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

In a hexagonal primitive unit cell of a close‑packed lattice, the voids are the empty spaces that can accommodate smaller atoms. To determine how many tetrahedral and octahedral voids are present, we examine the geometry of the cell.

**Step 1: Identify the positions of octahedral voids** In a hexagonal cell there are 12 edges. An octahedral void sits at the centre of each edge. Each edge is shared by two adjacent cells, so only half of an octahedral void belongs to a given cell.

Number of octahedral voids per cell = (12 edges × ½) = $6$.

**Step 2: Identify the positions of tetrahedral voids** Tetrahedral voids are located at the corners of the tetrahedra formed by three atoms in one layer and one atom in the adjacent layer. In a hexagonal primitive cell there are 24 such positions (four per each of the six faces that connect the two basal planes). Each tetrahedral void is completely contained within the cell.

Number of tetrahedral voids per cell = $24 ÷ 2\; =\; 12$ (since each void is counted twice when considering the two adjacent cells).

**Step 3: Summarise the counts** Thus the hexagonal primitive unit cell contains:

• Tetrahedral voids: $12$
• Octahedral voids: $6$

Therefore, the correct option is **12, 6**.