Question

Niobium crystallizes in hcp structure. All the tetrahedral voids present on the edge of unit cell (shown in the figure) are occupied by boron atoms while carbon atoms fill half of the octahedral voids present. The formula of 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

Answer: (A)

$N{b}_6{B}_4{C}_3$

Niobium crystallizes in hcp structure.

Number of atoms of $Nb=\frac{1}{6}\times 12+3+\frac{1}{2}\times 2=6$

Number of atoms of $B=\frac{1}{3}\times 12=4$

Number of atoms of $C=3$

Therefore, the formula is $N{b}_6{B}_4{C}_3$.

Hence, the correct answer is option $\text{A}$.