A compound XY crystallizes in BCC lattice with a unit cell edge length of 480 pm. If the radius of Y⁻ is 225 pm. What will be the radius of X⁺

The correct option is D

$190.68\mathrm{pm}$

The explanation for the correct answer:-

Option (D) $190.68$ pm

Step 1: Given data

As we know, the sum of the ionic radii is the nearest neighbour distance ‘d’.

In the case of a body-centred cubic unit cell, where a is the unit cell length.

Unit cell edge length of $480$ pm

Radius of $Y^{-}is225$pm

Step 2: Formula used

Neighbour distance, $d=\frac{\sqrt{3}}{2}a$

Step 3: Calculating neighbour distance

From 1) we get, d = $415.68$pm.

Step 4: Deriving radius of $X^{+}$

As $X^{+}+Y^{-}$ = $415.68$ pm

Therefore, $X^{+}$ = $415.68-225.00$

$=190.68pm$.

Hence, the radius of $X^{+}$ is $190.68\mathrm{pm}$, option (D) is the correct answer.