Question

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

Answer 1

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 `"LOSS of Electrons"`. Reduction involves the `"GAIN of Electrons"`. You know the old story `"LEO says GER"`, `"Loss of Electrons, Oxidation; Gain of electrons, Reduction"`

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

`C(s) + O_2(g) rarr CO_2(g)`

Both reactants are ZEROVALENT, i.e. a `0` oxidation state. During the reaction, `C` loses 4 electrons to give `C^(IV+)`, and oxygen gains 2 electrons to give `O^(-II)`.

Alternatively we could represent the oxidation of elemental iodine to iodate, `IO_3^-`:

`1/2I_2 + 3H_2O rarr IO_3^(-) +5e^(-) +6H^+`

Both mass and charge are balanced as is required.

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

`CararrCa^(2+) + 2e^-`

`1/2I_2 +e^(-) rarr I^(-)`

Again this is rationalized in terms of electron transfer.