How do I find the oxidation number(s) on the periodic table?

Explanation:
For metals on the left side of the periodic table the family number often gives the oxidation number
$\begin{array}{l}\text{ I }A=+1\\ \text{ II }A=+2\\ \text{ II }A=+3\\ \text{ IV }A=+4\text{ but can also be }+2,-2\text{ and }-4\text{. }\end{array}$
$\mathrm{V}A=+5$ but can also be +3 , and -3 (count to the right to find the number of negative electrons -e needed to become an inert gas.)
$\mathrm{VI}A=+6$ can also be $+4,+2$ but is more usually -2
$\mathrm{VII}\mathrm{A}=+7$ can also be $+5,+3+1$ but is usually -1
VIII $A=0$ This family is the inert gases that do not want to lose or gain electrons,
The transition metals can be very confusing in terms of oxidation number because of movement between the $s$ and d orbitals.
IB $=+1$ but also can be +2
II $B=+2$ almost always +2
III $B=+3$
IV $B=+4$ but also +2
$\vee B=+5$ but also +3 and +2
VI $B=+6$ but also $+4,+3$ and +2 ( Chromium means color because each of the
different oxidation numbers creates a different color.)
VII $B=+7$ but also $+5+3,+2$
VIII $B=+3$ and +2
IX $B=+3$ and +2
$XB=+3$ and +2
IB $=+1$ but also can be +2
II $\mathrm{B}=+2$ almost always +2
III $B=+3$
IV $B=+4$ but also +2
$\mathrm{V}=+5$ but also +3 and +2
VI B $=+6$ but also $+4,+3$ and +2 ( Chromium means color because each of the
different oxidation numbers creates a different color.)
VII $B=+7$ but also $+5+3,+2$
VIII $B=+3$ and +2
$\mathrm{IX}B=+3$ and +2
$XB=+3$ and +2