The number of P $$-$$ OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H$$_4$$P$$_2$$O$$_7$$), respectively are:

We start by recalling that pyrophosphoric acid, whose molecular formula is $$H_4P_2O_7$$, is obtained by the condensation of two molecules of orthophosphoric acid $$H_3PO_4$$ with the elimination of one molecule of water:

$$2\,H_3PO_4 \;\rightarrow\; H_4P_2O_7 \;+\; H_2O$$

Each orthophosphoric acid molecule possesses a tetrahedral $$PO_4$$ unit in which the phosphorus is linked to one $$P=O$$ bond and three $$P\!-\!OH$$ bonds. When two such units condense, one $$OH$$ from the first unit and one $$H$$ from the second unit combine to give $$H_2O$$, and the remaining oxygen from the first unit forms a bridge between the two phosphorus atoms. After this condensation the structure of $$H_4P_2O_7$$ can be written schematically as

$$HO{-}P(=O)(OH){-}O{-}P(=O)(OH){-}OH$$

We see clearly that each phosphorus atom now carries two $$P\!-\!OH$$ groups. Because there are two phosphorus atoms, the total number of $$P\!-\!OH$$ bonds in one molecule of pyrophosphoric acid is

$$2 \text{ (from first }P) + 2 \text{ (from second }P) = 4$$

Now we determine the oxidation state of phosphorus in $$H_4P_2O_7$$. Let the oxidation state of each phosphorus atom be $$x$$. The rules we shall employ are:

• The oxidation number of hydrogen is $$+1$$.
• The oxidation number of oxygen is $$-2$$.
• The algebraic sum of oxidation numbers of all atoms in a neutral molecule is zero.

Applying these rules to $$H_4P_2O_7$$, we write

$$4(+1) + 2x + 7(-2) = 0$$

Simplifying step by step, we have

$$+4 + 2x - 14 = 0$$

$$2x - 10 = 0$$

$$2x = 10$$

$$x = +5$$

Thus each phosphorus atom is in the $$+5$$ oxidation state.

We have found that pyrophosphoric acid contains four $$P\!-\!OH$$ bonds and that the oxidation state of phosphorus is $$+5$$.

Hence, the correct answer is Option B.