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Q.
$AB$ crystallizes in a body centred cubic lattice with edge length $'a'$ equal to $387\, pm$ . What is the distance between two oppositely charged ions in the lattice?
The Solid State
Solution:
For $bcc$ lattice body diagonal $=a\sqrt{3}$
The distance between the two oppositely charged ions
$=\frac{a}{2} \sqrt{3}$
$=\frac{387\times1.732}{2}$
$=335\,pm$