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(B) The ionic radius depends on the number of shells and the effective nuclear charge.$1$. Comparing $Na^{+}$ $(2, 8)$ and $Li^{+}$ $(2)$: $Na^{+}$ ha

Vedclass · Accepted Answer

$Na^{+} > Li^{+} > Mg^{2+} > Al^{3+} > Be^{2+}$

Vedclass · Answer

(B) The ionic radius depends on the number of shells and the effective nuclear charge.$1$. Comparing $Na^{+}$ $(2, 8)$ and $Li^{+}$ $(2)$: $Na^{+}$ has two shells while $Li^{+}$ has one,so $Na^{+} > Li^{+}$.$2$. Comparing $Mg^{2+}$ $(2, 8)$ and $Al^{3+}$ $(2, 8)$: Both have two shells,but $Al^{3+}$ has a higher nuclear charge,so $Mg^{2+} > Al^{3+}$.$3$. Comparing $Li^{+}$ $(2)$ and $Be^{2+}$ $(2)$: $Be^{2+}$ has a higher nuclear charge,so $Li^{+} > Be^{2+}$.$4$. Comparing $Na^{+}$ $(2, 8)$ and $Mg^{2+}$ $(2, 8)$: $Mg^{2+}$ has a higher nuclear charge,so $Na^{+} > Mg^{2+}$.Combining these,the correct order is $Na^{+} > Li^{+} > Mg^{2+} > Al^{3+} > Be^{2+}$.

The set representing the correct order of ionic radius is

Similar Questions

Which among the following isoelectronic species has the smallest size?
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Which of the following represents the correct order of ionic radii?

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