Editing Oxidation state (section)

== Determination ==
While introductory levels of chemistry teaching use [[postulate]]d oxidation states, the IUPAC recommendation<ref name="10.1515/pac-2015-1204" /> and the ''Gold Book'' entry<ref name="goldbookoxstate" /> list [[#Algorithm of assigning bonds|two entirely general algorithms for the calculation of the oxidation states]] of elements in chemical compounds.

=== Simple approach without bonding considerations ===
Introductory chemistry uses postulates: the oxidation state for an element in a chemical formula is calculated from the overall charge and postulated oxidation states for all the other atoms.

A simple example is based on two postulates,

# OS = +1 for [[hydrogen]]
# OS = −2 for [[oxygen]]

where OS stands for oxidation state. This approach yields correct oxidation states in oxides and hydroxides of any single element, and in acids such as [[sulfuric acid]] ({{chem2|H2SO4}}) or [[dichromic acid]] ({{chem2|H2Cr2O7}}). Its coverage can be extended either by a list of exceptions or by assigning priority to the postulates. The latter works for [[hydrogen peroxide]] ({{chem2|H2O2}}) where the priority of rule 1 leaves both oxygens with oxidation state −1.

Additional postulates and their ranking may expand the range of compounds to fit a textbook's scope. As an example, one postulatory algorithm from many possible; in a sequence of decreasing priority:

# An element in a free form has OS = 0.
# In a compound or ion, the sum of the oxidation states equals the total charge of the compound or ion.
# [[Fluorine]] in compounds has OS = −1; this extends to [[chlorine]] and [[bromine]] only when not bonded to a lighter halogen, oxygen or nitrogen.
# [[Group 1 element|Group 1]] and [[Group 2 element|group 2]] metals in compounds have OS = +1 and +2, respectively.
# Hydrogen has OS = +1 but adopts −1 when bonded as a [[hydride]] to metals or metalloids.
# Oxygen in compounds has OS = −2 but only when not bonded to oxygen (e.g. in peroxides) or fluorine.

This set of postulates covers oxidation states of fluorides, chlorides, bromides, oxides, hydroxides, and hydrides of any single element. It covers all [[oxoacids]] of any central atom (and all their fluoro-, chloro-, and bromo-relatives), as well as [[salt (chemistry)|salts]] of such acids with group 1 and 2 metals. It also covers [[iodide]]s, [[sulfide]]s, and similar simple salts of these metals.

=== Algorithm of assigning bonds ===
This algorithm is performed on a [[Lewis structure]] (a diagram that shows all [[valence electron]]s). Oxidation state equals the charge of an atom after each of its [[heteronuclear]] bonds has been assigned to the more [[Electronegativity#Methods of calculation|electronegative]] partner of the bond ([[#The algorithm's caveat|except when that partner is a reversibly bonded Lewis-acid ligand]]) and [[homonuclear]] bonds have been divided equally:

:[[File:1oxstate.svg|frameless|240px]]

where each "—" represents an electron pair (either shared between two atoms or solely on one atom), and "OS" is the oxidation state as a numerical variable.

After the electrons have been assigned according to the vertical red lines on the formula, the total number of valence electrons that now "belong" to each atom is subtracted from the number {{mvar|N}} of valence electrons of the neutral atom (such as 5 for nitrogen in [[Pnictogen|group 15]]) to yield that atom's oxidation state.

This example shows the importance of describing the bonding. Its summary formula, {{chem2|HNO3}}, corresponds to two [[structural isomer]]s; the [[peroxynitrous acid]] in the above figure and the more stable [[nitric acid]]. With the formula {{chem2|HNO3}}, the [[#Simple approach without bonding considerations|simple approach without bonding considerations]] yields −2 for all three oxygens and +5 for nitrogen, which is correct for nitric acid. For the peroxynitrous acid, however, both oxygens in the O–O bond have OS = −1, and the nitrogen has OS = +3, which requires a structure to understand.

[[Organic compound]]s are treated in a similar manner; exemplified here on [[functional group]]s occurring in between [[methane]] ({{chem2|CH4}}) and [[carbon dioxide]] ({{chem2|CO2}}):

:[[File:3oxstate.svg|frameless|500px]]

Analogously for [[transition-metal]] compounds; {{chem2|CrO(O2)2}} on the left has a total of 36 valence electrons (18 pairs to be distributed), and [[hexacarbonylchromium]] ({{chem2|Cr(CO)6}}) on the right has 66 valence electrons (33 pairs):

:[[File:2oxstate.svg|frameless|380px]]

A key step is drawing the Lewis structure of the molecule (neutral, cationic, anionic): Atom symbols are arranged so that pairs of atoms can be joined by single two-electron bonds as in the molecule (a sort of "skeletal" structure), and the remaining valence electrons are distributed such that sp atoms obtain an [[octet rule|octet]] (duet for hydrogen) with a priority that increases in proportion with electronegativity. In some cases, this leads to alternative formulae that differ in bond orders (the full set of which is called the [[Resonance (chemistry)|resonance formulas]]). Consider the [[sulfate]] anion ({{chem2|SO4(2-)}}) with 32 valence electrons; 24 from oxygens, 6 from sulfur, 2 of the anion charge obtained from the implied cation. The [[bond order]]s to the terminal oxygens do not affect the oxidation state so long as the oxygens have octets. Already the skeletal structure, top left, yields the correct oxidation states, as does the Lewis structure, top right (one of the resonance formulas):

:[[File:7oxstate.svg|frameless|450px]]

The bond-order formula at the bottom is closest to the reality of four equivalent oxygens each having a total bond order of 2. That total includes the bond of order {{sfrac|1|2}} to the implied cation and follows the 8&nbsp;−&nbsp;''N'' rule<ref name="10.1515/pac-2013-0505" /> requiring that the main-group atom's bond-order total equals 8&nbsp;−&nbsp;''N'' valence electrons of the neutral atom, enforced with a priority that proportionately increases with electronegativity.

This algorithm works equally for molecular cations composed of several atoms. An example is the [[ammonium]] cation of 8 valence electrons (5 from nitrogen, 4 from hydrogens, minus 1 electron for the cation's positive charge):

:[[File:5oxstate.svg|frameless|240px]]

Drawing Lewis structures with electron pairs as dashes emphasizes the essential equivalence of bond pairs and lone pairs when counting electrons and moving bonds onto atoms. Structures drawn with electron dot pairs are of course identical in every way:

:[[File:4oxstate.svg|frameless|200px]]

==== The algorithm's caveat ====
The algorithm contains a caveat, which concerns rare cases of [[transition-metal]] [[coordination complex|complexes]] with a type of [[ligand]] that is reversibly bonded as a [[Lewis acid]] (as an acceptor of the electron pair from the transition metal); termed a "Z-type" ligand in Green's [[covalent bond classification method]]. The caveat originates from the simplifying use of electronegativity instead of the [[molecular orbital|MO]]-based electron allegiance to decide the ionic sign.<ref name="10.1515/pac-2015-1204" /> One early example is the {{chem2|O2S\sRhCl(CO)([[triphenylphosphine|PPh3]])2}} complex<ref>{{cite journal|first1=K. W.|last1=Muir|first2=J. A.|last2=Ibers|title=The structure of chlorocarbonyl(sulfur dioxide)bis(triphenylphosphine)rhodium, (RhCl(CO)(SO2)(P(C6H5)3 2)|journal=Inorg. Chem.|volume=8|date=1969|issue=9|pages=1921–1928|doi=10.1021/ic50079a024}}</ref> with [[sulfur dioxide]] ({{chem2|SO2}}) as the reversibly-bonded acceptor ligand (released upon heating). The Rh−S bond is therefore extrapolated ionic against Allen electronegativities of [[rhodium]] and sulfur, yielding oxidation state +1 for rhodium:

:[[File:8oxstate.svg|frameless|450px]]

=== Algorithm of summing bond orders ===
This algorithm works on Lewis structures and bond graphs of extended (non-molecular) solids:
{{blockquote|Oxidation state is obtained by summing the heteronuclear-bond orders at the atom as positive if that atom is the electropositive partner in a particular bond and as negative if not, and the atom’s formal charge (if any) is added to that sum. The same caveat as above applies.}}

==== Applied to a Lewis structure ====
An example of a Lewis structure with no formal charge,
:[[File:9oxstate.svg|frameless|240px]]
illustrates that, in this algorithm, homonuclear bonds are simply ignored (the bond orders are in blue).

Carbon monoxide exemplifies a Lewis structure with [[formal charges]]:

:[[File:10oxstate.svg|frameless|240px]]

To obtain the oxidation states, the formal charges are summed with the bond-order value taken positively at the carbon and negatively at the oxygen.

Applied to molecular ions, this algorithm considers the actual location of the formal (ionic) charge, as drawn in the Lewis structure. As an example, summing bond orders in the [[ammonium]] cation yields −4 at the nitrogen of formal charge +1, with the two numbers adding to the oxidation state of −3:

:[[File:11oxstate.svg|frameless|240px]]

The sum of oxidation states in the ion equals its charge (as it equals zero for a neutral molecule).

Also in anions, the formal (ionic) charges have to be considered when nonzero. For sulfate this is exemplified with the skeletal or Lewis structures (top), compared with the bond-order formula of all oxygens equivalent and fulfilling the octet and 8&nbsp;−&nbsp;''N'' rules (bottom):

:[[File:13oxstate.svg|frameless|450px]]

==== Applied to bond graph ====
A [[bond graph]] in [[solid-state chemistry]] is a chemical formula of an extended structure, in which direct bonding connectivities are shown. An example is the {{chem2|AuORb3}} [[perovskite]], the unit cell of which is drawn on the left and the bond graph (with added numerical values) on the right:

:[[File:14oxstate.svg|frameless|360px]]

We see that the oxygen atom bonds to the six nearest [[rubidium]] cations, each of which has 4 bonds to the [[auride]] anion. The bond graph summarizes these connectivities. The bond orders (also called [[bond valence]]s) sum up to oxidation states according to the attached sign of the bond's ionic approximation (there are no formal charges in bond graphs).

Determination of oxidation states from a bond graph can be illustrated on [[ilmenite]], {{chem2|FeTiO3}}. We may ask whether the mineral contains {{chem2|Fe(2+)}} and {{chem2|Ti(4+)}}, or {{chem2|Fe(3+)}} and {{chem2|Ti(3+)}}. Its crystal structure has each metal atom bonded to six oxygens and each of the equivalent oxygens to two [[iron]]s and two [[titanium]]s, as in the bond graph below. Experimental data show that three metal-oxygen bonds in the octahedron are short and three are long (the metals are off-center). The bond orders (valences), obtained from the bond lengths by the [[bond valence method]], sum up to 2.01 at Fe and 3.99 at Ti; which can be rounded off to oxidation states +2 and +4, respectively:

:[[File:15oxstate.svg|frameless|200px]]

=== Balancing redox ===
Oxidation states can be useful for balancing chemical equations for oxidation-reduction (or [[redox]]) reactions, because the changes in the oxidized atoms have to be balanced by the changes in the reduced atoms. For example, in the reaction of [[acetaldehyde]] with [[Tollens' reagent]] to form [[acetic acid]] (shown below), the [[carbonyl]] carbon atom changes its oxidation state from +1 to +3 (loses two electrons). This oxidation is balanced by reducing two {{chem2|Ag+}} cations to {{chem2|Ag^{0} }} (gaining two electrons in total).

:[[File:Redox eqn 1.svg|600px]]

An inorganic example is the Bettendorf reaction using [[tin dichloride]] ({{chem2|SnCl2}}) to prove the presence of [[arsenite]] ions in a concentrated [[hydrochloric acid|HCl]] extract. When arsenic(III) is present, a brown coloration appears forming a dark precipitate of [[arsenic]], according to the following simplified reaction:
:{{chem2|2 As^{3+} + 3 Sn^{2+} -> 2 As^{0} + 3 Sn^{4+} }}
Here three [[tin]] atoms are oxidized from oxidation state +2 to +4, yielding six electrons that reduce two arsenic atoms from oxidation state +3 to 0. The simple one-line balancing goes as follows: the two redox couples are written down as they react;
:{{chem2|As^{3+} + Sn^{2+} <-> As^{0} + Sn^{4+} }}
One tin is oxidized from oxidation state +2 to +4, a two-electron step, hence 2 is written in front of the two arsenic partners. One arsenic is reduced from +3 to 0, a three-electron step, hence 3 goes in front of the two tin partners. An alternative three-line procedure is to write separately the [[half-reaction]]s for oxidation and reduction, each balanced with electrons, and then to sum them up such that the electrons cross out. In general, these redox balances (the one-line balance or each half-reaction) need to be checked for the ionic and electron charge sums on both sides of the equation being indeed equal. If they are not equal, suitable ions are added to balance the charges and the non-redox elemental balance.