The ease of adsorption of the hydrated alkali metal ions on an ion- exchange resins follow the order:
A.${K^ + } < N{a^ + } < R{b^ + } < L{i^ + }$
B.$N{a^ + } < L{i^ + } < {K^ + } < R{b^ + }$
C.$L{i^ + } < {K^ + } < N{a^ + } < R{b^ + }$
D.$R{b^ + } < {K^ + } < N{a^ + } < L{i^ + }$

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Hint:The ease of adsorption of hydrated metal ions depends on the size of the hydrated ion. As the size increases, the ease of adsorption also increases. To answer this question, you must recall the trend of change in the atomic radius of an atom in a periodic table. As we move from left to right in a period, the atomic size decreases and when we move down the group, an increase in the size is observed.

Complete answer:
In order to explain the behaviour of an atom or ion or molecule, we must have an idea about its size. Atomic radius is the most common way of expressing the size of an atom and similarly the size of an ion is given by the ionic radius.
As we move down in the group 1, the atomic size increases with the increase in the number of shells in the atoms. Similarly the ionic size also shows an increase as we move down the group. So, the order of size of the ions can be given as, $L{i^ + } < N{a^ + } < {K^ + } < R{b^ + }$
Smaller is the size of a cation, higher is the charge density on it and thus, higher is the number of water molecules that attach to it in an aqueous solution. As a result, the sizes of the hydrated ions follow the order, $R{b^ + } < {K^ + } < N{a^ + } < L{i^ + }$.
Thus, this will also be the order for the ease of adsorption of the ions on ion exchange resins.

Thus, the correct answer is D.

Cations are obtained when a neutral atom loses an electron from its valence shell and carries a positive charge. Since the number of electrons decreases, the effective pull on the valence shell and each electron increases. As a result, the size of a cation is smaller than its parent atom. The size of an ion is determined by the effective nuclear pull on the valence electrons.