# How to determine an oxidation number of an atom in the compound, that has at least two elements, which are not hydrogen or oxygen?

Wiki says:

[Oxidation state] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:

The oxidation state of a free element (uncombined element) is zero for a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion.

Hydrogen has an oxidation state of 1 and oxygen has an oxidation state of −2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of −1 in peroxides, e.g. H2O2 the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.

The same is written in my textbook. But how am I supposed to find the ox. number of an atom, which is in compound like $\ce{K2UO4}$?

• 1) Do not abbreviate Wikipedia.org as Wiki. 2) Do not use backticks unless there is some code somewhere. Commented Jul 20, 2015 at 19:37

In this case, the potassium ions are +1 each, and the oxygens are -2 as normal. That leaves uranium in a highly oxidized +6 state. The piece of information I think is missing is that Group 1 ions are almost always in the +1 state, because they so readily lose their outer valence electron. A molecule that is not an ion, like your example, must have an overall charge of zero.

• Group 1 ions are alkali metals? What about, say, $Mn_2U_2O_7$? How do you learn what oxidation number would some atom in such or more complex compound would have? How do you know which atom's ox. number is to be learned and which - to be determined by the learned one(s) and, optionally, oxygen and/or hydrogen? Commented Aug 15, 2013 at 18:46
• Good question! With rare exceptions, Group 1 & 2 elements always form ionic bonds and so have the +1 or +2 state, respectively. Same for halogens (Group 17) nearly always in the -1 state. Manganese and uranium can have many oxidation states, but transition metals like that tend to be unique in most inorganic molecules; i.e. I'm not certain your example exists! In most molecules, there is only one element that can have multiple oxidation states, and all the other elements will have consistent oxidation states like oxygen. There must be rare exceptions, but I can't think of one.
– user467
Commented Aug 15, 2013 at 19:16
• Yeah, I confused it for $Mg$. Does this apply to group 16? Commented Aug 15, 2013 at 19:43
• Not as much as you might think, because as you go down the periodic table, electronegativity decreases and you can have more oxidation states. This happen already at sulfur, which can have lots of oxidation states. If you look at a good period table, it will show the possible oxidation states. So does Wikipedia, which also shows the most common oxidation states. You start to learn what's possible and what's not.
– user467
Commented Aug 15, 2013 at 19:48

Technically (which may not be the answer you want) you compute the oxidation numbers in a molecule by first drawing its Lewis (dot) diagram showing all of the valence electrons. Then you assign the electron pair in each bond to the more electronegative atom. Then refigure the "charges" on each atom. The resulting "charge" will be the oxidation number.

Over the years folks have worked out shortcuts to this. For example, alkali metals easily lose an electron in almost all situations. Thus we have for example that $$\ce{Na}$$ has an oxidation number of +1.