Question

Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field.
A. $Li$
B. $Na$
C. $K$
D. $Rb$

Answer 1

Hint:Ionic mobility of an ion in solution is the speed of the ion per unit electric field applied to the solution.
i.e.$\mu = \dfrac{v}{E}$
where, v is the velocity of the ion and E is the applied electric field.
Units of ionic mobility are ${m^2}{V^{ - 1}}{s^{ - 1}}$.

Complete step-by-step answer:Ionic mobility depends upon the size of hydrated ions. The mobility of ions and conductivity in aqueous solution are inversely proportional to the size of hydrated ion,
i.e., Ionic mobility $ \propto \dfrac{1}{{Size{\text{ of hydrated ions}}{\text{.}}}}$
Therefore, smaller the size of hydrated ions, higher will be the ionic mobility and hydration power of ions depending upon charge density i.e., charge density $ = \dfrac{{Change}}{{Mass}}$. Therefore, higher the charge density of an ion, higher will be the hydration power. During hydration of an ion, the size of the ion increases.
In case of $L{i^ + },{\text{ N}}{{\text{a}}^ + },{\text{ }}{{\text{K}}^ + },{\text{ R}}{{\text{b}}^ + },{\text{ C}}{{\text{s}}^ + }$
$L{i^ + }$ion has high charge density and higher hydration power when dissolved in water because of the smallest size of lithium ion, so the effective size of lithium ion is increased due to presence of hydrated water molecules. Then the ionic mobility of the lithium ion decreases.
Hence, the order of ionic mobility is $L{i^ + }\left( {aq} \right) < N{a^ + }\left( {aq} \right) < {K^ + }\left( {aq} \right) < R{b^ + }\left( {aq} \right) < C{s^ + }\left( {aq} \right)$.
Thus, the ionic mobility of $L{i^ + }$ion is least as compared to other ions.

Hence, the correct option is (A).

Note:All alkali metal ions exist as hydrated ions in an aqueous solution. They form a complex with water. The degree of hydration decreases as the ionic size increases when we go down the group.