In a solid 'AB' having the NaCl structure, 'A' atom occupy the every CCP and 'B' atoms are km OHV. If all the face- centred atoms along one of the axes are removed, then what is the resultant stoichiometry of the solid?
Answers
Answer:'AB' has NaCl structure, otherwise known as Rock salt structure. Hence one type of atoms must be arranged in FCC pattern (at all the 8 corners and all the 6 face centers) and other type of atoms must be present on all the 12 edges as well as at the center of the unit cell.
Since, 'A' atoms are occupying the corners of unit cell, they must also be arranged on the face centers. The B atoms can be found at the 12 edges and at the center of the unit cell as shown below.
Therefore, before removing face centered atoms:
The number of 'A' atoms = (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 [there are 8 'A' atoms at the corners and 6 'A' atoms at the face centers]
The number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4 [there are 12 'B' atoms at the edges and 1 'B' atom at the center of the unit cell]
The number of 'A' atoms = (8 x 1/8) + (4 x 1/2) = 1 + 2 = 3 [there are now only 4 'A' atoms at the face centers]
The number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4
Therefore the stoichiometry after removing these atoms is: A3B4
Note that for Each atom at the corner contributes 1/8th, and each atom at the face center contributes 1/2nd and each atom along the edge contributes 1/4th to the unit cell.
So the final answer is A3B4
Hope this helps!'AB' has NaCl structure, otherwise known as Rock salt structure. Hence one type of atoms must be arranged in FCC pattern (at all the 8 corners and all the 6 face centers) and other type of atoms must be present on all the 12 edges as well as at the center of the unit cell.
Since, 'A' atoms are occupying the corners of unit cell, they must also be arranged on the face centers. The B atoms can be found at the 12 edges and at the center of the unit cell as shown below.
Therefore, before removing face centered atoms:
The number of 'A' atoms = (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 [there are 8 'A' atoms at the corners and 6 'A' atoms at the face centers]
The number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4 [there are 12 'B' atoms at the edges and 1 'B' atom at the center of the unit cell]
The number of 'A' atoms = (8 x 1/8) + (4 x 1/2) = 1 + 2 = 3 [there are now only 4 'A' atoms at the face centers]
The number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4
Therefore the stoichiometry after removing these atoms is: A3B4
Note that for Each atom at the corner contributes 1/8th, and each atom at the face center contributes 1/2nd and each atom along the edge contributes 1/4th to the unit cell.
So the final answer is A3B4
Hope this helps!
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