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Sir, last question in the video (Nov 9,'18)

H_{2}P_{2}O_{7}

In all cases P = 5+ (max. value for oxidation state)

H is always 1+ (except in hydrides = -1)

So in this case Oxidation Number of O = *-12/7*[2+10+7x = 0 so 7x = -12 therefore x = -12/7]

AM I RIGHT, SIR? DID I FOLLOW THE RULES OF OXIDATION STATE/NUMBER

**Dear Student,**

Check your process from below answer

**Oxidation number of P in H**the overall charge in this molecule is -2. The structure is given as,

_{2}P_{2}O_{7},Let the oxidation number of P be 'x'

The equation is,

2(oxidation number of H) + 2(oxidation number of P) + 7(oxidation number of O) = -2

2(1) + 2(x) + 7(-2) = -2

2 + 2x -14 = -2

x = +5

Hence, oxidation number of P in (H

_{2}P

_{2}O

_{7})

^{2- }is +5

Regards,

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