Rules for assigning oxidation states
Twenty Examples

Problems 1-10

Problems 11-30

Exampes and Problems with no answers

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I. Rule Number One

All free, uncombined elements have an oxidation state of zero.

This includes the seven diatomic elements (such as O2) and the other molecular elements (P4 and S8).
Ozone (O3) also has an oxidation state of zero.

II. Rule Number Two

Each hydrogen atom, in all its compounds except hydrides, has an oxidation state of +1 (positive one)

III. Rule Number Three

Each oxygen atom, in all its compounds except peroxides and superoxides, has an oxidation state of −2 (negative two).

IV. Rule Number Four

For single ions (in other words, not polyatomic), the charge on the ion is taken to be the oxidation state.

V. Rule Number Five

When no charge is indicated in a given formula, the total charge is taken to be zero.

VI. Rule Number Six

The oxidation state of fluorine is −1 in all of its compounds.

There are more extensive sets of rules (here is an example) and, for the most part, they derive from the six rules above. Here is another example. You might want to consider searching for redox rules to see how others have approached this topic.

There are compounds in which the oxidation state of a given atom appears to be fractional. There are referred to as "ambigious oxidation states." I will mention four below (from example #13 to example #15 and one in the bonus example). You may refer to the linked Wiki page for more information.

Also, please be aware that "oxidation state" is the term used currently. If you see older materials (or current materials prepared by an older person), you may see the older term "oxidation number."


Example #1: What is the oxidation state of Cl in HCl?

By rule #2, H equals +1. Therefore, the Cl must be −1 (minus one).

See KCl in Example #9. NaCl, BaCl2, AlCl3, etc. are more examples that have Cl equal −1.


Example #2: What is the oxidation state of Na in Na2O?

Since O equals −2 (rule #3), the two Na must each be +1. Notice that we treat each sodium as an individual unit, we do not address it as Na22+.

By the way, mercury in a +1 oxidation state exists, but it is written as follows:

Hg22+

There are resons for this which I will leave unexplained. As a consequence, that means writing the +1 cation like this:

Hg+

is always wrong.


Example #3: What is the oxidation state of Cl in ClO¯?

The O is −2, but since a −1 must be left over, then the Cl is +1.

This could also be asked with HClO, NaClO, KClO, Ba(ClO)2, etc.


Example #4: What is the oxidation state of Cl in ClO2¯?

Two O is −4 (from −2 x 2), but since a −1 must be left over, then the Cl is +3.

This could also be asked with HClO2, NaClO2, KClO2, Ba(ClO2)2, etc.


Example #5: What is the oxidation state of Cl in ClO3¯?

Three O is −6 (from −2 x 3), but since a −1 must be left over, then the Cl is +5.

This could also be asked with HClO3, NaClO3, KClO3, Ba(ClO3)2, etc.


Example #6: What is the oxidation state of Cl in ClO4¯?

Four O is −8 (from −2 x 4), but since a −1 must be left over, then the Cl is +7.

This could also be asked with NaClO4, KClO4, Ba(ClO4)2, etc.

Here's a comment I found associated with this question:

Please don't assume that chlorine has a charge of +7. It's not an "ionic" substance, and even if it was it couldn't possibly have a charge that large. The oxidation state of an element is an assigned number, only coincidentally may it agree with an actual charge, usually when the ion is in solution.

Example #7: What is the oxidation state of S in SO42¯?

O is −2. There are four oxygens for −8 total. Since −2 must be left over, the S must equal +6.

Example #8: What is the oxidation state of S in SO32¯

O is −2. There are three oxygens for −6 total. Since −2 must be left over, the S must equal +4.

Example #9: What are the oxidation states in KCl?

K is +1 because K2O exists. The O in K2O is −2 by defintion, therefore each K must be +1 in order to keep the K2O formula at zero charge.

Cl equals −1 because HCl exists. H is +1 by definition, therefore Cl must be −1. You can also say that Cl is −1 because K must be +1 and we need to have a zero charge for the formula.


Example #10: What is the oxidation state for each element in NaMnO4?

Na is +1 based on the fact that Na2O exists. We know the O = −2, so each Na is a +1. (We could also use this: we know NaCl exists and that Cl is a −1. We know this because HCl exists. Therefore, the Na is a +1.)

O equals −2 by definition

Mn equals +7. There are 4 oxygens for a total of −8, K is +1, so Mn must be the rest.


Example #11: What is the oxidation state of N and of P in NH4H2PO4?

Solution:

1) Let's split ammonium dihydrogen phosphate apart. It forms the following two polyatomic ions:

ammonium ---> NH4+
dihydrogen phosphate ---> H2PO4¯

2) Analysis of the ammonium ion:

We know that NH4+ takes on a +1 charge because we know that NH4Cl exists. Since the chloride ion is −1, therefore ammonium is a +1.

Remember that we know chloride is a −1 because HCl exists and H is defined as a +1 (except in hydrides).

In NH4+, the four hydrogens total to +4. (Ammonium is not a hydride.)

Therefore, the nitrogen is a −3, in order to leave a +1 for the overall charge on the ion.

2) Analysis of the dihydrogen phosphate ion:

Two H are present ---> +2
Four O are present ---> −8

In order for the entire ion to have a −1 charge, the P must be +5.


Example #12: For H2O2, is the oxidation state for O2, −2 or −4?

Solution:

An important skill in this area is being able to recognize a given formula as representing an ionic compound or a molecular one. In the previous example, the oxygens were ionically bonded to the Fe and were treated as individual oxide ions, not as an O4 polyatomic ion.

With H2O2, one needs to recognize it as the formula for hydrogen peroxide and that, in this formula, the O2 is seen to be a legitimate polyatomic ion, named peroxide. It is written O22¯ and the charge on the ion is −2. The oxidation state on each oxygen atom is, therefore, −1.

To me, the student that asked this question is aware of the peroxide rule, but is not sure if the oxygen in H2O2 should be dealt with as a peroxide and as separate atoms, as would be done in an oxide.


Example #13: (fractional oxidation state example) For Fe3O4, is the oxidation state for O4, −2 or −8?

Solution:

Notice the confusion shown by the student., in that he/she asks about the oxidation state of O4. This is improper. Each oxygen atom should be treated individually, not as a group. In the case, the correct answer is that each oxygen atom has an oxidation state of −2.

The fractional oxidation state is associated with the Fe. Note that a +8 (to counterbalance the −8 from the 4 oxygens) is required. That means that each Fe has an average oxidation state of +83.

However, the general rule is that each oxidation state must be an integer, which means that two of the Fe atoms are each +3 and one Fe atom is +2. This adds to +8, which offsets the −8 created by the four O atoms.


Example #14: (fractional oxidation state example) Determine the oxidation state for I in I3¯.

Solution:

The oxidation state of I in I3¯. is NOT −13. Fractional oxidation states do not exist.

However, some teachers may insist on an answer of −13.

This is when you use the term "average oxidation state," as in:

the average oxidation state of I in I3¯. is −13.

The best way to give an answer to this example is:

the oxidation state for the triiodide ion is −1.

Example #15: (fractional oxidation state example) What is the oxidation state of silver in Ag2F (silver subfluoride)?

Solution:

The oxidation state for F is always −1.

Therefore, the two silvers total up to +1.

That means the average oxidation state for each silver is +12

Since fractional oxidation states aren't really a thing, one of the two silvers is assigned an oxidation state of 0 and the other is assigned +1. Here's an example of that.


Example #16: What is the oxidation state of tungsten in tungsten disulfide, WS2?

Solution:

Sulfide is −2. He know that because H2S. Hydrogen is a +1 in this compound and, because there are two H, the S must be a −2 to counterbalance the +2 from the two hydrogens.

Therefore, W must be a +4 in order to counterbalance the 𕒸 total contributed by the two sulfide ions.


Example #17: What is the oxidation state of manganese in LiMnO2?

Solution:

Li is +1 because LiCl exists. We know the Cl to be a −1 because HCl also exists.

We know O to be a −2 because H2O exists. Two of them makes for −4.

The Mn is, therefore +3.


Example #18: Give the oxidation states of the elements in the following compounds.

(a) XeF2    (b) CO    (c) OF2    (d) Bi2O5    (e) MnO2

Solution:

(a) XeF2

F is always −1. Therefore, the Xe is +2.

By the way, XeF4 exists as well and XeF6 as well as a number of other xenon compounds.

(b) CO

O is −2. The carbon, therefore, is +2.

By the way, carbon is a +4 in CO2.

(c) OF2

This is an unusual one.

F is always −1. Since there are two of them, the O is +2.

O2F2 is another unusual one. You can read more about it at its Wikipedia page.

(d) Bi2O5
O is −2 and five of them makes for a total of −10.

Two Bi means each one has an oxidation state of +5.

(e) MnO2

This one would actually need a bit more context. This is because, from the formula alone, we do not know if the oxygen is present as two oxides (total ox. state = −4) or one peroxide (total ox. state = −2).

If oxide, then Mn is +4.

If peroxide, then Mn is +2.

Another example of this is PbO2.


Example #19:

(a) Se in SeO32¯     (b) I in H5IO6     (c) S in Al2(SO3)3     (d) N and C in HCN    (e) Cu in Na3[CuCl5]

Solution:

(a) Se in SeO32¯

Se + 3(−2) = −2

Se + (−6) = −2

Se = +4

(b) I in H5IO6
(5)(+1) + I + (6)(−2) = 0

5 + I + (−12) = 0

I = +7

(c) S in Al2(SO3)3

Al = +3, S = ?, O = −2

(2)(+3) + 3S + (9)(−2) = 0

Note the influence of the subscript outside the parenthesis.

6 + 3S + (−18) = 0

3S = +12

S = +4

(d) N and C in HCN

Since H is +1, we examine CN¯, the cyanide anion.

N is more electronegative than C, so it assumes a negative oxidation state.

Because NH3 exists, we know N will assume a −3 oxidation state.

In order to have a −1 charge on the cyanide anion, the carbon must be +2.

(e) Cu in Na3[CuCl5]

We know Na to be a +1, which means the polyatomic anion must be CuCl53¯

Each Cl contributes −1 for a total of −5.

The oxidation state of the Cu must be +2 in order for the anion to be an overall −3 charge


Example #20: Find the oxidation state of chlorine in:

(a) ClO    (e) ClF5    (i) Cl2O2 (chlorine peroxide)
(b) ClO2    (f) ClOF3    (j) Cl2O4 (chlorine perchlorate)
(c) ClF    (g) Cl2O6    (k) ClN3
(d) ClF3    (h) Cl2O7    (ℓ) Cl2

Solution:

(a) ClO
O = −2, Cl = ?
Cl + (−2) = 0
Cl = +2

(e) ClF5

F = −1, Cl = ?
Cl + (5)(−1) = 0
Cl = +5
(i) Cl2O2 (chlorine peroxide)
peroxide is O22¯
two Cl must total to +2
each Cl equals +1
(b) ClO2
O = −2, Cl = ?
Cl + (2)(−2) = 0
Cl = +4
(f) ClOF3
O = −2, F = −1, Cl = ?
Cl + (−2) + (3)(−1) = 0
Cl = +5
 
(j) Cl2O4 (chlorine perchlorate)
total from O is −8
each Cl is +4
The +4 is actually an average oxidation state.
one Cl is +7, the other is +1
(c) ClF
F = −1, Cl = ?
Cl + (−1) = 0
Cl = +1
 
(g) Cl2O6
O = −2, Cl = ?
2Cl + 6(−2) = 0
2Cl = +12
Cl = +6
(k) ClN3
recognize the anion as azide, N3¯
overall charge is zero
therefore, Cl = +1
 
(d) ClF3
F = −1
Three F = −3
One Cl = +3
 
(h) Cl2O7
O = −2, Cl = ?
2Cl + 7(−2) = 0
2Cl = +14
Cl = +7
(ℓ) Cl2
Cl2 is the element in its uncombined state.
The oxidation state of an uncombined element is 0 (zero).
 
 


Bonus Example: Find the oxidation state of nitrogen in:

(a) NO3¯    (e) N2O    (i) N2O4
(b) N2H4    (f) NO    (j) N2O3
(c) NO2+    (g) NO2¯    (k) N3¯
(d) N2O5    (h) NO2    (ℓ) N2

Solution:

(a) NO3¯
O = −2, N = ?
N + (3)(−2) = −1
N = +5

(e) NO

O = −2, N = ?
N + (−2) = 0
N = +2
 
(i) N2O4
O = −2, N = ?
2N + (4)(−2) = 0
2N = +8
N = +4
(b) N2H4
H = +1, N = ?
2N + (4)(+1) = 0
2N = −4
N = −2
(f) N2O
O = −2, N = ?
2N + (−2) = 0
2N = +2
N = +1
(j) N2O3
O = −2, N = ?
2N + (3)(−2) = 0
2N = +6
N = +3
(c) NO2+
O = −2, N = ?
N + (2)(−2) = +1
N = +5
 
(g) NO2¯
O = −2, N = ?
N + (2)(−2) = −1
N = +3
 
(k) N3¯
(fractional oxidation state example)
Each N has an average oxidation state of −13
It is usually just said that azide has an oxidation state of −1,
without any one particular N being associated with the −1.
(d) N2O5
Each O is −2
Five O is −10
Two N = +10
Each N = +5
(h) NO2
O = −2, N = ?
N + (2)(−2) = 0
N = +4
 
(ℓ) N2
N2 is the element in its uncombined state.
The oxidation state of an uncombined element is 0 (zero).
 
 


Problems 1-10

Problems 11-30

Exampes and Problems with no answers

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