A compound AB has an NaCl like structure. The radius of A+ and B− is 1.14∘A and 1.86∘A. What is the minimum distance between two A+ ions in angstroms? (Take √2= 1.41)
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Solution
The edge length is given by the expression, a=2(r++r−) = 2(1.14+1.86) = 6 ∘A The minimum distance between two cations is given by a√2 Substituting, 6√2 = 3×√2 = 4.23 ∘A