# Which oxidation state should I assume when balancing equations?

1. Nov 29, 2004

### wasteofo2

When dealing with an equation in which you combine 2 elements that have multiple oxidation states, let's say Magnesium and Nitrogen, which oxidation states do you assume you're working with? I know that in the instance of Magnesium and Nitrogen, it would have to be one of Nitrogen's negative oxidation states, and Magnesium would be +2, but aside from that, I'm not sure. Should you just assume it's the first listed on the periodic table for Nitrogen, or should I assume -2?

Last edited: Nov 29, 2004
2. Nov 29, 2004

### Staff: Mentor

Nitrogen in nitrides tends to have a valence -3 as in $$Mg_3 N_2$$.

Way back when, I remember looking at electronegativities, ionic radii, and thermodynamic considerations like heat of formation. Now its just easier to look up a compound in a table or database.

Transitional metals can be difficult because elements like Mo have valence states like (+2, +3, +4, +5, +6) and Mn has valence states of (+1 to +7, although +1 and +5 are uncommon). Which state depends on the anionic species.

Are you strictly looking at binary compounds.

I think Chem_tr is the one to answer this question.

3. Nov 29, 2004

### Gokul43201

Staff Emeritus
There are several things that determine the oxidation state that is right. In a crystalline solid, the neighbourhood of the atom often plays the most important role in determining the oxidation state. It's useful to know what are the most common oxidation states for the multivalent elements.

In the case of nitrogen, there is absolutely no reason for you to assume that it is -2 (did you pick this because Mg is +2 ?). More often than not, the oxidation state mentioned in a periodic table as the most common one will work. For nitrogen, the -3 and +5 states are the most common. Between these two, it's not hard to decide which state it takes in which compounds. It will have a positive oxidation state in a binary compound that involves a more electronegative element and a negative oxidation state if the other element is less electronegative.

In cases where the electronegativities are similar (or in crystalline solids) there is often more than one compound that may be formed, particularly if one of the elements is strongly multivalent. (eg : oxides of nitrogen, inter-halogen compounds, etc.)

A good way to determine the most likely oxidation state is by trying to draw the molecular (Lewis or other such type) structure, and picking the most stable configuration...but this may be a little advanced.

And, as astronuc mentioned, magnesium nitride is Mg(3)N(2).

4. Nov 29, 2004

### wasteofo2

Thanks alot, I did choose -2 because Mg is +2.

5. Nov 30, 2004

### chem_tr

Hello

As Gokul said, electronegativity is a principal option for decision. In many cases you can easily deduce for a metal and ametal, and even inter-ametallic compounds. You should first use electronegativity, I suppose.

In magnesium and nitrogen, if they form a compound, magnesium tends to give away electrons, and nitrogen does the opposite, both reach the most stable electronic orientation as in a noble gas.