Question

The number of tetrahedral voids present in bcc lattice is:
(a)- 4
(b)- 2
(c)- 1
(d)- 8

Answer 1

Hint: The bcc lattice is the structure known as a body-centered cubic. There are a total of 2 atoms in the bcc lattice. We can assign the number of octahedral voids as n and number of tetrahedral voids as 2n in the closed-pack structure.

Complete step by step answer:
In solids, the molecules are arranged in some order forming the crystal lattice. The crystal lattice is made up of a huge number of the unit cell. Unit cell the smallest portion which is repeated to form a structure.
Now, these unit cells have atoms at their corners, body center, face center, edge center, etc. and these are called a primitive cubic lattice, body-centered cubic lattice, face-centered cubic lattice, edge centered cubic lattice, etc. Each lattice has a different number of atoms.
In bcc i.e., body-centered cubic lattice there are 2 atoms.
So, between the atoms in the unit cell, there are spaces that are known as voids and these are of two types; octahedral and tetrahedral voids.
We can easily calculate the number of voids in the lattice by:
The total number of octahedral voids will be n and the total number of tetrahedral voids will be 2n, where n is the number of atoms in the lattice.
So, bcc has 2 atoms, then the number of octahedral voids will be 2 and the total number of tetrahedral voids will be = $2\text{ x 2 = 4}$.

So, the correct answer is an option is (a)- 4

Note: The total number of octahedral voids in fcc (face-centered cubic) will be 4 and the total number of tetrahedral voids will be 8 because the number of atoms in fcc is 4.