Question

Niobium crystallizes in HCP structure. If all the tetrahedral voids present on the edge of the unit cell (shown in the figure) are occupied by boron atoms and carbon atoms fill half of the octahedral voids present, then the formula of the resulting interstitial compound is:

• A. $N{b}_6{B}_4{C}_3$
• B. $N{b}_6{C}_4{B}_3$
• C. $N{b}_4{C}_6{B}_3$
• D. $N{b}_4{B}_3{C}_6$

Answer 1

The correct option is A $N{b}_6{B}_4{C}_3$

Boron atoms occupy all tetrahedral voids. All tetrahderal voids occur along the edges. This tells us that they are shared amongst unit cells. Since the internal angle of a hexagon is ${120}^o$, we can say that one void is shared among 3 unit cells equally.

Here, we can see that the highlighted part of the void is the unit cell we are concerned with. Hence, contribution of the void to one unit cell is number of voids$\times \frac{1}{3}$.
Number of Nb atoms = 6
Number of B atoms = Tetrahedral voids $\times \frac{1}{3}=\frac{12}{3}=4$
Number of C atoms = Octahedral voids$÷2=\frac{6}{2}=3$
Hence, the formula of the compound is: $N{b}_6{B}_4{C}_3$.