# What is oxidation state?

Jun 28, 2017

$\text{Oxidation state is the charge left.....}$

#### Explanation:

$\text{Oxidation state is the charge left on the central atom when}$  "all the bonding pairs of electrons are removed with the charge" $\text{assigned to the most electronegative atom.}$

Rules for assigning oxidation state are as follows.....

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

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

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

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

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

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

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

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

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

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

For a practical example of how oxidation numbers may be used to balance chemical equations, I include this old answer.....

Just to expand on the subject of oxidation of alcohols, the method of oxidation number, and gain or loss of electrons, can be utilized in the scenario, provided that we know how to assign oxidation numbers for the individual carbons in a given compound.

For the oxidation of ethanol to acetic acid, we could propose the given oxidation reaction as:

${H}_{3} \stackrel{- I I I}{C} - \stackrel{- I}{C} {H}_{2} O H + {H}_{2} O \rightarrow {H}_{3} \stackrel{- I I I}{C} - \stackrel{+ I I I}{C} {O}_{2} H + 4 {H}^{+} + 4 {e}^{-} \text{ (i)}$

And something, here $M n {O}_{4}^{-}$ is reduced down to $M {n}^{+ 2}$:

$\stackrel{V I I +}{\text{Mn"rarrstackrel(+II)"Mn}}$

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow \text{Mn"^(2+)+"4H"_2"O} \left(l\right)$ $\text{(ii)}$

Deep-red purple permanganate ion is reduced down to almost colourless $M {n}^{2 +}$. To give the final redox equation, we cross multiply $\left(i\right)$ and $\left(i i\right)$ so that electrons DO NOT appear in the final equation:

And thus $5 \times \left(i\right) + 4 \times \left(i i\right)$ gives...............

$4 \text{MnO"_4^(-) + "12H"^(+)+"5C"_2"H"_5"OH"rarr5"H"_3"CCO"_2"H"+"4Mn"^(2+)+"11H"_2"O}$

Which, if I have done my sums right, is balanced with respect to mass and charge; as indeed it must be if we purport to represent chemical reality.