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Chemistry Test 188

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Chemistry Test 188
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Weekly Quiz Competition
  • Question 1
    4 / -1

    Select the incorrect statement about the XeF6.

    Solution

    In tetramer, four square pyramidal XeF5+ ions are joined to two similar ions by means of two bridging F- ions. The Xe-F distances are 1.84 A on the square pyramidal units and 2.23 A and 2.60 A in the bridging groups. The solid has various crystalline forms, of which three are tetrameric and a fourth has both hexamers and tetramers. 

     

  • Question 2
    4 / -1

    Based on the Valence Shell Electron Pair Repulsion (VSEPR) model, what is the geometry around the sulphur, carbon and nitrogen in the thiourea – S , S – dioxide, O2SC(NH2)2 ? (consider the Lewis structure with zero formal charges on all atoms).

    Solution

     

  • Question 3
    4 / -1

    The following flow diagram represents the manufacturing of sodium carbonate

    Which of the following options describes the reagents, products and reaction conditions (given in parentheses)?

    Solution

     

  • Question 4
    4 / -1

    When boron is fused with potassium hydroxide which pair of species are formed ?

    Solution

     

  • Question 5
    4 / -1

    B2H+ NH3 → Addition compound  In the above sequence Y and Z are respectively :

    Solution

     

  • Question 6
    4 / -1

    Orthosilicate ions, (SiO44-) undergo polycondensation to froms pyrosilicate [O3Si - O - SiO3]6-, in presence of:

    Solution

    2SiO44- + H2O → [O3Si - O - SiO3]6- + 2OH- ; 2SiO44- + 2H+ → [O3Si - O - SiO3]6- + H2O.

     

  • Question 7
    4 / -1

    Which of the following statements is correct ?

    Solution

    Hydrogen atoms lie in the CSF2 axial plane. We know that the it bond involving a p-orbital on the carbon atom must lie in the equatorial plane of the molecule.

     

  • Question 8
    4 / -1

    The hybridisation and geometry of the anion of ICl3 in liquid phase is :

    Solution

     

  • Question 9
    4 / -1

    Which of the following orders is correct for the size ?

    (1) Mg2+ < Na+ < F < Al    

    (2) Al3+ < Mg2+ < Li+ < K+   

    (3) Fe4+ < Fe3+ < Fe2+ < Fe   

    (4) Mg > Al > Si > P

    Solution

    (1) Mg2+, Na+ and F- are isoelectronic and thus follows the order 12Mg2+ < 11Na+ < gF-. Al belongs to third period and has no charge so it is largest. Na+ = 102 pm; Mg2+ = 72 pm ; Al = 143 pm, F- = 133 pm.

    (2) K+ has more number of shells than Mg2+ and Al3+. Al3+ and Mg2+ are isoelectronic but Al3+ has higher nuclear charge so Al3+ < Mg2+. Mg2+ and Li+ has diagonal relationship. But due to +2 charge in Mg2+, the Mg2+ is smaller than Li+ Hence Al3+ is the smallest one. K+ =1.38 Å, Li2+ = 0.76 Å Mg2+ = 0.72 Å and Al3+ = 0.535 Å.

    (3) As the number of electrons are lost, the attraction between valence shell electrons and nucleus increases. As a consequence of this the electrons are pulled closer to the nucleus leading to the contraction in size of ions.

    (4) Across the period the nuclear charge increases and thus the size of atoms decreases. Mg = 160 pm; Al = 143 pm; Si = 118 ; P = 110 pm.

     

     

  • Question 10
    4 / -1

    Which reactions involves a change in the electron–pair geometry for the under lined atoms ?

    Solution

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