# What is the oxidation state of a free element? Of a monoatomic ion?

Sep 3, 2016

The oxidation state of a free element is $0$. The neutral element has neither gained nor donated electrons.

#### Explanation:

Redox reactions are considered to involve the formal TRANSFER of electrons. Oxidation involves the $\text{LOSS of Electrons}$. Reduction involves the $\text{GAIN of Electrons}$. You know the old story $\text{LEO says GER}$, $\text{Loss of Electrons, Oxidation; Gain of electrons, Reduction}$

When carbon reacts wth dioxygen this is certainly a formal redox process:

$C \left(s\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right)$

Both reactants are ZEROVALENT, i.e. a $0$ oxidation state. During the reaction, $C$ loses 4 electrons to give ${C}^{I V +}$, and oxygen gains 2 electrons to give ${O}^{- I I}$.

Alternatively we could represent the oxidation of elemental iodine to iodate, $I {O}_{3}^{-}$:

$\frac{1}{2} {I}_{2} + 3 {H}_{2} O \rightarrow I {O}_{3}^{-} + 5 {e}^{-} + 6 {H}^{+}$

Both mass and charge are balanced as is required.

For the monoatomic ion, the charge on the ion is simply the oxidation number:

$C a \rightarrow C {a}^{2 +} + 2 {e}^{-}$

$\frac{1}{2} {I}_{2} + {e}^{-} \rightarrow {I}^{-}$

Again this is rationalized in terms of electron transfer.