How to draw the structure and find the oxidation number of Phosphorous in H2P2O7 [closed]

Find the Oxidation number of Phosphorus in $$\ce{H2P2O7^2-}$$.

So I thought of drawing the lewis structure to find the oxidation numbers. Because I wasn't sure if Oxygen was bonded to a Hydrogen or not. Then I realised that the two Hydrogens could be ionised, so I'd be left with $$\ce{P2O7^4-}$$.

But then when I tried drawing this Lewis structure I got really confused, As I tried to draw it with the two Phosphorus as central atoms and I couldn't make it. (I know I can just use general rules to find oxidation numbers but I feel like lewis structures often give more accurate answers). So could you please tell me how to draw the structure and more importantly guide me through the process. Any help is appreciated, thanks!

• Please use $\LaTeX$ and also see How to ask a good question – Aniruddha Deb Jun 18 at 7:38
• You have O and P, both in their most common valence states. You start connecting them until there is nothing more to connect. What could possibly go wrong? – Ivan Neretin Jun 18 at 9:18
• Take 2 dihydrogenphosphate anions and condense 2 OH groups ( 1 from each ion ) into -O- bridge, eliminating water. – Poutnik Jun 18 at 13:36

You have realized your parent compound is $$\ce{H4P2O7}$$ (adding two $$\ce{H+}$$). Correct? If so, concentrate on following hypothetical reaction:
$$\ce{2H3PO4 -> H4P2O7 + H2O} \tag1$$
I believe you know what $$\ce{H3PO4}$$ is looks like. Now, draw two $$\ce{H3PO4}$$ side by side. Choose two closest $$\ce{OH}$$ groups (that would be $$\ce{-P-\color{red}{OH} \space \space \color{tequish}{HO}-P -}$$). Then, remove one $$\ce{\color{red}{-OH}}$$ and the $$\ce{\color{tequish}{-H}}$$ from the second $$\ce{\color{tequish}{-OH}}$$ you chose to eliminate a $$\ce{H2O}$$ molecule. Finally combine two remaining molecules together such that final structure forms $$\ce{-P\color{red}{-}\color{tequish}{O}-P -}$$ bond. This is your $$\ce{H4P2O7}$$. Eliminate any two hydrogens to get $$\ce{H2P2O7^2-}$$ structure.
Note: Since above hypothetical equation $$(1)$$ does not require oxidation or reduction to form $$\ce{H4P2O7}$$, it is safe to assume that oxidation state of $$\ce{P}$$ in both $$\ce{H3PO4}$$ and $$\ce{H4P2O7}$$ is same.