JEE Chemistry 2023 Question with Solution

Given: The compounds are $\mathrm{H_4P_2O_7}$, $(\mathrm{HPO_3})_3$ and $\mathrm{P_4O_{10}}$.

Find: The number of P–O–P bonds in each compound.

A P–O–P bond is counted only when an oxygen atom bridges two phosphorus atoms.

For $\mathrm{H_4P_2O_7}$, two $\mathrm{PO_4}$ tetrahedra are joined by one bridging oxygen atom. Therefore, it contains one P–O–P linkage.

For $(\mathrm{HPO_3})_3$, the structure is a cyclic ring of three $\mathrm{PO_3}$ units linked through oxygen atoms. Each linkage between two phosphorus atoms is one P–O–P bond, so the three-membered ring contains three such bonds.

For $\mathrm{P_4O_{10}}$, the cage structure contains bridging oxygen atoms connecting phosphorus atoms. The total number of such P–O–P linkages is six.

Thus, the required numbers are $1, 3, 6$. Therefore, the correct option is D.