Question

The mobility of metal ions in aqueous medium $(L{i}^{\oplus },N{a}^{\oplus },K^{\oplus },R{b}^{\oplus })$ in an electric field, follows the order:

• A. $L{i}^{\oplus }>N{a}^{\oplus }>K^{\oplus }>R{b}^{\oplus }$
• B. $R{b}^{\oplus }>N{a}^{\oplus }\equiv K^{\oplus }>L{i}^{\oplus }$
• C. $L{i}^{\oplus }<N{a}^{\oplus }<K^{\oplus }<R{b}^{\oplus }$
• D. $N{a}^{\oplus }\equiv K^{\oplus }>R{b}^{\oplus }>L{i}^{\oplus }$

Answer 1

The correct option is C $L{i}^{\oplus }<N{a}^{\oplus }<K^{\oplus }<R{b}^{\oplus }$

Due to their smaller sizes, lithium and sodium are surrounded by more number of water molecules, and hence it will be heavily hydrated. So the ionic mobility of those cations will be lesser than the larger cations like $R{b}^{\oplus }$ and $K^{\oplus }$ because they are hydrated to a lesser extent. If a cation can freely move in the solution, its ionic mobility will be high.

Correct order of ionic mobility as follows:
$L{i}^{\oplus }<N{a}^{\oplus }<K^{\oplus }<R{b}^{\oplus }$