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The density of NH(4)Cl is 1.534 g cm^(-3...

The density of `NH_(4)Cl` is `1.534 g cm^(-3)`. It crystallizes in the `CsCl` lattice.
a. Calculate the length of the length of the edge of `NH_(4)Cl` unit cell.
b. Calculate the shortest distance between a `NH_(4)^(o+)` ion and a `Cl^(ɵ)` ion.
c. Calculate the radius of `NH_(4)^(o+)` ion if the radius of the `Cl^(ɵ)` ion is `181` pm

Text Solution

Verified by Experts

The correct Answer is:
a. `387 pm` b. `335.1 pm` c. `154.1 pm`
a. `Mw` of `NH_(4)Cl = 53.5`
`Z_(eff) = 1`
`a^(3) = (53.5 xx 1)/(1.534 xx 6.023 xx 10^(23)) = 5.79 xx 10^(-23) cm`
`a = 3.87 xx 10^(8) xx 10^(10)` pm `= 387` pm
b. (`CsCl` type) (bcc type)
For atoms `[r = (sqrt3)/(4)a]`
For ionic compound `[2(r_(o+) + r_(ɵ))]^(2) = 3a^(2)`
`(r_(o+) + r_(ɵ)) = (sqrt3)/(2) a = sqrt3 xx 387//2 = 335.1` pm
c. `r_(o+) + r_(ɵ) = 335.1` pm
`r^(o+) = 335.1 - r_(ɵ)`
`= 335.1 - 181 = 154.1` pm
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