The correct order of ionic radii for the give | vedclass

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(B) The given species are $K^{+}$,$Na^{+}$,$Mg^{2+}$,and $Al^{3+}$.$K^{+}$ has electrons in the $n=3$ shell,while $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$ hav

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$K^{+} > Na^{+} > Mg^{2+} > Al^{3+}$

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(B) The given species are $K^{+}$,$Na^{+}$,$Mg^{2+}$,and $Al^{3+}$.$K^{+}$ has electrons in the $n=3$ shell,while $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$ have electrons in the $n=2$ shell.Therefore,$K^{+}$ has the largest ionic radius.For the isoelectronic species $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$,the ionic radius decreases as the nuclear charge (atomic number) increases.The atomic numbers are $Na (11)$,$Mg (12)$,and $Al (13)$.Thus,the order of ionic radii for these is $Na^{+} > Mg^{2+} > Al^{3+}$.Combining these,the correct order is $K^{+} > Na^{+} > Mg^{2+} > Al^{3+}$.

The correct order of ionic radii for the given species is:

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