atoms of the element B form hcp lattice and those of element A occupy 2by3rd part of tetrahydral voids what is the formula of the compound formed by the element A and B

Dear student!

In hcp lattice,

the no. of tetrahedral voids formed  is equal to twice the no. of atoms of element which forms the lattice.

i.e. Tetrahedral voids = 2N (No. of atoms of B)

Now, as element A occupies only 2/3rd part of the void for each atom of B

so the no. of atoms of A present in tetrahedral void = 2 x 2/3 = 4/3

Hence ratio of A:B = 4/3 :1 = 4:3

Hence formula = A4B3 

Regards

  • 29

There are totla 6 atoms in hcp lattice... so B = 6

Now , tetrahedral voids will be = 2N = 2*6 = 12 But given , A = 2/3 * 2N = 2/3 * 12 = 8

hence , A = 8 & B=6

so Formula of the compound = A8B6 = A4B3

  • 42
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