Solved question paper for CHEMISTRY Dec-2018 (B-TECH electronics and communication engineering 1st-2nd)

Question paper 1

1. 1. Answer briefly :
   a) What is fluorescence?

   Answer:

   When a substance is exposed to light or certain other radiations then the phenomenon of re-emitting the energy immediately after the absorption is known as fluorescence.

2. b) What do you understand by effective nuclear charge?

   Answer:

   It is the next charge experienced by an election in a poly electronic atom or in an atom having multiple electrons.

        ZeFF = Z-S
         Z=Atomic no.
         S=No. of shielding elections

3. c) What is optical activity?

   Answer:

   It is the ability of certain substance to rotate the plane of plane-polarized light as it passes through a crystal, liquid or solution. It occurs only when the substance is asymmetric in nature.

4. d) What is the essential condition for a molecule to be IR active?

   Answer:

   (i) Molecule must have finite dipole moment. If it does not have dipole moment, some of its vibrations must produce an induced dipole moment.
   (ii) Frequency of IR radiation must be equal to the vibrating frequency of a particular group or bond in the molecule.

6. e) Discuss entropy.

   Answer:

   It is the measure of the level of disorder in a closed but changing system in which energy can only be transferred in one direction. Higher the energy means higher the disorderliness of the system.

7. f) What is the usefulness of Ellingham diagrams?

   Answer:

   (i) It is mainly used in metallurgy. It tells us the conditions under which ore will be reduced to metal.
   (ii) Determine the partial pressure of oxygen that is in equilibrium with a metal oxide at a given temperature. 

8. g) What do you understand by polarizability?

   Answer:

    It is the ability to form instantaneous dipoles. It is a property of matter. In a solid, it is defined as the dipole moment per unit volume of crystal cell.

9. h) Write the electronic configurations for H2 and H2+ in term of molecular orbital theory.

   Answer:

   Electronic configuration of H2 and H2⁺ in terms of molecular orbital theory:

             H2 = (ϭ15)²

             H2⁺= (ϭ15)¹

11. i) How many signals would you expect to see in the ¹H NMR spectrum of the following :

               CH₃CH₂CH₂CH₃                                              

    Answer:

     CH₃CH₂CH₂CH₃   =  2 signals due to having two types of protons.

       =   Due to having two types of protons it has two signals in 1H NMR spectrum.

12. j) Indicate whether each of the following structures has the R or S configuration. Assign
    priorities to each group. What is the relationship between the two structures?

    

    Answer:

    

    Priorities are given as Br > O > Ethyl > Methyl.

    Both compounds are enantiomers in nature of each other.
     

13. SECTION-B

    Q2 a) Solve the Schrodinger wave equation for a particle in one-dimensional box.
    b) What will happen if the walls of the one-dimensional box are suddenly removed?
     

    Answer:

    (b) Y walls of one dimensional box are suddenly removed then the particle will be free and no external force is acting on it so potential energy of particle will be zero.
                  i.e. [u=0] in above equation.
     

14. Q3 a) With the help of a diagram, explain the splitting of d-orbital energy levels in tetrahedral ligand field? Account for the non-existence of tetrahedral complexes with low spin configurations.

    b) Discuss the relationship that exists between crystal field splitting and pairing energy in
    determining whether a given complex will be high or low spin.

    Answer:

    (a) In a tetrahedral complex, there are four ligands attached to control metal. The d orbitals also split into two different energy levels. The top three consist of dxy, dxz and dyz orbitals. The bottom of two consist of the dx2-y2 and dz2 orbitals. The reason for this is due to poor orbital overlap between the metal and the ligand orbitals. The orbitals are directed on the axis, while the ligands are not.

    

    Energy gap between the two energy levels i.e. ‘e’ and ‘t2’ in tetrahedral complex is very low. Almost 419 of that of octahedral complex, so the energy of promotion becomes less expensive than the election pairing energy. Hence electron will always go to higher states avoiding pairing, so low spin tetrahedral complex are not generally formed.

    (b) Crystal field splitting energy (▲):
                 Difference in energy between the two sets of d orbitals is called the crystal field splitting energy (▲0). Its magnitude depends upon change on the metal ion, position of metal in periodic table.

    Spin pairing energy (P):
                  Increase in energy of system when an electron is placed in an already occupied orbital is known as spin pairing energy.
    If ▲0 > p → low spin complex 
    If ▲0 < p → high spin complex

16. Q4 a) Discuss the principle of electronic spectroscopy. Explain with reference to CH₂ = CH₂, 1, 3- butadiende and carbonyl compounds.
    b) What is fluorescence? Discuss its applications in medicine.

    Answer:

    (a) Basically, electronic spectroscopy is an analytical technique to study electronic structure and dynamics in atoms and molecules. When any excitation source such as X-rays, or any other it will eject an electron from an inner shell of an atom. i.e. it is based upon the phenomenon of photo electric emission.
    In case of carbonyl compounds whose absorption range is 1760-1665 cm-1 due to stretching vibration of c = 0 band. It has characteristic high intensity and a few other functional groups in this region. Exact wave number of c=0 stretch can give us clues as to whether the compound is ketone, aldehyde or carboxylic acid.
    In case of ethane band intensities is 640-3260 cm-1. The probable transition occurs from π →π* at 170-190 nm.
    In case of 1,3- butadiene absorption range at 217 nm. Which is in u.v.ddz range. This corresponds to a π →π* transition.
    (b)  Fluorescence:
            When a substance is exposed to light or certain other radiations then the phenomenon of re-emitting the energy immediately after the absorption is known as fluorescence.
    Application in medicine:
    (i) The structure of protein can be studied by measuring the closeness of fluorescent groups in protein.
    (ii) It is used in fluorescent microscope and X-ray diagnosis.
    (iii) It is also used in determination of uranium in salts in field of nuclear research.

17. Q5 a) What are van der Waals forces? Discuss them briefly.
    b) What do you understand by potential energy surface? Explain with an example.

    Answer:

    (a) Vander wads forces:
             It is the force which accounts for mutual interaction between molecules or insert atoms. These forces are secondary in nature. Weak electrostatic attraction is due to unsymmetrical electrical changes in electrically neutral atoms or molecules. Basically, three types of Vander wad force are as follow.
    Dispersion bond:
              It fluctuating charge on one molecule tends to interact with the fluctuating charge on neighbouring molecule will result in a net attraction known as dispersion bond.
    Dipole-dipole interaction:
              It is a stronger force then dispersion bind. These force occur in molecule having polar in nature such as H2O. In this type of attraction of positive end and negative end of two polar molecules takes place.
    Hydrogen bond:
            It is a special type of strong dipole bond that occurs between the molecules in which one end is a hydrogen atom.
    (b) Potential energy surface:
            It is a mathematical function that gives the energy of a molecule as a function of its geometry.
    In case of H₂O the geometry is defined by two bond length and a bond angle. Bond length is the same then the graph of E versus two geometric parameters q1=O-H bond length q2= the H-O-H bond angle will represent the potential energy surface of water.

    

18. SECTION-C

    Q6 a) Calculate the solubility product of AgBr in water at 25°C from the cell :  
            Ag,Ag⁺ Br (sat.soln.) | AgBr_(s), Ag
    The standard potentials are  E⁰ _AgBr,Ag = 0.07 V;  E⁰ _Ag+,Ag  = 0.80 V

    b) What advantages does the use of “ion-exchange resin” provide over “zeolite process”
    for softening of hard water?

    Answer:

    (a)    Here
         Ag Br (s) + e^- → Ag (s) + Br^- (aq); E⁰R = 0.07 V
         Ag (s) + Ag⁺ (aq) + e^-; E⁰L= O.80 V

    Overall rxn:
         AgBr (s)     Ag⁺ (aq)  +  Br^- (aq)

       E⁰ = E⁰_R-E⁰_L =  -0.73V
      = 0.0591 log [(Ag⁺) (Br ^-)]
      = 0.0591 log Ksp

      = log Ksp = =

    =

    Ksp = 4.81 x 10^-11

    
    (b) Advantages of ion exchange method over zeolite method:
    (i)    Zeolite method replace ca2+ and mg2+ ions by Na+ ions but all acidic impurities eg: HCo3- and Co32- are present as such. But in case of ion exchange resin, even these impurities are also removed properly.
    (ii)    Soft water produced by zeolite method has 10-15 ppm hardness which can’t be used in high pressure boiler, but by ion exchange method water of hardness as low as 2 ppm is obtained which can be used in high pressure boilers.

19. Q7 a) Explain the concept of hard soft acids and bases.
    b) Discuss the geometry of the following : BF₃, H₂O

    Answer:

    (a) According to Hard soft acid and base concept, hard acids prefer binding to hard bases to give ionic complex, whereas soft acids prefer binding to soft base to give covalent complexes. It is sometimes known as hard soft interaction principle.

    •    Large electronegativity differences between hard acids and hard bases give rise to strong ionic interactions.
    •    Electronegativities of soft acids and soft bases are almost the same and hence have less ionic interactions.
    •    Interaction between hard acid, soft base or soft-acid hard base is mostly polar covalent.

    (b) 

    

    Geometry of BF₃ molecule is ‘trigonal planar’. Bond angle is 120⁰. Central atom boron has three valence electron. At the time of formation of BF₃ molecule, each electron in valence shell of B. atom forms a bond pair with F. atom. As a result, the control boron atom is surrounded by three bond pairs and molecule adopts trigonal planar geometry. 

    Geometry of H₂O molecule is bent or regular. The central atom O of water has six valence electrons. The bond angle in H₂O molecule is 104.5⁰.

21. Q8 a) What is optical activity? What is the essential condition for a compound to be optically
    active? Explain.
    b) Draw structural isomers for C₃H₈O and C₄H₁₀O?
     

    Answer:

    (a) Optional activity:        
         It is the ability of certain substance to rotate the plane of plane-polarized light as it passes through a crystal, liquid or solution. It occurs only when the substance is asymmetric in nature.

    Essential conditions for a compound to be optically active: 
    (i) It should have achiral carbon centre with all four groups different.
    (ii) Compound and its mirror should be non-super imposable.
    (iii) Molecule should not have any plane of symmetry.

    (b) Structural isomers of C₃H₈O:

    

    It has 3 isomers:

    1-   Methoxyethane CH₃-O-CH₂-CH₃

    2-    1- Propanol CH₃-CHOH-CH₃

    3-    2 - Propanol CH₃-CH₂-CH₂OH

    Structural isomers of C₄H₁₀O.

    

    Normal butyl alcohol: CH₂CH₂CH₂CH₂OH

    Isobutyl alcohol

    Secondary butyl alcohol

    Tertiary butyl alcohol.

22. Q9 a) Discuss the synthesis of a commonly used drug molecule by taking a suitable example.
    b) Discuss the S_N2 mechanism of alkyl halides in terms of kinetics, stereochemistry and
    reactivity of alkyl halides.

    Answer:

    (a) Benzodiazepines used as analgesic, hypnotic and anti-inflammatory. A simple and versatile method for synthesis for 1,5-benzodiazepines is via condensation of O-phenylenediamines and ketones in the presence of catalytic action of H-MCM- 22 using acetonitrile as solvent at room temperature. A mixture of OPDA, ketone, H.MCM-22 was stirred in acetonitrile at room temperature until their layer chromatography indicated that reaction was completed.
    (b) SN2 Mechanism of alkyl  halides: 
        Substitution Nucleophilic Biomolecularddz (SN2) Reaction follows second order kinetics. The rate determining step depends on both the concentrate of alkyl halides (R-X) and nucleophile present in the reaction. It is a one step process and there is no formation of intermediates. The basic mechanism of reaction is:

    

    Order of reactivity of alkyl halide towards this type of reaction is:
          1⁰ > 2⁰ > 3⁰.

    Order of reactivity of halide are:
      R-1 > R-Br > R-Cl > R-F.

    Sn² creates a product with an inverted stereo structure to that of the substrate. Essentially the nucleophile attaches to the opposite side form the leaving group results in the inversion of molecules original stereochemistry.