1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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+

A

$225\mathrm{pm}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

$255\mathrm{pm}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

$127.5\mathrm{pm}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

$190.68\mathrm{pm}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D $190.68\mathrm{pm}$The explanation for the correct answer:-Option (D) $190.68$ pmStep 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 ${Y}^{-}is225$pmStep 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.

Suggest Corrections
4
Join BYJU'S Learning Program
Related Videos
Density
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program