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.

Answer

282 pm

Explanation

Density = `(Z xx M)/(a^3 xx N_A)`
`2.165 = (4 xx 58.5)/(a^3 xx 6.022 xx 10^(23))`
`a^3 = (4 xx 58.5)/(2.165 xx 6.022 xx 10^(23))`
`a^3 = 179.48 xx 10^(-24) cm^3`
or `a = 5.64 xx 10^(-8) cm`
In NaCl structure, `a = 2 (r_(Na^+) + r_(Cl^-)`
`:.` Distance between `Na^(+) and Cl^(-)` ion,
`r_(Na^+) + r_(Cl^-) = a/2 = 2.82 xx 10^(-8) cm = 282` pm