TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Atom of B forms HCP lattice, so the atom at corners are shared by 6 unit cell.
So, its contribution is ‌

1

6

.
fcc atom contribute ‌

1

2

and middle layer atom
contribute 1 each.
Effective number of atoms in unit cell
=‌

1

6

×12+‌

1

2

×2+1×3
=2+1+3=6
Number of tetrahedral voids
=2× number of atoms forming the lattice
=2×6=12
Number of atoms of A=‌

2

3

×12=8.
Number of atoms of B=6.
Ratio of A:B is 8:6.
or, in simplest form =4:3
So, formula of the compound is A4B3.