How do you find oxidation numbers?

Dec 23, 2017

They are tattooed on the backs of the elements' collars....

Explanation:

To begin we note that oxidation number and oxidation state, are FORMALISMS....i.e. there is nothing fundamental about them...they are used to help us balance chemical equations....especially redox equations.

Rules of assignment (and clearly I have prepared these earlier!) are:

$1.$ $\text{The oxidation number of a free element is always 0.}$

$2.$ $\text{The oxidation number of a mono-atomic ion is equal}$ $\text{to the charge of the ion.}$

$3.$ $\text{For a given bond, X-Y, the bond is split to give } {X}^{+}$ $\text{and}$ ${Y}^{-}$, $\text{where Y is more electronegative than X.}$

$4.$ $\text{The oxidation number of H is +1, but it is -1 in when}$ $\text{combined with less electronegative elements.}$

$5.$ $\text{The oxidation number of O in its}$ compounds $\text{is usually -2, but it is -1 in peroxides.}$

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

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

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

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

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

SO I will go thru TWO examples.....the oxidation number of sulfur in sulfuric acid, and the oxidation number of nitrogen in nitric acid....

For ${H}_{2} S {O}_{4}$ the sum of the oxidation numbers is ZERO...(rule 9). From rule 4, and rule 5, $2 \times + 1 + 4 \times \left(- 2\right) + {S}_{\text{oxidation number}} = 0$...and so ${S}_{\text{oxidation number}} = + V I$...

For $H N {O}_{3}^{+}$, $+ 1 + 3 \times \left(- 2\right) + {N}_{\text{oxidation number}} = 0$...and so ${N}_{\text{oxidation number}} = + V$...

Can you tell us the oxidation number of chlorine in chloric, $H C l {O}_{3}$, and perchloric acids, $H C l {O}_{4}$? The hydrogen is BOUND to the oxygen....