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QUESTION

# What is the oxidation number of phosphorus in KH2PO4?

+5

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 from Group 1A has an oxidation state of +1 (e.g. Na^"+1", Li^"+1"). All Group 2A and 3A 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+".

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