What is the distance between `Na^(+)` and `Cl^(-)` ions in NaCl crystal if density is `2.165"g cm"^(-3)` ? NaCl crystallises in fcc lattice.

Explanation

Calculation of the edge length of the unit cell.
Let the edge length of unit cell = a
Volume of unit cel = `a^(3)`
Gram foumula mass of `NaCl =23+35.5=58.5"g mol"^(-1)`
No. of particles in fcc type unit cell (Z) = 4
`"Mass of unit cell"=(Zxx"Gram formula mass of NaCl")/("Avogadro's Number "(N_(0)))=(4xx(58.5"g mol"^(-1)))/((6.022xx10^(23)"mol"^(-1)))`
=`3.886xx10^(-22)g`
Density of unit cell `(rho)=2.165"g cm"^(-3)`
Density of unit cell `(rho)=2.165"g cm"^(-3)`
Now, `"Density of unit cell"=("Mass of unit cell")/("Volume of unit cell")`
`(2.165"g mol"^(-3))=((3.886xx10^(-22)g))/("Volume of unit cell"(a^(3)))`
Volume of unit cell`(a^(3))=((3.886xx10^(-22)g))/((2.165"g cm"^(-3)))=1.795xx10^(-22)cm^(3)=179.5xx10^(-24)cm^(3)`
Edge length (a) =`(179.5xx10^(-24)cm^(3))^(1//3)=5.64xx10^(-8)cm`
`=564xx10^(-10)cm=564 "pm"`
Step II. Calculation of distance between `Na^(+)` and `Cl^(-)` ions.
Edge length of `Na^(+)Cl^(-)` unit cell `=2(r_(Na^(+))r_(Cl^(-)))`
`therefore" "2(r_(Na^(+))+r_(Cl^(-)))="564 pm or "(r_(Na^(+))+r_(Cl^(-)))=(564)/(2)="282 pm"`
`therefore` Distance in `Na^(+)` and `Cl^(-)` ions = 282 pm