A compound of formula A _ 2 B _ 3 has the HCP lattice. Which atom...

Atom $B$ forms HCP as anions are larger in size.

From the formula ${A}_{2}{B}_{3}$,

Number of $B$ atoms$= 3$,

Then, number of tetrahedral voids $= 2 \times$Tetrahedral voids$= 2 \times 3 = 6$.

As $' A '$ is occupying tetrahedral voids,

$\dfrac{1}{3}$ of tetrahedral voids$= \dfrac{1}{3} \times 6 = 2$

Therefore, formula of the compound$= {A}_{2}{B}_{3}$