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Question
In a cubic arrangement of atoms of A, B and C, atoms of A are present at the corners of the unit cell, B atoms are at face centers and C at tetrahedral voids. If one of the atom from one corner is missing in the unit cell, then the simplest formula of the compound will be:
In a cubic arrangement of atoms of A, B and C, atoms of A are present at the corners of the unit cell, B atoms are at face centers and C at tetrahedral voids. If one of the atom from one corner is missing in the unit cell, then the simplest formula of the compound will be:
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Solution
The correct option is B A7B24C64
Atoms are present at the corners. and one of these is missing. so there are 7 corners at which A atoms are present in a cell. Thus the total number of A atoms per unit cell = 78.There are 6 face centers and each is occupied by B atoms. However, every B atom is shared by 2 unit cells. thus total number of B atoms = 12 x 6 = 3.Since it is a FCC arrangement there are 8 tetrahedral voids and each is occupied by C atoms. so total number of C atoms = 8thus the ratio of atoms of A:B:C is 78: 3: 8...multiplying by 8 on all sides, the ratio becomes 7: 24: 64thus the formula is A7B24C64
Atoms are present at the corners. and one of these is missing. so there are 7 corners at which A atoms are present in a cell. Thus the total number of A atoms per unit cell = 78.
There are 6 face centers and each is occupied by B atoms. However, every B atom is shared by 2 unit cells. thus total number of B atoms = 12 x 6 = 3.
Since it is a FCC arrangement there are 8 tetrahedral voids and each is occupied by C atoms. so total number of C atoms = 8
thus the ratio of atoms of A:B:C is 78: 3: 8...
multiplying by 8 on all sides, the ratio becomes 7: 24: 64
thus the formula is A7B24C64
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