
A crystal is made up of particles A, B and C. A forms FCC packing, B occupies all octahedral voids and C occupies all tetrahedral voids. If all the particles along one body diagonal are removed, then the formula of the crystal would be:
A.$AB{C_2}$
B.${A_2}B{C_2}$
C.${A_8}{B_4}{C_5}$
D.${A_5}{B_4}{C_8}$
Answer
495.9k+ views
Hint: In FCC packing which is also known as face centered cubic packing and HCP packing which is known as hexagonal close packed structure both have a packing factor of 0.74. They consist of closely packed planes of atoms, and have coordination number 12. The difference between them is that the hcp has a stacking sequence but not fcc.
Complete step by step answer:
As we know that in FCC unit cells the number of atoms present is 4.
The number of tetrahedral voids= 2 $\times$ of octahedral voids.
And we also know that the number of octahedral voids present in the unit cell is equal to the number of atoms present in a cubic unit cell. The number of atoms in a cubic cell is 4. So the tetrahedral voids number is 4 and octahedral voids number is 8.
In a unit cell on body diagonal there is only one octahedral void present at the body center and two tetrahedral voids which are present ¼ the distance from each corner. So after removing the atoms from body diagonals , we will have particle A with
$\dfrac{{15}}{4}$ $ = 6 \times \dfrac{1}{2} + \dfrac{{(8 - 2)}}{2}$
$ = 3 + \dfrac{3}{4}$
= $\dfrac{{15}}{4}$
So particle A will have $\dfrac{{15}}{4}$ atoms.
Particle B will have = 4-1=3 atoms
Particle C will have=8-2=6 atoms
The ratio of these particles is
$\dfrac{{15}}{4}$:3:6
On solving this ratio we get
5:4:8
So A=5
B=4 and
C=8
Hence the formula formed will be ${A_5}{B_4}{C_8}$ and the correct option is option’D’.
Additional information: The void which is surrounded by four spheres is called tetrahedral void.The void surrounded by eight spheres is called octahedral void.
Note:
Tetrahedral void is a simple triangular void in a crystal and is surrounded by four spheres arranged tetrahedrally around it. On the other hand, an octahedral void is a double triangular void with one triangle vertex upwards and the other triangle vertex downwards and is surrounded by six spheres.
Complete step by step answer:
As we know that in FCC unit cells the number of atoms present is 4.
The number of tetrahedral voids= 2 $\times$ of octahedral voids.
And we also know that the number of octahedral voids present in the unit cell is equal to the number of atoms present in a cubic unit cell. The number of atoms in a cubic cell is 4. So the tetrahedral voids number is 4 and octahedral voids number is 8.
In a unit cell on body diagonal there is only one octahedral void present at the body center and two tetrahedral voids which are present ¼ the distance from each corner. So after removing the atoms from body diagonals , we will have particle A with
$\dfrac{{15}}{4}$ $ = 6 \times \dfrac{1}{2} + \dfrac{{(8 - 2)}}{2}$
$ = 3 + \dfrac{3}{4}$
= $\dfrac{{15}}{4}$
So particle A will have $\dfrac{{15}}{4}$ atoms.
Particle B will have = 4-1=3 atoms
Particle C will have=8-2=6 atoms
The ratio of these particles is
$\dfrac{{15}}{4}$:3:6
On solving this ratio we get
5:4:8
So A=5
B=4 and
C=8
Hence the formula formed will be ${A_5}{B_4}{C_8}$ and the correct option is option’D’.
Additional information: The void which is surrounded by four spheres is called tetrahedral void.The void surrounded by eight spheres is called octahedral void.
Note:
Tetrahedral void is a simple triangular void in a crystal and is surrounded by four spheres arranged tetrahedrally around it. On the other hand, an octahedral void is a double triangular void with one triangle vertex upwards and the other triangle vertex downwards and is surrounded by six spheres.
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