The ionic radius of Na^{+} ions is 1.02 math | vedclass

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Question

(B) The ions $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$ are isoelectronic species,all having $10$ electrons.For isoelectronic species,the ionic radius decreases

Vedclass · Accepted Answer

$0.72$ and $0.54$

Vedclass · Answer

(B) The ions $Na^{+}$,$Mg^{2+}$,and $Al^{3+}$ are isoelectronic species,all having $10$ electrons.For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.The atomic numbers are $Na$ $(Z=11)$,$Mg$ $(Z=12)$,and $Al$ $(Z=13)$.Since the nuclear charge increases from $Na^{+}$ to $Al^{3+}$,the ionic radius decreases in the order $Na^{+} > Mg^{2+} > Al^{3+}$.Given the radius of $Na^{+}$ is $1.02 \ \mathring{A}$,the values for $Mg^{2+}$ and $Al^{3+}$ must be smaller than $1.02 \ \mathring{A}$ and follow the decreasing trend.Among the options,$0.72 \ \mathring{A}$ and $0.54 \ \mathring{A}$ fit this trend correctly.

The ionic radius of $Na^{+}$ ions is $1.02 \ \mathring{A}$. The ionic radii (in $\mathring{A}$) of $Mg^{2+}$ and $Al^{3+}$,respectively,are

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