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# Arrange the following ions in order of their increasing radii.${{K}^{+}}$$L{{i}^{+}},\text{ }M{{g}^{2+}},\text{ }{{K}^{+}},\text{ }A{{l}^{3+}}$:A. $A{{l}^{3+}}$ < $M{{g}^{2+}}$ < $L{{i}^{+}}$ < ${{K}^{+}}$B. $M{{g}^{2+}}$ < $A{{l}^{3+}}$ < $L{{i}^{+}}$ < ${{K}^{+}}$C. $A{{l}^{3+}}$ < $M{{g}^{2+}}$ < ${{K}^{+}}$ < $L{{i}^{+}}$D. None of these

Last updated date: 24th Apr 2024
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Hint: Think about the effective nuclear charge to solve this problem. Most of the problems in periodic properties can be solved using the concept of effective nuclear charge. Also take note of the trends in the properties that are observed along a period and down a group in the periodic table to narrow down the options.

We know that according to the general trend of periodic tables, when we go down the group the ionic radius will increase because the number of shells increases. Here, potassium has a higher number of shells, so that ${{K}^{+}}$ have the higher ionic radius. Since, Lithium is placed in the second period it has the lowest number of shells. So, $L{{i}^{+}}$ will be the smallest.
Among the$A{{l}^{3+}}$and$M{{g}^{2+}}$, they are placed in the same period and so they have the same number of shells. Now we can look at the effective nuclear charge on them. The effective nuclear charge is the charge experienced by each electron due to the nucleus. If the positive charge is higher, the size of the ion will decrease. This happens since as the atom loses more electrons the number of protons remain the same so they attract the remaining electrons more strongly and result in the reduction of size of the ion. The effective nuclear energy of each electron increases as the atom loses more electrons. So, $A{{l}^{3+}}$ has a smaller size than$M{{g}^{2+}}$. Therefore, the correct order is as follows:
$L{{i}^{+}}$ < $A{{l}^{3+}}$ < $M{{g}^{2+}}$ < ${{K}^{+}}$