Atoms of elements B form hcp lattice and those of element A occupy 23rd of tetrahedral voids. The formula of the compound formed is

Since atoms of B.forms hcp. The number of O.V = 1∴ No. of T.V = 2Atoms of A occupied by 23rd of TV=23×2=43∴The ratio of the number of atoms of A and B is 43:1 and the formula is A4B3

Atoms of elements B form hcp lattice and those of element A occupy 23rd of tetrahedral voids. The formula of the compound formed is

Since atoms of B.forms hcp. The number of O.V = 1∴ No. of T.V = 2Atoms of A occupied by 23rd of TV=23×2=43∴The ratio of the number of atoms of A and B is 43:1 and the formula is A4B3

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Atoms of elements B form hcp lattice and those of element A occupy ${\frac{2}{3}}^{r d}$ of tetrahedral voids. The formula of the compound formed is

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$A_{3} B_{4}$

$A_{4} B_{3}$

$A_{2} B_{3}$

$A_{3} B_{2}$

answer is B.

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Detailed Solution

Since atoms of B.forms hcp. The number of O.V = 1
$∴$ No. of T.V = 2
Atoms of A occupied by ${\frac{2}{3}}^{r d}$ of $T V = \frac{2}{3} \times 2 = \frac{4}{3}$