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Chemistry 2007 Set 1.1 Sol

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					2/4/2011                                         Subjective Test Paper - Chemistry - Meri…




  Chemistry 2007 Set 1                                                                           Close


 Subjective Test




      (i)    All questions are compulsory.
     (ii)    This question paper consists of four sections A, B, C and D.
              Section A contains 5 questions of one mark each.
              Section B is of 7 questions of two marks each.
              Section C is of 12 questions of three marks each and
              Section D is of 3 questions of five marks each.
     (iii)   There is no overall choice. However, an internal choice has been provided.
     (iv)    Wherever necessary, the diagrams drawn should be neat and properly labelled.



   Section A


    Question 1 ( 1.0 marks)
   What is the coordination number in a rock-salt type structure?

   Solution:
   In rock-salt, every Na + ion is surrounded by 6 Cl− ions and each Cl− ion is surrounded by 6 Na + ions.
   Thus, the coordination number of each type of ion in rock salt is 6.



    Question 2 ( 1.0 marks)
   State Raoult’s law for a binary solution containing volatile components.

   Solution:
   Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each
   component in the solution is directly proportional to its mole fraction.



    Question 3 ( 1.0 marks)
   What is meant by order of a reaction being zero?

   Solution:
   For a zero-order reaction, the sum of the powers to which the concentration terms are raised is
   zero. Thus, the reaction is independent of concentration.



    Question 4 ( 1.0 marks)
   Write the IUPAC name of the following compound:




   Solution:




   The IUPAC name of the given compound is 3, 3-dimethyl-butanoic acid.



    Question 5 ( 1.0 marks)
   Mention one commercial use of N, N-Dimethyl aniline (DMA).
…meritnation.com/…/6HfQ3MmOhVMw…                                                                             1/12
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…

   Solution:
   DMA is used as a promoter in the curing of polyesters and vinyl ester resins. It is also used for the
   preparation of several organic compounds.



   Section B


    Question 6 ( 2.0 marks)
   State as a mathematical formula the de Broglie relationship for moving particles. W hat experimental
   evidence is available for this concept?

                                                      OR

   Specify the ranges of values for quantum numbers ml and ms for an electron in an atom when the n
   quantum number value for it is 2. What is the significance of these values for the orbitals?

   Solution:
   According to De Broglie relationship for moving particles,




   Where, λ is the wavelength of the wave associated with mass ‘m’ moving with velocity ‘v’

   This concept was experimentally verified by Nickel Crystal with the help of the diffraction pattern of
   electrons.

   OR

   When Principal Quantum number, n = 2,

   l = 0, 1

   When l = 0,




   This signifies that the electron is present in the s sub-shell, and can have two orientations.

   When l = 1,




   This signifies that the electron is present in the p sub-shell, which has 3 orbitals. Electrons in each
   orbital can have two orientations.



    Question 7 ( 2.0 marks)
   When can an endothermic process be spontaneous? Give an example of such a process.

   Solution:
   Endothermic reactions can be made spontaneous by increasing the entropy of the system.

   For example −




   It is an endothermic reaction, but spontaneous.



    Question 8 ( 2.0 marks)
   Write balanced chemical equations for the following reactions:

   (i)


   (ii)


   Solution:
   (i)
…meritnation.com/…/6HfQ3MmOhVMw…                                                                             2/12
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…
   (i)




   (ii)




    Question 9 ( 2.0 marks)
   Explain any one of the following statements:

   (i) The transition metals are well known for the formation of interstitial compounds.

   (ii) The largest number of oxidation states are exhibited by manganese in the first series of
   transition elements.

   Solution:
   (i) Small atoms like hydrogen, carbon, boron and nitrogen occupy the interstitial sites in the lattices
   of transition metals. This gives rise to their interstitial compounds. This is mainly due to the vacant d
   orbital in transition metals and variable oxidation states.

   (ii) In each group of transition metals, oxidation states increase with increase in atomic number,
   reach a maximum in the middle, and then, start decreasing. The increase in oxidation states is due
   to the increase in the number of valence electrons with the increase in atomic number. As
   manganese lies in the middle, it has the maximum number of valence electrons. Hence, it shows
   maximum oxidation states.



    Question 10 ( 2.0 marks)
   Draw the three-dimensional representations of (R) - and (S) - butan- 2- ol.

   Solution:




    Question 11 ( 2.0 marks)
   Write chemical reaction equations to illustrate the following reactions:

   (i) Williamson synthesis of ethers

   (ii) Reimer−Tiemann reaction

   Solution:
   (i) Williamson synthesis of ethers:

   It is used for preparing both symmetrical and unsymmetrical ethers. It involves the treatment of alkyl
   halide with sodium alkoxide or sodium phenoxide to give ether and a sodium salt of halide.




   (ii) Reimer−Tiemann reaction:

   It involves the reaction between phenol and chloroform in the presence of an aqueous alkali at 333
   K − 343 K to give salicylaldehyde.




…meritnation.com/…/6HfQ3MmOhVMw…                                                                               3/12
2/4/2011                                            Subjective Test Paper - Chemistry - Meri…




   If the reaction is carried out with carbon tetrachloride, salicylic acid is produced.



    Question 12 ( 2.0 marks)
   Distinguish between addition polymers and condensation polymers and give one example of each
   class.

   Solution:

                       Addition polymer                                        Condensation polymer

     (i)    It is formed when the monomer units              (i)     It is formed when two or more monomer
            are repeatedly added to form long                        units react together.
            chains.

     (ii)   No elimination of any by-product takes           (ii)    Elimination of simple compounds like H2 O, R
            place.                                                   −OH takes place.

    (iii)   For example:                                     (iii)   For example:

            Polythene is addition polymer of                         Nylon (6, 6) is a condensation polymer
            ethylene.                                                obtained from hexamethylene diamine and
                                                                     adipic acid.



   Section C


    Question 13 ( 3.0 marks)
   Answer the following in the light of MO theory:

   (a) Which has a higher bond order, C 2 or          ?

   (b) Which species is not likely to exist, Li2 or Be 2 ?

   OR

   (a) Compare the structural shapes of the following species:

   SF6 and SF4

   (b) What type of intermolecular forces exists between Cl2 and CBr4 present in a mutual solution?


   Solution:
   (a) In C 2 ,

   Number of electrons = 12

   Electronic configuration =


   Now, bond order =




   In       ,

   Number of electrons = 12+ 2 = 14

   Electronic configuration =




…meritnation.com/…/6HfQ3MmOhVMw…                                                                                    4/12
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…
   Therefore, C 2 2- has the higher bond order.

   (b) For Li2 :

   Number of electrons = 6

   Electronic configuration =




   For Be 2 :

   Number of electrons = 8
   Electronic configuration =




   Therefore, Be 2 is unstable and is not likely to exist.

   OR

   (a) SF6 molecule has perfect octahedral geometry, while SF4 has distorted trigonal bipyramidal
   geometry.




   (b) Cl2 and CBr4 are both non-polar molecules. These molecules have no permanent dipoles. Hence,
   dispersion forces are the only forces that operate between these molecules.



    Question 14 ( 3.0 marks)
   (a) Name an element w ith which silicon can be doped to give an n-type semiconductor.

   (b) Which type of crystals exhibits piezoelectricity?

   Solution:
   (a) When Si is doped with group-15 elements, i.e., phosphorus, arsenic, etc., n-type semi-conductor
   is formed.

   (b) Piezoelectricity is exhibited by the crystals which produce voltage when pressure or stress is
   applied.

   Example − Rochelle salt, Barium titanate



    Question 15 ( 3.0 marks)
   The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a one molal
   solution of a non-volatile non-ionic solute in water.

   Solution:
   The molality of the solution is 1.

   Therefore, 1 mol of solute is present in 1000 g of water.


   Hence, number of moles of water in 1000 g =


   = 55.55

   Total number of moles in the solution = 1 + 55.55 = 56.55


   ∴Mole fraction of H2 O =


   Now, according to Raoult’s law,



…meritnation.com/…/6HfQ3MmOhVMw…                                                                         5/12
2/4/2011                                             Subjective Test Paper - Chemistry - Meri…




    Question 16 ( 3.0 marks)

   Using the values of ∆f HΘ and ∆rSΘ , given herein, calculate the standard molar Gibbs energy of
   formation, (∆f GΘ ) for CS2 (l).

   Given:


   Solution:
   ∆H = 89.70 kJ mol−1 = 89.70 × 10 3 J mol−1




   T = 298 K

   ∴Standard molar Gibbs energy of formation,         G

   = 89.70 × 103 J mol−1 − 298 K × 151.34 J K−1 mol−1

   = 89.70 × 103 − 45.10 × 10 3 J mol−1

   = 44.6 × 10 3 J mol−1



    Question 17 ( 3.0 marks)
   The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate
   activation energy of such a reaction.

   (R = 8.314 J mol−1 K−1 , log 2 = 0.3010)

   Solution:




   Where,

   k1 , k2 are rate constants

   Ea is activation energy

   R is universal gas constant = 8.314 J mol−1 K−1

   T1 = 298 K, T2 = 308 K

   On substituting the above values, we get




   ⇒ E a = 52897.77

   = 52.9 × 10 3 J

     53 kJ



    Question 18 ( 3.0 marks)
   State what is observed when

   (i) the electrodes connected to a battery are dipped into a sol.

   (ii) an electrolyte solution is added to a sol.

   (iii) an emulsion is subjected to high-speed centrifugation.

  Solution:
…meritnation.com/…/6HfQ3MmOhVMw…                                                                        6/12
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…
    Solution:
   (i) When the electrodes connected to a battery are dipped in a solution, the ions produced by the
   dissociation of the electrolyte move towards the oppositely charged electrodes. The cations move
   towards the negatively charged electrode while the anions move towards the positively charged
   electrode.

   (ii) The addition of an electrolyte causes a colloidal solution to coagulate.

   (iii) On high-speed centrifugation, an emulsion decomposes into its constituent liquids.



    Question 19 ( 3.0 marks)
   Answer the following questions:

   (i) Which element in the first series of transition elements does not exhibit variable oxidation states
   and why?

   (ii) What happens when a solution of copper (II) sulphate is saturated with ammonia?

   (iii) Why do actinoids, in general, exhibit a greater range of oxidation states than lanthanoids?

   Solution:
   (i) Scandium, with the atomic number ‘21’, does not show variable oxidation states. After losing
   three electrons, all the orbitals in scandium are fully filled.

   Thus, scandium exhibits only +3 oxidation state.

   (ii) In the presence of ammonia, copper (II) sulphate splits into two sets having slightly different
   energies. Thus, d−d transition occurs and it acquires blue colour.

   In this reaction, [Cu(NH3 )4 ]2+ is produced.

   (iii) Actinoids exhibit greater range of oxidation states than lanthanoids. This is because 5f, 6d and
   7s sub-shells in actinoids are of comparable energies.



    Question 20 ( 3.0 marks)
   (a) Illustrate the following with an example each.

   (i) Linkage isomerism

   (ii) Coordination isomerism

   (b) Why is           paramagnetic? (Ni = 28)


   Solution:
   (a)

   (i) Linkage isomerism results when there are two ways in which a ligand can get attached to the
   central atom.

   For example, NO 2 group can bind to a metal, either through a nitrogen atom or an oxygen atom, as
   indicated:

   (ii) Coordination isomerism is shown by the compounds in which the cation and the anion are
   complexes.

   For example,


   (b) In [NiCl4 ]2−, the oxidation state of nickel is +2. Thus, Ni has 8 electrons in the 3d orbital and no
   electron in the 4s orbital. Chlorine, being a weak-field ligand, forms sp3 hybridisation. As there are 2
   unpaired electrons in the 3d orbital of Ni, the complex is paramagnetic.



    Question 21 ( 3.0 marks)
   Write the nuclear reactions for the following radioactive changes:

   (i)      undergoes α-decay


   (ii)     undergoes β-decay


   (iii)     undergoes K-decay

   (You can put ‘X’ for the symbol which is not correctly known)

  Solution:
…meritnation.com/…/6HfQ3MmOhVMw…                                                                               7/12
2/4/2011                                        Subjective Test Paper - Chemistry - Meri…
    Solution:
   (i)


   (ii)


   (iii)



    Question 22 ( 3.0 marks)
   Explain the mechanism of nucleophilic addition to a carbonyl group and give one example of such
   addition reactions.

   Solution:
   The nucleophilic addition to a carbonyl groups proceeds via the following mechanism.




   Example:

   The addition of hydrogen cyanide proceeds as follows:




    Question 23 ( 3.0 marks)
   Write the chemical equations for the following chemical reactions:

   (a) 1-Nitropropene is prepared from acetaldehyde

   (b) Benzonitrile is converted into acetophenone

   (c) A primary amine is prepared from a primary alkyl halide

   Solution:
   (a )




   (b)




   (c)




    Question 24 ( 3.0 marks)
   Mention one important use of each of the following:
…meritnation.com/…/6HfQ3MmOhVMw…                                                                     8/12
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…
   Mention one important use of each of the following:

   (i) Equanil

   (ii) Sucralose

   (iii) Carbon fibres

   Solution:
   (i) Equanil is used in the treatment of depression & hypertension.




   (ii) Sucralose is used as an artificial sweetening agent.

   (iii) Carbon fibres are used for preparing strong fibres such as carbon fibre-reinforced plastics.



   Section D


    Question 25 ( 5.0 marks)
   (a) Write the formulation for the galvanic cell in which the reaction,

                                             takes place.

   Identify the cathode and the anode reactions in it.

   (b) Write Nernst equation and calculate the emf of the follow ing cell:




   OR

   (a) Explain with one example each the terms weak and strong electrolytes.

   (b) Write the Nernst equation and calculate the emf of the following cell:




   Solution:
   (a) The galvanic cell is represented as




   The reaction at the anode is given by



   The reaction at the cathode is given by




   (b) The Nernst equation for the given cell is




   Here, Eº ce ll = Eºcathode − Eºa node

   = 0 − (−0.14)

   = 0.14 V




…meritnation.com/…/6HfQ3MmOhVMw…                                                                        9/12
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…

   = 0.14 − 0.059

   = 0.081 V

   OR

   (a) Strong electrolytes completely ionise when dissolved in water.

   For example, sodium chloride dissociates as Na + & Cl−.

   Weak electrolytes do not ionise completely when dissolved in water.

   For example, carbonic acid dissociates as H+ &

   (b) The Nernst equation for the given cell is




   Here,




    Question 26 ( 5.0 marks)
   (a) How would you account for any two of the following:

   (i) PbO 2 is a stronger oxidising agent than SnO2

   (ii) H3 PO 2 acts as a monobasic acid.

   (iii) The pKa value for HOCl is higher than that of HOClO

   (b) Draw the structures of the following species:

   (i) Peroxodisulphuric acid, H2 S2 O 8

   (ii) Xenon tetrafluoride, XeF4

   OR

   (a) Assign reasons for any two of the following observations:

   (i) The lower oxidation state becomes more stable with increasing atomic number in Group 13.

   (ii) Hydrogen iodide is a stronger acid than hydrogen fluoride in aqueous solution.

   (iii) The basic character among the hydrides of Group-15 elements decreases with increasing atomic
   numbers.

   (b) Draw the structural formula for XeOF4 .


   Solution:
   (a)

   (i) PbO 2 is stronger oxidising agent than SnO 2 . This is because of the presence of poor shielding f
   orbitals in Pb. Thus, the effective nuclear charge on the outer shell electrons increases, and this
   makes PbO 2 a strong oxidising agent.

   (ii) The structure of H3 PO 2 is




…meritnation.com/…/6HfQ3MmOhVMw…                                                                            10/12
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…

   Though it has three hydrogen atoms, only one of these is attached to an oxygen atom. Only those
   hydrogen atoms are acidic which are connected to oxygen atoms; therefore, H3 PO 2 acts as a
   monobasic acid.

   (iii) The pKa value is determined by acidic strength. The conjugate base of HOCl is less stable than
   the conjugate base of HOClO. Thus, HO 2 Cl is a stronger acid. Also, increase in the number of oxygen
   atoms increases the stability of the conjugate base.

   (b)

   (i)




   (ii)




   OR

   (a)

   (i) The outer electronic configuration of group-13 elements is ns 2 np1. On moving down the group,
   due to inert pair effect, there is a decrease in the tendency of the s electrons of the valence shell to
   participate in bond formation. This is due to the poor shielding of ns 2 electrons by the intervening d
   orbital. Thus, the lower oxidation state becomes more stable with increasing atomic number.

   (ii) Hydrogen iodide is a strong acid. This is because the small size of hydrogen and the big size of
   iodine make the H−I bond very weak. On the other hand, due to the similar sizes of H and F, the H−F
   bond is quite strong. Thus, HI is a stronger acid than HF in aqueous solution.

   (iii) Among the group-15 elements, nitrogen has a small size and its density per unit volume is more.
   On moving down the group, the size of the central metal increases; thus, electron density
   decreases. As a result, the electron-donor capacity or the basic strength of group-15 hydrides
   decreases with increasing atomic numbers.

   (b)




    Question 27 ( 5.0 marks)
   (a) Name the three major classes of carbohydrates and give the distinctive characteristic of each
   class.

   (b) What are nucleotides? Name two classes of nitrogen-containing bases found amongst
   nucleotides.

                                                      OR

   (a) Describe the classification of lipids based on their chemical compositions. Mention the chief
   chemical characteristic of each class.

   (b) Explain the term ‘mutarotation’.

   Solution:
   (a) Based on their behaviour on hydrolysis, carbohydrates are classified into three major groups.

   (i) Monosaccharides − They cannot be hydrolysed to still simpler carbohydrates.

   Example: Glucose

   (ii) Oligosaccharides − They give two to ten units of monosaccharides on hydrolysis.

   Example: Sucrose

…meritnation.com/…/6HfQ3MmOhVMw…                                                                              11/12
2/4/2011                                        Subjective Test Paper - Chemistry - Meri…
   (iii) Polysaccharides − They are polymeric molecules and give large number of monosaccharide units
   on hydrolysis.

   Example: Starch

   (b) The monomeric units of nucleic acids are called nucleotides. It consists of three parts − a
   phosphate group, a five-carbon sugar and a nitrogen heterocyclic base.

   The nitrogen-containing bases are derivatives of two classes − purine and pyrimidine.

   OR

   (a) Out of current syllabus

   (b) Out of current syllabus




…meritnation.com/…/6HfQ3MmOhVMw…                                                                        12/12

				
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