Determinar el numero de oxidacion de p en H3PO4, Na3p, (PH4) + , POCI3?

₊₁ ₊ₓ ₋₂ H₃ P O₄

¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨ + 3 + x - 8 = 0 x = + 8 - 3 x = + 5 ( P = + 5 ) ₊₁ ₊ₓ Na₃ P ¨¨¨¨¨¨¨¨¨ + 3 + x = 0 x = - 3 ( P = - 3) ₓ ₊₁ (P H₄)⁺¹ ¨¨¨¨¨¨¨¨¨¨ + x + 4 = + 1 x = - 4 + 1 x = - 3 (P = - 3) ₊ₓ ₋₂ ₋₁ P O Cl₃ ¨¨¨¨¨¨¨¨¨¨¨¨ + x - 2 - 3 = 0 x = + 2 + 3 x = + 5 (P = + 5) ₊ₓ ₋₁ Cl F₃ ¨¨¨¨¨¨¨¨¨¨ + x - 3 = 0 x = + 3 ₊ₓ ₋₂ Cl₂O ¨¨¨¨¨¨¨¨¨¨ + 2x - 2 = 0 X = 2 / 2 = + 1 ₊ₓ ₋₂ Cl₂O₇ ¨¨¨¨¨¨¨¨¨¨ 2x - 14 = 0 x = 14 / 2 = + 7 ₊₁ ₊ₓ H Cl ¨¨¨¨¨¨¨¨ + 1 + x = 0 x = - 1.