The number of Octahedral voids in a hexagonal primitive unit cell are:

Step 1: Finding the number of atoms in a hexagonal primitive unit cell:

• It is a repetitive unit cell of a hexagonal crystal lattice.
• It is a unit cell which is abbreviated as an hcp unit cell.
• It has a hexagonal structure consisting of $6$atoms.

Step 2: Finding the relation between the no. of atoms in the unit cell and the no. of octahedral voids:

• Octahedral void is the empty unoccupied space having three atoms above the hole and three atoms below the hole.

• It is equal to the number of atoms in the crystal structure.
• So, No. of octahedral voids $=$ No. of atoms in the crystal structure.

Step 3: Calculating the no. of octahedral voids:

• No. of octahedral voids $=$ No. of atoms in the crystal structure
• No. of atoms in the crystal structure$=$$6$
• No. of octahedral voids $=$$6$

Hence, there are $6$octahedral voids in a hexagonal primitive unit cell.