The oxidation states of P in H_{4}P_{2}O_{7}, | vedclass

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(A) To find the oxidation state of $P$ (let it be $x$): For $H_{4}P_{2}O_{7}$: $4(+1) + 2x + 7(-2) = 0 \implies 4 + 2x - 14 = 0 \implies 2x = 10 \impl

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$5, 3$ and $4$

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(A) To find the oxidation state of $P$ (let it be $x$): For $H_{4}P_{2}O_{7}$: $4(+1) + 2x + 7(-2) = 0 \implies 4 + 2x - 14 = 0 \implies 2x = 10 \implies x = +5$. For $H_{4}P_{2}O_{5}$: $4(+1) + 2x + 5(-2) = 0 \implies 4 + 2x - 10 = 0 \implies 2x = 6 \implies x = +3$. For $H_{4}P_{2}O_{6}$: $4(+1) + 2x + 6(-2) = 0 \implies 4 + 2x - 12 = 0 \implies 2x = 8 \implies x = +4$. Thus,the oxidation states are $5, 3$ and $4$ respectively.

The oxidation states of $P$ in $H_{4}P_{2}O_{7}$,$H_{4}P_{2}O_{5}$ and $H_{4}P_{2}O_{6}$,respectively,are:

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