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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atom B is missing from one of the face centered points, the formula of the ionic compound is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) \[A{{B}_{2}}\]                                   

    B) \[{{A}_{5}}{{B}_{2}}\]

    C) \[{{A}_{2}}{{B}_{3}}\]                                    

    D) \[{{A}_{2}}{{B}_{5}}\]

    Correct Answer: D

    Solution :

                    Number of atoms\[A=\frac{1}{8}\times 8=1\] Number of atoms \[B=\frac{1}{2}\times 5=\frac{5}{2}\] (Since atom B is missing from one of the face centered point) \[\therefore \]Formula of ionic compound \[=A{{B}_{5/2}}={{A}_{2}}{{B}_{5}}\]


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