The Aluminium chloride crystallizes in BCC lattice with an edge length of unit cell equal to $387 p m$ . If the size of $C l^{-}$ ion is $181 p m$ , the size of $N H_{4}^{+}$ ion would be:

116 p m

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154 p m

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174 p m

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206 p m

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Solution

The correct option is C $154 p m$
In BCC lattice,

$\sqrt{3} \times e d g e l e n g t h = 2 (r_{+} + r_{-})$

$\Rightarrow \frac{\sqrt{3} \times 387}{2} =^{r} N H_{4}^{+} +$ $^{r} C l^{-}$

$\Rightarrow^{r} N H_{4}^{+} = \frac{387 \times \sqrt{3}}{2} - 181 p m$

$\Rightarrow^{r} N H_{4}^{+} = 154 p m$

Hence, option $(2)$ is correct.

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$N H_{4} C l$ 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

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