Question

why is the oxidation state of P in H3PO4 +5?

Answer 1

-8 + 3 = -5

Answer 2

I mean
P + H +O = 0
P + 3 -8 = 0
->P = +5

Answer 3

H is usually +1 (for only one electron in orbital)
O is usually -2 (for need in 2 electrons to satisfy the octect rule)
Now you can get it, 1*3 + (-2)*4 = -5 this is the oxidation state of P in H3PO4.
Hope this can help you.

Answer 4

If you dont want to do the math, you can just draw the structure, and believe it, it helps where the mathematical way may not.

See in the drawing, that P forms 5 bonds with O, a highly electronegative element. Each bond pair of electrons is attracted more to O, as a result of which, P gets a partial positive charge for each of the 5 bonds.
Thus, oxidation state of P here is +5

Answer 5

one more way u can by using formula
first mark x of which u have to find oxidation
as here u have to find oxidation of P so take it as x
now u know H has +1 charge on it and oxygen has -2 charge on it
so
here are 3molecules of H and 4 molecules of O so just multiply the molecule by its charge and here netcharge will be zero on the compound so equate it by zero
3(charge on H) + x + 4(charge on O)=compound charge
3(1)+x+4(-2)=0
3+x-8=0
x=+5