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KVPY Sample Paper KVPY Stream-SX Model Paper-1

  • question_answer
    The hybridisation and geometry of \[{{X}_{e}}{{F}_{4}}\] is

    A) \[ds{{p}^{2}},\] square planar   

    B) \[s{{p}^{3}}{{d}^{2}},\] octahedral

    C) \[s{{p}^{3}},\] tetrahedral         

    D) \[s{{p}^{3}}{{d}^{2}},\] square planar

    Correct Answer: D

    Solution :

    [d]
    The hybridisation of a compound can be calculated using formula.
    \[H=\frac{1}{2}[V+MO\pm \text{anion}/\text{cation}]\]                              
    where, H= Hybridised orbitals
    V= Valence
    MO = Number of monoatomic atoms
    For \[Xe{{F}_{4}}\]
    \[H=\frac{1}{2}[8+4]=6\]
    \[\Rightarrow \]   \[s{{p}^{3}}{{d}^{2}}\]
    It has 4 bond pairs and 2 lone pair, where 2 lone pair occupies axial positions and four F occupies equatorial positions. So, its geometry is square planar and not octahedral.


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