Class 12 Chemistry Sample Paper

Sample Paper Class 12

Please refer to the Class 12 Chemistry Sample Paper for the current academic year given below. We have provided the latest CBSE Sample Papers for Term 1 and Term 2 for Chemistry Class 12. All guess sample papers have been prepared based on the latest blueprint and examination pattern for the current year. All sample papers for Chemistry Class 12 Term 1 and 2 have been given with solutions. Students can access the multiple guess papers given below. Practicing more Class 12 Chemistry Sample Papers will help you to get more marks in upcoming exams.

CBSE Sample Papers for Class 12 Chemistry

Class 12 Chemistry Sample Paper Set A
Class 12 Chemistry Sample Paper Set B
Class 12 Chemistry Sample Paper Set C
Term 1 Sample Papers for Class 12 Chemistry
Class 12 Chemistry Sample Paper Term 1 Set A
Class 12 Chemistry Sample Paper Term 1 Set B
Class 12 Chemistry Sample Paper Term 1 Set C

Class 12 Chemistry Sample Paper Term 1 Set A

1. Name the non stoichiometric point defect responsible for colour in alkali metal halides.
Answer. Metal excess or anionic vacancies or F-centres

2. What is shape selective catalysis?
Answer. Catalysis which depends on the shape and size of the reactants and the products and to those of the pores and cavities of catalyst.

3. Amongst the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields a single monochloride.
Answer. Neopentane or 2,2-Dimethylpropane

4. Give the IUPAC name and structure of the amine obtained when 3-chlorobutanamide undergoes
Hoffmann–bromamide reaction.
Answer. 2-Chloropropanamine : CH3CH(Cl)CH2NH2

5. How many ions are produced from the complex,
[Co(NH3)6]Cl2 in solution?
Answer. Three ions : [Co(NH3)6]2+ and 2Cl

6. Will the elevation in boiling point be same if 0.1 mol of sodium chloride or 0.1 mol of sugar is dissolved in 1L of water? Explain.
Answer. No, the elevation in boiling point is not the same. Elevation in boiling point is a colligative property which depends on the number of particles. NaCl is an ionic compound which dissociates in solution to give more number of particles whereas sugar is made up of molecules and thus does not dissociate.

7. The following curve is obtained when molar conductivity (Λm) is plotted against the square root of concentration, C1/2 for two electrolytes A and B

(a) How do you account for the increase in the molar conductivity of the electrolyte A on dilution.
(b) As seen from the graph, the value of limiting molar conductivity (Λ°m) for electrolyte B cannot be obtained graphically. How can this value be obtained?

Answer. (a) As seen from the graph, electrolyte A is a strong electrolyte which is completely ionised in solution. With dilution, the ions are far apart from each other and hence the molar conductivity increases.
(b) To determine the value of limiting molar conductivity for electrolyte B, indirect method based upon Kohlrausch’s law of independent migration of ions is used.

8. Name the following :
(a) A transition metal which does not exhibit variation in oxidation state in its compounds.
(b) A compound where the transition metal is in the +7 oxidation state.
(c) A member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(d) Ore used in the preparation of potassium dichromate.
(a) Scandium (Sc)
(b) KMnO4
(c) Cerium (Ce)
(d) Chromite ore

9. Arrange the following in order of property indicated for each set :
(a) F2, Cl2, Br2, I2 – increasing bond dissociation enthalpy
(b) PH3, AsH3, BiH3, SbH3, NH3 – increasing base strength
(a) I2 < F2 < Br2 < Cl2 [1]
(b) BiH3 < SbH3 < AsH3 < PH3 < NH3

10. (a) Predict the major product of acid catalysed dehydration of 1-Methylcyclohexanol.
(b) You are given benzene, conc.H2SO4, NaOH and dil.HCl. Write the preparation of phenol using these reagents.
Draw the structures of any two isomeric alcohols (other than 1° alcohols) having molecular formula C5H12O and give their IUPAC names.
Answer. (a) 1-Methylcyclohexene

(i) CH3–CH2–CH2–CH(OH)–CH3 : Pentan-2-ol
(ii) CH3–CH2–CH(OH)–CH2–CH3 : Pentan-3-ol

11. An element occurs in the bcc structure with cell edge of 288 pm. The density of the element is 7.2 g cm–3. How many atoms of the element does 208 g of the element contain?
Answer. For the bcc structure, Z = 2

Or M = 51.8 g mol–1
By mole concept,
51.8 g of the element contains 6.022 × 1023 atoms
208 g of the element will contain

12. Calculate the boiling point of a 1 M aqueous solution (density 1.04 g mL–1) of potassium chloride (Kb for water = 0.52 K kg mol–1, Atomic masses
: K = 39 u, Cl = 35.5 u. Assume, potassium chloride is completely dissociated in solution.
Answer. Molar mass of KCl = 39 + 35.5 = 74.5 g mol–1
As KCl dissociates completely, number of ions produced are 2.
Therefore, van’t Ho factor, i = 2
Mass of KCl solution = 1000 × 1.04 = 1040 g
Mass of solvent = 1040 – 74.5 = 965.5 g = 0.9655 kg
Molality of the solution :

= 2 × 0.52 × 1.0357 = 1.078°C
Therefore, boiling point of solution
= 100 + 1.078 = 101.078°C

13. A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution.
Calculate the emf of the cell. Write the chemical equaiton for the electrode reactions and represent the cell.
(Given : E°Zn2+/Zn = –0.76V; E°Pb2+/Pb = –0.13V)
Answer. node reaction : Zn(s) → Zn2+(aq) + 2e
Cathode reaction : Pb2+(aq) + 2e– → Pb(s)
Cell representation : Zn(s) |Zn2+(aq) || Pb2+(aq)|Pb(s)
According to Nernst equation :

= 0.63 – 0.0295 × log 5
= 0.63 – 0.0295 × 0.6989
= 0.63 – 0.0206 = 0.6094 V

14. Answer the following questions :
(a) What happens when a freshly precipitated Fe(OH)3 is shaken with a little amount of dilute solution of FeCl3?
(b) Why are lyophilic colloidal sols more stable than lyophobic colloidal sols ?
(c) What form Freundlich adsorption equation will take at high pressure ?
Answer. (a) A reddish brown positively charged colloidal sol is obtained.
(b) Stability of lyophilic sols is due to :
(i) same charge on all the colloidal particles.
(ii) solvation of the colloidal particles.
(c) At high pressures, amount of gas adsorbed (x/m) becomes independent of pressure (p).

15. What chemical principle is involved in choosing a reducing agent for getting the metal from its oxide ore? Consider the metal oxides, Al2O3 and FeO and justify the choice of reducing agent in each case.

Account for the following facts :
(a) The reduction of a metal oxide is easier if the metal formed is in the liquid state at the temperature of reduction.
(b) Limestone is used in the manufacture of pig iron from haematite.
(c) Pine oil is used in the froth floatation process used to concentrate sulphide ores.
Answer. The feasibility of thermal reduction can be predicted on the basis of Ellingham diagram. Metals for which the standard free energy of formation (ΔfG°) is more negative can reduce those metals for which ΔfG° is less negative. At a given temperature, any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram. 
(a) Below the temperature approx 1623 K, corresponding to the point of intersection of Al2O3 and MgO curves, Mg can reduce alumina.
(b) At temperatures between 673 K and 1073 K, the CO , CO2 line lies below Fe, FeO line, thus CO can reduce FeO to Fe. At temperatures above this temperature, coke will reduce FeO and itself get oxidised to CO.
(a) Entropy is higher when a metal is in the liquid state than when it is in the solid state. Thus, TΔS° increases, thus, ΔG° becomes more negative and the
reduction becomes easier. (ΔG° = ΔH° – TΔS°) (b) Limestone provides the flux (CaO) which combines with the impurities (SiO2) to form slag (CaSiO3). Thus, it helps in the removal of impurities.
(c) Pine oil (collector) enhances the non wettability of the ore particles, which become lighter and hence rise to the surface along with the froth.

16. (i) For M2+/M and M3+/M2+ systems, E° values for some metals are as follows :
Cr2+/Cr = –0.9 V Cr3+/Cr2+ = –0.4 V
Mn2+/Mn = –1.2 V Mn3+/Mn2+ = +1.5 V
Fe2+/Fe = –0.4 V Fe3+/Fe2+ = +0.8 V
Use this data to comment upon
(a) The stability of Fe3+ in acid solution as compared to that of Cr3+ and Mn3+
(b) The ease with which iron can be oxidised as compared to the similar process for either Cr or Mn metals.
(ii) What can be inferred from the magnetic moment of the complex K4[Mn(CN)6]
Magnetic moment : 2.2 BM?
Answer. (i) (a) Cr3+/Cr2+ has a negative reduction potential. Hence, Cr3+ cannot be reduced to Cr2+. Cr3+ is most stable. Mn3+/Mn2+ have large positive E° values. Hence, Mn3+ can be easily reduced to Mn2+. Thus, Mn3+ is least stable. Fe3+/Fe2+ couple has a positive E° value but smaller than Mn3+/Mn2+. Thus, the stability of Fe3+ is more than Mn3+ but less stable than Cr3+
(b) If we compare the reduction potential values, Mn2+/Mn has the most negative value i.e., its oxidation potential value is most positive. Thus, it is most easily oxidised. us, the decreasing order for
their ease of oxidation is Mn > Cr > Fe.
(ii) K4[Mn(CN)6] Mn is in +2 oxidation state. Magnetic moment 1.4 indicates that it has one unpaired electron and hence forms inner orbital or low spin complex. In presence of CN(a strong ligand), hybridisation involved is d2sp3 (octahedral complex).

17. (i) Describe the type of hybridisation for the complex ion [Fe(H2O)6]2+.
(ii) Write the IUPAC name of the ionisation isomer of the coordination compound
[Co(NH3)5Br]SO4. Give one chemical test to distinguish between the two compounds.
Answer. (i) Fe exists as Fe2+. There are 4 unpaired electrons.

Water is a weak ligand. Thus, the hybridation involved is sp3d2. It is an octahedral outer orbital complex.
(ii) The ionisation isomer is [Co(NH3)5SO4]Br. The IUPAC name is Pentaamminesulphatocobalt (III) bromide.
The isomer [Co(NH3)5Br]SO4 gives a white precipitate of BaSO4 with BaCl2 solution whereas the isomer [Co(NH3)5SO4]Br does not form this precipitate.

18. (a) Explain why the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(b) An optically active compound having molecular formula C7H15Br reacts with aqueous KOH to give a racemic mixture of products. Write the mechanism involved in this reaction.
Answer. (a) Due to greater s-character, a sp2 hybrid carbon is more electronegative than a sp3 hybrid carbon. Therefore, the sp2 hybrid carbon of C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than a sp3 hybrid carbon of cyclohexyl chloride.

(b) Since the alkyl halide reacts with KOH to form a racemic mixture, it must be a 3° alkyl halide and the reaction will follow SNl mechanism.

19. Give the structures of A, B and C in the following reactions :

Answer. (a) A – C6H5NH2         B – C6H5N2+Cl
C – C6H5 – N2 – C6H4– OH
(b) A – C6H5CN          B – C6H5COOH

20. (a) A non reducing disaccharide ‘A’ on hydrolysis with dilute acid gives an equimolar mixture of D-(+)-glucose and D-(–)-Fructose.

Identify A. What is the mixture of D-(+)-glucose and D-(–)-fructose known as? Name the linkage that holds the two units in the disaccharide.
(b) a-amino acids have relatively higher melting points than the corresponding halo acids. Explain.
Answer. (a) A–Sucrose (C12H22O11)
The mixture of D-(+)-glucose and D-(–)-fructose is known as invert sugar.
The linkage which holds the two monosaccharide units through oxygen atom is called glycosidic
(b) The amino acids exist as dipolar zwitter ion. Due to this dipolar salt like character they have strong dipole attractions. Thus, their melting points are higher than the corresponding haloacids which do not exist as zwitter ions.

21. (a) Pick out the odd one from among the following on the basis of their medicinal properties mentioning the reasion : Luminal, Seconal, Phenacetin, Equanil.
(b) Give an example of a substance that can act as a disinfectant as well as antiseptic depending upon its concentration. (Specifiy
(c) Name any two macromolecules chosen as drug targets.
Answer. (a) Phenacetin is an antipyretic, while the rest are tranquilizers.
(b) 0.2% solution of phenol acts as antiseptic whereas 1% solution of phenol acts as disinfectant.
(c) Carbohydrates and proteins

22. The following is not an appropriate reaction for the preparation of tert-butyl ethyl ether :
C2H5ONa + (CH3)3C Cl → (CH3)3C OC2H5
(i) What would be the major product of the given reaction?
(ii) Write a suitable reaction for the preparation of tert-butyl ethyl ether, specifying the names of reagents used.
Justify your answer in both cases.
Answer. (i) Since the alkyl halide is a 3° halide and C2H5ONa is strong base, therefore, elimination occurs preferably. The product obtained is 2-methylprop-1-ene. CH3C(CH3) CH2
(ii) To prepare tert-butyl ethyl ether, the alkyl halide should be 1° i.e., chloroethane and the nucleophile should be sodium tert-butoxide because the 3° nucleophile is able to attack 1° alkyl halide.

23. Study the given passage carefully and answer the questions that follow :
Shalini studied a chapter on polymers in school and came across the following paragraph :
The durability, strength, low cost, water and chemicals resistance and light weight are advantages of plastic bags.
Shalini is confused as she has been reading in the newspaper about the ban on the usage of plastic substances. She further finds that despite the durability, the use of these materials has presented mankind with serious waste disposal problem as these materials do not disintegrate by themselves. In view of this, certain polymers are being developed which are broken down rapidly by microorganisms. Shalini feels relaxed that such kinds of biomaterials are being developed.
(a) Name the class of these useful polymers which do not harm the environment.
(b) Give any one example of these polymers and name its monomers.
(c) Comment on the qualities of Shalini.
Answer. (a) The class of polymers is biodegradable polymers.
(b) One example of biodegradable polymers is PHβV (poly-β-hydroxybutyrate-co-β-hydroxyvalerate) The names of its monomers are : 3-hydroxybutanoic acid and 3-hydroxypentanoic acid.
(c) Care for environment, concern for the health of the people.

24. (a) Give a plausible explanation for each one of the following :
(i) Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol.
(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(b) Carry out the following conversions in not more than two steps :
(i) Phenyl magnesium bromide to benzoic acid.
(ii) Acetaldehyde to but-2-enal.
(iii) Benzene to m-nitroacetophenone
(a) Give a simple chemical test to distinguish between the pair of organic compounds :
Ethanal and Propanal
(b) Name and complete the following chemical reaction :

(c) Draw the structures of the following derivatives :
(i) The 2,4-Dintitrophenylhydrazone of benzaldehyde
(ii) Acetaldehydedimethyl acetal
(iii) Cyclopropanone oxime

In case of phenoxide ion, strucutres (II-IV) carry a negative charge on the less electronegatice carbon atom.
In structure I and V, the negative charge on the oxygen atom remains localized while the electrons of the benzene rign only are delocalized.
While the negative charge on the carboxylate ion is delocalized over two oxygen atoms and in structure III and VII, the negative charge on the oxygen atom remians localized while the electrons of the benzene ring only are delocalized.
Thus, carboxylic acids are stronger acids than phenols.
(ii) Semicarbazide has two –NH2 groups. One oft hem, which is directly attached to C = O is involved in resonance. Thus, electron density on this group decreases and it does not act as a nucleophile. In contrast, the lone pair of electrons on the other –NH2 group is available for nucleophilic attack.
(b) (i) PhMgBr + O = C = O → PhCOOMgBr

(a) Ethanal and propanal can be distinghished by iodofrom test.
Ethanal gives a yellow precipitate of iodoform with an alkaline solution of NaOH. Propanal does not give this test.
CH3CHO + 4NaOH + 3I2 → CHI3 + HCOONa + 3H2O + 3NaI

25. (a) Write the rate law for a first order reaction. Justify the statement that half life for a first order reaction is independent of the initial concentration of the reactant.
(b) For a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%.
(a) For the reaction A→B, the rate of reaction becomes twenty seven times when the What is the order of the reaction?
(b) The activation energy of a reaction is 75.2 kJ mol–1 in the absence of a catalyst and it lowers to 50.14 kJ mol–1 with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25°C?
Answer. (a) For a first order reaction

[R] = conc. after time t
When half of the reaction is completed, [R] = [R]0/2.
Representing, the time taken for half of the reaction to be completed, by t½, equation becomes :

The above equation shows that half life first order reaction is independent of the initial concentration of the reactant.
(b) For a first order reaction,

26. (a) Write balanced chemical equations for the following :
(i) Complete hydrolysis of XeF6.
(ii) Disproportionation reaction of orthophosphorus acid.
(b) Draw the structure of a noble gas species which is isostructural with BrO3.
(c) Considering the parameters such as bond dissociation enthalpy, electron gain entalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2.
(d) Why is Ka2 << Ka1 for H2SO4 in water?
Explain the following :
(a) Hydrogen fluoride is a weaker acid than hydrogen chloride in aqueous solution.
(b) PCl5 is ionic in nature in the solid state.
(c) SF6 is inert towards hydrolysis.
(d) H3PO3 is diprotic.
(e) Out of noble gases only Xenon is known to form established chemical compounds.

(c) The bond dissociation enthalpy of F — F bond is lower than that of Cl — Cl bond and hydration enthalpy of F– ion is much higher than that of Cl– ion.
These two factors more than compensate the less negative electron gain enthalpy of F2. Thus, F2 is a stronger oxidizing agent than Cl2.
(d) H2SO4 ionises in two stages and hence has two dissociation constants.
Ka2 << Ka1.
This is because the negatively charged HSO4 ion has much less tendency to donate a proton to H2O as compared to neutral H2SO4.
(a) Due to stronger H-F bond than HCl bond, HF ionises less readily than HCl in aqueous solution to give H+ ions. Therefore, HF is a weaker acid than HCl.
(b) In solid state, PCl5 consists of ions [PCl4]+[PCl6]– On melting, these ions becomes free to move and
hence conducts electricity in the molten state.
(c) In SF6, S is sterically protected by six F atoms and hence does not allow H2O molecules to attack the S molecule. Also, F does not have d-orbitals to accept the electrons donated by H2O molecules.
(d) In the structure of H3PO3, it contains only two ionisable H-atoms which are present as –OH groups thus, it behaves as a dibasic acid.
(e) Except radon which is radiocative, Xenon has least ionisation energy among noble gases and hence it forms chemical compounds particularly with O2 and F2.

Class 12 Chemistry Sample Paper