In a solid MN , N forms hcp lattice and M is present in 2/3 rd tetrah…

The correct answer is Option (4) → \(M_4N_3\)

In any close packed arrangement (hcp or ccp) of N spheres there are N octahedral voids and 2N tetrahedral voids: every sphere has two tetrahedral voids, one above and one below.

Here the N atoms form the hcp lattice. Take n atoms of N:

    \( \displaystyle \text{tetrahedral voids} = 2n \)

    \( \displaystyle \text{atoms of M} = \frac{2}{3} \times 2n = \frac{4n}{3} \)

    \( \displaystyle M : N = \frac{4n}{3} : n = 4 : 3 \)

So the formula is M₄N₃.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Formula of a Compound and Number of Voids Filled – “… Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B?
Solution: The number of tetrahedral 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.”