# Oxidation Numbers

## Key Questions

• The valence electrons determine how many electrons an atom is willing to give up or how many spaces need to be filled in order to satisfy the rule of octet.

Lithium (Li), Sodium (Na) and Potassium (K) all have an electron configuration that ends as ${s}^{1}$. Each of these atoms would readily release this electron to have a filled valence shell and become stable as $L {i}^{+} 1$, $N {a}^{+} 1$ and ${K}^{+} 1$. Each element having an oxidation state of +1.

Oxygen (O) and Sulfur (S) all have an electron configuration that ends as ${s}^{2} {p}^{4}$. Each of these atoms would readily take on two electrons to have a filled valence shell and become stable as ${O}^{-} 2$, and ${S}^{-} 2$. Each element having an oxidation state of -2.

There are exceptions to the rules and the transition metals usually have more than one oxidation state.

SMARTERTEACHER • Oxidation numbers are related to the polarity of the bond between two atoms.

The basic rule is:
The more electronegative atom gets ALL of the shared electrons. For example, in CO₂, the more electronegative O atoms get all of the shared electrons. They end up with eight valence electrons each. Since they have each gained two electrons, their oxidation numbers are -2. The C atom has no valence electrons. It has lost four electrons, so its oxidation number is +4. The sum of the oxidation numbers is zero.

The same principle holds for ions. In SO₄²⁻, the more electronegative O atoms all get the shared pairs to the S atom. This gives them eight valence electrons each, so their oxidation numbers are each -2. The S atom is left with no valence electrons. It has lost six electrons, so its oxidation number is +6. The sum of the oxidation numbers adds up to the overall charge on the ion (-2).

This is formally the charge left on a central atom when all the bonding pairs of electrons are REMOVED, with the charge assigned to most electronegative atom......

#### Explanation:

The rules for assignment of oxidation numbers are given here.

As given, $\text{oxidation number}$ and $\text{oxidation state}$ are FORMALISMS, they are convenient fictions that may nevertheless have a practical use. And when we assign oxidation numbers, we use them to solve, i.e. balance, a chemical equation, with respect to mass and charge, $\text{redox equations}$, straightforwardly and systematically (says he, $\text{modelling a modern major general}$.)

And so we can interpret a given chemical reaction with the use of oxidation numbers. Consider the oxidation of ammonia to give nitrate ion..........in terms of formal oxidation state this is the transition, $\stackrel{- I I I}{N}$ to $\stackrel{+ V}{N}$, an 8 electron oxidation, which we formally represent in the equation.......

$N {H}_{3} \left(a q\right) + 3 {H}_{2} O \rightarrow N {O}_{3}^{-} + 9 {H}^{+} + 8 {e}^{-}$ $\left(i\right)$

Is this balanced with respect to mass and charge? It must be if we purport to represent physical reality.

And, inevitably, something must be reduced to effect the oxidation; let's say it is oxygen.

${\stackrel{0}{O}}_{2} + 4 {e}^{-} \rightarrow 2 {O}^{2 -}$ $\left(i i\right)$

We add the individual redox equations together in a way to eliminate the electrons, which are particles of convenience......And so we take $\left(i\right) + 2 \times \left(i i\right)$:

$N {H}_{3} \left(a q\right) + 3 {H}_{2} O + 2 {O}_{2} + 8 {e}^{-} \rightarrow N {O}_{3}^{-} + {\underbrace{9 {H}^{+} + 4 {O}^{2 -}}}_{4 {H}_{2} O + {H}^{+}} + 8 {e}^{-}$

And so we cancel out what we can.....

$N {H}_{3} + \cancel{3 {H}_{2} O} + 2 {O}_{2} + \cancel{8 {e}^{-}} \rightarrow N {O}_{3}^{-} + {\underbrace{9 {H}^{+} + 4 {O}^{2 -}}}_{\cancel{4} {H}_{2} O + {H}^{+}} + \cancel{8 {e}^{-}}$

....to give finally.........

$N {H}_{3} + 2 {O}_{2} \rightarrow N {O}_{3}^{-} + {H}_{2} O + {H}^{+}$

....or........

$\stackrel{- I I I}{N} {H}_{3} + 2 {\stackrel{0}{O}}_{2} \rightarrow H \stackrel{+ V}{N} {O}_{3} + {H}_{2} \stackrel{- I I}{O}$

I acknowledge that is a lot of pfaff.......but your question was rather open-ended.