# How do oxidation states work?

Aug 30, 2017

Oxidation number is the charge left on the central atom when all the bonding pairs of electrons are broken, with the charge assigned to the most electronegative atom.

#### Explanation:

$1.$ $\text{The oxidation number of a free element is always 0.}$

$2.$ $\text{The oxidation number of a mono-atomic ion is equal}$ $\text{to the charge of the ion.}$

$3.$ $\text{For a given bond, X-Y, the bond is split to give } {X}^{+}$ $\text{and}$ ${Y}^{-}$, $\text{where Y is more electronegative than X.}$

$4.$ $\text{The oxidation number of H is +1, but it is -1 in when}$ $\text{combined with less electronegative elements.}$

$5.$ $\text{The oxidation number of O in its}$ compounds $\text{is usually -2, but it is -1 in peroxides.}$

$6.$ $\text{The oxidation number of a Group 1 element}$ $\text{in a compound is +1.}$

$7.$ $\text{The oxidation number of a Group 2 element in}$ $\text{a compound is +2.}$

$8.$ $\text{The oxidation number of a Group 17 element in a binary compound is -1.}$

$9.$ $\text{The sum of the oxidation numbers of all of the atoms}$ $\text{in a neutral compound is 0.}$

$10.$ $\text{The sum of the oxidation numbers in a polyatomic ion}$ $\text{is equal to the charge of the ion.}$

Now this might be more than you want to know. But the maximum oxidation number of an element is equal to its Group number. Typically we assign oxygen (in its oxides) a $- I I$ charge, and fluorine, in it compounds has $- I$ charge.

For potassium permanganate, we got ${K}^{+} M n {O}_{4}^{-}$, i.e. $K \left(+ I\right) , M n \left(+ V I I\right) , 4 \times O \left(- I I\right)$; the sum of the numbers is ZERO. Why so?