Question

The hydration energy of $\text{M}{{\text{g}}^{\text{+2}}}$is higher than that of:
A.$\text{A}{{\text{l}}^{\text{+3}}}$
B.$\text{N}{{\text{a}}^{\text{+}}}$
C.$\text{B}{{\text{e}}^{\text{+2}}}$
D.$\text{Na}$

Answer 1

The amount of energy released when water molecules attach themselves to the metal ions to undergo hydration is called hydration energy. It is directly proportional to the positive charge present on the metal.

Complete step by step answer:
The hydration energy depends on a number of factors and the nuclear charge is one of them, the other being the size of the atom or the ion. When an ion is formed by loss of electrons then the number of electrons in the valence shell is less than the number of protons in the nucleus so there is an increased attraction between the nucleus and the electrons. Due to this increased nuclear pull, the size of the ion shrinks.
Now as the water molecules approach the metal ions in a solution, the increased nuclear pull attracts the negative dipole of the water, i.e., the oxygen atoms and hydration energy is released in the process. This hydration energy is higher for ions with higher positive charge. Now, in the question, the order of size of the ions is:
$\text{A}{{\text{l}}^{\text{+3}}}$>$\text{B}{{\text{e}}^{\text{+2}}}$>$\text{M}{{\text{g}}^{\text{+2}}}$>$\text{N}{{\text{a}}^{\text{+}}}$>$\text{Na}$
Hence the hydration energy of $\text{A}{{\text{l}}^{\text{+3}}}$should be the highest while that of $\text{Na}$should be lowest. And, the hydration energy of $\text{M}{{\text{g}}^{\text{+2}}}$higher than that$\text{N}{{\text{a}}^{\text{+}}}$.

So, the correct answer is option B.

Note:
In case of ionic solids, the solubility of a solute depends on the hydration energy of the constituent ions and the lattice energy of the solid. If the hydration energy is more than the lattice energy then only the solid dissolves to form the solution.