Zinc blende (ZnS) has a cubic cell of edge 541 pm. Each Zn^2+ is at the centre of a regular tetrahedron of S^2- ions. Using sqrt3 = 1.732, find the Zn-S distance in pm (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3ss175

Zinc blende ( $ZnS$ ) has a cubic cell of edge $541 pm$ . Each $Zn X^{2 +}$ is at the centre of a regular tetrahedron of $S X^{2 -}$ ions. Using $3 = 1.732$ , find the $Zn$ - $S$ distance in pm (nearest integer).

Numerical answer type, answer in pm, tolerance 0.5.

A tetrahedral-void ion lies at

(\frac{a}{4}, \frac{a}{4}, \frac{a}{4})

from a corner, so the nearest-neighbour distance is one quarter of the body diagonal:

d = \frac{3 a}{4} = \frac{1.732 \times 541}{4} = 234.25 pm

, i.e. about 234 pm.