CBSE Previous Year Solved Papers Class 12 Chemistry Outside Delhi 2009
Time allowed: 3 hours Maximum Marks: 70
- All questions are compulsory.
- Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
- Questions number 6 to 10 are short-answer questions and carry 2 marks each.
- Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
- Questions number 23 is a value based question and carry 4 marks.
- Questions number 24 to 26 are long-answer questions and carry 5 marks each.
- Use log tables, if necessary. Use of calculators is not allowed.
Questin.1. How do metallic and ionic substances differ in conducting
Answer : Metals conduct electricity through free electrons while ionic substances conduct electricity in aqueous form through ions.
Questin.2. What is the ‘coagulation’ process?
Answer : The process of setting of colloidal particles in a colloid is called coagulation. It is carried out by electrophoresis, persistent dialysis on addition of an electrolyte.
Questin.3. What is meant by the term ‘pyrometallurgy’?
Answer: Pyrometallurgy in the process of extracting metal by converting a metal oxide to metallic form on strong heating with a reducing agent.
Questin.4. Why is red phosphorus less reactive than white phosphorus?
Answer : Due to angular strain weak Yander weals forces are present in P4 molecules of white phosphorus whereas they are tetrahedrally attached by strong covalent bonds in red phosphorus. Therefore, red phosphorus is less reactive than white phosphorus.
Questin.5. Give the IUPAC name of the following compound:
Questin.6. Write the structural formula of 1-phenylpentan-l-one.
Questin.7. Arrange the following compounds in an increasing order of the basic strength in their aqueous solutions :
Questin.8. What does ‘6,6’ indicate in the name nylon-6,6?
Answer : 6,6 indicates the number of carbon atoms in the each of two monomers of nylon-6,6 and is desired from the two monomers hexamethylene diamine and adipic acid.
Questin.9. What type of cell is a lead storage battery? Write the anode and the cathode reactions and the overall cell reaction occurring in the use of a lead storage battery?
Two half cell reactions of an electrochemical cell are given below:
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation.
Answer: Lead storage battery is a secondary cell. The cathode, anode and overall cell reaction as follows :
- Elementary step in a reaction
- Rate of reaction
- If a reaction takes place in a single step with a single transition state then it is called elementary reaction.
- The change in concentration of the reactants or products per unit time is called rate of reaction.
Questin.11. Describe the underlying principle of each of the following metal refining methods:
- Electrolytic refining of metals
- Vapour phase refining of metals
- In electrolytic refining, impure metal is made to act as an anode and a thin sheet of pure metal as cathode, and a soluble salt solution of the metal is used as electrolyte. On passing electric current, the metal from anode passes into the solution and pure metal is deposited on cathode with impurities going into the solution or setting down as anode mud.
- The crude metal is first converted into its volatile compound which is then decomposed to give back pure metal.
Questin.12. Complete the following chemical reaction equations :
(i) XeF2+ H2O→
(ii) PH3 + HgCl2→
Questin.13. Complete the following chemical reaction equations
Questin.14. Which one in the following pairs undergoes SNl substitution reaction faster and why?
Questin.18. Differentiate between molecular structures and behaviours of thermoplastic and thermosetting polymers. Give one example of each type.
Questin.15. Complete the following reaction equation :
Questin.19. A first order reaction has a rate constant of 0.0051 min-1. If we begin with 0.10 M concentration of reactant what concentration of the reactant will be left after 3 hours?
Questin.16. Name the four bases present in DNA. Which one of these is not present in RNA?
Answer: Adenine, thymine, guanine and cytosine are present in DNA. Thymine is not present in RNA.
Questin.17. Name the two fat soluble vitamins, their sources and the diseases caused due to their deficiency in diet.
Answer:Vitamin A and’Vitamin K are fat soluble vitamins.
Source of Vitamin A – Carrot, papaya, Disease (Vitamin A) – Night blindness .
Sources of Vitamin K – Cereals, leafy vegetables, Diseases (Vitamin K) – Haemorrhagic conditions.
Questin.20. Silver crystallizes with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face atom is touching the four comer atoms.)
Questin.21. A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential measured is 0.422 V. Determine die concentration of the silver ion in the cell.
Questin.22. What happens in die following activities and why?
- An electrolyte is added to a hydrated ferric oxide sol in water.
- A beam of light is passed through a colloidal solution.
- An electric current is passed through a colloidal solution.
- Coagulation of sol takes place by absorbing OH” ions.
- The beam of light is scattered by the colloidal particles and the path of light becomes visible. This phenomenon is known as Tyndall effect.
- Electrophoresis takes place. The positively charged ions move towards cathode and negatively charged ions move towards anode and by loosing their charges get coagulated.
Questin.23. Giving a suitable example for each, explain the following:
- Crystal field splitting
- Linkage isomerism
- Ambidentate ligand
Compare the following complexes with respect to structural shapes of units, Magnetic behaviour and hybrid orbitals involved in units:
- When a transition metal ion is involved in complex formation, the degenerate orbitals split into two sets, one with lower energy and the other with higher energy. This is called crystal field splitting.
- Linkage isomerism occurs in compounds containing ambidentate ligands, i.e., when more than one atom may function as a donor and get linked with central metal ion.
- A ligand that has more than one donor atom but only one donor atom is attached to the neutral ion at a time is called ambidentate ligand eg. SCN–, NO–2
Questin.24. Explain the following:
(i) The boiling point of ethanol is higher than that of methoxymethane.
(ii)Phenol is more acidic than methanol.
(iii)o-and p-nitrophenols are more acidic than phenol.
- Due to intermolecular hydrogen bonding in ethanol.
- Phenol looses H+ ions and forms phenoxide ion. Ethanol looses H+ ion and forms ethoxide ion. Phenoxide ion is more stable than-ethoxide ion due to resonance. Therefore, phenol is more acidic than ethanol.
- Due to electron withdrawing character of NO2 group,
electron density in the OH bond of the substituted phenol decreases and hence loss of proton becomes easy, so it is more acidic.
Questin.25. How would you account for the following:
- Many of the transition elements and their compounds can act as good catalysts.
- The metallic radii of the third (5 d) series of transitions elements are virtually the same as those of the corresponding member of the second series.
- There is a greater range of oxidation states among the actinoids than among the lanthanoids.
- Due to presence of unpaired electrons in their incomplete -orbitals and variable oxidation states.
- Due to lanthanoid contraction in the second series, the atomic radii of the elements of second and third series become almost same, therefore, show same properties.
- Due to very small energy gap between 5f 6d and 7s – subshells all the electrons take part in bonding and show variable oxidation states.
Questin.26. Complete the following reaction equations :
Questin.27. Describe the following substances with one suitable example of each type :
- Non-ionic detergents
- Food preservatives
- Non-ionic detergents do not contain any ions. They are esters of high molecular mass prepared by reaction of polyethylene glycol with stearic acid e.g. Polyethylene glycol stearate.
- Preservatives are chemical substances used to protect food from bacteria, yeasts and moulds and avoid its spoilage, e.g. Sodium benzoate, salt-sugar etc.
- Disinfectants are chemicals which kill microbes but are applied only to non-living objects like floors and drains eg. Phenol, chlorine.
Questin.28. (a) Define the following terms:
- Mole fraction
- van’t Hoff factor
(b) 100 mg of a protein is dissolved in enough water to make 10.0 mL of the solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of protein?
(R = 0.0821 L atm mol-1 K-1 and 760 mm Hg = 1 atm.) OR
(a) What is meant by:
1. Colligative properties
2. Molality of a solution
(b) What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of 25°C, a total pressure of 1 atmosphere and mole fraction of nitrogen in air of 0.78
(KH for nitrogen = 8.42 x 10-7 M/mm Hg)
Answer : (a) (i) Mole fraction is the ratio of number of moles of a component to the total number of moles present in the solution.
(a) (1) Colligative, properties are those properties which depend on the number of particles of solute dissolved in a definite amount of solvent.
(2) Molality in the number of moles of solute dissolved in one
Questin.29. (a) Draw the structure of the following:
(b) How would you account for the following?
- NH3 is a stronger base than PH3
- Sulphur has a greater tendency for catenation than oxygen.
- F2 is a stronger oxidizing agent than Cl2.
(a) Draw the structure of the following:
(b) Explain the following observation :
- In the structure of HNO3, the N-0 bond (121 pm) is shorter than the N-OH bond (140 pm)
- All the P-Cl bond in PCl5 are not equivalent
(b) (i) Due to small size and high electronegativity of nitrogen ‘compound to phosphorus, electron density on nitrogen is more therefore, it can easily donate electrons and hence it is a stronger base than PH3.
(ii)Due to small size of oxygen, lone pair of oxygen repel (O- O) bond more than lone pair on sulphur atoms in S-S bond, hence S-S forms strong bond.
(iii) Due to small size and high electronegativity of electron- electron repulsion between lone pair of electrons are very large, therefore, bond dissociation energy of F2 is less than Cl2.
(b) (i) The N-O bond has partial double bond character while the N-OH is a single bond in the structure of HNO3.
(ii) PCl5 has a trigonal bipyramidal structure with 3 equivalent equatorial and 2 equivalent axial bonds. Due to higher bond pair-bond pair repulsion in case of axial bonds, they are longer than equatorial bonds. .
(iii) IC1 bond is weaker than I-I bond due to which I-I bond breaks easily to form halogen atoms which can easily bring out the reaction, hence it is more reactive than I2 :
Questin.30. (a) Write chemical equations to illustrate the following name bearing reaction :
- Cannizzaro’s reaction
- Hell-Volhard-Zelinsky reaction
(b) Give chemical test to distinguish between the billowing pairs of compound.
- Propanal and propanone
- Acetophenone and Benzophenone
- Phenol and Benzoic acid
(a) How will you bring about the following conversions :
- Ethanol to 3-Hydroxybutanal
- Benzaldehyde to benzophenone
(b) An organic compound A has the molecular formula C6H1602. It gets hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid also produced B. C
on dehydration reaction gives but-l-ene. Write all equations for the reactions involved.
(a) (i) Aldehydes which donot contain a-hydrogen atom undergo self oxidation and reduction with cone, alkali to give alcohol and carboxylic salt.
Note : Except these following questions, all the remaining questions have been asked in previous set.
Questin.1. Which point defect of its crystals decreases the density of a solid?
Questin.8. What is a primary structural feature necessary for a molecule to make it useful in a condensation polymerization reaction?
Answer : The monomers have a specific functional group representing their functionality and they combine through these functional groups to form condensation polymers.
Questin.19. Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cmn-3. Use this information to calculate Avogadro’s number. (At mass of iron = 56 g mol-1).
Questin.20. For a decomposition reaction the value of rate constant K at two different temperatures, are given below:
Questin.27. Complete the following reaction equations :
Questin.28. (a) Give chemical tests to distinguish between compounds in the following pairs of substances :
- Ethanol and propanal
- Benzoic acid and ethyl benzoate
(b) An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acids. Derive the structure of the compound.
(a) Arrange the following compounds in an increasing order of their indicated property.
- Benzoic acid, 4-nitrobenzoic acid, 3,4-dinitrobenzoic acid, 4-methoxybenzoic acid (acid strength)
- CH3CH2CH(Br)COOH,CH3CH (Br) CH2COOH (CH3)2 CHCOOH, CH3CH2CH2COOH (acid strength)
(b) how would you bring about he following conversions:
- Propanone to propene
- Benzoic acid to benzaldefiyde
- Bromobenzene to 1-phenylethanol
- Ethanol does not form silver mirror with Tollen’s reagent but propanal gives silver mirror with Tollen’s reagent due to presence of aldehyde group.
- Benzoic acid produce brisk effervescene with NaHCC>3 Solution while ethyl benzoate does not
Determination of structure : Since the compound doesnot reduce Tollen’s reagent and gives a positive iodoform test. So the given compound is Ketone not aldehyde. Since the given compound on vigrous oxidation gives a mixture of ethanoic acid and propanoic acid. Therefore, the given organic compound is methyl Ketone and its structure would be CH3COCH2CH2CH3
Questin.29. (a) Draw the structures of the following:
(b) How would you account for the following observations:
(i) Phosphorous has a greater tendency for catenation than nitrogen.
(ii)Bond dissociation energy of fluorine is less than that of chlorine.
(iii)No chemical compound of helium is known.
(a) Draw the structure of the following
(b) Explain the following observations:
(i) The electron gain enthalpy of sulphur atom has a greater negative value than that the oxygen atom.
(ii)Nitrogen does not form pentahalides.
(iii)In aqueous solution HI is a stronger acid than HCl.
(b) (i) This is because p—p Single bond is stronger than n—n Single bond.
(ii) Due to small size of fluorine, lone pairs of electrons. Create more repulsion than chlorine. Bond dissociation energy of fluorine is less than Cl2.
(iii) Valence shell electrons of He can’t be promoted to higher energy level due to lack of p or d orbitals.
(b) (i) This is due to smaller size of oxygen the electron cloud is distributed over a small region of space making electron density high which repels incoming electron.
(ii)Due to small size and absence of empty d-orbitals, Nitrogen does not form pentahalides.
(iii)The strength of acid depend up on its bond strength. Since bond dissociation energy of HCl is greater than HI. Therefore, HI is stronger than HCl.