Question

A compound of formula A₂B₃ has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms :
(A) hcp lattice - A, 2/3 Tetrahedral voids - B (B) hcp lattice - A, 1/3
Tetrahedral voids - B
(C) hcp lattice - B, 2/3
Tetrahedral voids - A (D) hcp lattice - B, 1/3 Tetrahedral voids – A

Answer 1

Answer:

The correct option is (4) hcp lattice - B, 1/3 Tetrahedral voids - A

Explanation:

Let B form hcp lattice no. of B = 6

tetrahedral void occupied by

A = 1/3 x 12

= 4 A4B6 or A2B3

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Answer 2

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• ❤.. The number of tetralhedral voids formed is equal to twice the number of atoms of element B and only 2/3rd of these are occupied by the atoms of element A.
• ❤..Hence the ratio of the number of atoms of A and B is ${2×(2/3):1 or 4:3}$ and the formula of the compound is ${A4B3}$.
• ❤.. Option 4.