The order of magnitude of ionic radii of ions | vedclass

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Question

(D) The ions $Na^{+}$,$Mg^{2+}$,$Al^{3+}$,and $Si^{4+}$ are isoelectronic species,meaning they all have the same number of electrons ($10$ electrons).

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The ions $Na^{+}$,$Mg^{2+}$,$Al^{3+}$,and $Si^{4+}$ are isoelectronic species,meaning they all have the same number of electrons ($10$ electrons).For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.The atomic numbers are $Na$ $(11)$,$Mg$ $(12)$,$Al$ $(13)$,and $Si$ $(14)$.As the nuclear charge increases,the electrostatic attraction between the nucleus and the electrons increases,pulling the electron cloud closer to the nucleus.Therefore,the order of ionic radii is $Na^{+} > Mg^{2+} > Al^{3+} > Si^{4+}$.

The order of magnitude of ionic radii of ions $Na^{+}$,$Mg^{2+}$,$Al^{3+}$ and $Si^{4+}$ is

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