Question

Anhydrous ${\text{AlC}}{{\text{l}}_{\text{3}}}$ is a covalent compound. From the data given below, predict whether it would remain covalent or become ionic in an aqueous solution. (Ionization energy of ${\text{Al = 5137kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ )
$\Delta {\text{H}}$ hydration for ${\text{A}}{{\text{l}}^{{\text{3 + }}}}$ ${\text{ = - 4665kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
$\Delta {\text{H}}$ hydration for ${\text{C}}{{\text{l}}^ - }$ ${\text{ = - 381kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
(A) Ionic
(B) Covalent
(C) Partially ionic
(D) Partially covalent

Answer 1

The third ionization energy of aluminium is very high due to which a lot of energy is necessary to convert aluminium into ${\text{A}}{{\text{l}}^{{\text{3 + }}}}$ ion. So, it forms covalent bonds preferably with chloride ions.
When dissolved in water, ${\text{AlC}}{{\text{l}}_{\text{3}}}$ will undergo hydration which will release a lot of energy which is enough to remove 3 electrons from aluminium.
If the ionization energy is less than the total hydration energy, the compound will be ionic in aqueous solution.

Complete step by step solution:
Given that anhydrous ${\text{AlC}}{{\text{l}}_{\text{3}}}$ is a covalent compound.
Also given, ionization energy of ${\text{Al = 5137kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
$\Delta {\text{H}}$ hydration for ${\text{A}}{{\text{l}}^{{\text{3 + }}}}$ ${\text{ = - 4665kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
$\Delta {\text{H}}$ hydration for ${\text{C}}{{\text{l}}^ - }$ ${\text{ = - 381kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
We need to find out whether aluminium chloride or ${\text{AlC}}{{\text{l}}_{\text{3}}}$ will remain covalent or become ionic in aqueous solution.
The total hydration energy when aluminium chloride is in an aqueous solution will be equal to the sum of the hydration energy of the aluminium ion and the hydration energy of the chloride ion. Thus, total hydration energy of ${\text{A}}{{\text{l}}^{{\text{3 + }}}}$ and ${\text{C}}{{\text{l}}^ - }$ ions of ${\text{AlC}}{{\text{l}}_{\text{3}}}$ is:
(Since there are one aluminium ion and three chloride ions)
$= \left[ { - 4665 + 3 \times \left( { - 381} \right)} \right]{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}} \\\ = - 5808{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}} \\\$
Now, this value of total hydration energy for aluminium chloride is higher than the ionization energy of aluminium which is ${\text{5137kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ . Thus, this is higher than the amount of energy required for the ionization of aluminium into ${\text{A}}{{\text{l}}^{{\text{3 + }}}}$ ion. Due to this reason, the aluminium chloride becomes ionic in aqueous solution. Its ionic form in aqueous solution is:
${\text{AlC}}{{\text{l}}_{\text{3}}}{\text{ + 6}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\left[ {{\text{Al}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]^{{\text{3 + }}}}{\text{ + 3C}}{{\text{l}}^{\text{ - }}}$
Thus, option A is correct.

Note:
Some of the factors affecting ionization energy are:
Size of atom: Smaller the size of the atom, more tightly held electrons and so greater IE.
Nuclear charge: Greater the nuclear charge, more energy needed to pull electrons and so greater IE.
Electronic configuration: Elements having half filled or fully filled valence orbitals have higher IE values than expected normally from their position in the periodic table.