Question

If all atoms along body diagonal of nacl are removed what is structure

Answer 1

Answer:

A type of atoms are at 8 corners, B type of atoms are at 6 face centres, C type of atoms occupy all tetrahedral voids and D type atoms occupy all octahedral voids.

Here, A and B together make a FCC(or CCP) unit cell structure.

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).

So, original formula will be AB3C8D4.

Now along a body diagonal, 2 corners, 2 tetrahedral voids and one octahedral void are present.

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.

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

Explanation: