A) \[AB{{O}_{2}}\]
B) \[{{A}_{2}}B{{O}_{2}}\]
C) \[{{A}_{2}}{{B}_{3}}{{O}_{4}}\]
D) \[A{{B}_{2}}{{O}_{2}}\]
Correct Answer: D
Solution :
| [d] According to ccp, |
| Number of \[{{\text{O}}^{\text{2-}}}\] ions = 4 |
| So, tetrahedral void = 8 |
| and octahedral void = 4 |
| Since, A ions occupied \[\frac{\text{1}}{\text{4}}\text{th}\] of tetrahedral void. |
| \[\therefore \]Number of A ions\[=\frac{1}{4}\times 8=2\] Again, B ions occupied all octahedral void. |
| \[\therefore \]Number of B ions = 4 |
| A : B : 0 = 2 : 4 : 4 |
| = 1 : 2 : 2 |
| \[\therefore \]Structure of oxide \[=A{{B}_{2}}{{O}_{2}}\] |
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