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

The correct answer is option 4. \(M_4N_3\).

To determine the molecular formula of the solid \(MN\) where \(N\) forms a hexagonal close-packed (hcp) lattice and \(M\) occupies \( \frac{2}{3} \) of the tetrahedral voids, follow these steps:

In an hcp lattice, each unit cell contains 2 atoms of \(N\).

The number of tetrahedral voids in the hcp lattice is equal to the number of atoms in the lattice. Therefore, there are 2 tetrahedral voids per unit cell.

Since \(M\) occupies \( \frac{2}{3} \) of the tetrahedral voids, for 2 tetrahedral voids, the number of tetrahedral voids occupied by \(M\) is \( \frac{2}{3} \times 2 = \frac{4}{3} \) voids.

Each unit cell has 2 atoms of \(N\) and \( \frac{4}{3} \) atoms of \(M\) (approximated for simplicity).

To get whole numbers, multiply by 3: \(M\) would be \( \frac{4}{3} \times 3 = 4\) and \(N\) would be \(2 \times 3 = 6\).

So, the molecular formula of the solid is: \(M_4N_3 \)

Therefore, the correct answer is: 4. \(M_4N_3\)