
In a solid of AB having NaCl structure, ‘A’ atoms occupy the corners of the cubic unit cell. If all the face- centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is:
a.) $A{{B}_{2}}$
b.) ${{A}_{2}}{{B}_{{}}}$
c.) ${{A}_{4}}{{B}_{3}}$
d.) ${{A}_{3}}{{B}_{4}}$
Answer
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Hint: The smallest unit which repeats and forms a crystal is known as unit cell. Unit cells are of many types: simple unit cell, body centered unit cell and face centered unit cell. In the face centered unit cell atoms are present at the center of all the faces and at all the corners of the cube.
Complete step by step answer:
In a face centered unit cell there are atoms at all the corners of the unit cell. Each corner atom is shared between 8 unit cells. The contribution of corner atoms are calculated as:
\[8\text{corner atoms X }\dfrac{1}{8}\text{atoms per unit cell}\]
\[8X\dfrac{1}{8}=1\]Atom
The number of atoms present at the face center of the cube = 6
And each face center atom is shared by two unit cells. The contribution of face center atoms is calculated as:
$\text{6 face-centered atoms X}\dfrac{1}{2}\text{ atoms per unit cell }$
\[6X\dfrac{1}{2}\]= 3 atoms.
The total number of atoms at the face centered cubic cell = 4 atoms
Here it is given that ‘A’ atoms occupy the corners of the cubic unit cell
From the above calculation we can say that total contribution of A is = 4
Now B occupies the octahedral void which will contribute $\dfrac{1}{4}$
Contribution of B= $1+12X\dfrac{1}{4}=4$
So the formula is ${{A}_{4}}{{B}_{4}} or AB$
Now, after removing all the face center atoms along one axis, total number of atoms removed
\[2(faces)X\dfrac{1}{2}(\text{per edge centre share)}\]= 1
Number of A atoms left $4-1=3$
So, the correct answer is “Option D”.
Note: You might get confused between the number of atoms in primitive and body center unit cells. The total number of atoms in the body centered unit cell is 2, one atom is present at the center of the unit cell and the total number of atoms in the primitive unit cell is 1.
Complete step by step answer:
In a face centered unit cell there are atoms at all the corners of the unit cell. Each corner atom is shared between 8 unit cells. The contribution of corner atoms are calculated as:
\[8\text{corner atoms X }\dfrac{1}{8}\text{atoms per unit cell}\]
\[8X\dfrac{1}{8}=1\]Atom
The number of atoms present at the face center of the cube = 6
And each face center atom is shared by two unit cells. The contribution of face center atoms is calculated as:
$\text{6 face-centered atoms X}\dfrac{1}{2}\text{ atoms per unit cell }$
\[6X\dfrac{1}{2}\]= 3 atoms.
The total number of atoms at the face centered cubic cell = 4 atoms
Here it is given that ‘A’ atoms occupy the corners of the cubic unit cell
From the above calculation we can say that total contribution of A is = 4
Now B occupies the octahedral void which will contribute $\dfrac{1}{4}$
Contribution of B= $1+12X\dfrac{1}{4}=4$
So the formula is ${{A}_{4}}{{B}_{4}} or AB$
Now, after removing all the face center atoms along one axis, total number of atoms removed
\[2(faces)X\dfrac{1}{2}(\text{per edge centre share)}\]= 1
Number of A atoms left $4-1=3$
So, the correct answer is “Option D”.
Note: You might get confused between the number of atoms in primitive and body center unit cells. The total number of atoms in the body centered unit cell is 2, one atom is present at the center of the unit cell and the total number of atoms in the primitive unit cell is 1.
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