CBSE Previous Year Solved  Papers  Class 12 Chemistry Delhi 2015

Time allowed: 3 hours                                                                                      Maximum Marks: 70

General Instructions:

  1. All questions are compulsory.
  2. Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
  3. Questions number 6 to 10 are short-answer questions and carry 2 marks each.
  4. Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
  5. Questions number 23 is a value based question and carry 4 marks.
  6. Questions number 24 to 26 are long-answer questions and carry 5 marks each.
  7. Use log tables, if necessary. Use of calculators is not allowed.


Question.1. What is the basicity of H3PO4 ?
Answer : H3POhas three ionisable hydrogen atoms. Hence, its basicity is 3.

Question.2. Write the IUPAC name of the given compound :

Question.3. Which would undergo SN2 reaction faster in the following pair and why ?
Answer : Primary alkyl halides will undergo SN2 reactions faster than tertiary alkyl halides because of less steric hindrance experienced by the approaching nucleophile. Hence, out of the given pair, primary alkyl halide (CH3CH2Br) would undergo SN2 reaction faster.

Question.4. Out of BaCl2 and KCl, which one is more effective in causing coagulation of a negatively charged colloidal Sol ? Give reason.
Answer : According to the Hardy-Schulze rule, greater the valency of a flocculating ion, the greater is its power to cause precipitation. Between Ba2+ (from BaCl2) and K+ (from KCl), Ba2+ has greater valency. Therefore, BaCl2 will be more effective in causing the coagulation of a negatively charged colloidal sol.

Question.5. What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1/3 rd of tetrahedral voids ?

Question.6. What are the transition elements ? Write two characteristics of the transition elements.
Answer : Elements which in their ground state or in any of their oxidation state have partially filled d-orbital. The name ‘transition given to the elements of d-block is only because of their position between d-block and p-block elements.
The two characteristics of transition elements are :
(i) They show variable oxidation states.
(ii) They generally form coloured compounds.

Question.7. (i) Write down the IUPAC name of the following complex: [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine)
(ii) Write the formula for the following complex : Pentaamminenitrito-o-Cobalt(III).
(i) Diamminedichloridethylenediaminchromium(III) chloride.
(ii) [Co(NH3)5(ONO)]2+.

Question.8. Name the reagents used in the following reactions :
(i) Sodium borohydride (NaBH4)
(ii) Alkaline potassium permanganate (KMnO4 – KOH).

Question.9. What is meant by positive deviations from Raoult’s law ? Give an example. What is the sign of ∆mixH for positive deviation ?
Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult’s law ? Give an example.
Answer : Positive deviation from Rauolt’s law means that the observed vapour pressure is greater than expected, and It occur A — B attractions are weaker than the average of the attractions in the pure component of the mixture. For example : A mixture of ethanol and acetone shows a positive deviation from Raoult s law.
In case of solutions showing positive deviations, absorption of heat takes palce; i.e., ∆mixH has a positive (+) sign.
Azeotropes are the binary mixtures which have the same composition in liquid and vapour phases and boil at a constant temperature.
A minimum-boiling azeotrope is formed by solutions showing a large positive deviation from Raoult s law at a specific composition.
Example : C2H5OH + H20 (An ethanol-water mixture)

Question.10. (a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution :
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why ? 
(b) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with decrease in concentration ?
Answer : The relationship between the standard free energy change and emf of a cell reaction is given by : ∆G° = -nFE°
Thus, more positive the standard reduction potential of a reaction, the more negative is the standard free energy change associated with the process and, consequently, the higher is the feasibility of the reaction.
Since E°Ag+/Ag has a greater positive value than E°H+/H, the reaction which is feasible at the cathode is given by
Ag+ (aq)+ e ——> Ag(s)
(b) The limiting molar conductivity of an electrolyte is defined as its molar conductivity when the concentration of the electrolyte in the solution approaches zero.
The conductivity of an electrolyte solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted or the concentration is decreased. As a result, the conductivity of an electrolyte solution decreases with the decrease in concentration.

Question.11. 3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the Van’t Hoff factor and predict the nature of solute (associated or dissociated). (Given s Molar mass of benzoic acid = 122 g mol-1, Kf for benzene = 4.9 K kg mol-1)

Question.12. (i) Indicate the principle behind the method used for the refining of zinc.
(ii) What is the role of silica in the extraction of copper ?
(iii) Which from of the iron is the purest form of commercial iron ?
Answer : (i) Zinc is a metal having low boiling point and it is refined by distillation method which is used for-metals having low boiling points.
(ii) During extraction of Copper, FeS or FeO is present as impurity hence Si02 is added as flux to form silicate (slag) which can be removed easily as it floats over molten copper (Cu)
(iii) Wrought iron is purest form of commercial iron.

Question.13. An element with molar mass 27 g mol-1 forms a cubic unit cell with edge length 4.05 x 10-8cm. If its density is 2.7 g cm-3, what is the nature of the cubic unit cell ?
Since the number of atoms in the unit cell is 4, the given cubic unit cell hats a face-centred cubic (fcc) or cubic-closed packed (ccp) structure.

Question.14. (a) How would you account for the following:
(i) Actinoid contraction is greater than lanthanoid contraction.
(ii) Transition metals form coloured compounds.
(b) Complete the following equation :
Answer : (a) (i) The 5 f-orbitals (in case of actinoids) have a poorer shielding effect than 4 f-orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more than that experienced by electrons in valence shells in case of lanthanoids. Hence, the size contraction in actinoids is greater than that in lanthanoids. (ii) In the presence of ligands, the d-orbitals of transition metal ions split up into two sets of orbitals having different energies. Thus, the transition of electrons takes place from one set to another. The energy required for these transitions is quite less and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

Question.15. (i) Draw the geometrical isomers of complex [Pt(NH3)2Cl2].
(ii) On the basis of crystal field theory, write the electronic configuration for ft ion if ∆0 < P.(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].
( of Ni = 28).
Carbonyl, CO being a strong field ligand causes the pairing of up valence electrons in the Ni atom against the Hund’s Rule of Maximum Multiplicity. This results in the formation of an inner orbital complex, [Ni(CO)4] . Since the complex [Ni(CO)4] has no unpaired electron, it is diamagnetic in nature and posses tetrahedral shape.

Question.16. Calculate emf of the following cell at 25°C :

Question.17. Give reasons for the following observations :
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process.
Answer : (i) Animal skin (hide) is colloidal in nature and has postively charged colloidal particles. When a hide is soaked in tanning, mutual coagulation takes place and as a result, leather get hardened.
(ii) The stability of lyophillic solution is depend on the result of two factors, the presence of a charge and the solvation of colloide particles. On the other hand, the stability of lyophobic solutions is only because of the presence of a charge. Thus, lyophilic solution is more stable than lyophobic solution due to the extensive solvation.
(iii) It is necessary to remove CO when ammnioa is prepared by Haber’s process because in this process the CO act as a poison and adversely affects the activity of iron catalyst, used in the process.

Question.18. Write the names and structures of the monomers of the following polymers:
(i) Nylon-6, 6
(ii) PHBV
(iii) Neoprene.

Question.19. Predict the products of the following reactions :

Question.20. How do you convert the following:
(i) Phenol to anisole
(ii) Propan 2-ol to 2-methylpropan-2-ol
(iii) Aniline to phenol
(a) Write the mechanism of the following reaction :
(b) Write the equation involved in the acetylation of Salicyclic acid.

Question.21. (i) Which one of the following is a disaccharide : Starch, Maltose, Fructose, Glucose ?
(ii) What is the difference between fibrous protein and globular protein ?
(iii) Write the name of vitamin whose deficiency causes
bone deformities in children.
Answer : (i) Maltose is a disaccharide, as it contains a-D- glucose units.
(iii) The deficiency that causes bone deformities in children is vitamin D.

Question.22. Give reasons:
(a) n-Butyl bromide has higher boiling point than t-butyl bromide.
(b) Racemic mixture is optically inactive.
(c) The presence of nitro group (-N02) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer: (a) n-butyl bromide is a straight chain molecule having larger surface area and therefore, has stronger intermolecular forces. On the other hand, butyl bromide (CH3)3 CBr is branched molecule, so it has a smaller surface area. Hence, it has weaker intermolecular force.
(b) Racemic mixture is an equimolar solution of two enantiomers (d and / forms) and thus, the rotation due to one isomer is cancelled by the rotation due to another. Therefore, it has zero optical rotation and hence, it is optically inactive.
(c) This is because nitro group (-N02) at o/p positions withdraw the electrons from the benzene ring. The reduced electron density at o/p positions for haloarenes facilitate the attack of nucleophile. The negative charge in the Carbanion formed, at o/p positions will respect to halogen atom is stabilised through and the presence of nitro groups (N02) respectively.

Question.23. Mr. Roy, the principal of one reputed school organized a seminar in which he invited parents and . principals to discuss the serious issue of diabetes and depression in students. They all resolved this issue by strictly banning the junk food in school and to introduce healthy snacks and drinks like soup, lassi, milk etc. in school canteens. They also decided to make compulsory half an hour physical activities for the students in the morning assembly daily. After six months, Mr. Roy conducted the health survey in most of the schools and discovered a tremendous improvement in the health of students.
After reading the above passage, answer the following:
(i) What are the values (at least two, displayed by Mr. Roy ?
(ii) As a student, how can you spread awareness about this issue ?
(iii) What are tranquilizers 1 Give an example.
(iv) Why is use of aspartame limited to cold foods and drinks ?
Answer : (i) Mr. Roy showed concern for the health of the student; and Caring in nature and aware towards the harmful effect of junk food.
(ii) As i student, I can spread awareness regarding diabetes and depression among students by conducting seminars, health camps, debates, distribution of pamphets, and workshops by doctors. To highlight the needs to follow healthy eating.
(iii) Tranquilizers are neurologically active drugs that induce a sense of well being and are used to treat stress, anxiety and mild or severe mental disease. They perform their function by inhibiting the message transfer mechanism from nerve to receptor e.g., equanil, meprobamate and iproniazid etc.
(iv) Because aspartame is unstable at cooking temperature.

Question.24. (a) Account for the the following :
(i) Adding character increases from HF to HI.
(ii) There is large difference between melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) ClF3
(ii) XeF4 ,
(i) Which allotrope of phosphorus is more reactive and why ?
(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone layers ?
(iii) F2 has lower bond dissociation enthalpy than Cl2. Why?
(iv Which noble gas is used in filling balloons for meteorological observations ?
(v) Complete the equation : XeF2 + PF5
Answer : (a) (i) The acidic strength of the hydrohalic acids increases from HF to HI because the stability of the acids decreases from HF to HI on account of decrease in bond dissociation enthalpy of H — X bond from HF to HI.
(ii) The difference in melting point and boiling point of oxygen and sulphur is due to the difference in their atomicities oxygen exists as a diatomic (02) molecule, while sulphur exists as a polyatomic (S8) molecule and also oxygen is small in size and have high electro negativities.
(iii) Being an element of second period, Nitrogen has no ‘d’ orbitals and its maximum covalency is restricted to four. Hence, due to the non-availability of d-orbitals, it can’t form pentahalides.
(i) White phosphorus is most reactive of all the allotropes of phosphorus. It is because it exists as P4 discrete tetrahedral units with 60° angle, which results in angular strain and makes it highly reactive.
(ii) Since supersonic jets fly in the stratosphere near the ozone layer, they are responsible for the depletion of ozone layer. The oxide emitted from the exhausts of supersonic jet aeroplanes readly combine with ozone to form nitrogen dioxide and diatomic oxygen.
NO(g) + 03 (g) ——> NO2 (g) + 02 (g)
(iii) The size of a fluorine atom is veiy small as compared to a chlorine atom. Therefore, the repulsion between electrons in the outer most shell of the two atoms in a fluorine molecule is much greater than that in a chlorine molecule. Hence, it requires less energy to break up the fluorine molecule, making its bond dissociation energy lesser than that of chlorine molecule.
(iv) Helium, gas is used for filling of balloons for meteorological observations.

Question.25. An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions.
(a) Write the structures of main products when aniline reacts with the following reagents:
(i) Br2 water
(ii) HCl
(iii) (CH3C0)20/pyridine.
(b) Arrange the following in the increasing order of their boiling point:
C2H5NH2, C2H50H, (CH3)3N
(c) Give a simple chemical test to distinguish between the following pair of compounds :
(CH3)2NH and (CH3)3N.
(b) Ethanol has high boiling point then ethylamine because oxygen, being more electronegative and form Strong extensive hydrogen bond as compared to that of nitrogen. In trimethylamine there is no hydrogen and hence has the lowest boiling point.

Question.26. For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
(a) For a reaction A + B —>P, the rate is given by
Rate = k[A] [B]2
(i) How is the rate of reaction affected if the concentration of B is doubled ?
(ii) What is the overall order of reaction if A is present in large excess ?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010)
(a) (i) Since the given reaction has order two with respect to reactant B, thus if the concentration of B is doubled in the given reaction, then the rate of reaction will become four times.
(ii) It the concentration of B is doubled i.e. , [B]2 the overall reaction will be two, because if A is present in large excess, then the reaction will be independent of the concentration of A and will be dependent only on the concentration of B. Order of reaction = 2.

One thought on “CBSE Previous Year Solved Papers Class 12 Chemistry Delhi 2015

  1. I go through whole paper .. it was quite easy to understand and also save my alot of time. Thank you

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