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Ammonium chloride crystallizes in a body centred cubic lattice with edge length of unit cell of 390 $\mathrm{pm}$. If the size of chloride ion is $180 \mathrm{pm}$, the size of ammonium ion would be
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Verified Answer
The correct answer is:
$158 \mathrm{pm}$
$158 \mathrm{pm}$
For bcc structure
$$
\begin{aligned}
\text { Interionic distance }= & r^{+}+r^{-}=\frac{\sqrt{3}}{2} a \\
& (a=\text { edge length })
\end{aligned}
$$
$$
\begin{aligned}
& r_{\mathrm{NH}_4^{+}}^{+}+r_{\mathrm{Cl}^{-}}^{-}=\frac{\sqrt{3}}{2} a \\
& r_{\mathrm{NH}_4^{+}}^{+}+180=\frac{\sqrt{3}}{2} \times 390
\end{aligned}
$$
$$
\mathrm{r}_{\mathrm{NH}_4^{+}}^{+}=338 \mathrm{pm}-180 \mathrm{pm}=158 \mathrm{pm}
$$
$$
\begin{aligned}
\text { Interionic distance }= & r^{+}+r^{-}=\frac{\sqrt{3}}{2} a \\
& (a=\text { edge length })
\end{aligned}
$$
$$
\begin{aligned}
& r_{\mathrm{NH}_4^{+}}^{+}+r_{\mathrm{Cl}^{-}}^{-}=\frac{\sqrt{3}}{2} a \\
& r_{\mathrm{NH}_4^{+}}^{+}+180=\frac{\sqrt{3}}{2} \times 390
\end{aligned}
$$
$$
\mathrm{r}_{\mathrm{NH}_4^{+}}^{+}=338 \mathrm{pm}-180 \mathrm{pm}=158 \mathrm{pm}
$$
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