White phosphorus reacts with thionyl chloride to give

We need to find the products when white phosphorus ($$P_4$$) reacts with thionyl chloride ($$SOCl_2$$).

Thionyl chloride ($$SOCl_2$$) is a chlorinating agent. When it reacts with white phosphorus, it converts phosphorus into a phosphorus chloride and undergoes reduction itself.

The balanced reaction is:

$$P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2$$

Let us verify this by analyzing the reaction. White phosphorus ($$P_4$$) has phosphorus in the 0 oxidation state. Thionyl chloride provides chlorine atoms to phosphorus. Each phosphorus atom accepts three chlorine atoms to form $$PCl_3$$ (phosphorus trichloride), where phosphorus goes to the +3 oxidation state.

Now, the sulfur in $$SOCl_2$$ is in the +4 oxidation state (since O is $$-2$$ and each Cl is $$-1$$, giving $$S + (-2) + 2(-1) = 0$$, so $$S = +4$$). In the products, sulfur appears in $$SO_2$$ (oxidation state +4) and $$S_2Cl_2$$ (oxidation state +1). The sulfur in $$S_2Cl_2$$ has been reduced from +4 to +1, while phosphorus has been oxidized from 0 to +3.

So the products of the reaction are $$PCl_3$$, $$SO_2$$, and $$S_2Cl_2$$.

Hence, the correct answer is Option B.