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In $\mathrm{HCP}$ of $A, \frac{1}{3}$ of tetrahedral voids are occupied by $B$. What is the formula of the compound?
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The correct answer is:
$A_3 B_2$
No. of atoms $(A)=6(h c p)$,
no. of $B$ atoms $=\frac{1}{3} \times 12=4$
Hence, the formula of the compound is $A_6 B_4$ or $A_3 B_2$.
no. of $B$ atoms $=\frac{1}{3} \times 12=4$
Hence, the formula of the compound is $A_6 B_4$ or $A_3 B_2$.
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