Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are :

The correct option is 12, 6.

To determine the number of tetrahedral and octahedral voids in a hexagonal primitive (or hexagonal close-packed, HCP) unit cell, we can follow these step-by-step logical relations:

Step 1: Find the number of atoms per hexagonal unit cell (Z)
In a hexagonal primitive unit cell (HCP structure):
- There are 12 corner atoms. Each corner atom is shared among 6 adjacent unit cells, contributing:
$12\times \frac{1}{6}=2$ atoms.
- There are 2 face-center atoms (one on the top face and one on the bottom face). Each is shared between 2 adjacent unit cells, contributing:
$2\times \frac{1}{2}=1$ atom.
- There are 3 completely unshared atoms located inside the body of the unit cell, contributing:
$3$ atoms.
Adding these contributions together, the total number of atoms in one hexagonal unit cell ($Z$) is:
$Z=2+1+3=6$ atoms.

Step 2: Calculate the number of octahedral voids
For a close-packed lattice containing $N$ atoms, the number of octahedral voids generated is equal to the number of atoms, $N$.
Since $N=Z=6$ for a hexagonal unit cell:
Number of octahedral voids = $6$.

Step 3: Calculate the number of tetrahedral voids
For a close-packed lattice containing $N$ atoms, the number of tetrahedral voids generated is twice the number of atoms, $2N$.
Using $N=6$:
Number of tetrahedral voids = $2\times 6=12$.

Thus, the number of tetrahedral and octahedral voids in a hexagonal primitive unit cell are 12 and 6 respectively.