A CsCl-type crystal has molar mass 168.5 g mol^-1 and density 3.99 g cm^-3. Using N_A = 6.022x10^23 mol^-1 and sqrt3 = 1.732, calculate the distance between the nearest cation and anion in pm (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss180

A CsCl-type crystal has molar mass $168.5 g mo l^{- 1}$ and density $3.99 g c m^{- 3}$ . Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ and $3 = 1.732$ , calculate the distance between the nearest cation and anion in pm (nearest integer).

Numerical answer type, answer in pm, tolerance 1.

With

Z = 1

a^{3} = \frac{M}{N _{A} ρ} = \frac{168.5}{6.022 \times 1 0 ^{23} \times 3.99} = 7.0127 e - 23 c m^{3}

, so

a = 412.4 pm

. The nearest cation-anion distance is half the body diagonal:

\frac{3}{2} a = 0.866 \times 412.4 = 357.1 pm

, about 357 pm.