Question

A compound formed by two elements m and n .the element n forms hcp lattice and atom m occupy 2/3rd of the tetrahedral void.what is the formula of this compound​

Answer 1

Answer:

According to the question, the atoms of element M occupy 1/3rd of the tetrahedral voids. Therefore, the number of atoms of M is equal to 2 1/3 = 2/3rd of the number of atoms of N. Therefore, ratio of the number of atoms of M to that of N is M : N = (2/3):1 = 2:3 Thus, the formula of the compound is M2 N3.

Explanation:

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Answer 2

Given that the elements form HCP lattice⇒

The contribution of atoms would be as follows:

Atoms at the corner are shared by 6 units cell.  

Hence, the contribution will be $\frac{1}{6}$.

Face-centered atoms contribution will be $\frac{1}{2}$.

Middle layer atom contribution will be$\\ 1$ .

The effective number of atoms  in the unit cell : $\\\frac{1}{6 } \times 12 + \frac{1}{2} \times 2 + 1\times 3$

                                                                      : $\\ 2 + 1 + 3 = 6$

Tetrahedral voids $\\ = 2 \times Effective\ number\ of \atoms \ in\ the \unit\ cell$

                            = $\\ 2 \times 6 = 12$

So, the number of atoms of M = $\\ \frac{2}{3} \times 12 = 8$

Number of atoms of N = $\\ 6$  

 $\\ N : M = 6 : 8 = 3 : 4$

So, the formula of the compound would be N₃M₄.