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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    The structure of \[\text{Xe}{{\text{F}}_{\text{6}}}\]is experimentally determined to be distorted octahedron. Its structure according to VSEPR theory is

    A)  octahedron

    B)  trigonal bipyramid

    C)  pentagonal bipyramid

    D)  tetragonal bipyramid

    Correct Answer: C

    Solution :

     In \[\text{Xe}{{\text{F}}_{\text{6}}}\text{,Xe,}\]the central atom contains, 8 valence electrons. Out   of which 6 are utilised with fluorine in bonding {i.e., it contains six bond pairs of electrons) while one pair    remains as lone pair.                    Thus, the total pairs \[=6+1=7\] Hence, its shape is pentagonal bipyramid  according to VSEPR theory.


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