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In a face centred cubic lattice, atoms of A form the corner points and atoms of B form the face centred points. If two atoms of $\mathrm{A}$ are missing from the corner points, the formula of the ionic compound is :
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The correct answer is:
$\mathrm{AB}_4$
$\mathrm{AB}_4$
A form corner points and two atoms of A are missing from corner
$$
\begin{aligned}
\therefore \quad & \text { Atoms at corner }(\mathrm{A})=6 \times \frac{1}{8}=\frac{3}{4} \\
& \text { Atoms at face centre }(\mathrm{B})=6 \times \frac{1}{2}=3 \\
\therefore & \mathrm{A}_{3 / 4} \mathrm{~B}_3 \text { i.e., } \mathrm{AB}_4
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad & \text { Atoms at corner }(\mathrm{A})=6 \times \frac{1}{8}=\frac{3}{4} \\
& \text { Atoms at face centre }(\mathrm{B})=6 \times \frac{1}{2}=3 \\
\therefore & \mathrm{A}_{3 / 4} \mathrm{~B}_3 \text { i.e., } \mathrm{AB}_4
\end{aligned}
$$
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