Oxidation states of $P$ in $H_{4} P_{2} O_{6}, H_{4} P_{2} O_{6}, H_{4} P_{2} O_{7}$ are respectively:

+3, +5, +4

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+5, +3, +4

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+5, +4, +3

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+3, +4, +5

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Solution

The correct option is D +3, +4, +5
$H_{4} P_{2} O_{5} : 4 (1) + 2 x + 5 (- 2) = 0$
$4 + 2 x - 10 = 0$
$x = + 3$
$H_{4} P_{2} O_{6} : 4 (1) + 2 x + 6 (- 2) = 0$
$x = + 4$
$H_{4} P_{2} O_{7} : 4 (1) + 2 x - 14 = 0$
$x = + 5$

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