JEE Main 2019 — Chemistry solution

JEE Main2019Chemistry

An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is :

A
$\frac{3}{2} a$
B
$\frac{a}{2}$
C
$a$
D
$2a$

Solution

In FCC, tetrahedral voids are located on the body diagonal at a distance of $\frac{3a}{4}$ from the corner. Together they form a smaller cube of edge length $\frac{a}{2} .$

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