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NEET Chemistry The Solid State / ठोस प्रावस्था Question Bank Self Evaluation Test - The Solid State

  • question_answer
    \[CsBr\] has bcc structure with edge length 4.3. The shortest interionic distance in between \[C{{s}^{+}}\] and \[B{{r}^{-}}\]is

    A) 3.72     

    B) 1.86  

    C) 7.44                 

    D) 4.3

    Correct Answer: A

    Solution :

    [a] For bcc structure, atomic radius, \[r=\frac{\sqrt{3}}{4}a\] \[=\frac{\sqrt{3}}{4}\times 4.3=1.86\] Since, r = half the distance between two nearest neighbouring atoms. \[\therefore \] Shortest interionic distance \[=2\times 1.86=3.72\]


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