Oxidation states of P in $H_{4} P_{2} O_{5}, H_{4} P_{2} O_{6}, H_{4} P_{2} O_{7},$ 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
$H_{4} P_{2} O_{5}$ $H_{4} P_{2} O_{6}$ $H_{4} P_{2} O_{7}$

$4 + 2 x - 10 = 0$ $4 + 2 x - 12 = 0$ $4 + 2 x - 14 = 0$
$2 x = 6$ $2 x = 8$ $2 x = 10$
$x = + 3$ $x = + 4$ $x = + 5$

Hence, the correct option is $C$

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