increasing order of basic strength :
C6H5NH2, NH3, C2H5NH2, (C2H5)2NH
Answer : C6H5NH2 (aniline) < NH3 (ammonia) < C2H5NH2 (ethylamine) < (C2H5)2NH (diethylamine)
(iii) Give one chemical test to distinguish between aniline and benzylamine.
Answer – Nitrous Acid Test and Hinsberg Test
SECTION – D (4 Marks : CASE STUDY)
31. Transition elements have partly filled d-orbitals in their normal oxidation states or in their common oxidation states. There are four series in transition elements and these are 3d-series (first transition series), 4d-series (second transition series), 5d-series (third transition series) and 6d-series (fourth transition series). Each series contains ten elements. Elements of group 12 are not considered as transition elements as these have fully-filled d-subshells. Transition elements show all the characteristics of metals. These elements show variable oxidation states, form coloured ions, form complexes from alloys and interstitial compounds. They show high enthalpies of atomisation, show catalytic properties and magnetic moment.
Questions :
(i) Why Zn, Cd and Hg are not considered as transition elements?
Answer – Transition metals are defined as the elements having incompletely filled d-orbitals. Since Zn, Cd and Hg have completely filled d-orbital, they are not regarded as transition elements.
(ii) Which element of the d-block shows maximum oxidation state in the first transition series?
Answer – Manganese
(iii) Why transition elements show variable oxidation states?
Answer – Transition elements show variable state oxidation in their compounds because there is a very small energy difference between (n-1)d and ns orbitals. As a result, electrons of (n-1)d orbitals as well as ns-orbitals take part in bond formation. Thus, transition elements have variable oxidation states.
(iv) Calculate the magnetic moment of a divalent ion in its aqueous solution if its atomic number is 25.
Answer : Magnetic moment (µ) = √n(n+2)
= √5(5+2) = √35 = 5.92 BM
32. An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of concentration. The solutions for which vapour pressure is either higher or lower than predicted by Raoult’s law are called non ideal solutions. Non ideal solutions can show either positive or negative deviations from Raoult’s law depending on whether the A-B interactions in solution are stronger or weaker than A-A and B-B interactions.
Questions :
(i) What type of solution is formed when chloroform is mixed with acetone and why?
Answer – Negative deviation (chloroform + acetone form H-bonded A-B interactions stronger than A-A or B-B).
(ii) What are the signs of ∆H and ∆V for a solution with positive deviation from ideal behaviour?
Answer : ΔΗmix > 0, ∆Vmix > for positive deviation.
(iii) Name two factors on which the vapour pressure of liquid depends?
Answer – Vapour pressure depends on temperature and nature / intermolecular forces of the liquid.
(iv) Give one example of a solution showing positive deviation from Raoult’s law.
Answer – Ethanol and acetone show positive deviation from Raoult’s law.
SECTION – E (5 Marks)
33. (i) Arrange the following according to their increasing acidic strength:
Acetic acid, formic acid, fluoroacetic acid, propanoic acid.
Answer – Acidic strength of carboxylic acids increases with electron-withdrawing effect and decreases with electron-releasing alkyl groups.
Propanoic acid < Acetic acid < Formic acid < Fluoroacetic acid
(ii) Describe the following :
(a) Cannizarro Reaction
Answer – Aldehydes that do not contain α-hydrogen atoms undergo disproportionation in the presence of concentrated alkali to give an alcohol and a carboxylic acid (or its salt). Example:
2HCHO + NaOH → HCOONa + CH3OH
(b) Esterification Reaction
Answer – When a carboxylic acid reacts with an alcohol in the presence of concentrated sulphuric acid, an ester and water are formed. Example:
CH3COOH + C2H5OH + conc.H2SO4 → CH3COOC2H5 + H2O
(c) Cross Aldol Condensation
Answer – When two different aldehydes or ketones containing α-hydrogen atoms react together in the presence of a dilute base, the reaction is called cross aldol condensation, producing a mixture of aldol products. Example:
CH3CHO + CH3CH2CHO + dil.NaOH → CH3CH(OH)CH2CHO
34. Explain giving reasons :
(i) Transition elements and many of their compounds show paramagnetic behaviour.
Answer – Transition elements and their compounds show paramagnetic behaviour due to the presence of unpaired electrons in the (n-1)d orbitals, which causes attraction towards a magnetic field.
(ii) The enthalpy of atomization of transition metals is very high.
Answer – Transition metals have high enthalpy of atomization because of strong metallic bonding arising from the participation of both ns and (n-1)c electrons in bond formation.
(iii) The transition metals generally form coloured compounds.
Answer – Transition metal compounds are coloured due to d → d electronic transitions in partially filled d-orbitals when they absorb visible light.
(iv) The transition metals from large number of complexes.
Answer – Transition metals form a large number of complexes due to their small size, high charge density, availability of vacant d-orbitals, and ability to accept lone pairs from ligands.
(v) When we move from left to right in a transition series, the decrease in atomic size is small.
Answer – Across a transition series, electrons are added to the (n-1)c subshell, which partially shields the increasing nuclear charge. As a result, the decrease in atomic size is small.
35. (i) Why is salt bridge used in electrochemical cell?
Answer – A salt bridge is used to complete the electrical circuit and maintain electrical neutrality in the electrochemical cell by allowing the migration of ions. It also prevents mixing of the two solutions and minimises liquid junction potential, while electrons flow only through the external circuit.
(ii) Calculate the EMF of Cu-Ag cell when Cu rod is dipped in 0.1 M CuSO4 solution and Ag rod is dipped in 0.01 M AgNO3 solution at 298 K. (Eo Cu2+/Cu = 0.34 V, Eo Ag+/Ag = 0.80 V)
Answer – Here,
Ag+/Ag act as cathode
Cu/Cu2+ act as anode
Eocell = Eocathode – Eoanode = 0.80 – 0.34 = 0.46 V
The given reaction is
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag
From here n = 2
We can calculate the emf of cell by Nernst equation
Ecell = Eocell – 0.0591/n log[Cu2+] / [Ag+]2
= 0.46 – 0.0591/2 log [0.1] / [0.01]2
= 0.46 – 0.02995 log(103)
= 0.46 – 0.02995 × 3
= 0.46 – 0.08985
= 0.37V