QUES. In a closed packed arrangement ,A type of atoms are at corners , B type of atoms are at face corners , C type of atoms are at tetrahedral voids and D type of atoms are at octahedral voids. If all the atoms along any one body diagonal are removed. Then what will be the formula of compound?
QUES. In a closed packed arrangement ,A type of atoms are at corners , B type of atoms are at face corners , C type of atoms are at tetrahedral voids and D type of atoms are at octahedral voids. If all the atoms along any one body diagonal are removed. Then what will be the formula of compound?
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
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
