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Step-by-Step Tutorial to Determine Oxidation Numbers
First, let’s define oxidation number (ON)
Oxidation Number – is a positive (+) or negative (-) number that indicates which indicates how
many electrons an element has gained, lost or shared when bonded to another element.
Second, let’s review the oxidation rules. These will make more sense as we use them, but you
need to have them handy for reference. The ON of:
1. A pure elements is 0 (individual atoms and covalently bonded elements)
2. The ON of an ion is equivalent to the ion's charge
3. Group 1 - Alkali metals = +1
4. Group 2 - Alkaline metals = +2
5. Group 7 - Halogens = -1
6. Hydrogen is +1 when bonded with nonmetals, -1 when bonded with another metal
7. Oxygen is -2 unless it is bonded with florine, -1 in peroxide H2O2
8. Florine is -1
9. For compounds the sum of the ON = 0
10. For polyatomic ions the sum = the charge of the ion
Now let’s go through the steps you can take to determine the ON.
Step 1 – Determine if the substance is elemental – meaning it is only comprised of one element. If
it is then Rule #1 applies and the oxidation number is 0.
Example 1: H2 = 0
In this example two hydrogen atoms have bonded with eachother. No other element has bonded
to it. It is a pure element. Therefore Rule #1 applies and the ON is 0
Example 2: Al = 0
Al represents a single atom of aluminum and is a pure element. Rule #1 applies again and the ON
is 0.
Step 2 – Determine if the substance is an ion – does it have a charge? If it does, then Rule #2
applies and the ON is equal to the charge. (Just be sure to write the ON correctly – ONs are
written with the sign first and then the number. Ionic charges are written with the number first and
then the charge.)
Example 1: Na+
Na = +1
In this example, sodium has a charge of 1+, meaning it has lost an electron and has more protons
than electrons. Therefore Rule #2 applies and the ON is +1.
Example 2: N3N = -3
Nitrogen has a charge of 3-. It has gained 3 electrons and has more electrons than protons. Rule
#2 applies and the ON is -3.
Step 3 – Apply Rules #3 - 8 as appropriate
Example 1: KCl
K=+1 Cl=-1
In this example there are two elements and there is no charge to the compound, so Rules #1 and
#2 do not apply.
Potassium (K) is an alkali metal – it is in Group 1 of the periodic table. Therefore, it has a +1 ON
according to rule #3.
Chlorine (Cl) is a halogen – it is in Group 7 of the periodic table. Therefore it has a -1 ON
according to rule #5.
Example 2: H2O
H=+1 O=-2
Two atoms of hydrogen are bonded with one atom of oxygen.
Rule #6 tells us that hydrogen has an ON of +1 when bonded with non-metals and -1 when bonded
with a metal. Oxygen is a non-metal, so in this compound the hydrogen atoms have an ON of +1.
Rule #7 tells us that oxygen has an ON of -2 unless it is bonded with florine, which it isn’t. So, the
O
Example 2: SnO
Here, tin is bonded with oxygen. It is not a single element and it has no charge, so Rules #1 and
#2 do not apply.
Let’s look at oxygen first. According to Rule #7, oxygen has an ON of -2, unless bonded with
florine. In this case it is bonded with tin, so its ON is -2.
None of the other rules help us determine the ON for Sn, so let’s continue on to Rule #9…
Step 4 – Balance the ONs of all the elements, so the sum is 0. (Rule #9) To use this rule, we will
need to do some math.
Example 1: Let’s go back to our previous example SnO
We know that oxygen has a ON of -2, but we don’t have any rules to tell us directly the ON for Sn.
However, the sum of all ONs for a compound must equal 0.
Sn
+
O
=
0
Sn
+
-2
=
0
Sn
(+2) =
0
(+2) bring the -2 to the other side
Sn
=
2
So, the ON of tin is +2
Step 4 (continued) – when there are more than one atoms of an element in a molecule, you need
to multiply the oxidation number of that element by the number of atoms.
Example 2: H2CO4
H= +1
C = +6
O= -2
This molecule has subscripts which indicate that there are two hydrogen atoms and four oxygen
atoms. So we will need to multiply the number of atoms by the oxidation number of the element.
Using Rules #6 and #7, we know that the oxidation number for H is +1 and O is -2.
2(H)
+
C
+
4(O) =
0
2(+1)
+
C
+
4(-2) =
0
2
+
C
+
(-8)
=
0
C
+
(-6)
=
0 add 2 and -8
C
=
6 bring -6 to the other side
Step 5 – When the compound contains a polyatomic ion – meaning several elements that act as a
single unit - you can just use the charge of the ion to balance the oxidation numbers. (Use your
polyatomic ion sheet to identify common ones.)
Example 1: K3PO4
Phosphate (PO4) is a polyatomic ion with a charge of 3-. Using Rule #10, we know that its ON is
-3. To find the ON for potassium (K) we use Rule #9.
3(K)
+
(PO4)
=
0
3(K)
+
(-3)
=
0
3(K)
=
3 bring -3 to the other side
(K)
=
1 divide each side by 3 (# of K atoms)