Question

How to find oxidation number?

Answer 1

Divide a number by 5, add 37, then take away the number you first thought of....

Explanation:

Now oxidation number is a FORMALISM, a construct that has little physical significance. Nevertheless, we can use this formalism to approach redox reactions systematically.

I give the rules of assignment here...

`1.` `"The oxidation number of a free element is always 0."`

`2.` `"The oxidation number of a mono-atomic ion is equal"` `"to the charge of the ion."`

`3.` `"For a given bond, X-Y, the bond is split to give "X^+` `"and"` `Y^-`, `"where Y is more electronegative than X,"`
`"to give formal oxidation numbers of X(-I), and Y(+I)."`

`4.` `"The oxidation number of H is +I, but it is -I in when"` `"combined with less electronegative elements."`

`5.` `"The oxidation number of 0 in its"` compounds `"is usually -II, but it is -I in peroxides."`

`6.` `"The oxidation number of a Group 1 element"` `"in a compound is +I."`

`7.` `"The oxidation number of a Group 2 element in"` `"a compound is +II."`

`"8. The oxidation number of a Group 17 element in a binary compound is -I."`

`9.` `"The sum of the oxidation numbers of all of the atoms"` `"in a neutral compound is 0."`

`10.` `"The sum of the oxidation numbers in a polyatomic ion"` `"is equal to the charge of the ion."`

And I give an example of its use here...

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_3stackrel(-III)C-stackrel(-I)CH_2OH+H_2OrarrH_3stackrel(-III)C-stackrel(+III)CO_2H +4H^(+) + 4e^(-) " (i)"`

And something, here `MnO_4^(-)` is reduced down to `Mn^(+2)`:

`stackrel(VII+)"Mn"rarrstackrel(+II)"Mn"`

`"MnO"_4^(-) + "8H"^(+)+ "5"e^(-) rarr"Mn"^(2+)+"4H"_2"O"(l)` `"(ii)"`

Deep-red purple permanganate ion is reduced down to almost colourless `Mn^(2+)`. to give the final redox equation, we cross multiply `(i)` and `(ii)` so that electrons DO NOT appear in the final equation:

And thus `5xx(i)+4xx(ii)` gives...............

`4MnO_4^(-) + 12H^(+)+5H_3C-CH_2OHrarr5H_3C-CO_2H+4Mn^(2+)+11H_2O(l)`

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.

Answer 2

Elements always have oxidation number 0.
When they form compound, oxidation number indicates the electrons' tendency in the bond.
Total oxidation number in a compound must sum to zero.

Explanation:

Hydrogen is always +1 when forming compound with nonmetal; -1 when forming compound with metal.

Fluorine is always -1 , other halogens are usually -1 when they're not binding with a more powerful oxidizing agent.

Oxygen is usually -2 , but exceptions do occur. Like in OF2, O is +2 (it's extremely unstable!); in H2O2, O is -1.(thus it's a strong oxidizing agent)

They're most common elements used in organic chemistry.(carbon's oxidation number can vary from compound to compound)

By the way, ions' oxidation number equal to their charge.
If you still have any question after reading this, just ask and I'll answer you later.