The arrangement of X– ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of X

Engineering

Chemistry

Voids or Interstitial Site and Radius Ratio

Question

The arrangement of X^– ions around A⁺ ion in solid AX is given in the figure (not drawn to scale). If the radius of X^– is 250 pm, the radius of A⁺ is

57 pm

125 pm

104 pm

183 pm

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

It is an octahedral void

$\frac{r_{A^{+}}}{r_{X^{-}}}$ = 0.414

$r_{A^{+}}$ = 0.414 × 250 = 104 pm

Please subscribe our Youtube channel to unlock this solution.