Question

Give reasons when ${\rm{C}}{{\rm{l}}_{\rm{2}}}$ reacts with excess of ${{\rm{F}}_{\rm{2}}}$, ${\rm{Cl}}{{\rm{F}}_{\rm{3}}}$ is formed and not ${\rm{FC}}{{\rm{l}}_{\rm{3}}}$.

Answer 1

Hint: Fluorine and chlorine both present in the group 17 (group of halogens). Fluorine is the only halogen which possesses valency of -1. Other halogens such as chlorine, bromine etc shows variable valency.

Complete step by step answer:
Let’s first discuss the d-orbital present in chlorine. The atomic number of chlorine is 17. So, its electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}$. Here, we have observed that no d-orbitals present. But some elements like phosphorus, sulphur, chlorine etc. contain vacant d-orbitals and result in different values of covalency in different covalent compounds. This phenomenon is termed as variable valency.
Here, we are asked to identify the product when chlorine reacts with excess of fluorine. Chlorine can promote its 3p electrons to empty 3d orbitals so that it can have more half filled orbitals to accommodate three fluorine atoms around it. But fluorine does not have vacant d-orbital, so it cannot accommodate three chlorine atoms. So, the compound formed is ${\rm{Cl}}{{\rm{F}}_{\rm{3}}}$ not ${\rm{FC}}{{\rm{l}}_{\rm{3}}}$.
More, the product can be identified comparing the size of the atoms. We know that the size of fluorine is smallest among the halogen that means chlorine is larger than fluorine. As fluorine is smaller in size it cannot accommodate three larger chlorine atoms. So, the formation of compound ${\rm{FC}}{{\rm{l}}_{\rm{3}}}$ is not possible.
Therefore, when excess ${\rm{C}}{{\rm{l}}_{\rm{2}}}$ reacts with excess of ${{\rm{F}}_{\rm{2}}}$
formation of ${\rm{Cl}}{{\rm{F}}_{\rm{3}}}$ occur.

Note: Always remember that fluorine is the most electronegative element and thus it possesses valency of -1. But other halogens such as bromine, chlorine etc. possess d-orbital and therefore can expand their octets and show +1,+3, +5 and +7 oxidation states.