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NH(4)Cl crystallizes in a body centred c...

`NH_(4)Cl` crystallizes in a body centred cubic lattice , with a unit cell distance of 387 pm. Calculate (a) the distance between the oppositely charged ions in the lattice and (b) the radius of the `NH_(4)^(+)` ion if the radius of the `Cl^(-)` ion is 181 pm.

A

`335.15` pm

B

`387` pm

C

181 pm

D

`154.51` pm

Text Solution

Verified by Experts

The correct Answer is:
D
In a body centred cubic lattice, oppositely charged ions touch each other along the cross-diagonal of the cube, So, we can write,
`2r^(+)+2r^(-)=sqrt(3)a`
`r^(+)+r^(-)=sqrt(3)/(2)a=(387"pm")=335.15"pm"b`
Now, since `r^(-)=181"pm"`
We have, `r^(+)=(335.15-181)"pm"=154.15"pm"`
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