CBSE Previous Year Question Papers Class 12 Chemistry 2014 Delhi
Time allowed: 3 hours
Maximum Marks: 70
CBSE Previous Year Question Papers Class 12 Chemistry 2014 Delhi Set I
Give one example each of ‘oil in the water’ and ‘water in oil’ emulsion.  Answer:
Oil in water – Milk
Water in oil – Butter
Which reducing agent is employed to get copper from the leached low-grade copper ore?  Answer:
Scrap iron is used as a reducing agent to obtain copper metal from the solution containing copper.
Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq)
Which of the following is more stable complex and why?
[Co(NH3)6]3+ and [Co(en)3]3+  Answer:
NH3 is a unidentate ligand and H2NCH2CH2NH2(en) is a bidentate ligand. Chelating ligands form more stable complexes compared to non-chelating ligands. Thus, [Co(en)3]3+ is more stable.
Write the IUPAC name of the compound. 
Which of the following isomers is more volatile:
o-nitrophenol or ph-nitrophenol?  Answer:
o-nitrophenol is more volatile because of the presence of intramolecular hydrogen bonding.
What are isotonic solutions?  Answer:
Two or more solutions having the same osmotic pressure are called isotonic solutions.
e.g. 0.5 M NaCl, 0.5 M KCl and 1 M glucose are isotonic.
Arrange the following compounds in increasing order of solubility in water:  C6H5NH2, (C2H5)2NH, C2H5NH2
C6H5NH2 < (C2H5)2NH < C2H5NH2
Which of the two components of starch is water-soluble?  Answer:
Amylose is the water-soluble content of starch between Amylose and Amylopectin.
An element with density 11.2 gm cm-3 forms f.c.c. lattice with edge length 4 × 10-8 cm. Calculate the atomic mass of the element. (Given NA = 6.022 × 1023 mol-1) 
Examine the given defective crystal 
Answer the following questions:
(i) What type of stoichiometric defect is shown by crystal?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances shows such a defect?
Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (Kf = 5.12 K kg mol-1).  Answer:
Given, MB = 256 g mol-1, WA = 75 g
Kf = 5.12 kg mol-1, ΔTf = 0.48K
From the formula,
Define an ideal solution and write one of its characteristics.  Answer:
A solution which obeys Raoult’s law at all temperatures and concentrations is called an ideal solution.
For an ideal solution,
P = PA + PB
PA = PA0 × A
PB = PB0 × B
P = PA0 × A + PB0 × B
Where, P = Total pressure and PA, PB, PA0, PB0 and XA, XB are partial pressure, the pressure of pure component and mole fraction of component A and B respectively.
An ideal solution, no volume or enthalpy change takes place while mixing to form such a solution,
i.e., ΔHmix = 0 and ΔVmix = 0
Write two differences between ‘order of reaction’ and ‘molecularity of reaction’.  Answer:
|S.No.||Molecularity||Order of Reaction|
|1.||It is the number of reacting species taking part in an elementary reaction.||It is equal to the sum of all the exponents of all the reactants present in the rate law expression.|
|2.||It is a theoretical concept and cannot be zero or fractional.||It is determined experimentally and can be equal to zero or can have fractional value.|
Outline the principles behind the refining of metals by the following methods:  (i) Zone refining method
(ii) Chromatographic method
(i) Zone refining Method: This method is employed when impurities are more soluble in the melt than in the solid form of the metal. It is used to obtain metals such as Germanium (Ge), Silicon (Si), Gallium (Ga), etc in their purest form.
(ii) Chromatographic Method: This method is used when the impurities are not very different in chemical properties from the element to be purified. The components of the mixture show different mobility on the stationary phase, i.e., the components are adsorbed differently on the adsorbent.
Complete the following chemical equations:  (i) Ca3P2 + H2O →
(ii) Cu + H2SO4 (conc.) →
Arrange the following in order of property indicated against each set:
(i) HF, HCl, HBr, HI: increasing bond dissociation enthalpy
(ii) H2O, H2S, H2Se, H2Te: increasing acidic character.
Write the IUPAC name of the complex [Cr(NH3)4Cl2]+. What type of isomerism does it exhibit?  Answer:
Tetraamminedichloridochromium(III) ion. The complex exhibits geometric (cis-trans) isomerism.
(i) Which alkyl halide from the following pair is chiral and undergoes faster SN2 reaction? 
(ii) Out of SN1 and SN2, which reaction occurs with
(a) Inversion of configuration
(i) Compound (b), i.e., CH3CHBrCH2CH3 possesses chiral centre and undergoes faster SN2 reaction.
(ii) (a) Inversion of configuration results in SN2 reaction as there is the formation of the intermediate transition state in which there are simultaneous attack and migration of the leaving the group.
(b) Racemisation in SN1 reaction is due to an attack of the nucleophile on both sides of the planar carbocation.
Draw the structure of major moon halo product in each of the following reactions: 
(a) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.  (b) Write an important characteristic of lyophilic sols.
(c) Based on the type of particles of the dispersed phase, give one example each of associated colloid and multimolecular colloid.
(a) The expression for Freundlich adsorption isotherm
Where x = mass of the gas adsorbed (adsorbate)
m = mass of the adsorbent (solid)
p = pressure of the gas
n and k are constants, which depend on the nature of the adsorbate and adsorbent.
(b) Lyophilic sols are stable and reversible in nature
(c) Associated colloid: Soap solution or detergent solution.
Multimolecular colloid: Sulphur sol or Gold sol.
(a) Draw the structures of the following molecules:  (i) XeOF4
(b) Write the structural difference between white phosphorus and red phosphorus
Account for the following:  (i) PCl5 is more covalent than PCl3.
(ii) Iron on reaction with HCl forms FeCl2 and not FeCl3.
(iii) The two O-O bond lengths in the ozone molecule are equal.
(ii) As HCl is a mild oxidising agent and thus the hydrogen gas produced in the reaction prevents further oxidation of Fe2+ to Fe3+.
Fe + 2HCl → FeCl2 + H2.
(iii) An ozone molecule is a resonance hybrid of the molecule in which there is a single bond with one terminal oxygen and double bond with other terminal oxygen of the central oxygen atom.
So, neither a single or double bond is pure. Thus both O-O bond lengths are equal.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume:  SO2Cl2 → SO2 + Cl2
Calculate the rate constant.
(Given: log 4 = 0.6021, log 2 = 0.3010)
(i) Give two examples of macromolecules that are chosen as drug targets.  (ii) What are antiseptics ? Give an example.
(iii) Why is the use of aspartame limited to cold foods and soft drinks?
(i) Proteins, enzymes, nucleic acids and lipids are called drug targets as drug interacts with these macromolecules.
(ii) Antiseptics are the chemical substances which prevent the growth of microorganism and are capable of killing them without harming the human tissues. These are applied on wounds, ulcer, cuts and diseased skin surfaces, e.g. savlon, 0.2% solution of phenol, Dettol, iodine tincture etc.
(iii) Aspartame decomposes on heating hence it is used as an artificial sweetener for foods and soft drinks at low temperatures.
(i) Deficiency of which vitamin causes night blindness?
(ii) Name the base that is found in the nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane.
What does it suggest about the structure of glucose?  Answer:
(i) Vitamin A.
(iii) Glucose exists in the acyclic straight six-membered carbon chain
⇒ the open structure of Glucose.
CHO – (CHOH)4 – CH2OH
After the ban on plastic bags, students of a school decided to make the people aware of the harmful effects of plastic bags on the environment and Yamuna River. To make the awareness more impactful, they organized rally by joining hands with other schools and distributed paper bags to vegetable vendors, shopkeepers and departmental stores. All the students pledged not to use polythene bags in the future to save the Yamuna River.  After reading the above passage, answer the following questions:
(i) What values are shown by the students?
(ii) What are bio-degradable polymers? Give one example.
(iii) Is polythene a condensation or the addition polymer?
(ii) Polymers that are decomposed over a period of time either by itself or by the action of micro-organisms are called biodegradable polymers. PHBV (Poly β-hydroxybutyrate Co-β-hydroxy valerate)
(iii) Polythene is an addition polymer.
(a) Write the mechanism of the following reaction: 
(b) Write the equation involved in Reimer Tiemann reaction.
(a) Mechanism of the reaction
HBr → H+ + Br–
Give the structures of A, B and C in the following reactions: 
How will you convert the following?
(i) Nitrobenzene into aniline.
(ii) Ethanoic acid into methenamine.
(iii) Aniline into N-phenylethanamide.
(Write the chemical equations involved.)
(a) Define the following terms:
(i) Limiting molar conductivity
(ii) Fuel cell
(b) The resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L-1 KCl solution. The conductivity of 0.1 mol L-1 KCl solution is 1.29 × 10-1 Ω-1 cm-1.
(a) State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
(b) Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+ (0.1M) | | Cu2+ (0.01 M) | Cu(s)
[Given E0cell = +2.71 V, 1F = 96500 C mol-1]  Answer:
(a) (i) The limiting molar conductivity is the molar conductivity of a solution when the concentration approaches zero (infinite dilution, C → 0).
(ii) A fuel cell is a galvanic cell in which the heat energy of the fuel is directly converted into electrical energy.
(b) Given, R1 = 100 Ω
C1 (Concentration of KCl solution) = 0.1 mol L-1
K1 = 1.29 × 10-2 Ω-1 cm-1
R2 = 520 Ω
C2 (Concentration of KCl solution) = 0.02 mol L-1
Cell constant (G) = conductivity × resistance
= 1.29 × 10-2 Ω-1 cm-1 × 100 Ω
= 1.29 cm-1
(a) The mass of substance deposited or liberated at an electrode is proportional to the quantity of electric charge passed through an electrolysis cell.
Reduction of mol of Cu2+ to 2 moles Cu
One mole of electrons carries 1 F charge. Since 2 moles of electrons are required for reduction of 1 mol of Cu2+ So charge required is 2F.
(a) How do you prepare: [2, 3] (i) K2MnO4 from MnO2
(ii) Na2Cr2O7 from Na2CrO4
(b) Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(ii) The enthalpy of atomization is lowest for Zn in 3d series of the transition elements.
(iii) Actinoid elements show a wide range of oxidation states.
(i) Name the elements of 3d transition series which shows the maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of the 3d series has positive E°(M2+/M) value and why?
(iii) Out Of Cr3+ and Mn3+, which is a stronger oxidizing agent and why?
(iv) Name a member of the Lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation
MnO4– + 8H+ + 5e– →
(a) (i) Pyrolusite is fused with KOH in the presence of atmospheric oxygen or an oxidizing agent (KNO3 or KClO3) to give potassium permanganate.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
(ii) Acidification of yellow solution of sodium chromate with green core H2SO4 gives an orange solution of Na2Cr2O7.
(b) (i) The electronic configuration of Mn2+ is [Ar] 3d5 and Fe2+ is [Ar] 3d6. As it is clear from the electronic configuration, Mn2+ is already in half-filled stable d5 configuration whereas Fe2+ ion by losing one electron it can attain stable d5 configuration. Thus, Mn2+ shows stability while Fe2+ can be easily oxidized to Fe3+ state.
(ii) Unpaired electrons account for the stronger metallic bond. Zinc lacks impaired electrons as it has electronic configuration of [Ar] 3d104s2, thus metallic bonding is the weakest. So, Zn has the lowest enthalpy of atomization.
(iii) Due to comparable energies of 5f, 6d and 7s orbitals, actinoids show a wide range of oxidation states.
(i) Manganese shows the maximum number of oxidation states in the 3d transition series. This is because all the five d-orbital electrons are unpaired i.e [Ar] 3d54s2. Thus, Mn shows oxidation states from +2 to +7.
(ii) Cu has positive E (Cu2+/Cu) value i.e. +0.34. This is because Cu has a high value of ΔaH and low value of Δhyd H. ΔiH required is not compensated by the energy released. Therefore, E0(CU2+/CU) is positive.
(iii) E0 (Cr3+/Cr2+) = -0.4V and E0(M3+/Mn2+) = +1.5V
The negative value of E0 shows the stability of Cr3+ and high positive value of E0 in case of manganese shows that Mn3+ is easily converted to Mn2+. Thus, Mn2+ is a stronger oxidizing agent than Cr3+.
(iv) Europium shows +2 oxidation state owing to its half-filled electronic configuration.
(v) MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
(a) Write the products of the following reactions: [2, 3]
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzaldehyde and Benzoic acid
(ii) Propanal and Propanone
(a) Account for the following:
(i) CH3CHO is more reactive than CH3COCH3 towards reaction with HCN.
(ii) Carboxylic acid is a stronger acid than phenol.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Wolff-Kishner reduction
(ii) Aldol condensation
(iii) Cannizzaro reaction
(b) (i) Benzoic acid reacts with NaHCO3 to produce effervescence of CO2 gas but benzaldehyde does not
(ii) Propanone gives iodoform test but propanal does not because propanone has CH3CO- group
(a) (i) CH3CHO is more reactive than CH3COCH3 because the reactivity of the compound depends on the steric hindrance due to the groups present around the carbonyl group. More the steric hindrance less will be the reactivity of the compound. Therefore, due to the presence of more electrophilic carbonyl carbon in CH3CHO, it is more reactive than CH3COCH3.
(ii) Carboxylic acid is stronger than phenol because of resonance, stabilisation of more electronegative oxygen atom in carboxylate ion than carbon atom in phenoxide ion.
The reason for greater stability of carboxylate ion is that it involves equivalent resonance in which negative charge is present on oxygen atom all the time whereas, in phenoxide ion, out of total four resonating structures, in three structures, negative charge is present on a carbon atom which makes it less stable.
(b) (i) Reduction of aldehyde or ketone to respective hydrocarbon.
CBSE Previous Year Question Papers Class 12 Chemistry 2014 Delhi Set II
Note: Except for the following questions, all the remaining questions have been asked in the previous set.
Give one example each of sol and gel.  Answer:
Write the IUPAC name of the compound 
Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’?  Answer:
Azeotropes are mixtures with fixed concentrations of components such that they boil at constant temperatures, e.g. 95% ethanol and 5% water by the mass mixture.
Which component of starch is a branched polymer of a-glucose and insoluble in water?  Answer:
Amylopectin is a branched polymer of α-glucose and insoluble in water.
State Henry’s law. What is the effect of temperature on the solubility of a gas in a liquid?  Answer:
Henry’s law states that “the partial pressure of the gas in the vapour phase (p) is directly proportional to the mole fraction of gas (x) in the solution.”
P = KHX
(KH – Henry’s law constant)
The solubility of a gas in liquid decreases with rising in temperature as dissolution of a gas in a liquid is an exothermic process.
Define the following terms:  (i) Pseudo-first-order reaction
(ii) The half-life period of reaction (t1/2).
(i) Reactions which are actually not first order but behave as first-order under certain conditions like excess of one of the reactants is a pseudo-first-order reaction, e.g. Acid hydrolysis of ethyl acetate.
(ii) The time taken for the concentration of reactants to be reduced to half of its initial concentration is called the half-life of a reaction.
Write the principle behind the following methods of refining:  (i) Hydraulic washing
(ii) Vapour-phase refining
(i) Hydraulic washing is based on the differences in densities of gravities of the ore and the gangue particles. The lighter gangue particles are washed away and the heavier ores are left behind.
(ii) In Vapour phase refining method the metal is converted into its volatile compound and is collected elsewhere. It is then thermally decomposed to get the pure metal, e.g. Mond’s process.
(a) Draw the structures of the following:  (i) XeF2
(b) Write the structural difference between white phosphorus and red phosphorus.
(b) White phosphorus consists of discrete tetrahedral P4 molecule with six P-P covalent bonds.
Red Phosphorus has a polymeric structure in which P4 tetrahedral are linked together through P-P covalent bond to form a chain.
Account for the following:  (i) Bi(V) is a stronger oxidizing agent than Sb(V).
(ii) N – N single bond is weaker than the P – P single bond.
(iii) Noble gases have very low boiling points.
(i) Due to the inert pair effect +3 oxidation state of Bi is more stable than its +5 oxidation state while the oxidation state of Sb is more stable than its +3 oxidation state. Therefore, Bi (V) can accept a pair of electrons to form more stable Bi (III) more easily than Sb (V).
(ii) N has the small size so lone pair is more concentrated over N hence repulsion takes place and bond become weak but P has large size, therefore, no repulsion takes place and Hence, the bond becomes strong.
(iii) Noble gas are monoatomic with weak Vander Waals forces of attraction, Hence, noble gases have a very low boiling point.
(i) Name the sweetening agents used in the preparation of sweets for a diabetic patient.  (ii) What are antibiotics? Give an example.
(iii) Give two examples of macromolecules that are chosen as drug targets.
(iii) Carbohydrate, lipid, proteins, enzymes, nucleic acid.
(i) Deficiency of which vitamin causes rickets?  (ii) Give an example for each of fibrous protein and globular protein.
(iii) Write the product formed on reaction of D-Glucose with Br2 water.
(i) Vitamin D.
(ii) Fibrous protein: Keratin or myosin
Globular protein: Insulin or albumin.
(iii) Gluconic acid i.e. HOOC (CH2OH)4.CH2OH.
CBSE Previous Year Question Papers Class 12 Chemistry 2014 Delhi Set III
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Give one example each of lyophobic sol and lyophilic sol.  Answer:
Lyophobic sol: Metal sol or metal sulphide.
Lyophilic sol: Gum, Starch, gelatin.
Write the IUPAC name of the compound. 
What type of intermolecular attractive interaction exists in the pair of methanol and acetone?  Answer:
Hydrogen bonding (intermolecular)
Name the products of hydrolysis of sucrose.  Answer:
Glucose and fructose.
State Raoult’s law for the solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law?  Answer:
Raoult’s law states that the partial pressure of the vapour of a volatile component in a solution is directly proportional to its mole fraction in the solution.
Raoult’s law is a special case of Henry’s law.
Explain the following terms:  (i) The rate constant (k)
(ii) The half-life period of reaction (t1/2).
(i) The rate constant (k) is the rate of reaction when the concentration of reactants is unity.
(ii) Half-life period (t1/2) of reaction is the time in which the concentration of reactant is reduced to half of its initial concentration.
Write the principles of the following methods:  (i) Froth floatation method
(ii) Electrolytic refining
(i) It is based on the difference in wetting qualities of gangue and the sulphide ore particles with water and oil. Whereas the ore particles are wetted by oil, the gangue or the earthy particles are wetted by water.
(ii) In this process, a slab of impure copper is used as anode and a thin sheet of pure copper as the cathode. The copper sulphate is used as an electrolytic solution. Bypassing electricity through the cell copper is dissolved from the anode and deposited on the cathode. The impurities either remain in solution or collect as an insoluble gangue.
(a) Draw structure of the following compounds:
(b) Write the structural difference between white phosphorus and red phosphorus.  Answer:
Account for the following:
(i) Sulphur in vapour form exhibit paramagnetic behaviour.
(ii) SnCl4 is more covalent than SnCl2.
(iii) H3PO2 is a stronger reducing agent than H3PO3.  Answer:
(i) In vapour form sulphur partly exits as S2 molecule which has two unpaired electrons in the antibonding p molecular orbitals like O2 molecule and hence exhibits paramagnetism.
(ii) It is due to higher oxidation state (+4) of Sn in SnCl4 or because of its high polarising power which increases the covalent character of bond formation between the central atom and the atoms around it.
(iii) It is because of two P-H bonds in H3PO2 whereas there is only one P-H bond in H3PO3.
(i) What are disinfectants? Give an example.  (ii) Give two examples of macromolecules that are chosen as drug targets.
(iii) What are anionic detergents? Give an example.
(i) Chemicals which are used to kill microorganisms and applied on non-living objects like floors and drains are called disinfectants e.g. 1% phenol solution.
(ii) Proteins, amino acids and enzymes.
(iii) Detergents in which the anionic part of the molecule is responsible for cleansing action are called anionic detergents eg. Sodium lauryl sulphate.
(i) Deficiency of which vitamin causes scurvy?
(ii) What type of linkage is responsible for the formation of proteins?
(iii) Write the product formed when glucose is treated with HI.  Answer:
(ii) Peptide linkage
(iii) n-hexane: CH3CH2CH2CH2CH2CH3