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Arrange the following ions in the correct order with respect to their ionic radii.
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The correct answer is:
$\mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{Na}^{+}>\mathrm{Be}^{2+}$
The electronic configuration of $\mathrm{Be}^{2+}$ is $\left[{ }_2 \mathrm{He}\right], \mathrm{Na}^{+}$is $\left[{ }_{10} \mathrm{Ne}\right], \mathrm{Cl}^{-}$is $\left[{ }_{18} \mathrm{Ar}\right]$ and $\mathrm{Br}^{-}$ is $\left[{ }_{36} \mathrm{Kr}\right]$.
We know, site of group 18 elements follow the order $\mathrm{He} < \mathrm{Ne} < \mathrm{Ar} < \mathrm{Kr} < \mathrm{Xe}$.
So, the order of ionic radii of the given ions will be
$\mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{Na}^{+}>\mathrm{Be}^{2+}$
We know, site of group 18 elements follow the order $\mathrm{He} < \mathrm{Ne} < \mathrm{Ar} < \mathrm{Kr} < \mathrm{Xe}$.
So, the order of ionic radii of the given ions will be
$\mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{Na}^{+}>\mathrm{Be}^{2+}$
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