Question

The ionic radii of alkali metal ions in water (hydrated radii) are in the order

• A. $L{i}^{+}>N{a}^{+}>K^{+}>R{b}^{+}<C{s}^{+}$
• B. $L{i}^{+}(aq.)>N{a}^{+}(aq.)>K^{+}(aq.)>R{b}^{+}(aq.)>C{s}^{+}(aq.)$
• C. $L{i}^{+}<N{a}^{+}>K^{+}>R{b}^{+}>C{s}^{+}$
• D. $L{i}^{+}(aq.)>N{a}^{+}(aq.)<K^{+}(aq.)<R{b}^{+}(aq.)<C{s}^{+}(aq.)$

Answer 1

The correct option is B $L{i}^{+}(aq.)>N{a}^{+}(aq.)>K^{+}(aq.)>R{b}^{+}(aq.)>C{s}^{+}(aq.)$

If we just look at only the ionic radii - then clearly $L{i}^{+}$ should be the smallest and common sense might compel us to think that the small ion should exhibit the best conductivity. However, among alkali metal cations - $L{i}^{+}$ has the highest charge density and is thus hydrated to the largest extent. Thus $L{i}^{+}$ has the highest hydrated ion radius and least conductivity in water.