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Oxidation numbers assigned by rules with a priority order, all adding up to the species' charge; the bookkeeping that shows where electrons go.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to assign oxidation numbers to every atom in a molecule or ion using the rules and the charge-sum rule, handle the exceptions for peroxides and ions, read and write Roman numerals in names, order species of one element by oxidation number, and interpret a fractional average.
From Chemistry 1 you know that ions carry charges because atoms have gained or lost electrons, and you have written half equations with electrons in them. The acids unit tracked one kind of particle moving between species — the proton. This unit tracks electrons, and oxidation numbers are how you see them move when an equation does not show them.
An oxidation number (or oxidation state) is the charge an atom would have if every bond it made were fully ionic, with the shared electrons given to the more electronegative atom. It is written with its sign first: $+3$, $-2$. Electronegativity is how strongly an atom pulls shared electrons towards itself; fluorine and oxygen are the most electronegative elements. A Roman numeral in a name, as in iron(III) chloride, gives a metal's oxidation number.
Oxidation numbers are assigned by rules, applied in this order of priority:
The last rule does the work. The others fix the atoms that are always the same; the last one finds the atom that varies. Take the manganate(VII) ion, $\mathrm{MnO_4^-}$:
| Atom | How many | Each | Total |
|---|---|---|---|
| O | 4 | $-2$ | $-8$ |
| Mn | 1 | $x$ | $x$ |
| whole ion | $-1$ |
$x - 8 = -1$, so manganese is $+7$ — which is what the (VII) in its name says.
Oxidation numbers are a model, not a measurement. Manganese in manganate(VII) does not really carry a $+7$ charge; its bonds to oxygen are covalent. But the numbers do something real: when they change during a reaction, electrons have moved, and the size of the change is how many.
Another way: picture
Think of splitting a restaurant bill. Some diners have fixed orders — oxygen always pays $-2$, hydrogen $+1$ — and the bill's total is fixed by the charge on the species. Whoever is left pays whatever makes the total come out right. If two people share the leftover, each pays half.
Another way: steps
To find an unknown oxidation number:
Transition metals and non-metals in the middle of the periodic table take many oxidation numbers, and their chemistry follows them. Nitrogen runs the whole range:
| Species | Nitrogen's oxidation number |
|---|---|
| ammonia, $\mathrm{NH_3}$ | $-3$ |
| hydrazine, $\mathrm{N_2H_4}$ | $-2$ |
| nitrogen gas, $\mathrm{N_2}$ | $0$ |
| dinitrogen oxide, $\mathrm{N_2O}$ | $+1$ |
| nitrogen monoxide, $\mathrm{NO}$ | $+2$ |
| nitrogen dioxide, $\mathrm{NO_2}$ | $+4$ |
| nitric acid, $\mathrm{HNO_3}$ | $+5$ |
The lowest value, $-3$, is nitrogen with three more electrons than a neutral atom's share: it can only lose electrons. The highest, $+5$, is nitrogen with all five outer electrons given away: it can only gain. Everything in between can go either way. That is why nitric acid is an oxidising agent and ammonia can be a reducing agent — a connection the next lesson makes precise.
The Roman numerals in names come from the same idea. Iron(II) sulfate contains iron at $+2$ and iron(III) oxide iron at $+3$; the numeral is never the number of atoms.
Where do the rules themselves come from? From electronegativity. In a bond between two different atoms the shared electrons are pulled towards the more electronegative one, and the oxidation number pretends that pull is complete. Oxygen, second only to fluorine, wins against almost everything, so it is counted as having gained two electrons: $-2$. Hydrogen loses against every non-metal, so it is counted as having given its one electron away: $+1$. Against a metal such as sodium, hydrogen is the more electronegative partner and becomes $-1$. Two atoms of the same element share their electrons equally, which is why an element on its own is $0$ and why the two oxygens bonded to each other in a peroxide split the difference.
In magnetite, $\mathrm{Fe_3O_4}$, four oxygens at $-2$ make $-8$, so the three irons share $+8$: an average of $+\tfrac{8}{3}$. No atom has a fractional charge — magnetite contains one iron(II) and two iron(III) for every four oxides, and $\tfrac{2 + 3 + 3}{3} = \tfrac{8}{3}$. A fraction from the rules is a sign that atoms of the same element are sitting in different environments, and the rules can only report their average.
The same happens with carbon in organic compounds: in propane, $\mathrm{C_3H_8}$, eight hydrogens at $+1$ make the three carbons share $-8$, an average of $-\tfrac{8}{3}$, though the end carbons are $-3$ and the middle one $-2$.
Every atom in a compound has oxidation number zero. The sum is zero for a neutral compound. Individual atoms are positive or negative.
The sum is zero for an ion too. For an ion the sum is the ion's charge: $-1$ for manganate(VII), $+1$ for ammonium.
Oxygen is always $-2$. In peroxides it is $-1$, because hydrogen's $+1$ takes priority.
The subscript is the oxidation number. In $\mathrm{KMnO_4}$ the 4 counts oxygen atoms; manganese is $+7$.
Only one atom of the unknown element counts. In dichromate, $\mathrm{Cr_2O_7^{2-}}$, the two chromiums share $+12$, so each is $+6$.
$\mathrm{Cr_2O_7^{2-}}$: overall charge $-2$.
An ion, so the total is its charge.
Seven oxygens at $-2$ give $-14$; two chromiums give $2x$. So $2x - 14 = -2$.
Count every atom.
$2x = 12$, so each chromium is $+6$: this is the dichromate(VI) ion.
Share the total between the two atoms.
$\mathrm{NH_4^+}$: overall charge $+1$.
Positive this time.
Four hydrogens at $+1$ give $+4$, so $x + 4 = +1$.
Hydrogen is $+1$ bonded to a non-metal.
Nitrogen is $-3$, its lowest state — the same as in ammonia, because adding a proton moves no electrons.
Proton transfer is not electron transfer.
Three oxygens at $-2$ give $-6$, and the ion's charge is $-1$.
Fixed atoms and the total.
So $x - 6 = -1$, and $x = \ldots$
…$+5$, which is why the ion is named chlorate(V); check that $+5 - 6 = -1$.
What is the oxidation number of chlorine in $\mathrm{ClO_3^-}$?
Answer:
Give the oxidation number of each element in $\mathrm{H_2SO_4}$.
| oxidation number | |
|---|---|
| hydrogen | |
| sulfur | |
| oxygen |
Nitrogen's oxidation number runs from $-3$ to $+5$. Put the marker at nitrogen's oxidation number in $\mathrm{NO}$, nitrogen monoxide.
-3 |——————————| 5
Mark the position with a cross, then write the value:
Hydrogen peroxide, $\mathrm{H_2O_2}$, is used as a rocket propellant. What is the oxidation number of oxygen in it?
Match each name to its formula.
| $\mathrm{FeCl_2}$ | $\mathrm{FeCl_3}$ | $\mathrm{Cu_2O}$ | $\mathrm{CuO}$ | |
|---|---|---|---|---|
| iron(II) chloride | ||||
| iron(III) chloride | ||||
| copper(I) oxide | ||||
| copper(II) oxide |
Put these sulfur species in order of sulfur's oxidation number, lowest first.
Number the steps in order (write the number in the box):
$\mathrm{Fe_3O_4}$ is magnetite, the magnetic iron ore. Using the usual rules, what is the average oxidation number of iron in it? A fraction is fine.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Manganese forms compounds in many oxidation states. In which of these is manganese $+4$?
You can assign oxidation numbers so that they add up to the charge. Say out loud why manganese in $\mathrm{MnO_4^-}$ is $+7$ and not $+8$. Next: using changes in oxidation number to spot oxidation and reduction in a reaction.
10. Your turn: what is chlorine's oxidation number in the chlorate ion, $\mathrm{ClO_3^-}$?, step 3