# The order of magnitude of ionic radii of ions  and ${\mathrm{Si}}^{4+}$ is :

1. A

$\begin{array}{r}{\mathrm{Na}}^{+}<{\mathrm{Mg}}^{2+}<{\mathrm{Al}}^{3+}<{\mathrm{Si}}^{4+}\end{array}$

2. B

$\begin{array}{r}{\mathrm{Mg}}^{2+}>{\mathrm{Na}}^{+}>{\mathrm{Al}}^{3+}>{\mathrm{Si}}^{4+}\end{array}$

3. C

$\begin{array}{r}{\mathrm{Al}}^{3+}>{\mathrm{Na}}^{+}>{\mathrm{Si}}^{4+}>{\mathrm{Mg}}^{2+}\end{array}$

4. D

${\mathrm{Na}}^{+}>{\mathrm{Mg}}^{2+}>{\mathrm{Al}}^{3+}>{\mathrm{Si}}^{4+}$

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### Solution:

The order of magnitude of ionic radii of ions ${\mathrm{Na}}^{+}>{\mathrm{Mg}}^{2+}>{\mathrm{Al}}^{3+}>{\mathrm{Si}}^{4+}$. All are isoelectronic, but ${\mathrm{Si}}^{4+}$ has the highest nuclear charge per electron. As a result, ${\mathrm{Na}}^{+}$ has the smallest size and nuclear charge per electron.

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