Question

In a crystal lattice p atoms occupy at the body centre

Answer 1

If all the atoms along any one body diagonal are removed, what is the formula of the compound?

If all the atoms along any one body diagonal are removed, remaining atoms are 6 corners, 6 face centers, 6 total tetrahedral voids and 3 total octahedral voids.

Each Corner and face-centered atom contribute 1/8 and 1/2 respectively towards one unit cell.

For one FCC unit cell, there are total 8 tetrahedral voids (2 along each body diagonal at 1/4 from eah corners,, all completely inside FCC unit cell so each contribute complete 1).

For one FCC unit cell, there are total 4 octahedral voids (12 along each edge centre which contribute 1/3 each towards one unit cell and one complete octahedral void at the body centre of FCC).

Contribution by A atoms at corners = 6 x (1/8) = 3/4

Contribution by B atoms at face-centre positions = 6 x (1/2) = 3

So, new formula is A(3/4) B3 C6 D3 = A3 B12 C24 D12 = A B4 C8 D4