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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    The radius of divalent cation \[{{M}^{2+}}\] is \[94\,\,pm\] and that of divalent anion \[{{X}^{2-}}\] is\[146\,\,pm\]. Thus \[{{M}^{2+}}{{X}^{2-}}\] has

    A) \[NaCl\] structure           

    B)  linear structure

    C) \[CsCl\] structure            

    D)  \[ZnS\] structure

    Correct Answer: A

    Solution :

    For\[{{M}^{2+}}{{X}^{-2}}\]\[\frac{{{r}_{+}}}{{{r}_{-}}}=\frac{94}{146}=0.6438\] Since, the radius ratio is in between \[0.414\] to\[0.732\], it has \[NaCl\] structure.


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