JEE Chemistry 2023 Question with Solution

Given: One mole of $P_4$ reacts with $8$ moles of $SOCl_2$ to form $4$ moles of $A$, $x$ mole of $SO_2$ and $2$ moles of $B$.

Find: The identities of $A$, $B$ and the value of $x$.

The balanced chemical equation given in the solution is:

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

From this equation:

• $A = PCl_3$
• $B = S_2Cl_2$
• $x = 4$

Therefore, the correct option is C.

Given: The reaction of $P_4$ with $SOCl_2$.

Find: Which product corresponds to $A$, which corresponds to $B$, and how many moles of $SO_2$ are formed.

Using the balanced reaction:

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

Compare this directly with the form stated in the question:

$$P_4 + 8SOCl_2 \rightarrow 4A + xSO_2 + 2B$$

Matching coefficients and products term by term:

$$4A = 4PCl_3 \Rightarrow A = PCl_3$$ $$2B = 2S_2Cl_2 \Rightarrow B = S_2Cl_2$$ $$xSO_2 = 4SO_2 \Rightarrow x = 4$$

Hence, $A = PCl_3$, $B = S_2Cl_2$, and $x = 4$. The correct option is C.

The solution's lists answer key as option (1), but the extracted solution clearly concludes option C. Since the solution is the primary source, C is taken as correct.