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The radius of a divaent cation `A^(2+)` is 94pm and of divalent anion `B^(2-)` is 146pm. The compound AB has:A. NaCl structureB. Linear structureC. CsCl structureD. ZnS structure

Answer» Given , radius of cation of 94 pm, radius of anion = 146pm
For `M^(2+)X^(2-)`, radius ratio , `(r_+)/(r_-)=94/146=0.6438`
Since , the radius ratio is in between 0.41 to 0.732 , it has NaCl structure.


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