Answer
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Hint: The solids are believed to be arranged in a closed packing of unit cells, these unit cells are arranged in different kinds of packing. One of which is ccp that is cubic close packing. The chemical formula can be found out by first finding out the number of octahedral and tetrahedral voids.
Complete Step by step answer: We know that the close packing system involves two kinds of voids-tetrahedral voids and octahedral voids. The former has four spheres adjacent to it while the latter has six spheres adjacent to it. These voids are only found in either FCC or hexagonal primitive unit cells. Let us first consider a FCC unit cell, such that there is an atom at the centre of an edge. Let it be big enough to touch one of the corner atoms of the fcc.
The closed packed structures have both octahedral and tetrahedral voids. In a ccp structure, there is 1 octahedral void in the centre of the body and 12 octahedral void on the edges. Each one of which is common to four other unit cells. Thus in ccp structure,
Octahedral voids in the centre of the cube=1
Effective number of octahedral voids located at the 12 edge of = $12 \times \dfrac{1}{4} = 3$
Total number of octahedral voids = 4
In ccp structure, there are 8 tetrahedral voids. In a closed packed structure, there are eight spheres in the corners of the unit cell and each sphere is in contact with three groups giving rise to eight tetrahedral voids. Therefore from the question we can conclude that
Effective number of A = 4
Effective number of B = $4 \times \dfrac{1}{2} = 2$
Effective number of O = 8
The formula is ${A_4}{B_2}{O_8} \;or {A_2}B{O_4}$
Hence the correct option is A.
Note: In this case, it can be easily understood that it would also touch six other atoms at the same distance. Such voids in an FCC unit cell in which if we place an atom it would be in contact with six spheres at equal distance(in the form of an octahedron) are called octahedral voids.
Complete Step by step answer: We know that the close packing system involves two kinds of voids-tetrahedral voids and octahedral voids. The former has four spheres adjacent to it while the latter has six spheres adjacent to it. These voids are only found in either FCC or hexagonal primitive unit cells. Let us first consider a FCC unit cell, such that there is an atom at the centre of an edge. Let it be big enough to touch one of the corner atoms of the fcc.
The closed packed structures have both octahedral and tetrahedral voids. In a ccp structure, there is 1 octahedral void in the centre of the body and 12 octahedral void on the edges. Each one of which is common to four other unit cells. Thus in ccp structure,
Octahedral voids in the centre of the cube=1
Effective number of octahedral voids located at the 12 edge of = $12 \times \dfrac{1}{4} = 3$
Total number of octahedral voids = 4
In ccp structure, there are 8 tetrahedral voids. In a closed packed structure, there are eight spheres in the corners of the unit cell and each sphere is in contact with three groups giving rise to eight tetrahedral voids. Therefore from the question we can conclude that
Effective number of A = 4
Effective number of B = $4 \times \dfrac{1}{2} = 2$
Effective number of O = 8
The formula is ${A_4}{B_2}{O_8} \;or {A_2}B{O_4}$
Hence the correct option is A.
Note: In this case, it can be easily understood that it would also touch six other atoms at the same distance. Such voids in an FCC unit cell in which if we place an atom it would be in contact with six spheres at equal distance(in the form of an octahedron) are called octahedral voids.
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