how do you figure out ocidation numbers in specific elements?

Answer 1

If it is an element, its oxidation number is 0 in its elementtal form:
H2 = 0, Cl2 = 0, O2 = 0

If it is an atom which forms and ion, the oxidation number it equal to its ionic charge:
Mg = 2+, Na = 1+, O = 2-

If it is in a ternary compound, you must determine the oxidation number by a little calculation:
MgSO4
Mg = 2+
O = 2- x 4 = 8-
These add up to a total of 6-
That means the S must be 6+ to cancel out the 6-:
Mg = 2+
O = 2-
S = 6+

Rules to remember:
H is almost always a 1+
alkali metals are always 1+
alkaline earth metal are always 2+
halogens (as the last element written in the formula) are 1-
O is almost always 2-
Start with these rules and work your way to the center of the formula. The atoms in the center are the ones which can usually have a different charge.