Question

Question #6d6d4

Answer 1

`P(V)`; the maximum oxidation number.

Explanation:

We have `H_2PO_4^-`. The sum of the oxidation numbers must give the charge on the ion, which is `-1`. Oxygen generally has an oxidation state of `-II` in its compounds, and it does here; hydrogen generally has an oxidation state of `+1`. Is the answer correct? Don't assume I have done my arithmetic correctly; I have trouble tying my shoelaces!

Would the answer change if I had the parent acid, phosphoric acid, `H_3PO_4`? Why or why not?

Answer 2

+5

Explanation:

Okay, with determining the oxidation states there are some rules. This would help you to not get confuse.

(1) The total charge of a stable compound is always equal to zero (meaning no charge).

For example, the `H_2O` molecule exists as a neutrally charged substance.

(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the ion is the charge (i.e. oxidation state of `NO_3^"-1"` ion is -1).

(3) All elements from Group 1A has an oxidation state of +1 (e.g. `Na^"+1"`, `Li^"+1"`). All Group 2A and 3A elements have an oxidation state of +2 and +3, respectively. (e.g. `Ca^"2+"`, `Mg^"2+"`, `Al^"3+"`)

(4) Oxygen always have a charge -2 except for peroxide ion (`O_2^"2-"`) which has a charge of -1.

(5) Hydrogen always have a charge of +1 if it is bonded with a non-metal (as in the case of `HCl`) and always have a -1 charge if it is bonded with a metal (as in `AlH_3`).

So let's try solving your problem, the oxidation state of `P` in the substance `KH_2PO_4`.

Based on rule 1, the whole substance has an oxidation state of zero.

`KH_2PO_4` = 0

Based on rules 3 and 5, the oxidation states of `K` and `H` atoms are both +1 (since there no metal atom in this substance).

(+1) + [(+1)(2)] + `PO_4` = 0

Notice that since `H` atoms have a subscript, I multiply its oxidation state by 2.

Next, based on rule 4 the `O` atom has a charge of -2.

(+1) + [(+1)(2)] + `color (red) x` + [(-2) (4)] = 0 where `color (red) x` is the oxidation state of `P` atom

Notice again that since `O` atom has a subscript, I multiplied the oxidation state by 4. Now, we are ready to solve for `x`.

(+1) + [(+1)(2)] + `color (red) x` + [(-2) (4)] = 0

(+1) + (+2) + `color (red) x` + (-8) = 0

(+3) + `color (red) x` + (-8) = 0

`color (red) x` + (-5) = 0

`color (red) x` = +5

Therefore, the oxidation state of `P` atom is +5 or `P^"5+"`.