HBSE Class 12th Chemistry Sample Paper 2024 Answer

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HBSE Class 12th Chemistry Sample / Model Paper 2024 Answer

Class – 12th               Subject – Chemistry
Time : 3 Hours            Maximum Marks : 70

General Instructions –
(i) This question paper is divided into five sections A, B, C, D and E.
(ii) This question paper contains 35 questions. All questions are compulsory.
(iii) In Section A – Question No. 1 to 18 are multiple choice (MCQ) type questions carrying 1 mark each.
(iv) In Section B – Question No. 19 to 25 are very short answer (VSA) type questions carrying 2 marks each.
(v) In Section C – Question No. 26 to 30 are short answer (SA) type questions carrying 3 marks each.
(vi) In Section D – Question No. 31 and 32 are case based questions carrying 4 marks each.
(vii) In Section E – Question No. 33 to 35 are long answer (LA) type questions carrying 5 marks each.
(viii) There is no overall choice. However an internal choice has been provided in two questions in Section B, two questions in Section C, two questions in Section D and two questions in Section E.
(ix) Use of calculators is not allowed.

SECTION – A (18 Marks)
(18 MCQ of 1 Mark Each)

1. The deficiency of which of the following Vitamins causes pernicious anemia.
(a) Vitamin B1
(b) Vitamin B2
(c) Vitamin B6
(d) Vitamin B12
Answer – (d) Vitamin B12

2. In comparison to 0.01 M solution of glucose, the depression in freezing point of 0.01 M MgCl2 solution is :
(a) the same
(b) about twice
(c) about three times
(d) about six times
Answer – (c) about three times

3. Electrode potential for Cu electrode varies according to the equation is :
ECu²+/Cu = E°Cu²+/Cu – 0.0591/2 log1/[Cu²+]
The graph of ECu²+/Cu vs log [Cu²+] is

Answer – Option (b)

4. For a first order reaction, the time taken to reduce the initial concentration by a factor of 1/4th is 20 minutes. The time required to reduce initial concentration by a factor of 1/16th is
(a) 20 min
(b) 10 min
(c) 80 min
(d) 40 min
Answer – (d) 40 min
The time required to reduce the initial concentration to a factor 1/4th is 20 min.
So, half life of the reaction is (t½) = 10 min.
The time required to reduce the initial concentration to a factor 1/16th i.e. 4 half lives is (10 x 4) min = 40 min

5. Glucose on treatment with sodium amalgam gives :
(a) n-heptanoic acid
(b) Sorbitol
(c) Gluconic acid
(d) Glucaric acid
Answer – (b) Sorbitol

6. The reaction of chloroform with alcoholic KOH and p-toluidine forms.

Answer – Option (a)

7. When 1 mol CrCl3.6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The Formula of Complex is
(a) [CrCl3(H2O)3].3H2O
(b) [CrCl3(H2O)4]Cl.2H2O
(c) [CrCl(H2O)5]Cl2.H2O
(d) [Cr(H2O)6]Cl3
Answer – (d) [Cr(H2O)6]Cl3

8. Monochlorination of toluene in sunlight followed by hydrolysis with aqueous NaOH yield.
(a) o-cresol
(b) m-cresol
(c) 2,4-dihydroxy toluene
(d) Benzyl alcohol
Answer – (d) Benzyl alcohol

9. At pH = 11, Cr2O7²- ion changes to
(a) CrO3
(b) CrO4²-
(c) Cr³+
(d) CrO2²+
Answer – (b) CrO4²-

10. Ethanol on warming with conc. H2SO4 at 413 K gives
(a) Ethene
(b) Diethyl ether
(c) Dimethyl ether
(d) Ethyl hydrogen sulphate
Answer – (b) Diethyl ether

11. Which of the following compound will dissolve in an alkali solution after it undergoes reaction with Hinsberg’s reagent.
(a) CH3NH2
(b) (CH3)2NH
(c) C6H5NHC6H5
(d) (CH3)3N
Answer – (a) CH3NH2

12. The end product (D) in the following sequence is

(a) Salicylic Acid
(b) Salicylaldehyde
(c) Phenyl acetate
(d) Aspirin
Answer – (d) Aspirin

13. Aniline on oxidation with Na2Cr2O7 and H2SO4 gives
(a) Benzoic acid
(b) m-amino benzoic acid
(c) Schiff’s base
(d) p-Benzoquinone
Answer – (d) p-Benzoquinone

14. We are having three aqueous solutions of K2SO4 labelled as ‘X’, ‘Y’ and ‘Z’ with the concentration of 0.001 M, 0.01 M, 0.1 M respectively. The value of Van’t Hoff factor for these solutions will be in the order
(a) ix < iy < iz
(b) ix > iy > iz
(c) ix = iy = iz
(d) ix < iy > iz
Answer – (c) ix = iy = iz

For Questions 15 to 18 two statements are given-one labeled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

15. Assertion (A) : For CH3COOH the Molar conductance of 0.1 M CH3COOH and equivalent conductance of 0.1 N CH3COOH is same.
Reason (R) : These do not depend upon concentration.
Answer – (c) Assertion (A) is true, but Reason (R) is false.

16. Assertion (A) : For an exothermic reaction, activation energy for the backward reaction is more than the activation energy of the forward reaction.
Reason (R) : If the activation energy is high, the reaction is slow.
Answer – (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A).

17. Assertion (A) : Lanthanoids show a limited number of oxidation states where as Actinoids show a large number of oxidation states.
Reason (R) : Energy gap between 4f, 5d and 6s subshell is small where as that between 5f, 6d and 7s subshell in large.
Answer – (c) Assertion (A) is true, but Reason (R) is false.

18. Assertion (A) : Aniline is more basic than ethylamine.
Reason (R) : The lone pair on N atom is present in Conjugation with benzene ring and becomes less available for protonation because of resonance.
Answer – (d) Assertion (A) is false, but Reason (R) is true.

SECTION – B (14 Marks)
(7 Questions of 2 Marks Each)

19. Account for the following :
(a) Aniline does not undergo Friedal craft reaction.
Answer – Aniline being lewis base react with Anhydrous AlCl3 which is lewis acid to form salt.

(b) Methylamine is water reacts with ferric chloride to precipitate hydrated ferric oxide.
Answer – Methylamine accept proton from water and liberate OH- ion which combine with Fe³+ ion to form hydrated ferric oxide Fe(OH)3 or Fe2O3.3H2O

20. Among the isomeric alkanes of molecular formula C5H12, identify the one which on photochemical chlorination yields
(a) a single monochloride
Answer : 2,2-dimethyl propane (neopentane)

(b) four isomeric monochlorides
Answer : 2-methyl butane


Explain :
(a) Why Grignard reagent should be prepared under anhydrous conditions.
Answer – Because Grignard reagent reacts with moisture and form Alkane.

(b) Why the dipole moment of chloro benzene is lower than that of cyclohexyl chloride.
Answer : C-Cl bond in chloro benzene acquire some double bond character due to delocalization of ions pair on chlorine so bond length decreases.

21. (a) Give one structural difference between amylose and amylopectin.
Answer – Amylose is water soluble linear polymer of α-D glucose whereas amylopectin is water insoluble branched (C1-C6) glycosidic linkage carrying branched polymer.

(b) What type of bonding helps in stabilising the α-helix structure of proteins.
Answer – Intra molecular H-Bonding

22. Draw all the isomers (geometrical and optical) of [COCl2(en)2]+
Answer –

23. How many electrons flow through a metallic wire if a current of 0.5 A is passed for 4 hours.
Answer : Current (I) = 0.5 A, Time (t) = 4 × 60 × 60 seconds
Charge = Current × Time
Q = I × t = 0.5 × 4 × 60 × 60 = 7200 C
We also know that charge on one electron = 1.6 × 10-¹⁹ C
No. of electrons = Total Charge ÷ charge on one electron = 7200 ÷ (1.6×10-¹⁹) = 4.5 × 10²²
No. of electrons = 4.5 × 10²² e-

24. (a) Explain the following through example
(i) Azeotropic mixture
Answer – Azeotropic mixture is type of liquid mixture having definite composition and boiling like a pure liquid.
eg. 95.37% C2H5OH + 4.63% H2O

(ii) Isotonic solutions
Answer – Solutions which have the same osmotic pressure at same temperature.
eg. 0.9% solution of pure NaCl is isotonic with RBC


(b) Derive relationship between mole fraction and vapour pressure of components of an ideal solution in liquid phase and vapour phase.
Answer – If we have two completely miscible volatile liquid A and B having mole fraction XA and XB. Then at certain temperature partial pressures PA and PB and vapour pressure in pure state PA° and PB° are expressed as
PT = PA + PB
PT = PA°.XA + PB°.XB
PT = PA°.(1-XB) + PB°.XB
when XA = 1,  PT = PA°.XA
when XB = 1,  PT = PB°.X

25. What is the effect of adding a catalyst on
(i) Activation Energy (Ea) of the reaction
Answer – Ea decrease

(ii) Gibbs free energy (∆G) of the reaction.
Answer – No effect on ∆G

SECTION – C (15 Marks)
(5 Questions of 3 Marks Each)

26. Define the following as related to protein
(a) Peptide linkage
Answer – It is the amide linkage present between -COOH group of one α-amino acid and NH2 group of other amino acid.

(b) Denaturation
Answer – When protein in native form is subjected to physical changes like change in temperature or pH then hydrogen bonds are broken, it looses its biological activity and all structures are destroyed and only primary structure remain intact.

(c) Primary structure
Answer – It is the sequence in which various α-amino acids present in a protein are linked to one another.


How do you explain the amphoteric behaviour of amino acid.
Answer – Amino acids contain acidic and basic group within same molecule. In aqueous solution they neutralize each other, carboxyl group loses a proton and amino group accept it.

27. Give equations of the following reactions :
(a) Oxidation of propanol with alkaline KMnO4 solution
(b) Bromine in CS2 with phenol
(c) Propanone with methyl magnesium bromide
Answer –

28. For a certain chemical reaction, variation in the concentration, ln[R] vs time(min) plot is shown below.

For this reaction :
(a) What is the order of reaction ?
Answer : 1st order

(b) What are the units of rate constant K for this reaction ?
Answer : min-¹

(c) Give the relationship between K and t½ of this reaction.
Answer : t½ = 0.693/K

29. Two elements A and B form compounds having formula AB2 and AB4 when dissolved in 20 gm of benzene (C6H6), 1 gm of AB2 lowers the freezing point by 2.3 K where as 1.0 gm of AB4 lowers it by 1.3 K. The molal depression constant for benzene is 5.1 Kkg/mol. Calculate atomic masses of A and B.
Answer : For AB2
MAB2 = (Kf × WB × 1000) ÷ (WA × ∆Tf) = (5.1 × 1 × 1000) ÷ (20 × 2.3) = 110.87 u
MAB4 = (5.1 × 1 × 1000) ÷ (20 × 1.3) = 196.50 u
Atomic mass of A = a and Atomic mass of B = b.
a + 2b = 110.87 …….(i)
a + 4b = 196.50 ……..(ii)
Solve eq.(i) and (ii), we get
a = 25.29u and b = 42.64u
So, Atomic mass of A is 25.29 u and Atomic mass of B is 42.64 u

30. Write the product of the following reactions :

Answer –


Do the following conversions :
(a) Ethanal to But-2-enal
(b) Benzaldehyde to 3-phenyl propan-1-ol
(c) Benzone acid to Benzaldehyde
Answer –

SECTION – D (8 Marks)
(2 Case Study Questions of 4 Marks Each)

31. One of the latest theory to explain bonding in coordination compounds is crystal field theory which consider metal ligand bond to be ionic unlike valence bond theory. It considers the effect of ligands on the relative energy of d orbitals of the central metal atom/ion. In free transition metal ion, all the d-orbitals are degenerate. However this degeneracy split in the presence of ligands which may be anions or polar molecules like H2O, NH3 etc. The pattern of splitting depends upon the nature of crystal field generated by the ligands around the metal ion.
Answer the following questions :
(a) What is the crystal field splitting energy ?
Answer – The difference of energy between the two sets of a orbitals is called as crystal field splitting energy.

(b) Draw crystal field splitting diagram for Octahedral field.
Answer –


(b) The hexaaqua manganese (II) ion contains five unpaired electrons while the hexacyano ion contain only one unpaired electron. Explain using crystal field theory.
Answer : Mn²+ = 4s°3d⁵
H2O being weak ligand, don’t cause pairing 5 unpaired electron. i.e. t2g³ eg²
CN- strong ligand, cause pairing so there is 1 unpaired electron. i.e. t2g⁵ eg°

(c) How does the magnitude of ∆o decide the actual configuration of d-orbital in a coordination entity.
Answer : ∆o > P (pairing occurs)
o < P (no pairing occurs)

32. Alkyl halides undergo two types of nucleophillic substitution reactions SN¹ and SN². SN¹ reactions are two step reactions which proceed through the formation of carbocation while SN² are one step reactions proceed through the formation of transition state.
Answer the following questions :
(a) Out of C6H5CH2Cl and C6H5CHClC6H5 which is more easily hydrolysed by aqueous KOH following SN¹ mechanism.
Answer : C6H5CHClC6H5

(b) Arrange the following compounds in the order of reactivity towards SN² displacement
2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
Answer : 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane

(c) Why allyl chloride is hydrolysed through SN¹ mechanism.
Answer – Allylic carbocation is stable

(d) Why 1-Jodobutane is more reactive than 1-Chlorobutane towards nucleophilic substitution reaction.
Answer : I- is better leaving group than Cl-.


Answer –

SECTION – E (15 Marks)
(3 Questions of 5 Marks Each)

33. (a) Describe the preparation of potassium dichromate from iron chromite ore. Write reactions involved in the preparation.
Answer : (i) 4FeOCr2O3 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2
(ii) 2Na2CrO4 + conc. H2SO4 → Na2Cr2O7 + Na2SO4 + H2O
(iii) Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

(b) Describe the oxidizing action of potassium dichromate on (Write ionic equation for the reactions)
(i) iodide ion
Answer : Cr2O7²- + 14H+ + 6I- → 2Cr³+ + 7H2O + 3I2

(ii) iron (II) solution
Answer : Cr2O7²- + 6Fe²+ + 14H+ → 2Cr³+ + 6Fe³ + 7H2O


(a) Describe the preparation of potassium permanganate.
Answer : (i) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
(ii) 2K2MnO4 + Cl2 → 2KMnO4 + 2KCl

(b) Compare the chemistry of Lanthanoids with that of Actinoids with special reference to :
(i) Electronic Configuration
Answer – Lanthanoids have general electronic configuration [Xe] 4f¹-¹⁴ 5d⁰-¹ 6s²
Actinoids have general electronic configuration [Rn] 5f¹-¹⁴ 6d⁰-¹ 7s²

(ii) Atomic and Ionic size
Answer – Regular decrease in size from left to right is known as Lanthanoid contraction.
Regular decrease in size from left to right known as Actinoid contraction.

(iii) Chemical reactivity
Answer – Highly electropositive lanthanoids have almost similar chemical reactivity. Actinoids (electropostive and highly reactive) are more reactive (specially in finely divided state) than lanthanoids.

34. (a) Write the Nernst equation and calculate emf of the following cells at 298 K
Mg²+ (0.001M)/Mg(s) || Cu²+ (0.0001M)/Cu(s)
E°Mg²+/Mg = -2.36V and E°Cu²+/Cu = 0.34V
Answer : E°cell = E°cathode – E°anode = E°Cu²+/Cu – E°Mg²+/Mg = 0.34 – (-2.36) = 2.70
Mg(s) + Cu²+ → Mg²+ + Cu(s)
Ecell = E°cell – 0.0591/2 Log[Mg²+]/[Cu²+]
= 2.70 – 0.0591/2 Log[0.001]/[0.0001]
= 2.70 – 0.0295 Log10
= 2.70 – 0.0295×1 = 2.6705
Ecell = 2.6705 V

(b) Why does the conductivity of a solution decrease with dilution ?
Answer – Because the number of ions per unit volume decreases.


(a) Write the chemistry of recharging of the lead storage battery highlighting all the material that are involved during recharging.
Answer – (i) During recharging, cell is operated like electrolytic cell.
(ii) Electrical energy is supplied to it from external source.
(iii) Electrode reactions are reverse of that of discharging.
(iv) At cathode (Reduction) :
PbSO4(s) + 2e- → Pb(s) + SO4²-(aq)
At Anode (oxidation) :
PbSO4(s) + 2H2O → PbO2(s) + SO4²-(aq) + 4H+ + 2e-
Overall reaction :
2PbSO4 + 2H2O → Pb(s) + PbO2(s) + 4H(aq)+ + 2SO4²-(aq)

(b) Can we store copper sulphate solution in a zinc pot ?
Answer – We can’t store CuSO4 in zinc pot because E°cell +ve means reaction is spontaneous and in this reaction zinc is oxidised.
cell = E°Cu²+/Cu – E°Zn²+/Zn = 0.34 – (-0.76) = + 1.10 V

35. Write the structures of product of the following reactions :

Answer –


(a) Give chemical test to distinguish between phenol and benzoic acid.
Answer – Phenol gives violet colouration with neutral FeCl3 solution but benzoic acid does not.

(b) Out of acetaldehyde and acetone which is more reactive towards nucleophillic addition reaction and why ?
Answer – Acetaldehyde is more reactive towards nucleophillic addition reaction.

(c) An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces tollen’s reagent and undergoes eannizaro reaction. On vigorous oxidation it gives 1,2-benzene dicarboxylic acid. Identify the compound. Write the reactions involved.
Answer –


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