A) AB
B) \[{{A}_{2}}B\]
C) \[A{{B}_{2}}\]
D) \[A{{B}_{4}}\]
Correct Answer: C
Solution :
[c] Suppose no. of \[{{B}^{-}}\] ions = N \[\therefore \] no. of tetrahedral holes = 2N As 25% tetrahedral holes are occupied by \[{{A}^{+}}\] ions. \[\therefore \] no. of \[{{A}^{+}}\] inos \[=\frac{2N}{4}=\frac{N}{2}\] \[\therefore \] Ratio of \[{{A}^{+}}:{{B}^{-}}=\frac{N}{2}:N\Rightarrow 1:2\] \[\therefore \] the formula is \[A{{B}_{2}}\]You need to login to perform this action.
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