Use standard oxidation numbers

For these neutral molecules:

• Hydrogen has oxidation state $+1$
• Oxygen has oxidation state $-2$
• Let the oxidation state of each phosphorus atom be $x$ on average

Since the molecule is neutral, the sum of oxidation states is zero.

For $H_4P_2O_7$

Let oxidation state of P be $x$.

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

$4 + 2x - 14 = 0$

$2x - 10 = 0$

$2x = 10$

$x = 5$

So, in $H_4P_2O_7$, phosphorus has oxidation state $+5$.

For $H_4P_2O_5$

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

$4 + 2x - 10 = 0$

$2x - 6 = 0$

$2x = 6$

$x = 3$

So, in $H_4P_2O_5$, phosphorus has oxidation state $+3$.

For $H_4P_2O_6$

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

$4 + 2x - 12 = 0$

$2x - 8 = 0$

$2x = 8$

$x = 4$

So, in $H_4P_2O_6$, phosphorus has oxidation state $+4$.

Final sequence

The oxidation states are:

5,3,4

This matches Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

Hence, they agree.