The pair in which phosphorous atoms have a formal oxidation state of +3 is:

To decide which pair contains phosphorus atoms with a formal oxidation state of $$+3$$, we first recall the definition: the algebraic sum of oxidation states of all atoms in a neutral molecule is zero. Hydrogen in acids is taken as $$+1$$ and oxygen as $$-2$$. We now examine every acid that appears in the four options, writing and solving the oxidation-state equation each time.

We have orthophosphorous acid, whose molecular formula is $$H_3PO_3$$. Writing the oxidation-state balance,

$$3(+1) + (\,\text{O.S. of }P\,) + 3(-2) = 0.$$

So $$3 + x - 6 = 0 \Longrightarrow x = +3.$$ Hence in $$H_3PO_3$$ phosphorus is in the $$+3$$ state.

Next comes hypophosphoric (also written hypophospheric) acid, formula $$H_4P_2O_6$$. The balance is

$$4(+1) + 2x + 6(-2) = 0.$$

This gives $$4 + 2x - 12 = 0 \Longrightarrow 2x = 8 \Longrightarrow x = +4.$$ Therefore here phosphorus is $$+4$$; not $$+3$$.

Now consider pyrophosphorous acid, whose formula is $$H_4P_2O_5$$. Setting up the equation,

$$4(+1) + 2x + 5(-2) = 0.$$

We obtain $$4 + 2x - 10 = 0 \Longrightarrow 2x = 6 \Longrightarrow x = +3.$$ Each phosphorus atom in $$H_4P_2O_5$$ is therefore in the $$+3$$ state.

Pyrophosphoric acid has the formula $$H_4P_2O_7$$. Writing its balance,

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

Thus $$4 + 2x - 14 = 0 \Longrightarrow 2x = 10 \Longrightarrow x = +5.$$ Phosphorus is $$+5$$ in this acid.

Summarising the results:

$$H_3PO_3$$ (orthophosphorous)   $$\rightarrow$$  $$+3$$

$$H_4P_2O_5$$ (pyrophosphorous)   $$\rightarrow$$  $$+3$$

$$H_4P_2O_6$$ (hypophosphoric)   $$\rightarrow$$  $$+4$$

$$H_4P_2O_7$$ (pyrophosphoric)   $$\rightarrow$$  $$+5$$

Only the pair “orthophosphorous acid and pyrophosphorous acid” gives phosphorus in the same oxidation state, namely $$+3$$.

Option C lists exactly this pair. Hence, the correct answer is Option C.