Why is the number of valence electrons associated with the oxidation number of the element?

Jun 12, 2017

Why, because we conceive of the oxidation number as the number of electrons GAINED or RECEIVED by an atom upon chemical reaction.

Explanation:

And oxygen, which as an element has a high nuclear charge (i.e. it comes to the right of the Periodic Table as we face it), tends to accept electrons.........

Guidelines for assigning oxidation numbers 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 $\text{is usually -2, but it is -1 in peroxides.}$

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

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

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

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

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

And so when elemental oxygen reacts with say carbon, we conceive that the 2 oxygen atoms accept 2 electrons, and that carbon atom donates two electrons...............

$\stackrel{0}{C} \left(s\right) + {\stackrel{0}{O}}_{2} \left(g\right) \rightarrow \stackrel{- I I}{O} = \stackrel{+ I V}{C} = \stackrel{- I I}{O} \left(g\right)$

As always the sum of the oxidation numbers of the elements in a compound or ion, is equal to the charge of the compound or ion, and here the $C {O}_{2}$ product is neutral.