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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • question_answer
    The number of paired electrons in \[Xe\] atom of \[Xe{{F}_{2}}\] is

    A) \[3\]                                     

    B) \[5\]                     

    C) \[4\]                     

    D) \[2\]

    Correct Answer: A

    Solution :

    The dot structure of \[Xe{{F}_{2}}\] is as follows                 \[F-:\underset{\centerdot \,\,\centerdot }{\overset{\centerdot \,\,\centerdot }{\mathop{Xe}}}\,-F\] Number of molecular hybrid orbitals = number of sigma bonds + number of lone pair of electrons. \[\therefore \]Number of molecular hybrid orbitals \[=2+3=5\] So, hybridization is\[s{{p}^{3}}d\].


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