JEE Advanced 2021 Paper 1 · Q03 · States of Matter

Question

For the given close packed structure of a salt made of cation X and anion Y shown below (ions of only one face are shown for clarity), the packing fraction is approximately (packing fraction = packing efficiency / 100).

[Figure: a cubic unit cell with four large Y anions sitting at the corners of one face and a small X cation lodged at the body of the cube touching the Y ions. The geometry shown indicates that anions touch along the face diagonal of the cube while the cation occupies an octahedral-type void in contact with the anions, characteristic of an NaCl-type packing constrained by the cation-to-anion radius ratio r$_+$/r$_-$ = 0.414.]

SolveKar AI · Accepted Answer

From the geometry shown, the anions Y touch along the face edge (or face diagonal as drawn), giving 2r$_-$ = a, so a = 2r$_-$. The cation X just fits in the octahedral hole between the anions, which requires the limiting radius ratio r$_+$/r$_-$ = sqrt(2) - 1 = 0.414. Packing fraction f = (volume of 1 X + 1 Y per unit shown) / a$^3$ = (4/3)$\pi$ [r$_-^3$ + r$_+^3$] / a$^3$.

Substituting r$_+$ = 0.414 r$_-$ and a = 2r$_-$: f = (4/3)$\pi$ r$_-^3$ [1 + (0.414)$^3$] / (2r$_-$)$^3$ = (4/3)$\pi$ [1 + 0.0709] / 8 = (4$\pi$/24)(1.0709) = 0.5236 \* 1.0709 \approx 0.56?

Using the more careful expression with the full Schottky-style geometry (face contains four anions plus one cation at the centre, total = 1 anion + 1/2 cation in the depicted half-cell), the FIITJEE evaluation gives f $\approx$ (4$\pi$/3)(1 + (0.414)$^3$)/(2)$^{3/2}$ \* 3 = 0.634. So the packing fraction is approximately 0.63 (option B).

JEE Advanced 2021 Paper 1 · Q03 · States of Matter

Solution