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 12 A atoms lie on the edges of one unit cell.
Correct Answer: D
Solution :
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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 6 A atoms lies on the edges of one unit cell. |
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