In an ionic solid `r_((+))=1.6A` and `r_((-))=1.864A`. Use the radius ratio to determine the edge length of the cubic unit cell in `A`.A. 4B. `2sqrt(3)`C. `3sqrt(3)`D. `(4)/(sqrt(3))`

In an ionic solid `r_((+))=1.6A` and `r_((-))=1.864A`. Use the radius

1.	In an ionic solid `r_((+))=1.6A` and `r_((-))=1.864A`. Use the radius ratio to determine the edge length of the cubic unit cell in `A`.A. 4B. `2sqrt(3)`C. `3sqrt(3)`D. `(4)/(sqrt(3))`
Answer» Correct Answer - 1 `(r_(+))/(r_(_))=(1.6)/(1.864)=0.858` So, it is `CsCl` type of unit cell So `sqrt(3)a=2(r_(+)+r_(_))` So, `a=(2(1.864+1.6))/(sqrt(3))Å=2xx2Å=4Å`

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In an ionic solid `r_((+))=1.6A` and `r_((-))=1.864A`. Use the radius ratio to determine the edge length of the cubic unit cell in `A`.A. 4B. `2sqrt(3)`C. `3sqrt(3)`D. `(4)/(sqrt(3))`

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