The electrical conductivity in aqueous solution is high for:
(A) $ M{g^{ + + }} $
(B) $ N{a^ + } $
(C) $ C{a^{ + + }} $
(D) $ {K^ + } $

Hint: The conductivity of water is determined by the concentration of ions present in it. It is the ability of water to flow current in it. The extent of dissociation of ions decides the conductivity of the solution, more the dissociation, more will be the conductivity.

Complete step by step solution:
Compounds that are dissolved in water are called electrolytes. These electrolytes contain ions in them, more the number of ions in the solution, more will be the conductivity.
We are given ions of alkali metals and alkaline earth metals. They all conduct electricity in an aqueous solution. So, now we will arrange the electrical conductivity of ions according to their hydration energy. The electrical conductivity is inversely proportional to the hydration energy of the ion.
 $ Electrical\,Conductivity\,\alpha \,\dfrac{1}{{Hydration\,Energy}} $
It means that the ions that will have the minimum hydration energy will have the highest electrical conductivity in the aqueous solution. Now, we will look for the trends of hydration energy down the group and across the period. On moving from left to right in a period, the hydration energy increases, and on moving down the group the hydration energy decreases. It means that the element that lies in the bottom of the group will have the least hydration energy. In the given option, $ {K^ + } $ lies at the bottom so it will have the least hydration energy and it will conduct electricity to the maximum extent. Hence, the electrical conductivity in an aqueous solution will be high for $ {K^ + } $ .
Therefore, option (D) is correct.

Note:
Electrical conductivity is the measure of the solution which decides its ability to conduct electricity through it. The electrical conductivity of water is directly proportional to the ion concentration. The extent of ions in the solution decides the current that will flow in the solution.