# 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. – It's Over Jul 20 '15 at 19:37

• 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? – Anonymous Aug 15 '13 at 18:46
• Yeah, I confused it for $Mg$. Does this apply to group 16? – Anonymous Aug 15 '13 at 19:43
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.