Oxidation states of P in H_4 P_2 O_5, H_4 P_2 O_6 , H_4 P_2 O_7 ,...

Key Idea Oxidation state of $\mathrm{H}$ is +1 and that of $\mathrm{O}$ is -2. Let the oxidation state of $\mathrm{P}$ in the given compounds is $\mathrm{x}$. In $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5$, $$\begin{aligned} (+1) \times 4+2 \times x+(-2) \times 5 & =0 \\ 4+2 x-10 & =0 \\ 2 x & =6 \\ \therefore x & =+3 \end{aligned}$$$$\begin{array}{l} \text{In }\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6, \\ (+1) \times 4+2 \times \mathrm{x}+(-2) \times 6 =0 \\ 4+2 \mathrm{x}-12 =0 \\ 2 \mathrm{x} =8 \\ \therefore \mathrm{x} =+4 \end{array}$$ In \mathrm{H}_4\mathrm{P}_2\mathrm{O}_7 $$\begin{aligned} (+1) \times 4+2 \times x+(-2) \times 7 & =0 \\ 4+2 x-14 & =0 \\ 2 x & =10 \\ \mathrm{x} & =+5 \end{aligned}$$$Thus, the oxidation states of$\mathrm{P} in \mathrm{H}_4\mathrm{P}_2\mathrm{O}_5, \mathrm{H}_4 \mathrm{P}_2\mathrm{O}_6$and$\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7a$re$+3,+4$ and +5 respectively.