Q9 / 90 · JEE Main 2019 (10 Jan Shift 2) · Chemistry

A compound of formula A2B3 has the HCP lattice. Which atom forms the HCP lattice and what fraction of the tetrahedral voids are occupied by the other atoms?

  • A

    HCP lattice - B  

    23tetrahedral voids - A

  • B

    HCP lattice - A

    13 tetrahedral voids - B 

  • Correct answer
    C

    HCP lattice - B

    13 tetrahedral voids - A

  • D

    HCP lattice - A

    23 tetrahedral voids - B

Solution

Atom B forms HCP as anions are larger in size.

From the formula A2B3,

Number of B atoms=3,

Then, number of tetrahedral voids =2×Tetrahedral voids=2×3=6.

As 'A' is occupying tetrahedral voids,

  1 3  of tetrahedral voids=13×6=2

Therefore, formula of the compound=A2B3

Hence, the correct answer is option: 'C'.

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