in addition to previous Q:

Question

for eg:
why is the oxidation state (#) of Ca in CaH2 "+2"? According to a p.table I have here +2 is the only oxidation # for Ca...but how is that figured?
One other eg:
the oxidation state (#) of the maganese atom in KMnO4.6H2O is "+7". How is this so & not: 2,3,or4?

If you assign charges to oxygen, it's (-2). There are 4 of these in the molecule. So 4*(-2) = -8. potassium is a +1 charge. So now you have (-8) +1 = (-7). Therefore, since there's only one Mn, that one atom is a +7 charge so that the -7 is balanced out to be neutral (since it's not an ion).

I hope this helps.. Just assign the charges for what you know, then calculate based upon the overall charge and the amount of that element in the molecule.

For the Ca ion, it is in group IIA (or 2 depending upon the system you are using). Alkali metals (group IA) and alkaline earth metals (IIA) are assigned oxidation states of +1 and {2 respectively. The other way you can do it is to know that hydrogen, in hydrides, (and this is a hydride) is
-1. Since there are two of them it makes the charge on H -2 for two H atoms; therefore, Ca must be +2 for the compound of CaH2 to be neutral.