Reaction of thionyl chloride with white phosphorus forms a compound [A], which on hydrolysis gives [B], a dibasic acid. [A] and [B] are respectively

Thionyl chloride ($$SOCl_2$$) reacts with white phosphorus ($$P_4$$).

When $$SOCl_2$$ reacts with $$P_4$$, it chlorinates the phosphorus. The reaction produces phosphorus trichloride ($$PCl_3$$) as compound [A]:

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

On hydrolysis, $$PCl_3$$ gives phosphorous acid ($$H_3PO_3$$) as compound [B]:

$$PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$$

$$H_3PO_3$$ (phosphorous acid) is a dibasic acid because it has only two ionizable O-H bonds (the third hydrogen is directly bonded to phosphorus as a P-H bond and is not ionizable).

Therefore [A] = $$PCl_3$$ and [B] = $$H_3PO_3$$.

The answer is Option B.