Figure 2-110. Determination of the oxidation state of O atoms in H2O2.
The oxidation state of -1 for oxygen is rare. The most common oxidation state of oxygen is -2 because in most of its compounds, oxygen forms covalent bonds to atoms of lower electronegativity.
Many texts and chemical education websites such as this one
provide a set of rules for determination of the oxidation state of an atom in a compound. Although everyone is free to memorize and use rules, I strongly believe in the benefit of understanding and logical thinking over straight memory work. Now that we are aware of the idea behind the oxidation state concept, let us quickly go over the rules and make some comments.
1. "The oxidation number of a free element is always 0."
(Here the author means a simple substance, one of those that are composed of just one element, such as O2
, etc.) Yes, it is 0 (zero) because all shared electron pairs in a molecule of a simple substance are shared equally by identical atoms.
2. "The oxidation number of a monatomic ion equals the charge of the ion."
A monoatomic ion does not have shared electron pairs. There are neither shared electrons nor other atoms to shift them to.
3. "The oxidation number of H is +1, but it is -1 in [sic] when combined with less electronegative elements."
Hydrogen conventionally forms covalent bonds with nonmetals. As hydrogen is less electronegative than other nonmetals, a covalent bond between an H atom and an atom of another nonmetal is polarized away from the H, meaning the oxidation state of +1 for the H. Compounds where a hydrogen atom is bonded to a metal, called metal hydrides
, are well-known, albeit less frequently encountered. Since most metals are less electronegative than hydrogen, the oxidation state of the H atom in a metal hydride is -1. The oxidation state of H in H2
is obviously zero.
4. "The oxidation number of O in compounds is usually -2, but it is -1 in peroxides."
Not only do we know that, but we also understand why the oxidation state of oxygen is -1 in peroxides, such as H2
(Figure 2-110). In O2
, the oxidation state of O is zero. In all other oxygen compounds, the oxidation state of oxygen is invariably -2, except
for those featuring an O-F bond. Fluorine is more electronegative than oxygen, and is, in fact, the only element that is more electronegative than oxygen. For example, in oxygen difluoride (OF2
), the oxidation state of the oxygen atom is +2.
5. "The oxidation number of a Group 1 element in a compound is +1. The oxidation number of a Group 2 element in a compound is +2."
The alkali (Group 1) and alkaline-earth (Group 2) metals are the least electronegative elements that willingly donate their one (Group 1) or two (Group 2) valence electrons. Hence the indicated oxidation states.
6. "The oxidation number of a Group 17 element in a binary compound is -1."
(A binary compound is a chemical compound composed of only two elements.) That is true but not always. All halogen atoms need one electron to attain the stable octet configuration of the outermost shell. This and the high electronegativity of the halogens explain the oxidation state of -1. However, there are binary halogen compounds where the oxidation state of a halogen is other than -1. For instance, the oxidation state of iodine in ICl is +1, of Cl in Cl2
is +7, of Br in BrF3
is +3, and of I in I2
is +5. These and many more examples, which I can provide, are yet another piece of evidence for the benefit of understanding rules rather than simply memorizing them.
7. "The sum of the oxidation numbers of all of the atoms in a neutral compound is 0."
This is self-explanatory if you understand the oxidation state idea rather than memorize the rules. When, in our mind, we shift the shared electron pairs in a molecule (molecules are always neutral) to more electronegative atoms, electrons are neither added nor taken away.
8. "The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion."
Same as item 7 above. No matter how we move electrons around within an ion, the overall charge remains the same.
Although oxidation states are hypothetical
numbers, they are broadly used in chemistry for a variety of purposes, including the balancing of redox reaction equations. We will learn that in the next subsection. 2.9.2. Exercises.
1. Determine the oxidation states of all elements in the following molecules and ions: (a) SO2
; (b) K+
; (c) F2
; (d) F-
; (e) H2
; (f) HNO3
; (g) Na2
; (h) KHSO4
; (i) P4
; (j) KMnO4
; (k) MnO2
; (l) CuI; (m) CO2
; (n) CO; (o) CuSO4
; (p) K2
; (q) CrO3
; (r) CrO42-
; (s) KClO3
; (t) Ca2+
; (u) SO32-
; (v) Fe2
; (w) FeO; (x) Fe3
; (y) Li2
S; (z) HOCl (structure: H-O-Cl). (I would give you more but there are only 26 letters in the English alphabet.) Answer