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The face diagonal length of FCC cubic cell is `660sqrt2` pm.If the radius of the cation is 110 pm, What should be the radius of the anion following radius ratio rules |
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Answer» Correct Answer - 220pm `sqrt2a=660sqrt2` pm so a=660 pm Now if tetrahedral void is occupied by cations than `sqrt3/4 a=(r_(+)+r_(-)),r_(-)=((sqrt3xx660)/4-110)=110[3/2sqrt3-1]=1.598xx110` so, `r_(+)/r_(-)=1/1.598=1/1.6=10/16=0.625` but `r_(+)/r_(-) gt 0.414` so it must not be occupying tetrahedral void then `a=2r_(+)+r_(-) " " implies 330=r_(+)+r_(-)` `r_(-)=220` pm `{r_(+)/r_(-)=0.5" it can occupy octahedral void"}` |
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