Question

Atoms of elemants B from hcp lattice and those of the element A occupy 2/3rd of the octahedral voids.what is the formula of the compound formed by those element A and B

Answer 1

If B forms HCP lattice then no of B ATMs present will be 6.

We know that number of octahedral voids in HCP lattice will be 6 and A occupies 2/3 of it.

So number of A atoms will be

2/3x6 = 4

So the formula is A4B6 or A2B3.

I hope this helps you.

Incase you still have doubt feel free to ask.

Answer 2

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\purple{R}}}}$

• ❤.. 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 ${A4B3}$.