Question

Why is $FeC{l_3}$ the product, when $Fe$ and $C{l_2}$ reacts and not $FeC{l_2}$?

Answer 1

Hint: Stability of the compound depends upon the number of electrons filled in the orbitals, whether it is completely filled or half-filled. If the element is completely filled then it is in the most stable state whereas in a half-filled state, it is partially stable, less than completely filled but more the atoms whose orbital are not either completely filled or half-filled.

Complete answer:
When iron heated in the stream of dry chlorine then iron(III) chloride is produced. This is happened due because chlorine is the powerful oxidising agent thus it brings out the higher oxidation state of iron:
$2Fe(s) + 3C{l_2}(g) \to 2FeC{l_3}(s)$
Moreover,$FeC{l_3}$ is more stable than $FeC{l_2}$ relative to the elements, $Fe + C{l_2}$, thus the energy of the electrons is lower in the system if rearranging in this way.
The extra stability of $FeC{l_3}$ can be explained with help of the central $Fe$ atom, the oxidation state is $ + 3$ and therefore it has $5$ electrons in its $3d$ orbitals. This is exactly half of the maximum number, i.e. $10$ thus it has extra stability due to its half-filled $d$ orbital . Now again $Fe$ atom with oxidation number $ + 2$ , it has $3$ electrons in its $3d$orbitals. Which is less stable than $F{e^{ + 3}}$. Thus, $Fe$ and $C{l_2}$ form $FeC{l_3}$ in place of $FeC{l_2}$ because it is more stable.

Note:
In some atoms, the binding energy is great enough to hold the nucleus together. This type of atom is known as stable. In some the binding energy is not strong enough to hold the nucleus together thus termed as unstable atoms. Atoms are in their most stable state when their outermost energy level is either filled with electrons or partially filled with electrons.