Question bank/JEE Main/Chemistry

Consider a Body Centered Cubic(bcc) arrangement, let de, dfd,
dbd be the distances between successive atoms located along the edge,
the face-diagonal, the body diagonal respectively in a unit cell.Their
order is given by

Ade< dfd < dbdBdfd > dbd > deCdfd> de> dbdDdbd > de> dfd,

Show answer and solution▾

$d_{fd} = \sqrt{2}a$
$d_{bd} = \frac{\sqrt{3}a}{2}$
$\therefore d_{fd} > d_{e} > d_{bd}$

Source: Grass-G / CH-19 Chemistry Paper 28 Nov. Fourth.docx