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Chemistry Test 217

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Chemistry Test 217
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Weekly Quiz Competition
  • Question 1
    4 / -1

    1s orbital of H is represented by,  where a0 is Bohr's.

    The probability of finding the electron at a distance r, from the nucleus in the region dV is radius.

    Solution

    P(r) = ∫∣ψ2∣4πr2dr

    4πr2 dr = dv = volume of thin spherical shell around nucleus at distance (r).

     

  • Question 2
    4 / -1

    Amongst the following, the most basic compound is

    Solution

    Lone pair is not involved in resonance, most basic. In all other cases, lone-pair of nitrogen is involved in resonance, less basic.

     

  • Question 3
    4 / -1

    A real gas has an equation of state  (Vm is molar volume). Find a and b Vc and Tc. where Vc and Tc are critical constants of gas.

    Solution

    At critical point, i.e., point of Inflection.

     

  • Question 4
    4 / -1

    Find out the process by which  can undergo radioactive decay to attain a more stable isotope.

    Solution

    In stable isotope of Na, there are 11 protons and 12 neutrons. In the given radioactive isotope of sodium (Na24) there are 13 neutrons, one neutron more than that required for stability. A neutron-rich isotope always decay by β−emission as

     

  • Question 5
    4 / -1

    The correct order of reducing power for the four successive elements Cr, Mn, Fe and Co is: (Their E°red. values are given below.)

    Solution

    The values of E°red. gives us an idea of the ease with which metals get reduced. A higher negative value implies that it has greater tendency to get oxidised.

    For a metal to act as a reducing agent, it should itself get oxidised. Hence, the order of reducing ability is proportional to ease of oxidation which is inversely proportional to E°red.

    Hence, the answer will be:

    Mn > Cr > Fe > Co

     

  • Question 6
    4 / -1

    Electrolysis of a solution of  Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mol of  per hour ?

    Solution

     

  • Question 7
    4 / -1

    The catalyst used in the manufacture of polyethylene by Ziegler-Natta's method is:

    Solution

    Ziegler - Natta's catalyst which is used in polymerisation of ethene is (C2H5)3 Al + TiCl4.

     

  • Question 8
    4 / -1

    An unknown alkyl halide (A) reacts with alcoholic KOH to produce a hydrocarbon (C4H8) as the major product. Ozonolysis of the hydrocarbon affords one mole of propanaldehyde and one mole of formaldehyde. Suggest which organic compound among the following has the correct structure of the above alkyl halide (A)?

    Solution

    Since the product on ozonolysis is propanaldehyde and formaldehyde, it implies that the the double bond is shared between C1 and C2.

    If the compound were option 1, the result of such elimination would have been 2 butene whose ozonolysis would give ethanals.

    1−butene on ozonolysis gives propanaldehyde and formaldehyde thus, the hydrocarbon is

     

  • Question 9
    4 / -1

    The shape of XeF4 is

    Solution

    Due to the presence of two lone pairs of electrons, it is of square planar shape.

     

  • Question 10
    4 / -1

    Which of the following reactions of Xenon compounds is not feasible?

    Solution

    XeO3 + 6HF ⟶ XeF6 + 3H,O

    Above reaction is not feasible because XeF6 formed will further produce XeO3 by getting hydrolysed.

    XeF6 + H2O → XeOF4 + 2HF

    XeOF4 + H2O → XeO2 F2 + 2HF

    XeO2 F2 + H2O → XeO3 + 2HF

     

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