Atoms of element B form hcp lattice and those of element A occupy two-third of tetrahedral voids. The formula of compound is :

A_{4} B_{3}

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A B_{2}

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A B_{3}

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Solution

The correct option is B $A_{4} B_{3}$
Total no of atoms $= n = 6$ (B)

Tetrahedral voids $= 2 n = 2 \times 6 = 12$

Given,

$A = \frac{2}{3} \times 2 \times N = \frac{2}{3} \times 12 = 8$

So formula of compound $= A_{8} B_{6}$ reduced to $A_{4} B_{3}$

Hence, option $A$ is correct.

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Atoms of element B form hcp lattice and those of element A occupy two-third of tetrahedral voids. The formula of compound is :

The correct option is B A4B3Total no of atoms =n=6 (B) Tetrahedral voids =2n=2×6=12 Given, A=23×2×N=23×12=8 So formula of compound =A8B6 reduced to A4B3 Hence, option A is correct.

The correct option is B $A_{4} B_{3}$
Total no of atoms $= n = 6$ (B)

Tetrahedral voids $= 2 n = 2 \times 6 = 12$

Given,

$A = \frac{2}{3} \times 2 \times N = \frac{2}{3} \times 12 = 8$

So formula of compound $= A_{8} B_{6}$ reduced to $A_{4} B_{3}$

Hence, option $A$ is correct.