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# An ionic compound made up of atoms A and B has a face-centred cubic arrangement in which atoms A are at the corners and atoms B are at the face-centres. If one of the atoms is missing from the corner, what is the simplest formula of the compound?

$\begin{array}{1 1}(A)\;A_6B_{23}&(B)\;A_7B_{27}\\(C)\;A_7B_{24}&(D)\;A_6B_{24}\end{array}$

Number of atoms of A at the corners = 7 (because one A is missing)
$\therefore$ Contribution atoms of A towards unit cell=$7\times \large\frac{1}{8}=\frac{7}{8}$
Number of atoms B at face-centres = 6
$\therefore$ Contribution of atom B towards unit cell =$6\times \large\frac{1}{2}$$=3 Ratio of A : B =\large\frac{7}{8}$$: 3=7 : 24$
$\therefore$ Formula is $A_7B_{24}$
Hence (C) is the correct answer.