Sir, last question in the video (Nov 9,'18)
H₂P₂O₇
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 H₂P₂O₇, the overall charge in this molecule is -2. The structure is given as,

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 (H₂P₂O₇)^2- is +5

Regards,