Question

Explain why, iron dissolves in HCl to form $FeC{{l}_{2}}$ not $FeC{{l}_{3}}$?

Answer 1

When two chemicals react with each other the formation of the product is going to depend on the energy required to convert chemical reactants into products. If the energy to convert reactants into products is too high then the reaction is not going to happen.

Complete answer:
In the question it is given that iron (Fe) reacts with HCl and forms $FeC{{l}_{2}}$(Ferrous Chloride) not $FeC{{l}_{3}}$ (Ferric chloride).
$Fe+HCl\to FeC{{l}_{2}}$
To know about the above reaction we should know that the electronic configuration of Iron.
Electronic configuration of Iron is as follows.
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}4{{s}^{2}}$
The outermost electrons in $4{{s}^{2}}$ are readily given electrons to Chlorine atoms.
But to form $FeC{{l}_{3}}$ it requires a catalyst or oxidant like ${{H}_{2}}{{O}_{2}}$etc., to donate it electrons in $3{{d}^{6}}$ orbitals.
To donate the electron from $3{{d}^{6}}$of iron, it requires more energy, also called third ionization energy.
First and second ionization energy of iron are less so that the first two electrons are donated easily to two chlorine atoms.
Coming to the third ionization energy of iron, it is very high because the next electron should come from $3{{d}^{6}}$.
In $3{{d}^{6}}$ the last electron (6th electron) is paired in 3d orbital. So, it is very difficult to donate a third electron from Iron to chlorine.
So, iron forms only $FeC{{l}_{2}}$(Ferrous Chloride) not $FeC{{l}_{3}}$ (Ferric chloride).

Note:
The formation of the products in a chemical reaction is going to depend on ionization energy of the chemicals. If the reaction needs the highest amount of ionization energy we have to use a catalyst to complete the reaction.