Calcium fluoride `(CaF_(2))` crystallises in a cubic crystal structure. Unit cell of this structure contains four `Ca^(2+)` ions and eight `F^(-)` ions. If density of `CaF_(2)` is `3.2g*cm^(-3)`, then calculate the edge length of the unit cell in pm.

Explanation

According to the question, the unit cell of `CaF_(2)` contains `4Ca^(2+)` ions and `8F^(-)` ions. Hence, number of formula units of `CaF_(2)` per unit cell = 4.
Therefore, mass of a unit cell of `CaF_(2)`
`=(4xx"formula mass of " CaF_(2))/N=(4xx78g*mol^(-1))/(6.022xx10^(23)mol^(-1))`
`=5.18xx10^(-22)g`
Density of unit cell = `"Mass"/"Volume"=(5.18xx10^(-22)g)/a^(3)=3.2g*cm^(-3)`
where a = edge length of the unit cell
`therefore" "a^(3)=(5.18xx10^(-22))/3.2cm^(3)=1.618xx10^(-22)cm^(3)`
`therefore" "a=5.45xx10^(-8)cm` = 545pm
`therefore" "`Edge length of the unit cell of `CaF_(2)` = 545pm.