A) Formula of the compound is \[{{A}_{2}}{{B}_{2}}X\]
B) Coordination number of A can be 4.
C) Coordination number of X can be 6.
D) Centres of 12A atoms lie on the edges of one unit cell.
Correct Answer: D
Solution :
B like \[{{S}^{2-}}\] (making hcp) |
A like \[Z{{n}^{2+}}\](alternate tetrahedral void) |
X (alternate octahedral void) |
no. of B in one unit cell =6 |
no. of A in one unit cell \[=\frac{1}{2}\times 12=6\] |
no. of X in one unit cell \[=\frac{1}{2}\times 6=3\] |
\[\Rightarrow \]Formula \[{{A}_{6}}{{B}_{6}}{{X}_{3}}\Rightarrow {{A}_{2}}{{B}_{2}}X\] |
\[\because \] A is in tetrahedral void, so, coordination number is 4. |
\[\because \] X is in octahedral void so, coordination number is 6. |
when alternate tetrahedral voids are occupied, then centres of 6A atoms lies on the edges of one unit cell. |
You need to login to perform this action.
You will be redirected in
3 sec