Answer
354.9k+ views
Hint: To solve this question we first need to know what is a crystal cubic system. A crystal system in which the shape of the unit cell is a cube is known as a crystal cubic system. The three main types of these crystals are face-centered, body-centered, and primitive cubic.
Complete answer:
Now, in a body-centered cubic system (cl), there is one lattice point on the corner of each unit along with a lattice point at the center of the cube.
So the number of atoms in the unit cell will be
\[(\dfrac{1}{8}\times 8)+1=2\]
It has 6 net octahedral voids and 12 net tetrahedral voids.
In a face-centered cubic system (cF), there is one lattice point on the corner of each unit along with a lattice point at the center of the faces of the cube (which given $\dfrac{1}{2}$ atom contribution).
So, the number of atoms in the unit cell will be
\[(\dfrac{1}{8}\times 8)+(\dfrac{1}{2}\times 6)=4\]
It has 4 net octahedral voids and 4 net tetrahedral voids.
So, the number of atoms of A = 4 as it occupies lattice points in a face-centered unit.
And the number of atoms of B = 4 as it occupies octahedral voids in a body-centered unit.
So, the formula of the compound will be ${{A}_{4}}{{B}_{4}}$ or option (A) AB.
Note:
It should be noted that in a primitive cubic system (cP), there is one lattice point on the corner of each unit. Since the atom at the lattice point is shared by all the 8 adjacent sides of the cube.
So, the number of atoms in the unit cell will be
\[\dfrac{1}{8}\times 8=1\]
It has a single cubic void in the center.
Complete answer:
Now, in a body-centered cubic system (cl), there is one lattice point on the corner of each unit along with a lattice point at the center of the cube.
![seo images](https://www.vedantu.com/question-sets/cce8919a-47d4-4333-af39-8b130375fa043313281544643031853.png)
So the number of atoms in the unit cell will be
\[(\dfrac{1}{8}\times 8)+1=2\]
It has 6 net octahedral voids and 12 net tetrahedral voids.
In a face-centered cubic system (cF), there is one lattice point on the corner of each unit along with a lattice point at the center of the faces of the cube (which given $\dfrac{1}{2}$ atom contribution).
![seo images](https://www.vedantu.com/question-sets/bed7065c-0a01-4b8b-b133-b0dc23ee21d92133032060297763053.png)
So, the number of atoms in the unit cell will be
\[(\dfrac{1}{8}\times 8)+(\dfrac{1}{2}\times 6)=4\]
It has 4 net octahedral voids and 4 net tetrahedral voids.
So, the number of atoms of A = 4 as it occupies lattice points in a face-centered unit.
And the number of atoms of B = 4 as it occupies octahedral voids in a body-centered unit.
So, the formula of the compound will be ${{A}_{4}}{{B}_{4}}$ or option (A) AB.
Note:
It should be noted that in a primitive cubic system (cP), there is one lattice point on the corner of each unit. Since the atom at the lattice point is shared by all the 8 adjacent sides of the cube.
![seo images](https://www.vedantu.com/question-sets/572c9007-9eeb-499f-8c2b-bb579db677e98596786990205986496.png)
So, the number of atoms in the unit cell will be
\[\dfrac{1}{8}\times 8=1\]
It has a single cubic void in the center.
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