JEE Main 2019ChemistrySolid StateMediumMCQ

JEE Main 2019 — Solid State Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

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?

Choose an option

Show full solutionCorrect option: C
Correct answer
C

HCP lattice - B

13 tetrahedral voids - A

Step-by-step explanation

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

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About this question

This is a previous-year question from JEE Main 2019, covering the Solid State chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.