Ammonium chloride cystallise in a body-centred cubic lattice with a unit distance of $387 p m$ . Calculate
(a) the distance between oppositely charged ion in the lattice and
(b) the radius of the ${N H}_{4}^{+}$ ion if the radius of ${C l}^{-}$ ion is $181 p m$

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Solution

$1.$ $N H_{4}^{+}$ is at body centre and $C l^{-}$ at body corners. Distance between opposite charged ion $= \frac{a \sqrt{3}}{2} = \frac{387 \sqrt{3}}{2} = 335.15$ $p m$
$2.$ For BCC, Ideal ration of $\frac{r_{+}}{r_{-}} = 0.732$
$\frac{r_{N H_{4}^{+}}}{r_{C l^{-}}} = 0.732$
$⟹ r_{N H_{4}^{+}} = 0.732 \times 181 = 132.49 p m$

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Q. Ammonium chloride crystallises in a body-centred cubic lattice with a unit distance of 387 pm. Calculate (a) the distance between oppositely charged ions in the lattice and (b) the radius of the

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crystallises in a body-centered cubic lattice with an unit cell edge length of 387 pm. The distance between the oppositely charged ions in the lattice is:

$N H_{4} C l$ crystallizes in a body-centered cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the $C l^{-}$ ion is 181 pm, radius for $N H_{4}^{+}$ ion is

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Ammonium chloride cystallise in a body-centred cubic lattice with a unit distance of 387pm. Calculate(a) the distance between oppositely charged ion in the lattice and(b) the radius of the NH+4 ion if the radius of Cl− ion is 181pm

1. NH+4 is at body centre and Cl− at body corners. Distance between opposite charged ion =a√32=387√32=335.15 pm2. For BCC, Ideal ration of r+r−=0.732rNH+4rCl−=0.732⟹rNH+4=0.732×181=132.49 pm

Ammonium chloride cystallise in a body-centred cubic lattice with a unit distance of $387 p m$ . Calculate
(a) the distance between oppositely charged ion in the lattice and
(b) the radius of the ${N H}_{4}^{+}$ ion if the radius of ${C l}^{-}$ ion is $181 p m$