Calcium fluoride (molar mass 78 g mol^-1) has a density of 3.15 g cm^-3 and the fluorite structure. Use N_A = 6.022x10^23 mol^-1. Using sqrt3=1.732, calculate the shortest Ca^2+-F- distance in pm (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss144

Calcium fluoride (molar mass $78 g mo l^{- 1}$ ) has a density of $3.15 g c m^{- 3}$ and the fluorite structure. Use $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ . Using $3 = 1.732$ , calculate the shortest $Ca X^{2 +}$ - $F X^{-}$ distance in pm (nearest integer).

Numerical answer type, answer in pm, tolerance 1.

Z = 4

a^{3} = \frac{4 \times 78}{3.15 \times 6.022 \times 1 0 ^{23}} = 1.645 \times 1 0^{- 22} c m^{3}

and

a = 547.9 pm

. The

Ca X^{2 +}

F X^{-}

contact lies along the quarter body diagonal:

d = \frac{3 a}{4} = \frac{1.732 \times 547.9}{4} = 237.2 pm \approx 237 pm