Oxidation number examples working

H₃PO₄

Phosphorus is not on our oxidation number rule list, so we set the other two elements to their normal values, H = +1 and O = -2. There is no charge on this molecule (otherwise it would be written at the top right of the formula), so all the oxidation numbers must add up to zero. 3 x +1 (the hydrogen) and 4 x -2 (the oxygen) makes -5, so the phosphorus must be +5.

HSO₄^-

Sulfuris not on our oxidation number rule list, so we set the other two elements to their normal values, H = +1 and O = -2. This ion has a -1 charge, so the oxidation numbers must all add up to give -1. +1 (the hydrogen) and 4 x -2 (the oxygen) makes -7. If we make the sulfur +6, +6 and -7 leaves the -1 charge of the ion.

N₂

This is an element and the oxidation number of all elements is zero. Alternatively, this molecule has no charge. So we are looking for an oxidation number which when multiplied by two (the two nitrogens) gives zero. It's got to be N = 0, as 2 x 0 = 0.

NH₄⁺

Nitrogen is not on our oxidation number rule list, so we set the hydrogen to its usual oxidation number of +1. This ion has a +1 charge, so all the oxidation numbers must add up to +1. The four hydrogens add up to +4 (4 x +1), so we set N = -3 to leave a +1 charge left over.

F₂O

In this molecule both the elements appear in our oxidation number list, but fluorine is always -1 in its compounds. This means that we set the oxidation number of the oxygen to give the correct overall charge. When the oxygen's oxidation number is forced by a more important rule it may not have its normal value of -2. This molecule has no charge. The two fluorines give a charge of -2 (2 x -1) so the oxygen must be +2 (unusual) to give no charge.

Na₂S₄O₆

Sulfuris not on our oxidation number rule list, so we set the other two elements to their normal values, Na = +1 and O = -2. There is no charge on this molecule, so all the oxidation numbers must add up to zero. The sodium and oxygen adds up to -10 (2 x +1 and 6 x -2). This means that the four sulfurs must add up to +10 (to give no overall charge). Each sulfur must have an oxidation number of +10/4 = +2.5. This can also be written as +2½. Fractional oxidation numbers are unusual, but they do appear on the exam.

H₂O₂

In this example both the elements appear on our oxidation number rule list, but the hydrogen is higher up, so we set it to its normal value of +1. The molecule is neutral, so all the oxidation numbers must add up to zero. The two hydrogens add up to +2, so the two oxygens must be -2. Therefore, each oxygen must be -2/2 = -1. -1 is an unusual oxidation number for oxygen.

KH

In this example both the elements appear on our oxidation number rule list, but potassium is always +1. This means that the hydrogen must have an oxidation number of -1 to give no overall charge. This compound is ionic and contains the hydride ion, H^-, as opposed to the more common hydrogen ion, H⁺.

VO²⁺

Vanadium is not on our oxidation number rule list, so we set the oxygen to its normal value of -2. This ion has a 2+ charge, so the oxidation numbers must add up to give +2. This means that the vanadium is +4, with one oxygen at -2 these two values add up to give the +2 charge on the ion.

ICl

This can't be done with the rules provided, but we have already learnt from the displacement reactions that the smaller chlorine atom is better at attracting electrons to itself than the larger iodine atom. This means that it is the chlorine which has the oxidation state of -1. As the molecule is neutral the oxidation numbers must add up to zero. So the iodine must be +1.

NaHCO₃

Carbon is not on our oxidation number rule list, so we set the other three elements to their normal values, Na = +1, H = +1 and O = -2. The compound has no charge so the oxidation numbers must add up to zero. One sodium (+1), one hydrogen (+1) and three oxygens (3 x -2) add up to -4, so the carbon must be +4.

UF₆

This is a nice easy one, straight off the oxidation number rule list, fluorine is always -1. You should be able to see that the uranium must have an oxidation number of +6.

KAuO₂

Gold is not on our oxidation number rule list, so we set the other two elements to their normal values, K = +1 and O = -2. The compound has no charge so the oxidation numbers must add up to zero. One potassium (+1) and two oxygens (2 x -2) add up to -3, so the gold must be +3.

K₄Fe(CN)₆

With iron, carbon and nitrogen not on our oxidation number rule list this could be quite difficult, unless you recognize the cyanide ion, CN^-. This has a charge of -1 so we don't need to worry about the individual oxidation numbers of the carbon and the nitrogen. We now proceed in the same way ensuring that all the charges add up to zero as this compound is uncharged. Four potassiums (4 x +1) and six cyanide ions (6 x -1) add up to -2, so the iron must be +2.

NH₄VO₃

Another difficult one, unless you recognize the ammonium ion, NH4+. Again, we don't worry about the individual oxidation numbers of the nitrogen and hydrogen but treat the ion as a whole. One ammonium ion (+1) and three oxygens (3 x -2) add up to -5. As the compound has no charge the oxidation numbers must add up to give zero, so the vanadium is +5.

Cr₂O₇^2-

Here we have an ion with a 2- charge, so the oxidation numbers must add up to -2. Chromium is not on our oxidation number rule list, so we set the oxygen to its normal oxidation state of -2. Seven oxygens give a value of -14 (7 x -2), so the two chromiums must be +12 to leave the 2- charge on the ion. A common error is to leave it at that, but we need to remember that oxidation number is always the value for a single atom. So +12 needs dividing by two to give the oxidation number of the chromium as +6.