A CsCl-type solid has $r_{+} / r_{-} = 0.732$ (the limiting value), so anions touch each other along the cell edge and the cation touches the anions along the body diagonal. Taking $π = 3.14$ and $3 = 1.732$ , the fraction of the cell volume occupied by the ions is closest to

Answer: B

Set

r_{-} = 1

, so

r_{+} = 0.732

. Along the body diagonal

3 a = 2 (1 + 0.732)

, giving

a = 2.000

and

a^{3} = 8.000

(consistent with

a = 2 r_{-}

). Ion volume

= \frac{4}{3} π (1 + 0.73 2^{3}) = \frac{4}{3} (3.14) (1.392) = 5.829

. Fraction

= 0.729 \approx 0.73