NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer The ionic radii of \[{{\text{A}}^{\text{+}}}\]and \[{{\text{B}}^{}}\]ions are \[0.98\text{ }\times \text{ }10{{}^{10}}\text{m}\]and \[1.81\text{ }\times \text{ }{{10}^{10}}\text{ m}.\] The coordination number of each ion in AB is :-                                                                                                                                                 

    A)  6                                            

    B)   4

    C)   8                                           

    D)   2

    Correct Answer: A

    Solution :

                     \[\text{radio ratio = }\frac{{{r}_{+}}}{{{r}_{-}}}=\frac{0.98\times {{10}^{-10}}}{1.81\times {{10}^{-10}}}=0.54\]radii ratio is in between 0.414 to 0.732 so, coordination number is 6

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