Question

A compound $XY$ crystallizes in BCC lattice with unit cell edge length of $480$ pm. If the radius of $Y^{-}$ is $225$ pm, then the radius of $X^{+}$ is:-

• A. $127.5$ pm
• B. $190.68$ pm
• C. $225$ pm
• D. $255$ pm

Answer 1

The correct option is B $190.68$ pm

We know that the sum of the ionic radii is the nearest neighbour distance $d$.
In case of a body centred cube, $d=\frac{\sqrt{3}}{2}a$, where $a$ is the unit cell length.
Calculating, we get $d=415.68$ pm
Since, $X^{+}+Y^{-}=415.68$ pm
Therefore, $X^{+}=415.68-225.00=190.68$ pm