How can you determine what the oxidation number of an ion is?

Answer 1

This is an EXTREMELY complex area, but there is a fairly basic rule, but there are MANY rule breakers. Let's look at it this way, an ion forms so that an element can achieve 8 electrons, they want to achieve this with the least amount of energy possible. (Keep in mind we are only working with s and p sublevel electrons, or the valence electrons, not the d's and f's). First you need to undestand this (this is a rough chart, as I am ignoring all of the rule breakers):

Group 1: 1 valence electron

Groups 2-12: 2 valence electrons

Group 13: 3 valence electrons

Group 14: 4 valence electrons

Group 15: 5 valence electrons

Group 16: 6

Group 17: 7

Group 18: 8 (noble gases)

Everything left of the metalloid line will LOSE valence electrons to achieve the state of the noble gas on the period before it, for example:

Calcium (Ca #20) has 2 valence electrons, so it will lose those two to achieve the electron configuration (E.C.) of Argon (Ar #18), and since it has two more positively charged protons than it does negatively charged electrons, it forms the ion Ca2+.

Everything to the right of the metalloid line will GAIN valence electrons to achieve the E.C. of the noble gas on its period, for example:

Selenium (Se #34) has 6 valence electrons, so instead of losing 6 electrons to achieve the state of Argon, like what Calcium did, it will gain two electrons to achieve the state of Krypton (Kr #36), and since its negatively charged electrons outnumber its positively charged protons it gets the oxidation number 2-, making it Se2-.

Just remember that, metals (to the left of the zigzag/metalloid line) lose electrons to form positive ions, and nonmetals (to the right of the zigzag/metalloid line) gain electrons to form negative ions. Hope this helped, and I hope you do not mind that I disregarded the rule breakers.