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 ra
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
5 views
1 Answers
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"}`
5 views
Answered