CBSE Previous Year Question Papers Class 12 Chemistry 2011 Outside Delhi

Time allowed: 3 hours
Maximum Marks: 70

CBSE Previous Year Question Papers Class 12 Chemistry 2011 Outside Delhi Set I

Question 1.
Define ‘order of a reaction’. [1]
Answer:
The sum of powers of the concentrations of the reactants in the rate law expression is called the order of a reaction.

Question 2.
What is meant by ‘shape-selective catalysis’? [1]
Answer:
Catalysis using selective absorbents like zeolites as a catalyst is called shape-selective catalysis. In this catalysis, small-sized molecules are absorbed in the pores and cavities of zeolites.

Question 3.
Differentiate between a mineral and an ore. [1]
Answer:

S.No.	Mineral	Ore
1.	These are naturally occurring chemical substances obtained from earth crust by mining.	Ores are those minerals from which metals can be extracted profitably and conveniently.
2.	All minerals are not ores.	All ores are minerals.

Question 4.
What is meant by ‘lanthanoid contraction’? [1]
Answer:
The lanthanoid contraction refers to the steady and regular decrease in atomic radii along the period from La⁺³ to Lu⁺³.

Question 5.
Write the IUPAC name of the following compound:
CH₂=CHCH₂Br [1]
Answer:
3-Bromo-1-propene

Question 6.
Draw the structure of 4-chloropentan-2-one. [1]
Answer:

Question 7.
How would you convert ethanol to ethene? [1]
Answer:

Question 8.
Rearrange the following in increasing order of their basic strengths: [1]
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂.
Answer:
C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH

Question 9.
Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of the unit cell of its crystal. [2]

Question 10.
Calculate the packing efficiency of a metal crystal for a simple cubic lattice. [2]
Answer:
Edge length = a

Question 11.
State the following: [2]
(i) Raoult’s law in its general form in reference to solutions.
(ii) Henry’s law about partial pressure of a gas in a mixture.
Answer:
(i) Raoult’s Law: The partial vapour pressure of each component of a solution is equal to the vapour pressure of pure component multiplied by its mole fraction in the solution.
P = P_Ax_A+P_Bx_B
Where
P = Total vapour pressure of a solution.
P_A and P_B = vapour pressure of components A and B.
x_A and x_B = mole fractions of components A and B.

(ii) Henry’s Law: It states that the partial pressure of a gas in the vapour phase (P) is directly proportional to the mole fraction of gas (x) in the solution and is expressed as
P = K_Hx
where K_H is Henry constant.

Question 12.
What do you understand by the rate law and rate constant of a reaction? Identify the order of a reaction if the units of its rate constant are: [2]
(i) L^-1 mol s^-1
(ii) L mol^-1 s^-1
Answer:
Rate law of a chemical reaction is the expression relating the rate of reaction to the concentrations or pressures of various reactants taking part in the reaction.
The rate of reaction at a unit concentration of all reactants is known as the rate constant (K)

Zero-order
Second-order

Question 13.
The thermal decomposition of HCO₂H is a first-order reaction with a rate constant of 2.4 × 10^-3 s^-1 at a certain temperature. Calculate how long will it take for three-fourths of the initial quantity of HCO₂H to decompose, (log 0.25 = -0.6021) [2]
Answer:

Question 14.
Describe the principle controlling each of the following processes: [2]

Vapour phase refining of titanium metal
Froth floatation method of concentration of a sulphide ore

Answer:

Titanium is converted to its volatile form which is evaporated and then decomposed to give pure titanium.
The ore particles get adsorbed on oil droplets and come to the surface where they can be collected as froth gangue is wetted by water and gets settle down.

Question 15.
How would you account for the following: [2]

Cr²⁺ is reducing in nature while with the same d-orbital configuration (d⁴) Mn³⁺ is an oxidizing agent.
In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series.

Answer:

Cr²⁺ has configuration d⁴ which easily changes to d³ due to stable half-filled 12g orbitals. Therefore, Cr²⁺ is reducing agent and Mn³⁺ easily changes to Mn²⁺ and acts as an oxidizing agent.
Due to the presence of more unpaired electrons and more number of partially filled orbitals in the middle of a transition series, these metals exhibit the greatest number of oxidation states.

Question 16.
Complete the following chemical equations: [2]

OR
State reasons for the following:
(i) Cu (i) ion is not stable in an aqueous solution.
(ii) Unlike Cr³⁺, Mn²⁺, Fe³⁺ and the subsequent other M²⁺ ions of the 3d series of elements, the 4d and the 5d series metals generally do not form stable cationic species.
Answer:

OR
(i) Cu²⁺ is more stable than Cu⁺ because hydration energy of Cu²⁺ is high and it is, therefore, stable in aqueous solution. Therefore, Cu⁺ disproportionates to Cu²⁺ and Cu.
2Cu⁺ → Cu²⁺ + Cu
(ii) Because of Lanthanoid contraction, an expected increase in size does not occur. That is why they do not form stable cations.

Question 17.
Explain what is meant by the following: [2]
(i) Peptide linkage
(ii) Pyranose structure of glucose
Answer:
(i) Peptide linkage is the peptide bond formed between amino acids. It is a covalent bond formed between an amino group of one molecule and a carboxylic acid group of another molecule, causing the release of one molecule of water.

(ii) The cyclic structure of glucose is called pyranose because it resembles the pyran ring. Its structure includes a six-membered ring with 5-carbon atoms and one oxygen atom having no double bonds.

Question 18.
Write the main structural difference between DNA and RNA. Of the four bases, name those which are common to both DNA and RNA. [2]
Answer:

S.No.	DNA	RNA
1.	Sugar present in DNA, is 2-deoxy D-ribose	In RNA, sugar is D-ribose
2.	DNA is a double stranded molecule	RNA is a single-stranded molecule
3.	DNA can replicate	RNA can not replicate

Adenine, cytokine and guanine are bases present in both DNA and RNA

Question 19.
A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torrs at 25°C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass. [3]
Answer:

Question 20.
Classify colloids where the dispersion medium is water. State their characteristics and write an example of each of these classes. [3]
OR
Explain what is observed when
(i) an electric current is passed through a Solution
(ii) a beam of light is passed through a Solution
(iii) an electrolyte (say NaCl) is added to ferric hydroxide sol
Answer:
Colloids can be classified into two types where the dispersion medium is water. They are as follows:
1. Hydrophilic Colloids or Lyophilic: The substances when mixed with dispersion medium form colloidal solution directly are called hydrophilic colloids. They are quite stable, reversible solutions and can’t get precipitated easily, e.g., Gum starch, etc.
2. Hydrophobic Colloids or Lyophobic: The substances which do not form a colloidal solution with dispersion medium are called hydrophobic colloids. They are unstable, irreversible solutions and can be easily precipitated, e.g., Metals and their sulphides.
OR
(i) When an electric current is passed through a solution, then positively charged ions move towards the cathode and negatively charged ions move towards the anode. Then they get coagulate.
This is known as Electrophoresis.

(ii) When a beam of strong light is passed through the solution, the light gets scattered by the colloidal particles and the path of light becomes visible. This is known as the Tyndall effect.

(iii) When NaCl is added to ferric hydroxide sol then a negatively charged solution is obtained with the absorption of OH^– ion.

Question 21.
How would you account for the following: [3]

H₂S is more acidic than H₂O.
The N-O bond in NO₂^– is shorter than the N-O bond in NO₃^–
Both O₂ and F₂ stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine.

Answer:

Size of sulphur is larger than oxygen due to which S-H bond length increases and hence the bond dissociation energy of S-H is less than O-H. Therefore S-H easily loses H+ and is more acidic than H₂O.
Due to the tendency of nitrogen to form pπ-pπ multiple bonds, there is a difference in N-O bond lengths of NO₂ and NO₃.
Due to the property of oxygen to form double bonds with the metal atoms, oxygen stabilizes the higher oxidation state even more than fluorine.

Question 22.
Explain the following terms giving a suitable example in each case: [3]

Ambident ligand
Denticity of a ligand
Crystal field splitting in an octahedral field

Answer:

Ligands which can ligate to the central atom in two places ligands such as SCN which can attach at either S atom or N atom are called ambidentate ligand.
The number of donor atoms of a ligand, when bound with the central atom in a coordination complex, is called its denticity.
The splitting of five degenerated d-orbitals in two sets, one with three orbitals and another with two orbitals is known as crystal field splitting.

Question 23.
Rearrange the compounds of each of the following sets in order of reactivity towards S_N2 displacement: [3]

2-Bromo-2-methyl butane, 1-Bromopentane, 2-Bro-neopentane
1-Bromo-3-methyl butane, 2-Bromo-2- methyl-butane, 2-Bromo-3-methyl butane
1-Bromobutane, 1-Bromo-2, 2-dimethylpropane, 1- Bromo-2- methyl butane.

Answer:

1-bromopentane > 2-bromopentane > 2-bromo-2-methylbutane.
1-bromo-3 methylbutane > 2-bromo-3- methylbutane > 2-bromo-2-methyl butane.
1-bromobutane > 1-bromo-3-methylbutane > 1-bromo-2, 2-dimethylpropane.

Question 24.
How would you obtain the following: [3]
(i) Benzoquinone from phenol
(ii) 2-Methylpropan-2-ol from methyl magnesium bromide
(iii) Propan-2-ol from propene
Answer:

Question 25.
State reasons for the following: [3]
(i) pK_b value for aniline is more than that for methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not soluble in water.
(iii) Primary amines have higher boiling points than tertiary amines.
Answer:
(i) A higher value of pK_b means lower basicity, therefore, aniline is less basic than methylamine because in aniline the lone pair of electrons on N atom gets delocalized over the benzene ring and remains unavailable for protonation due to resonance but this is absent in methylamine.

(ii) Ethylamine forms H-bonds with water therefore, it is soluble in water but aniline does not form H-bonds with water due to its larger hydrocarbon part and is insoluble in water.

(iii) In primary amines, two H-atoms are attached to N-atom and they undergo intermolecular H-bonding, but tertiary amines due to the absence of H-atom on N-atom do not undergo H-bonding. Therefore, primary amines have a higher boiling point than tertiary amines.

Question 26.
Draw the structures of the monomers of the following polymers: [3]
(i) Polythene
(ii) PVC
(iii) Teflon
Answer:
(i) Ethene —CH₂=CH₂—
(ii) Vinyl chloride —HCH₂=CHCl—
(iii) Tetrafluoroethylene —CF₂=CF₂—

Question 27.
What are the following substances? Give one example of each. [3]

Food preservatives
Synthetic detergents
Antacids

Answer:

Chemicals added to food to prevent its spoilage by killing or preventing the growth of microorganisms like bacteria, yeasts and moulds, e.g., sodium benzoate.
Synthetic detergents are sodium or potassium salts of long-chain sulphonic acids. They don’t precipitate in hard water, e.g., Sodium lauryl sulphate.
Antacids are chemicals consumed to get relief from acidity in the stomach by neutralizing excess acid, e.g., milk of magnesia.

Question 28.
(a) What type of battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery. [5]
(b) Calculate the potential for half-cell containing 0.10 M K₂Cr₂O₇ (aq), 0.20 M Cr³⁺ (aq) and 1.0 × 10^-4 M H⁺ (aq). The half-cell reaction is and the standard electrode potential is given as E⁰ = 1.33 V.

OR
(a) How many moles of mercury will be produced by electrolyzing 1.0 M Hg(NO₃)₂ solution with a current of 2.00 A for 3 hours? [Hg(NO₃)₂ = 200.6 g mol^-1]
(b) A voltaic cell is set up at 25°C with the following half-cells Al³⁺ (0.001 M) and Ni²⁺ (0.50 M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.

Answer:
(a) The lead storage battery is a secondary cell (rechargeable). The electrode reaction is as follows:

Question 29.
(a) Draw the structures of the following molecules: [5]
(i) (HPO₃)₃
(ii) BrF₃
(b) Complete the following chemical equations:
(i) HgCl₂ + PH₃ →
(ii) SO₃ + H₂SO₄ →
(iii) XeF₄ + H₂O →
OR
(a) What happens when
(i) Chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) Sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt?
(b) Answer the following:
(i) What is the basicity of H₃PO₃ and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do noble gases have very low boiling points?
Answer:
(a) (ii) BrF₃

(b) (ii) Since fluorine is the most electro-negative halogen, it does not acts as a central atom in interhalogen compounds due to the absence of d-orbital.
(iii) Noble gases are monoatomic with weak Vander Waals forces of attraction and hence have low boiling points.

Question 30.
(a) Illustrate the following name reactions: [5]
(i) Cannizzaro’s reaction
(ii) Clemmensen reduction
(b) How would you obtain the following:
(i) But-2-enal from ethanal
(ii) Butanoic acid from butanol
(iii) Benzoic acid from ethylbenzene
OR
(a) Give chemical tests to distinguish between the following:
(i) Benzoic acid and ethyl benzoate
(ii) Benzaldehyde and acetophenone
(b) Complete each synthesis by giving missing reagents or products in the following:

Answer:
(a) (i) Aldehydes having no a -hydrogen atom undergoes self oxidation and reduction on treatment with concentrated alkali and produces alcohol and carboxylic acid salt.

(iii) Reduction of aldehydes and ketones to their respective hydrocarbons. On treating with zinc amalgam and concentrated hydrochloric acid

OR
(a) (i) Benzoic acid gives CO₂ gas on reacting with NaHCO₃ but ethyl benzoate does not
(ii) Acetophenone gives iodoform test but

CBSE Previous Year Question Papers Class 12 Chemistry 2011 Outside Delhi Set II

Note: Except for the following questions, all the remaining questions have been asked in the previous set.

Question 2.
What are lyophobic colloids? Given one example for them. [1]
Answer:
Colloids in which the colloidal particles have no affinity for the dispersion medium and they do not form a colloidal solution are lyophobic colloids, like, Al(OH)₃ and AS₂S₃ sols.

Question 3.
Why is it that only sulphide ores are concentrated by ‘froth floatation process’? [1]
Answer:
Due to the affinity of heavy oil droplets to adsorb sulphide particles. The ore particles come on the surface as froth, gangue particles are wetted by water and get settle down.

Question 5.
Write the IUPAC name of the following compound: [1]

Answer:
3-Bromo-2-methyl prop-1-ene

Question 6.
Draw the structure of 2, 6-Dimethylphenol. [1]
Answer:

Question 9.
Define the following terms in relation to crystalline solids: [2]
(i) Unit Cell
(ii) Coordination number

Question 12.
A reaction is of second order with respect to a reactant How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of the rate constant for such a reaction? [2]
Answer:

Question 14.
Describe the principle controlling each of the following processes: [2]
(i) Zone refining of metals
(ii) Electrolytic refining of metals
Answer:
(i) Zone refining of Metals: It is based on the principle that impurities are more soluble in the molten state of metal than in the solid-state

(ii) Electrolytic refining: Impure metal is made of an anode, a thin sheet of pure metal is made of a cathode and a salt of the metal is used as an electrolyte. On passing current, metal from anode goes into the solution and ions in the solution reduce on cathode leading to deposition of pure metal.

Question 15.
Explain giving a suitable reason for each of the following: [2]
(i) Transition metals and their compounds are generally found to be good catalysts.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series.
Answer:
(i) Due to the presence of vacant orbitals and tendency to form a large number of oxidation states, transition metals have a high tendency to form complexes and hence acts as catalysts.

(ii) In transition metals of 4d and 5d series, the 4d and 5d electrons are at a greater distance from the nucleus, therefore, they are less tightly held to the atom by the nucleus and hence contribute more to metallic bonding as compared to transition metals of 3d series.

Question 19.
What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50°C? The freezing point depression constant (K_f) for water is 1.86°C/m. Assume Van’t Hoff factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g) [3]
Answer:

Question 22.
Write the structures and names of all the stereoisomers of the following compounds: [3]
(i) [Co(en₃)]Cl₃
(ii) [Pt(NH₃)₂Cl₂]
(iii) [Fe(NH₃)₄Cl₂]Cl
Answer:
(i) Stereoisomerism is of two types:
(1) Geometrical isomerism
(2) Optical isomerism
[Co(en)₃]Cl₃: Tris (ethylenediamine) cobalt (III) chloride

Question 27.
(a) Differentiate between a disinfectant and an antiseptic. Give one example of each.
(b) What is a tincture of iodine and what is it used for? [3]
Answer:

S.No.	Disinfectant	Antiseptic
1.	Chemical substances used to kill microorganisms	Chemical substances which prevent the growth of microorganisms and may even loll them.
2.	They are applied on non-living objects	They are safe to apply to live tissues.
3.	They are used in drains, toilets, floors etc.	They are used on wounds, cut, ulcers etc.
4.	Example: Phenol (1%)	Example: Soframycin

(b) 2-3% solution of iodine in alcohol and water is called tincture iodine and is widely used as an antiseptic.

CBSE Previous Year Question Papers Class 12 Chemistry 2011 Outside Delhi Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Question 1.
Define ‘activation energy’ of a reaction. [1]
Answer:
Activation energy is the minimum energy that is to be provided to the reactants for the reaction to take place.

Question 2.
What is meant by ‘reverse osmosis’? [1]
Answer:
When a pressure higher than the osmotic pressure is applied the solvent will flow from the solution into the pure solvent through the semipermeable membrane, called reverse osmosis.

Question 3.
What type of ores can be concentrated by magnetic separation method? [1]
Answer:
Ores with different magnetic properties and impurities which are magnetic in nature can be concentrated by magnetic separation method. e.g. Chromite (FeO.Cr₂O₃), Magnetite (Fe₃O₄)

Question 14.
Describe the principle controlling each of the following processes: [2]
(i) Preparation of cast iron from pig iron.
(ii) Preparation of pure alumina (Al₂O₃) from bauxite ore.
Answer:
(i) Preparation of cast iron from pig iron: The iron obtained from a blast furnace is called pig iron. Cast iron is prepared by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%). It is extremely hard and brittle.

(ii) Preparation of pure alumina from bauxite ore: The principal ore of aluminium is bauxite Al₂O₃xH₂O. Bauxite is concentrated by digesting the powdered ore with a concentrated solution of NaOH at 473-523K. The Al₂O₃ is leached out as sodium aluminate. The sodium aluminate is neutralized by passing CO₂ gas and hydrated Al₂O₃ is precipitated which is filtered, dried and heated to give pure Al₂O₃.

Question 15.
Explain giving reasons: [2]
(i) Transition metals and their compounds generally exhibit paramagnetic behaviour.
(ii) The chemistry of actinoids is not so smooth as that of lanthanoids.
Answer:
(i) Transition metals and their compounds exhibit a paramagnetic behaviour due to the presence of unpaired electrons in the Penultimate shell of d-orbital.

(ii) Lanthanoids show a limited number of oxidation states, +2, +3, +4 because of the large energy gap between 4f and 5d subshells. Actinoids show a number of oxidation states +4, +5, +6, +7 due to the small energy difference between 5f, 6d and 7s subshells.

Question 18.
Write such reactions and facts about glucose which cannot be explained by its open-chain structure. [2]
Answer:
Limitations of open chain structure of glucose:

Glucose does not form NaHSO₃ as an additional product.
Glucose penta-acetate does not react with NH₂OH due to the absence of the aldehyde group.

Question 21.
How would you account for the following: [3]
(i) NF₃ is an exothermic compound but NCl₃ is not.
(ii) The acidic strength of compounds increases in the order PH₃ < H₂S < HCl
(ii) As electronegativity increases in the same period from left to right so their electronegativity is in the increasing order P < S < Cl. Therefore, acidic strength increases in the order PH₃ < H₂S < HCl
(iii) SF6 is protected by six F atoms and hence does not allow to attack sulphur atom.

Question 22.
Write the state of hybridization, the shape and the magnetic behaviour of the following complex entitles: [3]

[Cr(NH₃)₄Cl₂]Cl
[Co(en)₃]Cl₃
K₂[Ni(CN)₄]

Answer:

d²sp³, octahedral, diamagnetic
d²sp³, octahedral, diamagnetic
dsp², square planar, diamagnetic

Question 26.
Write the names and structures of the monomers of the following polymers: [3]
(i) Buna-S
(ii) Dacron
(iii) Neoprene
Answer:
(i) Buna-S: 1, 3-Butadiene
CH₂=CH-CH-CH₂ and styrene C₆H₅CH=CH₂
(ii) Dacron: Ethylene glycol HOCH₂CH₂OH, Terephthalic acid

(iii) Neoprene: Chloroprene