\(Na^+\), \(Mg^{2+}\), \(Al^{3+}\), \(Si^{4+}\) are isoelectronic. Their ionic size follows the order:

\(Na^+ > Mg^{2+} > Al^{3+} > Si^{4+}\)

The correct answer is option 4. \(Na^+ > Mg^{2+} > Al^{3+} > Si^{4+}\).

Let us explain why the ionic sizes of \(Na^+\), \(Mg^{2+}\), \(Al^{3+}\), and \(Si^{4+}\) follow the order \(Na^+ > Mg^{2+} > Al^{3+} > Si^{4+}\).

Isoelectronic ions are ions that have the same number of electrons but different numbers of protons. In this case:

\(Na^+\) (11 protons, 10 electrons)

\(Mg^{2+}\) (12 protons, 10 electrons)

\(Al^{3+}\) (13 protons, 10 electrons)

\(Si^{4+}\) (14 protons, 10 electrons)

Each of these ions has the same electron configuration as neon (\([Ne]\), 10 electrons), but the number of protons varies.

Effective Nuclear Charge

The effective nuclear charge (\(Z_{\text{eff}}\)) is the net positive charge experienced by electrons in an atom or ion. It accounts for the actual nuclear charge (number of protons) minus the shielding effect of inner-shell electrons. In this case, because there are no inner-shell electrons to provide significant shielding beyond the 1s² and 2s²2p⁶ configuration, the \(Z_{\text{eff}}\) increases directly with the number of protons.

Relationship Between Nuclear Charge and Ionic Size

Higher Nuclear Charge: As the number of protons increases, the effective nuclear charge increases.

Stronger Attraction: With a higher effective nuclear charge, the electrons are pulled more strongly towards the nucleus.

Smaller Ionic Radius: This increased attraction causes the electron cloud to contract, resulting in a smaller ionic radius.

Ionic Size Order

\(Na^+\) (11 protons): The nucleus exerts a certain amount of attraction on the 10 electrons. Because it has the fewest protons among the ions listed, the attractive force is the weakest, resulting in the largest ionic radius.

\(Mg^{2+}\) (12 protons): The nucleus has one more proton than \(Na^+\), leading to a stronger attraction on the same 10 electrons. This results in a smaller ionic radius compared to \(Na^+\).

\(Al^{3+}\) (13 protons): With yet another proton, the nucleus attracts the electrons even more strongly, resulting in an even smaller ionic radius compared to \(Mg^{2+}\).

\(Si^{4+}\) (14 protons): The highest number of protons in this series means the strongest attraction on the 10 electrons, leading to the smallest ionic radius among the ions listed.

The increasing nuclear charge as we move from \(Na^+\) to \(Si^{4+}\) leads to a progressively stronger attraction on the electron cloud, causing the ionic radius to decrease. Therefore, the correct order of ionic sizes is:\(Na^+ > Mg^{2+} > Al^{3+} > Si^{4+} \)

This explains why \(Na^+\) has the largest radius and \(Si^{4+}\) has the smallest radius among these isoelectronic ions.