Chemistry 2007 Set 1 Sol

Document Sample
Chemistry 2007 Set 1 Sol Powered By Docstoc
					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 number of atoms per unit cell in a body-centred cubic structure?

   Solution:

   The number of atoms per unit cell in a body-centred cubic structure is 2 (8 corner atoms ×     atom

   per unit cell + 1 body-centre atom).



    Question 2 ( 1.0 marks)
   Define osmotic pressure.

   Solution:
   Osmotic pressure is defined as the equilibrium hydrostatic pressure set up by a difference between
   the concentrations of solutions on the two sides of a semi-permeable membrane. Experimentally, it
   has been found that for a dilute solution, osmotic pressure (π) is proportional to the molarity (C) of
   the solution at a given temperature (T).

   π = CRT

   Where, R is the gas constant


   Or,                [        n → Moles of the solute, V → Volume of solution]



   Or,            [           , w2 → Mass of solute, M2 → Molar mass of solute]




    Question 3 ( 1.0 marks)
   For the reaction




   the rate law is expressed as

   rate = k[Cl2 ][NO]2

   What is the overall order of this reaction?

   Solution:
   Overall order of this reaction = 1 + 2

   =3

…meritnation.com/…/NiakQOGIPNGTBO…                                                                          1/14
2/4/2011                                            Subjective Test Paper - Chemistry - Meri…


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




   Solution:




    Question 5 ( 1.0 marks)
   Why do nitro compounds have high boiling points in comparison with other compounds of same
   molecular mass?

   Solution:
   Nitro compounds are highly polar in nature; therefore, there is strong electrostatic attraction
   between the nitrogen and oxygen atoms of a nitro group. As a result, a large amount of energy is
   required to break the nitrogen and oxygen bonds. Hence, nitro compounds have high boiling points
   in comparison with other compounds of same molecular mass.




   Section B


    Question 6 ( 2.0 marks)
   State ‘Pauli’s exclusion principle’. Explain giving an example how this principle limits the maximum
   occupancy of an energy level in an atom.

                                                         OR

   State ‘Aufbau principle’ and give the order in which the energies of orbitals increase and hence they
   are filled in that order.

   Solution:
   Pauli’s exclusion principle states that only two electrons may exist in the same orbital and those two
   electrons must have opposite spins.

   The atomic number of Na is 11. Therefore, a neutral atom of sodium has 11 electrons.

   There is one 1s orbital. So, the first two electrons occupy the 1s orbital. Similarly, there is one 2s
   orbital, so the next two electrons occupy the 2s orbital. The next six electrons occupy the 2p orbitals
   as there are three 2p orbitals viz.,the 2px , 2py and 2pz orbitals. Then, the remaining one electron
   occupies the 3s orbital.




   It can be observed that a maximum of two electrons can occupy the s subshell and a maximum of six
   electrons can occupy the p subshell.

                                                         OR

   Aufbau principle states that in the ground state of atoms, the orbitals are filled in the order of their
   increasing energies, i.e., electrons occupy the higher energy orbitals only after the lower energy
   orbitals are filled.

   The order in which the energies of orbitals increase, and hence, the orbitals are filled is

   1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p,..




…meritnation.com/…/NiakQOGIPNGTBO…                                                                            2/14
2/4/2011                                         Subjective Test Paper - Chemistry - Meri…




    Question 7 ( 2.0 marks)

   A reaction with             , alw ays has an equilibrium constant value greater than 1. Why?


   Solution:
   For B, the reaction is spontaneous.

   Mathematically,

   K=

   If            ,


   Then −∆G° /RT > 0

   So,           >1

   Therefore, K > 1

   For a spontaneous reaction, the value of equilibrium constant (K) is greater than 1. This is why a
   reaction with           always has an equilibrium constant value greater than 1.



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

   (i) Aluminium dissolves in aqueous hydrochloric acid

   (ii) Tin reacts with a hot alkali solution

   Solution:
   (i) Aluminium reacts with aqueous hydrochloric acid to give aluminium chloride and hydrogen gas.




   (ii) Tin reacts with hot sodium hydroxide solution to give sodium stannite and hydrogen gas.




    Question 9 ( 2.0 marks)
   Write the structures of the following species:

   (i) H3 PO 2 (ii) H2 SO 5


   Solution:
   (i)




   (ii)


…meritnation.com/…/NiakQOGIPNGTBO…                                                                      3/14
2/4/2011                                        Subjective Test Paper - Chemistry - Meri…




    Question 10 ( 2.0 marks)
   Identify whether the following pairs of compounds are structural or geometrical isomers:

   (i)




   (ii)




   Solution:
   (i) The given two compounds are geometrical isomers as the relative positions of their atoms or
   groups in space are different.

   (ii) The given two compounds are structural isomers as they have different structures.



    Question 11 ( 2.0 marks)
   How would you account for the following:

   (i) Phenols are much more acidic than alcohols.

   (ii) The boiling points of ethers are much lower than those of the alcohols of comparable molar
   masses.

   Solution:
   (i)




   Phenoxide ion is more stable than alkoxide ion due to the delocalisation of negative charge in the
   former. Hence, phenols are more acidic than alcohols.

   (ii) Alcohols undergo intermolecular H-bonding while ethers do not.




   Therefore, extra energy is required to break the H-bonds in alcohol. Hence, the boiling point of
   ethers is much lower than those of the alcohols of comparable molar masses.



    Question 12 ( 2.0 marks)
   Draw the structure of the monomer of each of the following polymers:

   (i) Polyvinylchloride (PVC) (ii) Nylon-6
…meritnation.com/…/NiakQOGIPNGTBO…                                                                      4/14
2/4/2011                                            Subjective Test Paper - Chemistry - Meri…
   (i) Polyvinylchloride (PVC) (ii) Nylon-6

   Solution:
   (i) The monomer of polyvinylchloride (PVC) is




   (ii) The monomer of nylon-6 is




   Section C


    Question 13 ( 3.0 marks)
   Write the molecular orbital configurations of the following species and rearrange them in the
   increasing order of their bond lengths:

           and O 2 .


   Solution:
   The molecular orbital configuration of      is




   The molecular orbital configuration of     is




   The molecular orbital configuration of O 2 is




   Bond order of



        (9 − 4)


   = 2.5


   Bond order of



        (7 − 4)


   = 1.5


   Bond order of O 2



        × (10 − 6)


   =2

…meritnation.com/…/NiakQOGIPNGTBO…                                                                 5/14
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…
   Hence,            and O 2 are arranged in the increasing order of their bond lengths as


       < O2 <



    Question 14 ( 3.0 marks)
   Explain each of the following with a suitable example:

   (i) Paramagnetism

   (ii) Piezoelectric effect

   (iii) Frenkel defect in crystals

   Solution:
   (i) Paramagnetism:

   The phenomenon due to which a substance gets attracted towards a magnetic field is called
   paramagnetism. The substances attracted by a magnetic field are called paramagnetic substances.
   Some examples of paramagnetic substances are O 2 , Cu 2t, Fe 3t and Cr3t.

   Paramagnetic substances get magnetised in a magnetic field in the same direction, but lose
   magnetism w hen the magnetic field is removed. To undergo paramagnetism, a substance must have
   one or more unpaired electrons. This is because the unpaired electrons are attracted by a magnetic
   field, thereby causing paramagnetism.

   (ii) Piezoelectric effect:

   The production of electricity due to the displacement of ions, on the application of mechanical stress,
   or the production of mechanical stress and/or strain due to atomic displacement, on the application
   of an electric field is known as piezoelectric effect. Piezoelectric materials are used in transducers −
   devices that convert electrical energy into mechanical stress/strain or vice-versa. Some piezoelectric
   materials are lead-zirconate (PbZrO 3 ), ammonium dihydrogen phosphate (NH4 H2 PO 4 ), quartz, etc.

   (iii) Frenkel defect in crystals:

   Frenkel defect is observed when an ion is absent from its lattice site (causing a vacancy or hole in its
   stead), and occupies the interstitial sites. Ionic solids containing large differences in the sizes of ions
   show this type of defect. Electrical neutrality as well as stoichiometry of the compound is maintained.
   Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI and ZnS show
   this type of defect.




    Question 15 ( 3.0 marks)
   In the production of water gas the reaction involved is:

   C(s) + H2 O(g) → CO(g) + H2 (g), ∆rH° = +131.4 kJ mol−1

   For this reaction, ∆rS° is +134 JK−1 mol−1 . Find out the spontaneous feasibility of this reaction at (i)
   25°C and (ii) 1000°C.

   Solution:
   We know that

   ∆rG° = ∆rH° − T∆rS°

   (i) ∆rG° = 131.4 kJ mol−1 − (25 + 273)K × 134 J K−1 mol−1

   = 131400 J mol−1 − 298 × 134 J mol−1

   = 91468 J mol−1

   Since ∆rG° is positive, the reaction is non-spontaneous at 25°C.

   (ii) ∆rG° = 131.4 kJ mol−1 − (1000 + 273) K × 134 J K−1 mol−1

…meritnation.com/…/NiakQOGIPNGTBO…                                                                               6/14
2/4/2011                                                  Subjective Test Paper - Chemistry - Meri…
   = 131400 J mol−1 − 1273 × 134 J mol−1

   = −39182 J mol−1

   Since ∆rG° is negative, the reaction is spontaneous at 1000°C.



    Question 16 ( 3.0 marks)
   An antifreeze solution is prepared from 222.6 g of ethylene glycol [C 2 H4 (OH)2 ] and 200 g of water.
   Calculate the molality of the solution. If the density of this solution be 1.072 g mL−1 , what will be the
   molarity of the solution?

   Solution:
   Molar mass of ethylene glycol [C 2 H4 (OH)2 ] = 62 g mol−1


   So, number of moles of ethylene glycol =


   = 3.6 mol


   Mass of water in kg


   = 0.2 kg


   Therefore, molality of the solution




   = 18 mol kg −1


   Given density of the solution




   = 1.072 kg L−1

   Hence, molarity of the solution = 18 mol kg −1 × 1.072 kg L−1

   = 19.296 mol L−1



    Question 17 ( 3.0 marks)

   The decomposition of NH3 on platinum surface,                                                    is a zero order

   reaction with k = 2.5 ×   10 −4   M   s −1 .   What are the rates of production of N2 and H2 ?


   Solution:
   Rate for the given reaction is given by

   k = 2.5 × 10 −4 M s −1

   Therefore, for the given reaction, rate is given by




   Hence, the rate of production of N2 is


           = 2.5 × 10 −4 M s −1


   And, the rate of production of H2 is

…meritnation.com/…/NiakQOGIPNGTBO…                                                                                    7/14
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…

            = 3 × 2.5 × 10 −4 M s −1


   = 7.5 × 10 −4 M s −1



    Question 18 ( 3.0 marks)
   Explain the following terms giving a suitable example in each case:

   (i) Emulsification

   (ii) Homogeneous catalysis

                                                       OR

   Define adsorption. Write any two features which distinguish physisorption from chemisorption.

   Solution:
   (i) Emulsification is the process of making an emulsion. Emulsions are liquid−liquid colloidal systems.
   These are unstable, and sometimes form two separate layers when left undisturbed. To stabilise an
   emulsion, a small quantity of a third substance known as emulsifying agent or emulsifier is added.
   Soaps and detergents are frequently used as emulsifying agents. Some other emulsifying agents are
   proteins, gums, heavy metal salts of fatty acids, long-chain alcohols, etc.

   (ii) The reaction in which the catalyst is present in the same phase as the reactants is said to be
   homogeneous catalysis.

   An example of this reaction is the oxidation of sulphur dioxide into sulphur trioxide in the presence of
   oxides of nitrogen in a lead chamber.




                                                       OR

   Adsorption is a surface phenomenon of accumulation of molecules of a substance on the surface
   rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the
   ‘adsorbate’, and the substance on whose surface the adsorption takes place is called the
   ‘adsorbent’.

                                 Physisorption                               Chemisorption

           1.   In this, the adsorption is attached to the      In this, strong chemical bonds are
                surface of the adsorbent with weak van der      formed between the adsorbate and the
                Waals’ forces of attraction.                    surface of the adsorbent.

           2.   It is generally found to be reversible in       It is usually irreversible in nature.
                nature.



    Question 19 ( 3.0 marks)
   How would you account for the following?

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

   (ii) Hydrogen fluoride is much less volatile than hydrogen chloride.

   (iii) Interhalogen compounds are strong oxidising agents.

   Solution:
   (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 s electrons of the valence shell to
   participate in bond formation. This is due to the poor shielding of the ns 2 electrons of the valence
   shell by the intervening d or f electrons. So, only the p electrons can participate in bonding. Hence,
   the oxidation state is restricted to +1 only.

   (ii) Fluorine is more electronegative than chlorine. Therefore, HF undergoes stronger H-bonding than
   HCl. Hence, HF is much less volatile than HCl.

   (iii) Halogens are very electronegative, so they act as strong oxidising agents. In interhalogen
   compounds, X−X′ bond is very weak and can be easily broken. As a result, interhalogen compounds
   also act as strong oxidising agents.



    Question 20 ( 3.0 marks)
   Write the name and draw the structure of each of the following complex compounds:

   (i) [Co(NH3 )4 (H2 O)2 ]Cl3
…meritnation.com/…/NiakQOGIPNGTBO…                                                                            8/14
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…
   (i) [Co(NH3 )4 (H2 O)2 ]Cl3

   (ii) [Pt(NH3 )4 ] [NiCl4 ]


   Solution:
   (i) Tetraaminediaquocobalt(III)chloride




   (ii) Tetraamineplatinum(II)tetrachloronickelate(II)



    Question 21 ( 3.0 marks)
   The net nuclear reaction of a radioactive decay series is written as:




   Write three pieces of information that you get from the above equation.

   Solution:
   (i)       is an unstable nuclei.


   (ii)      undergoes radioactive decay and changes into a stable              nuclei.


   (iii) In the decay series     to        , 8 α particles and 6 β particles are emitted.



    Question 22 ( 3.0 marks)
   Give chemical tests to distinguish between the following pairs of compounds:

   (i) Propanal and propanone

   (ii) Methyl acetate and ethyl acetate

   (iii) Benzaldehyde and benzoic acid

   Solution:
   (i) Propanal is an aldehyde. Thus, it reduces Tollen’s reagent. Being a ketone, propanone does not
   reduce Tollen’s reagent.




   (ii) Methyl acetate and ethyl acetate can be distinguished by the iodoform test of their hydrolysis
   products.

   When ethyl acetate is boiled with excess of NaOH, ethyl alcohol and sodium acetate is formed. When
   this alkaline solution is heated with I2 , yellow precipitate of iodoform is formed.




   On hydrolysis, methyl acetate gives methyl alcohol that does not respond to the iodoform test.




…meritnation.com/…/NiakQOGIPNGTBO…                                                                       9/14
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…


   (iii) Benzaldehyde and benzoic acid can be distinguished by the NaHCO 3 test. Being an acid, benzoic
   acid responds to this test, but benzaldehyde does not.

   Benzoic acid reacts with NaHCO 3 solution to produce a brisk effervescence with the evolution of CO 2 .




    Question 23 ( 3.0 marks)
   How would you achieve the following conversions:

   (i) Nitrobenzene to aniline

   (ii) An alkyl halide to a quaternary ammonium salt

   (iii) Aniline to benzonitrile

   Write the chemical equation with reaction conditions in each case.

   Solution:
   (i)




   (ii)




   (iii)




    Question 24 ( 3.0 marks)
   (i) Give an example of a hybrid propellant.

   (ii) What are acid dyes?

   (iii) Name a food preservative which is most commonly used by food producers.

   Solution:
   (i) A propellant consisting of liquid N2 O 4 and acrylic rubber is a hybrid propellant.

…meritnation.com/…/NiakQOGIPNGTBO…                                                                           10/14
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…
   Hybrid propellants are mixtures of a solid fuel and a liquid oxidiser.

   (ii) Acid dyes are usually salts of sulphonic acids. For example, orange I is an acid dye.




   (iii) Sodium benzoate (C 6 H5 COONa) is most commonly used by food producers as a food
   preservative.



   Section D


    Question 25 ( 5.0 marks)
   (a) Describe the general trends in the following properties of the first series of the transition
   elements:

   (i) Stability of +2 oxidation state

   (ii) Formation of oxometal ions

   (b) Assign reason for each of the following:

   (i) Transition elements exhibit variable oxidation states

   (ii) Transition metal ions are usually coloured

                                                          OR

   (a) Write the steps involved in the preparation of:

   (i) K2 Cr2 O 7 from Na 2 CrO 4

   (ii) KMnO 4 from K2 MnO 4

   (iii) Calomel from corrosive sublimate

   (b) What is meant by lanthanoid contraction? What effect does it have on the chemistry of the
   elements which follow lanthanoids?

   Solution:
   (a)

   (i)

            Sc                 +3

            Ti          +2     +3   +4

            V           +2     +3   +4   +5

            Cr    +1    +2     +3   +4   +5   +6

            Mn          +2     +3   +4   +5   +6     +7

            Fe          +2     +3   +4   +5   +6

            Co          +2     +3   +4   +5

            Ni          +2     +3   +4

            Cu    +1    +2     +3

            Zn          +2

   It is evident from the above table that the maximum number of oxidation states is shown by Mn,
   varying from +2 to +7. The numbers of the oxidation states increase on moving from Sc to Mn. On
   moving from Mn to Zn, the numbers of the oxidation states decrease due to a decrease in the
   number of available unpaired electrons. The relative stability of the +2 oxidation state increases on
   moving from top to bottom. This is because on moving from top to bottom, it becomes more and
   more difficult to remove the third electron from the d orbital.

   (ii) The elements of the first series of transition metals form a variety of oxides of different oxidation
   states having general formulae MO, M2 O 3 , M3 O 6 , MO 2 , MO 3 , etc. All metals except Sc form MO oxides
   which are ionic in nature. Beyond group 7, no higher oxides except Fe 2 O 3 are known. The oxocations
…meritnation.com/…/NiakQOGIPNGTBO…                                                                               11/14
2/4/2011                                                Subjective Test Paper - Chemistry - Meri…
   which are ionic in nature. Beyond group 7, no higher oxides except Fe 2 O 3 are known. The oxocations
   stabilise as VO 2   +   , VO 2   +   and   TiO 2+.

   (b)

   (i) Transition metals have incompletely filled d orbitals. As a result, they exhibit variable oxidation
   states.

   (ii) Transition metal ions undergo d−d transition of electrons by absorbing light from the visible region
   of electromagnetic spectrum. Due to this d−d transition, colour is imparted to transition metal ions.

                                                            OR

   (a)

   (i)




   (ii) There are three methods for preparing KMnO 4 from K2 MnO 4 .

   (a) By bubbling carbon dioxide gas




   (b) By bubbling chlorine gas




   (c) By reaction with ozonide oxygen


   (iii)


   (b) Lanthanoid contraction is the regular decrease in the atomic and ionic radii from lanthanum to
   lutetium. The decrease in the ionic radii of M3+ ions is regular, but this is not the case with the
   atomic radii.

   As a result of the lanthanoid contraction, the radii of the members of the third transition series are
   very similar to those of the corresponding members of the second transition series. This is the
   reason why Zr and Hf have almost identical radii, and as a result, they occur together in nature and
   cannot be separated easily.



    Question 26 ( 5.0 marks)
   (a) Calculate the emf of the cell

   Mg(s) | Mg 2+ (0.1M) || Cu 2+ (1 × 10 −3 M) | Cu (s)

   Given: E° Cu 2+/Cu = +0.34V, E° Mg 2+/Mg = −2.37 V

   (b) Explain with examples the terms weak and strong electrolytes.

                                                            OR

   (a) The resistance of a conductivity cell containing 0.001 M KC1 solution at 298 K is 1500 . What is
   the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10−3 S cm−1 ?

   (b) Predict the products of electrolysis in the following:

   A solution of H2 SO 4 with platinum electrodes.


   Solution:
   (a) For the given cell,




   = +0.34 − (−2.37) V

   = +2.71 V

   Now, emf of the cell is given by




…meritnation.com/…/NiakQOGIPNGTBO…                                                                             12/14
2/4/2011                                          Subjective Test Paper - Chemistry - Meri…




   = 2.651 V

   (b) Strong electrolytes are substances which dissociate completely (100%) in solution. Strong acids
   (such as HCl, HNO 3 ), strong bases (such as NaOH, KOH) and salts that are not weak acids or weak
   bases (such as NaCl, KBr) are strong electrolytes.

   Weak electrolytes are substances which dissociate partially (1% − 10%) in solution. Weak acids
   (such as CH3 COOH, H2 CO 3 ) and weak bases (such as NH3 , C 5 H5 N) are weak electrolytes.

                                                       OR

   (a)

   Given,

   Conductivity, κ = 0.146 × 10−3 S cm−1

   Resistance, R = 1500

     Cell constant = κ × R

   = 0.146 × 10−3 × 1500

   = 0.219 cm−1

   (b) At the cathode, the following reduction reaction occurs to produce H2 gas.




   At the anode, the following processes are possible.




   For dilute sulphuric acid, reaction (i) is preferred to produce O 2 gas.

   For concentrated sulphuric acid, reaction (ii) occurs.



    Question 27 ( 5.0 marks)
   (a) Name the three major classes of carbohydrates and give an example of each of these classes.

   (b) Answer the following:

   (i) What type of linkage is responsible for the primary structure of proteins?

   (ii) Name the location where protein synthesis occurs in our body.

                                                       OR

   (a) How are lipids classified? Give an example of each class.

   (b) Explain the following terms:

   (i) Mutarotation

   (ii) Avitaminosis

   Solution:
   (a) The three major classes of carbohydrates are −

   (i) Monosaccharides

   Example: glucose

   (ii) Oligosaccharides

   Example: sucrose (disaccharide)

   (iii) Polysaccharides

   Example: starch

…meritnation.com/…/NiakQOGIPNGTBO…                                                                       13/14
2/4/2011                                           Subjective Test Paper - Chemistry - Meri…
   (b) (i) Peptide linkage is responsible for the primary structure of proteins. Peptide linkage is an amide
   formed between −COOH group and −NH2 group of two amino acids.

   (ii) Protein synthesis occurs in the nucleus of a cell.

                                                       OR

   (a) Out of current syllabus

   (b) (i) Out of current syllabus

   (ii) Out of current syllabus




…meritnation.com/…/NiakQOGIPNGTBO…                                                                             14/14

				
DOCUMENT INFO
Shared By:
Stats:
views:10
posted:9/14/2012
language:English
pages:14
Description: PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS