Animated Solution for Chemistry - s and p-Block Elements: The product formed in the reaction of ${\text{SOCl}}_2$ with white phosphorous is

The reaction of white phosphorus with thionyl chloride is a classic and highly important chemical transformation in inorganic chemistry, frequently tested in competitive exams like JEE. Let's dive deep into the mechanics and the nuances of this reaction.

We start with white phosphorus, which exists as discrete ${\text{P}}_4$ molecules. The structure of ${\text{P}}_4$ is a regular tetrahedron, meaning each phosphorus atom is bonded to three other phosphorus atoms. This geometry forces the bond angles to be $60^{\circ }$, which is significantly smaller than the ideal angle for $s{p}^3$ hybridization. This creates immense ring strain, making white phosphorus highly reactive.

Our other reactant is thionyl chloride (${\text{SOCl}}_2$). Thionyl chloride is a widely used chlorinating agent in both organic and inorganic chemistry. When it reacts with white phosphorus, it serves to transfer its chlorine atoms to the phosphorus.

When these two chemicals are brought together, a vigorous reaction ensues. The balanced chemical equation for this process is:

Let's break down the products formed: 1. Phosphorus trichloride (${\text{PCl}}_3$): This is the primary phosphorus-containing product. The phosphorus atom is oxidized from an oxidation state of $0$ in ${\text{P}}_4$ to $+3$ in ${\text{PCl}}_3$. 2. Sulfur dioxide (${\text{SO}}_2$): This is a gaseous byproduct. The evolution of ${\text{SO}}_2$ gas helps drive the reaction forward according to Le Chatelier's principle. 3. Sulfur monochloride (${\text{S}}_2{\text{Cl}}_2$): This is another byproduct, typically a yellow-red liquid.

A very common trap in JEE is confusing thionyl chloride (${\text{SOCl}}_2$) with sulfuryl chloride (${\text{SO}}_2{\text{Cl}}_2$).

While ${\text{SOCl}}_2$ is a milder chlorinating agent that yields ${\text{PCl}}_3$, reacting white phosphorus with sulfuryl chloride (${\text{SO}}_2{\text{Cl}}_2$) provides a more exhaustive chlorination, leading to the formation of phosphorus pentachloride (${\text{PCl}}_5$):

Always pay close attention to the specific reagent given in the problem!

Returning to our original question, the reaction of ${\text{SOCl}}_2$ with white phosphorus yields ${\text{PCl}}_3$ as the main product. Therefore, comparing this with the given options, Option (A) is the correct answer.