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Chemistry Test 224

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Chemistry Test 224
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  • Question 1
    4 / -1

    One gram equimolecular mixture of Na2CO3 and NaHCO3 is reacted with 0.1 N HCI. The millilitres of 0.1 N HCI required to react completely with the above mixture are

    Solution

     

  • Question 2
    4 / -1

    A compound of vanadium has a magnetic moment of 1.73 BM. What is the electronic configuration of vanadium ion in the compound?

    Solution

    The magnetic moment (μ) of a compound is given by

    Thus, there is only one unpaired electron in vanadium ion. Vanadium has atomic number 23. So, the electronic configuration of vanadium atom, V (Atomic number = 23) - 1s2 2s2 2p6 3s2 3p6 3d3 4s2

    Thus, it has 3 unpaired electrons in 3d orbitals. To be left with only one unpaired electron, vanadium atom should lose four electrons i.e. vanadium ion in the compound is V4+. The electronic configuration of V4+ ion is 1s2 2s2 2p6 3s2 3p6 3d1.

     

  • Question 3
    4 / -1

    2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?

    Solution

     

  • Question 4
    4 / -1

    Directions For Questions

    The standard reduction potentials at 298 K for the following half-cell reactions are given below:

    Zn2+ (aq) + 2e- ⇌ Zn (s); -0.762 V

    Cr3+ (aq) + 3e- ⇌ Cr (s); -0.74 V

    2H+ (aq) + 2e- ⇌ H2 (g); +0.00 V

    Fe3+ (aq) + e- ⇌ Fe2+ (aq); +0.77 V

    ...view full instructions

    Which one of the following is the strongest reducing agent?

    Solution

    The metal with the lowest value of reduction potential and consequently the highest value of the oxidation potential, among the given species, is the strongest reducing agent. Zn has a higher negative value of standard reduction potential. Therefore, it is the strongest reducing agent.

     

  • Question 5
    4 / -1

    A nuclide of Barium, an alkaline earth metal, undergoes radioactive decay by emission of three α-particles in succession. The resulting daughter element belongs to which of the following groups of the periodic table?

    Solution

    When a radioactive element emits an α-particle, its atomic number is reduced by 2 units. Hence, the daughter element after the emission of an α-particle from the nuclide of an alkaline earth metal (Group 2) will belong to Group 18. Since the emission of three α-particles is in succession, the ultimate product will be Lead, i.e. a Group 14 element.

     

  • Question 6
    4 / -1

    When CO2 is bubbled through a solution of barium peroxide in water,

    Solution

    When CO2 is bubbled through a cold pasty solution of barium peroxide in water, H2O2 is obtained.

    Barium carbonate, being insoluble, is filtered off. This is known as Merck's process.

     

  • Question 7
    4 / -1

    Select the correct order of initials T (True) and F (False) for the following statements:

    (I) Co(II) is stablised in the presence of strong field ligands, while Co(III) is stablised in the presence of weak field ligands.

    (II) [Cr(NH3)6]3+ is an inner orbital complex with crystal field stablisation energy equal to -1.2Δo.

    (III) In [CoF6]3-, F- is a weak field ligand so that Δo < P and it is a low spin complex.

    (IV) Crystal field splitting in ferrocyanide ion is greater than that in ferricyanide ion.

    Solution

    (I) Both Co2+ and Co3+ are stablised in the presence of strong field ligands due to higher CFSE values.

    (II) It is a correct statement.

    (III) In [CoF6]3-, Δo < P; thus it is a high spin complex.

    (IV) The greater the charge on the central atom, the greater will be the splitting.

     

  • Question 8
    4 / -1

    Which of the following statements is correct?

    Solution

    (a) Ethanol forms a lower boiling azeotropic mixture containing 95.6 and 4.4 by volume of alcohol and water, respectively. Water can be removed by adding Mg metal to azeotropic mixture, which reacts with water forming Mg(OH)2 and releasing H2.

    (b) The OH group of cyclohexanol is more exposed and available for H-bonding with water due to more compactness of the remaining alkyl group.

    (c) The product is 2-phenyl-2-butanol.

    (Note: Although phenyl is a better migrator than H, yet hydride migrates so as to give a more stable 3° benzylic carbocation)

    (d) The product is cis- and trans-2-methylcyclopentanol. In this reaction, addition of H+ to C1 gives a 2° carbocation, which rearranges by ring expansion to give substituted cyclopentanol.

     

  • Question 9
    4 / -1

    Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of

    Solution

    In an SN2 reaction, in the transition state, there will be five groups attached to the carbon atom at which reaction occurs. Thus, there will be crowding in the transition state and presence of bulky groups makes the reaction sterically hindered. Moreover, SN1 mechanism is preferred due to stability of the carbocation intermediate.

     

  • Question 10
    4 / -1

    The following reactions are carried out.

    The final product 'Z' is

    Solution

     

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