CBSE Previous Year Solved Papers Class 12 Chemistry Delhi 2012
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.
Note : Except for the following questions, all the remaining questions have been asked in previous sets.
Question.1. What is meant by ‘doping’ in a semiconductor ?
Answer : Addition of impurities (like phosphorous and arsenic) to a semiconductor (like silicon) to improve the
conductivity is called doping.
Question.2. What is the role of graphite in the electrometallurgy of aluminium?
Answer : Graphite is used as anode and useful for the reduction of Al2O3 into Al.
Question.3. Which one of PCl+4and PCl–4 is not likely to exist and why?
Answer: The oxidation state of P in PCl+14 is +5 while that PCl-14 is +3. As we move down the group, the stability of +5 oxidation decreases and +3 oxidation state increases. Chlorine belongs to third period. Hence, PCl+14 compound is more likely to exist.
Question.4. Give the IUPAC name of the following compound.
Question.5. Draw the structural formula of 2-methylpropan-2ol molecule.
Question.6.Arrange the following compounds in ait increasing order of their reactivity in nucleophilic addition reactions : ethanol, propanal, propanone, butanone.
Answer : Butanone < Propanone < Proponal < Ethanol
Question.7. Arrange the following in the decreasing order of their basic strength in aqueous solutions:
Question.8. Define the term, ‘homopolymerisation’ giving an example
Answer : Polymerisation involving the presence of one monomer is called homopolymerisation, e.g. polyethene is a homopolymer
Question.9. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3 COOH) is heated to its boiling point. The solution has the boiling point of 100.18 °C. Determine the Van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol-1)
Define the following terms :
- Mole fraction
- Isotonic solutions
- Van’t Hoff factor
- Ideal solution
- Ratio of the number of moles of a component in a mixture to the total number of moles in the mixture is called the mole fraction of that component. It is denoted by ‘x
- Two solutions having the same molar concentration are said to be isotonic solutions, eg : All intravenous injections must be isotonic with body’fluids.
- The ratio of observed colligative property to the calculated colligative property is called the Van’t Hoff factor. It is denoted by Y.
- Solutions that follow Raoult s law at all temperatures and concentrations are called ideal solutions.
Question.10. What do you understand by the ‘order of a reaction’ ?
Identify the reaction order from each of the following units of reaction rate constant:
- L-1 mol s-1
- L mol-1 s-1
Answer : The sum of the powers to which the concentration of reactants are raised in the rate law expression is called the order of a reaction.
- Zero order reaction
- Second order reaction.
Question.11. Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved.
Answer: Catalysis can be positive, that is, it increases the rate of the reaction or negative i.e. decreases the rate of reaction. Depending on the phase of the reactants and the catalyst, catalysis can be :
(i) Homogenous catalysis : The reactants and catalyst are in the same phase.
Question.12. What is meant by coagulation of a colloidal solution?
Describe briefly any three methods by which coagulation of lyophobic sols can be carried out.
Answer : The process of setting of colloidal particles is called coagulation of sol. Methods of coagulation are :
- Electrophoresis : In this process, the colloidal particles move towards opposite charged electrodes and get discharged and precipitated.
- Mixing two oppositely charged sols : Equal proportions of oppositely charged sols are mixed, they get neutralized and get precipitated.
- Dialysis : Electrolytes are- removed from the sol and colloid becomes unstable and gets coagulated.
Question.13. Describe the principle involved in each of the following processes.
- Mond process for refining of Nickel.
- Column chromatography for purification of rare elements.
- Nickel combined with carbon monoxide to form volatile complex within further be decomposed to get back pure nickel.
- The basic principle involved in column chromatography is that different elements present in a mixture are adsorbed on adsorbent at different extents.
Question.14. Explain the following giving an appropriate reason in each case.
- O2 and F2both stabilize higher oxidation states of metals but O2 exceeds F2 in doing so.
- Structures of xenon fluorides cannot be explained by Valence Bond approach.
- Due to the difference in atomic size of oxygen and fluorine and the property of oxygen to form multiple bonds with metals,O2 exceeds F2, stabilize higher oxidation states,
- For explaining the structures of xenon fluorides, we need to use VSEPR and hybridization theories because in VBT, covalent bonds are formed by overlapping of half filled atomic orbital. But xenon has fully filled electronic configuration.
Question.15. Complete the following chemical equations :
Question.16. What is meant by
- peptide linkage
- Peptides linkage is the amide bond that helps to connect amino acids to form proteins. It is formed between -COOH and -NH2 group of two amino acids with the loss of water molecule.
- Biocatalysts are enzymes that catalyses the biochemical reactions in the bodies of living organisms, e.g. Amylase.
Question.17. Write any two reactions of glucose which cannot be explained by the open chain structure of glucose molecule.
Answer : Two reactions which can’t be explained by open chain structure of glucose are :
- Despite having the aldehyde group, glucose does not give 2,4 —DNP test.
- The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free -CHO group.
Question.18. Draw the structure of the monomer for each of the following polymers:
(i) Nylon 6 (ii) Polypropene.
Question.19. Tungsten crystallizes in body centered cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom!
Iron has a body centered cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro’s number. (At mass of Fe = 55.845 u)
Question.20. Calculate the amount of KCl which must be added to 1kg of water so that the freezing point is depressed by 2K.
(Kf for water = 1.86 K kg mol-1)
Question.21.For the reaction
the following data were collected. All the measurements were taken at 263 K:
(a) Write the expression for rate law.
(b) Calculate the value of rate constant and specify its units.
(c) What is the inidal rate of disappearance of Cl2 in exp. 4?
Question.22. How would you account for the following?
- Many of the transition- elements are known to form interstitial compounds.
- The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.
- Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical.
- Transition metal lattice have voids and hence these voids can trap small atoms like H, C and N to form interstitial compounds.
- It is because electrons first fill the 4forbitals and than the 5d orbitals causing decline in radii of third (5d) series.
- This is because of the comparable energies of 5f 6d and Is orbitals so all of them can participate
Question.23. Give the formula of each of the following coordination
(i) CO3+ ion is bound to one Cl–one NH3 molecule and two bidentate ethylene diamine (en) molecules.
(ii)Ni2+ ion is bound to two water molecules and two oxalate ions.
Write the name and magnetic behavior of each of the above coordination entities.
Question.24. Although chlorine is an electron ‘withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why it is so?
Answer : Due to resonance, electron density is increased on ortho and para positions for electron donating (via resonance) groups like chlorine.
Question.25. Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidizing agent is used.
(i) CH3CH2CH2CH2OH (ii) 2-butenol
Question.26. Write chemical equations for the following conversions :
(i) Nitrobenzene to benzoic acid.
(ii)Benzyl chloride to 2-phenylethanamine.
(iii)Aniline to benzyl alcohol.
Question.27. What are the following substances? Give one example of each one of them.
- Food preservatives
- Synthetic detergents.
- Tranquilizers are drugs that act on the central nervous system to get relief from anxiety, stress etc. They are used in treatment of stress related mental disorders.
- These are chemicals used to preserve food by protecting it against microbial growth, eg. Sodium benzoate.
- Synthetic detergents have all the properties of soap but they do not- precipitate in hard water, eg. Sodium p-dodecyl benzenesulphonate.
Question.28. (a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reactions occurring in a lead storage battery when current is drawn from it.
(b) In the button cell, widely used in watches, the following reaction takes place.
(a) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte.
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell
constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 x 10-3S cm-1?
Question.29. (a) Complete the following chemical reactions. equations:
Question.30. (a) Illustrate the following name reactions giving suitable example in each case :
(i) Clemmensen reduction
(b) How are the following conversions carried out?
(i) Ethyicyanide to ethanoic acid
(ii)Butan-l-ol to butanoic acid
(iii)Benzoic acid to m-bromobenzoic acid
(a) Illustrate the following reactions giving a suitable example for each:
(i) Cross aldol condensation
(b) Give simple tests to distinguish between the following pairs of compounds :
(i) Pentan-2-one and Pentan-3-one
(ii)Benzaldehyde and Acetophenone
(iii)Phenol and Benzoic acid
Answer : (a)
(i) Aldehydes and ketones are reduced to CH2group on treatment with zinc-amalgam and cone. HCl.
(ii) Carboxylic acids containing a-hydrogen atom gives halo carboxylic acids on treatment with halogens in the presence of red phosphorous.
(ii) Carboxylic acid lose carbon dioxide to form hydrocarbon, when their salts are heated with sodium. The reaction is known as decarboxylation reaction
Note: Except for the following questions, all the remaining questions have been asked in previous set.
Question.1.Write a point of distinction between a metallic solid and an
ionic solid other than metallic luster.
Answer : Metallic solids do not ionize in aqueous state but ionic solids ionize immediately
Question.11.Describe a conspicuous change observed when :
- a solution of NaCl is added to a sol of hydrated ferric oxide.
- a beam of light is passed through a solution of NaCl and then through a sol.
- Coagulation of ferric hydroxide sol. would take place.
- NaCl solution is transparent so when beam of light is passed, no tyndall effect is produced. But on passing through soln the path of light becomes visible due to Tyndall effect.
Question.13. Describe the following:
- The role of cryolite in electro-metallurgy of aluminium.
- The role of carbon monoxide in the refining of crude nickel.
- Cryolite lowers the melting point of the mixture and brings conductivity. Therefore, it is mixed with alumina during metallurgy of aluminium.
- Carbon monoxide forms a volatile complex with nickel.
The volatile complex is then subjected to high temperature to get pure metal through decomposition.
Question.14. What is meant by
- peptide linkage
- Peptide linkage: It is the linkage between amino acids, formed due to loss of a water molecule and
- Biocatalysts : Biocatalysts are enzymes which are used to perform chemical reactions on organic compounds.
Question.18. Write the main structural difference between DNA and RNA. Of the two bases, thymine and uracil, which one is present in DNA?
Question.23. How would you account for the following?
- With the same d-orbital configuration (d4) Cr2+ is a reducing agent while Mn3+ is an oxidizing agent.
- The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
- Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.
- Cr2+ has d4 configuration. It gets oxidized , to Cr3+ with electronic configuration d3 which is more stable. Therefore, Cr2+ is a reducing agent. Mn3+ has d4 configuration. It gets reduced to Mn2+with d5 configuration. This is half-filled d-orbital and is stable. Therefore Mn3+is an oxidizing agent.
- Because 5f, 6d and 7s energy levels has small enery gap in the actinoid series. Due to these orbitals actinoids exhibit large number of oxidation states.
- Due to partial absorption of visible light the electron from one orbital gets promoted to another orbital of the d subshell. Due to presence of unpaired electrons transition metals are coloured.
Question.30. (a) Give a possible explanation for each one of the following:
(i) There are two -NH2 groups in semicarbazide. However, only one such group is involved in the formation of semicarbazones.
(ii) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcydohexanone does not.
(b) An organic compound with molecular formula C9H10O forms 2, 4, -DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzene-di-carboxylic acid. Identify the compound.
(a) Give chemical tests to distinguish between
(i) Phenol and Benzoic acid
(ii)Acetophenone and Benzpphenone
(b) Write the structures of the main products of following
Answer : (a)
(i) Semicarbazide show resonance involving one of the two -NH2 groups, which is attached to the carboxyl carbon atom. Due to which electron density on —NH2 group involved in resonance decreases. So it cannot act as nucleophile. Other -NH2 group can act as nucleophile to produce semicarbazones with aldehydes and ketones.
(ii) In cyclohexanone CN– can easily attack without any steric hindrance. But in 2, 4, 6 — Trimethylcyclohexanone due to presence of methyl groups steric hindrance is produced and CN– cannot attack effectively.
(iii) C9H10Ois aldehyde because it reduces Tollens reagent. It undergoes Cannizaro’s reaction therefore it is substituted benzaldehyde. It gives 1, 2, Benezene-di carboxylic acid. The compound is 2-Ethylbenzaldehyde.
Note : Except for the following questions, all the remaining questions have been asked in previous sets.
Question.3. Of PH3 and H2 S which is more acidic and why?
Answer : H2S is more acidic than PH3 due to smaller size and higher electronegativity of sulphur. Therefore S-H bond is polar than P-H bond and easy to remove.
Question.5. Draw the structure of hex-l-en-3-ol compound.
Question.12. Explain the following terms-giving one example for each :
- Micelles are aggregates which exhibit colloidal behavior at higher concentration, with the hydrophilic part outside and the hydrophobic part towards the oil and dirt particle, eg. soap.
- An aerosol is a colloid of fine solid particle or liquid drops in air or another gas. It can be natural or artificial. Example dust particle and smoke.
Question.20. 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at -0.34 °C. What is the molar mass of this material? (Kf for water = 1.86 K kg mol-1)
Question.22. Explain the following observations giving an appropriate reason for each :
- The enthalpies of atomization of transition elements are quite high.
- There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. 3rd series).
- Mn2+is much more resistant than Fe2+ towards oxidation.
- Due to the presence of metallic bonds as a result of large number of valence electrons.
- The presence of valence electrons and unpaired d-orbital electrons help heavy transition metals to form metallic bonds.
- Due to stability of Mn2+ because of half filled-subshell (3d5) it does not gets oxidized. But Fe2+ has 3d6configuration and it can lose one electron to become 3d5 which is stable. Therefore it is easily oxidized.
Question.23. Write the name, the structure and the magnetic behavior of each of the following complexes:
(i) Amminechloridonitrito—N-platinum (II),
trigonal planar, diamagnetic.
(ii)Tetra-ammine dichlorido cobalt (III) chloride, octahedral, diamagnetic
(iii)Tetracarbonyl nickel (0), Tetrahedral, diamagnetic
Question.27. Explain the following terms giving one example of each , type:
- Antacids, .
- Substances consumed to reduce acidity in the stomach by neutralizing excess HCl produced by the stomach, eg. Milk of magnesia.
- Disinfectants are chemicals used to kill microorganism, applied only to non-living objects like floors and drains, e.g. 1% phenol solution.
- Enzymes are structurally globular proteins that catalyse biochemical reactions in living organisms eg. Trypsin.
Question.30. Draw the molecular structures of following compounds:
(i) XeF6 (ii) H2S2O8
(b) Explain the following observations :
(i) The molecules NH3 and NF3 have dipole moments which are of opposite direction.
(ii)All the bonds of PCl5 molecule are not equivalent.
(iii)Sulphur in vapour state exhibits paramagnetism.
(b) (i) Fluorine is more electronegative than nitrogen while hydrogen is less electronegative than nitrogen resulting in opposite c(ipole moments of NH3 and NF3. Dipole points towards N in NH3 and towards F in NF3.
(ii) PCl5 has trigonal bipyramidal shape. The two axial bonds feel more repulsion from the three equatorial bond pairs and are hence longer. Therefore all bonds in PCl5 are not equivalent.
(iii)Sulphur in vapour state exists asS2 and hence likeO2 has two unpaired electrons in antibonding n orbitals leading to paramagnetism.
(b)(i) Due to the small size, absence of d-orbitals and presence of triple bond in nitrogen, it is less reactive than phosphorous.
(ii)Due to the increase in size and inert pair effect, +5 oxidation state decreases down the group.
(iii)Bond angles in NO–2 and NO+2 are not of same value because NO+2 is linear and NO–2 has bent shape.