We have provided Class 12 Chemistry Sample Paper as per the latest CBSE examination pattern for the current academic year. The following CBSE Sample Papers for Class 12 Chemistry has been prepared based on the guess papers issued recently. Students will be able to practice these papers and get good marks in upcoming Chemistry exams for Class 12.
CBSE Sample Papers for Class 12 Chemistry
Class 12 Chemistry Sample Paper Term 1 With Solutions Set A
1. Write the structure of 1-bromo-4-sec-butyl-2-methylbenzene.
2. For the reaction, X2 + 2Y2 → 2XY2, write the rate equation in terms of the rate of disappearance of Y2.
3. Pick out the odd one amongst the following on the basis of their medicinal properties mentioning the reason.
Chloroxylenol, phenol, chloramphenicol, bithional.
Answer. Chloramphenicol is an antibiotic while all the remaining are antiseptics.
4. Aniline is a weaker base than cyclohexylamine. Why?
Answer. In aniline there is delocalisation of lone pair of electrons of N atom in benzene ring therefore it is less basic than cyclohexylamine
5. Account for the following : o-nitrophenol has lower boiling point than p-nitrophenol.
Answer. o-Nitrophenol has intramolecular H-bonding which is weaker than intermolecular H-bonding present in p-nitrophenol.
6. A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Answer. No. of atoms in the close packing = 0.5 mol
= 0.5 × 6.022 × 1023 = 3.011 × 1023
No. of octahedral voids = 1 × No. of atoms in the packing = 3.011 × 1023
No. of tetrahedral voids = 2 × No. of atoms in
the packing = 2 × 3.011 × 1023 = 6.022 × 1023
Total no. of voids = 3.011 × 1023 + 6.022 × 1023
= 9.033 × 1023
7. Rate constant of the first order reaction is 6.93 × 10–3 s–1. Calculate
(ii) time of 75% completion of reaction.
Answer. (i) For first order reaction,
(ii) For first order reaction, half-life is related to time of 75% completion of reaction by t75 = 2 × half-life = 2 × 100 = 200 s
8. (i) Give the electronic configuration of the d-orbitals of Ti in [Ti(H2O)6]3+ ion in an octahedral crystal field.
(ii) Why is the given complex coloured? Explain on the basis of distribution of electrons in the d-orbitals.
Answer. (i) In [Ti(H2O)6]3+ ion, Ti is in +3 oxidation state. The electronic configuration of Ti3+ion is 3d1. In an octahedral field, it is t1 2ge0g.
(ii) Due to d-d transition, the electron present in t2g absorbs green and yellow radiation of white light for excitation to eg. The complementary colour is purple.
9. Predict the products of electrolysis obtained at the electrodes in each case when the electrodes used are of platinum :
(i) An aqueous solution of AgNO3.
(ii) An aqueous solution of H2SO4.
10. Explain how vacancies are introduced in an ionic solid when a cation of higher valency is added as an impurity in it.
Ionic solids, which have anionic vacancies due to metal-excess defect, develop colour. Explain with the help of a suitable example.
Answer. When a cation of higher valency is added as animpurity in the ionic solid, some of the sites of the original cations are occupied by the cations of higher valency. Each cation of higher valency replaces two or more original cations and occupies the site of one original cation and the other site(s) remains vacant.
Cationic vacancies produced = Number of cations of higher valency × Difference in valencies of the original cation and cation of higher valency
In ionic solids with anionic vacancies due to metal-excess defect, when the metal atoms deposit on the surface, they diffuse into the crystal and after ionization, the metal ion occupies cationic vacancy while electron occupies anionic vacancy. These electrons get excited to higher energy levels by absorption of suitable wavelengths from the visible white light and, therefore appear coloured.
Example : Zinc oxide is white in colour at room temperature. On heating it loses oxygen and turns yellow.
The excess Zn2+ ions move to interstitial sites and the electrons to neighbouring interstitial sites. These electrons absorb visible light and impart yellow colour to zinc oxide.
11. Give reasons for the following :
(i) Addition of Cl2 to KI solution gives it a brown colour, but excess of Cl2 turns it colourless.
(ii) Phosphinic acid behaves as a monoprotic acid.
(iii)White phosphorus is much more reactive than red phosphorus.
Answer. (i) Chlorine being stronger oxidising agent than iodine displaces iodine from KI which brings brown colour to the solution. In excess of chlorine, the liberated iodine is further oxidised to iodic acid and solution becomes colourless.
Cl2 + 2KI → 2KCl + I2;
I2 + 5Cl2 + 6H2O → 2HIO3 + 10HCl
(ii) Phosphinic acid has only one replaceable hydrogen atom.
(iii) White P4 is monomeric whereas red phosphorus is polymeric. White P4 has less bond dissociation energy than red P4.
12. (i) A given ore is a mixed sulphide of lead and zinc (PbS-ZnS). Suggest a method to separate the mixture.
(ii) Write the chemical reactions involved in the
extraction of gold by cyanide process.
Answer. (i) PbS and ZnS both form froth with pine oil on bubbling air inside the mixture. In such case NaCN is used as a depressant.
Before froth floatation process is used, NaCN is mixed (as a depressant) which forms a complex with ZnS.
ZnS + 4NaCN → Na2[Zn(CN)4] + Na2S PbS remains insoluble and is separated.
CuSO4 is then added to activate depressed ZnS and air is blown when ZnS floats. This method is called differential floatation.
(ii) Step I : 4Au(s) + 8CN–(aq) + O2(g) + 2H2O(aq) →
4[Au(CN)2]–( aq) + 4OH–(aq)
Step II : 2[Au(CN)2]–(aq) + Zn(s) →
2Au(s) + [Zn(CN)4]2–(aq)
13. Explain the following terms with a suitable example in each case :
(i) Shape-selective catalysis
Answer. (i) Shape-selective catalysis : The catalytic reaction which depends upon the pore size of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts because of their honey comb-like structures.
(ii) Dialysis : The process of separating a crystalloid from a colloid by flteration or diffusion through a membrane is called dialysis. The apparatus employed to affect such a separation is known as dialyser. Purification of blood in the artificial kidney machine.
(iii) Multimolecular colloids : A large number of atoms or smaller molecules of a substance on dissolution aggregate together to form species having size (diameter < 1 nm) in the colloidal range (1–1000 nm). Such species are known as multimolecular colloids. For example, a sulphur sol consists of particles containing a thousand or more of S8 sulphur molecules.
14. (i) Give the IUPAC name of
(ii) Compare the magnetic behaviour of the complex entities [Fe(CN)6]4– and [FeF6]3–.
[Fe = 26].
(i) Diamminechlorido(methylamine) platinum(II) chloride
(ii) (a) [Fe(CN)6]4– ion
15. A zinc rod is dipped in 0.1 M solution of ZnSO4.
the salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential (E°Zn2+/Zn = –0.76 V)
Answer. The electrode reaction written as reduction reaction is Zn2+ + 2e– → Zn(n = 2) Applying Nernst equation, we get
= – 0.76 – 0.02955 (log 1000 – log 95)
= – 0.76 – 0.02955 (3 – 1.9777)
= – 0.76 – 0.03021 = – 0.79021 V
16. Write the structures of A, B and C in the following reactions.
17. Write structures and names of the monomers of the following polymers :
(i) Nylon 6,6
(ii) Nylon 6
(iii) Urea formaldehyde resin
(i) How will you differentiate between low density and high density polythenes?
(ii) Arrange the following polymers in order of increasing intermolecular forces :
Bakelite, Nylon 6,6, Polythene, Neoprene.
Answer. (i) HOOC(CH2)4COOH and NH2(CH2)6NH2
Adipic acid Hexamethylene
18. State Henry’s law for solubility of a gas in a liquid and give its important applications.
Answer. Henry’s law : Mole fraction of a gas in the solution is proportional to the partial pressure of the gas over the solution.
Applications of Henry’s law :
(i) Soft drinks and soda water bottles are sealed under high pressure to increase the solubility of CO2.
(ii) The tanks used by scuba divers [persons swimming under water use underwater breathing apparatus known as scuba] are filled with air diluted with helium.
19. Give plausible explanation for each of the following :
(i) There are two —NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
(ii) Give the mechanism of esterification of carboxylic acids.
As electron density on one —NH2 group decreases due to resonance it does not act as a nucleophile while the lone pair of electrons on the other NH2 group (i.e., attached to NH) is not involved in resonance and hence is available for nucleophilic attack on the C = O group of aldehydes and ketones.
(ii) Mechanism of esterification : It is a nucleophilic acyl substitution.
(a) Protonation of carboxyl oxygen:
20. (i) What is the structural difference between a nucleoside and a nucleotide?
(ii) “The two strands of DNA are not identical but are complementary.” Explain.
(iii) Differentiate between flbrous and globular proteins.
Answer. (i) A unit formed by attachment of a base to C-1 position of sugar is known as nucleoside. When nucleoside is linked to phosphoric acid at C-5′ position of sugar moiety, resulting species is called nucleotide.
(ii) James Watson and Francis Crick gave a double strand helix structure for DNA. Two nucleic acid chains are bonded to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine (A = T) whereas cytosine forms hydrogen bonds with guanine (C ≡ G).
21. Mention the action of the following on the human body in bringing relief from a disease.
22. Outline the principles of refining of metals by the following methods :
(i) Zone refining
(ii) van Arkel method
(iii) Liquation method
Answer. (i) Zone refining : The method is based on the difference in solubility of impurities in molten and solid state of the metal.
(ii) van Arkel method : In this method, impure metal is converted to a volatile compound while the impurities are not affected. The volatile compound is then decomposed to get pure metal.
(iii) Liquation method : this process is based on the difference in fusibility of metal and impurities. Impure metal is gently heated on the sloping hearth of a furnace. The metal melts and flows down leaving behind the impurities on the hearth.
23. Dr. Rachna – a dietician conducted a seminar for the students of class XII. She emphasized the importance of balanced diet for good health. She also discussed with students that under special conditions like for pregnant women, growing children and HIV/AIDS patients or heart
patients, protein rich diet is recommended.
(i) What values are displayed by Dr. Rachna?
(ii) Why do HIV/AIDS or heart patients need a protein rich diet?
(iii) Name some rich sources of proteins.
(iv) What are the chief components of a balanced diet?
Answer. (i) Dr. Rachna, showed concern towards the health needs of students.
(ii) Protein rich diet helps to strengthen the immune system of HIV/AIDS patients and helps to reduce blood pressure as well as cholesterol of heart patients.
(iii) Eggs, sprouts, beans, lentils are rich sources of proteins.
(iv) Carbohydrates, fats and proteins are basic components of a balanced diet.
24. (i) Give one chemical test to distinguish between the following pairs of compounds:
(a) Methylamine and dimethylamine
(b) Ethanal and propanal
(c) Benzoic acid and ethyl benzoate
(ii) How will you bring about the following conversions in not more than two steps?
(a) Benzene to m-nitroacetophenone
(b) 2-Methylpropanol to 2-methylpropene
(i) An organic compound (A) [molecular formula C8H16O2] was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of
(C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.
(ii) Write the equations involved in the following reactions :
(a) Kolbe’s reaction
(b) Cannizzaro reaction
Answer. (i) (a) Methylamine and dimethylamine can be distinguished by carbylamine test. Methylamine, a primary amine, gives offensive smell on heating with chloroform and alcoholic solution of KOH whereas dimethylamine does not react.
(b) Ethanal and propanal can be distinguished by iodoform test. Yellow precipitate of iodoform will be formed from ethanal on heating with iodine and sodium hydroxide solution whereas propanal does not give iodoform test.
(c) Benzoic acid and ethyl benzoate can be distinguished by their reactions with sodium bicarbonate solution. Benzoic acid will give effervescence with NaHCO3 whereas ethyl benzoate does not react.
(i) Compound ‘A’ (C8H16O2) on hydrolysis gives an acid ‘B’ and an alcohol ‘C’. It shows that ‘A’ is an ester. Since the oxidation of alcohol ‘C’ also gives the acid ‘B’ indicates that ‘B’ and ‘C’ both contain same number of carbon atoms, i.e., four carbon atoms each and same arrangement of atoms. Formation of but-1-ene on dehydration of ‘C’ indicates it to be butan-1-ol, so the possible ester (A) could be butyl butanoate.
25. (i) Why is freezing point depression of 0.1 M sodium chloride solution nearly twice that of 0.1 M glucose solution?
(ii) At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If osmotic pressure of solution is 1.52 bar at the same temperature, what would be its concentration?
(iii) Osmotic pressure is more useful to calculate molecular mass than other colligative properties. Why?
(i) 15 g of an unknown molecular mass substance was dissolved in 450 g of water. The resulting solution freezes at – 0.34°C. What is the molar mass of the substance? (Kf for water = 1.86 K kg mol–1)
(ii) Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type.
Answer. (i) Sodium chloride being a strong electrolyte completely dissociates in the solution while glucose does not dissociate. The number of particles in sodium chloride solution becomes double than that of glucose solution. Hence freezing point depression of sodium chloride is nearly twice that of glucose solution of same molarity, because it is a colligative property..
(iii) Magnitude of osmotic pressure is large even for very dilute solutions and it can be measured at room temperature hence it is more useful for the calculation of molecular mass.
(i) Given: w2 = 15 g, w1 = 450 g
Kf = 1.86 K kg mol–1, M2 = ?
ΔTf = 0 – (– 0.34) = 0.34°C or 0.34 K
Now, ΔTf = Kf ⋅m
(ii) When the vapour pressure of a non-ideal solution is either higher or lower than that predicted by Raoult’s law, the solution exhibits deviation. These deviations are caused because of unequal intermolecular attractive forces between solute-solvent molecules and solute-solute or solvent-solvent molecules.
Examples of positive deviations : Mixture of ethanol and acetone, carbon disulphide and acetone. Examples of negative deviations : Chloroform and acetone, nitric acid and water.
26. Assign reason for the following :
(i) The enthalpies of atomisation of transition elements are high.
(ii) The transition metals and many of their compounds act as good catalyst.
(iii) Actinoid contraction is greater from element to element than the lanthanoid contraction.
(iv) The E° value for the Mn3+/Mn2+ couple
is much more positive than that for Cr3+/Cr2+.
(v) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as transition element.
(i) Write the steps involved in the preparation of
(a) K2Cr2O7 from Na2CrO4
(b) KMnO4 from K2MnO4.
(ii) What is meant by lanthanoid contraction?
What effect does it have on the chemistry of the elements which follow lanthanoid contraction?
Answer. (i) As transition metals have a large number of unpaired electrons in the d-orbitals of their atoms they have strong interatomic attractions or metallic bonds. Hence they have high enthalpy of atomization.
(ii) Many transition metals and their compounds act as catalysts. The catalytic activity is due to their ability to exhibit multiple oxidation states. For example, V2O5 in Contact process and finely divided iron in Haber process.
(iii) The actinoid contraction is more than lanthanoid contraction due to poor shielding by 5f-electrons than by 4f-electrons.
(iv) Much larger third ionisation energy of Mn (where change is d5 to d4) is mainly responsible for this. This also explains that +3 state of Mn is of little importance.
(v) Scandium (Z = 21) has incompletely filled 3d-orbitals in the ground state (3d1). Hence it is considered as a transition element.
(i) (a) 2Na2CrO4 + H2SO4 →
Na2Cr2O7 + Na2SO4 + H2O
Na2Cr2O7 + 2KCl →
K2Cr2O7 + 2NaCl
(b) The potassium manganate is oxidised to potassium permanganate by oxidation with chlorine.
2K2MnO4(aq) + Cl2(g) → 2KMnO4(aq) + 2KCl(g)
(ii) Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons. Lanthanum (La) has the largest ionic radius while lutetium (Lu) has the smallest among the 4f-series elements. Consequences of lanthanoid contraction :
(a) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(b) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g. Atomic radii of zirconium(Zr) is same as that of hafnium(Hf).