Question #6d6d4

Oct 31, 2015

$P \left(V\right)$; the maximum oxidation number.

Explanation:

We have ${H}_{2} P {O}_{4}^{-}$. The sum of the oxidation numbers must give the charge on the ion, which is $- 1$. Oxygen generally has an oxidation state of $- I I$ 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, ${H}_{3} P {O}_{4}$? Why or why not?

Oct 31, 2015

+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 ${H}_{2} O$ 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 $N {O}_{3}^{\text{-1}}$ ion is -1).

(3) All elements from Group 1A has an oxidation state of +1 (e.g. $N {a}^{\text{+1}}$, $L {i}^{\text{+1}}$). All Group 2A and 3A elements have an oxidation state of +2 and +3, respectively. (e.g. $C {a}^{\text{2+}}$, $M {g}^{\text{2+}}$, $A {l}^{\text{3+}}$)

(4) Oxygen always have a charge -2 except for peroxide ion (${O}_{2}^{\text{2-}}$) 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 $H C l$) and always have a -1 charge if it is bonded with a metal (as in $A l {H}_{3}$).

So let's try solving your problem, the oxidation state of $P$ in the substance $K {H}_{2} P {O}_{4}$.

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

$K {H}_{2} P {O}_{4}$ = 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)] + $P {O}_{4}$ = 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)] + $\textcolor{red}{x}$ + [(-2) (4)] = 0 where $\textcolor{red}{x}$ 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)] + $\textcolor{red}{x}$ + [(-2) (4)] = 0

(+1) + (+2) + $\textcolor{red}{x}$ + (-8) = 0

(+3) + $\textcolor{red}{x}$ + (-8) = 0

$\textcolor{red}{x}$ + (-5) = 0

$\textcolor{red}{x}$ = +5

Therefore, the oxidation state of $P$ atom is +5 or ${P}^{\text{5+}}$.