# What are the oxidation states of the individual atoms in hydrogen peroxide? How are oxidation numbers determined?

Jul 25, 2017

We gots peroxide $\stackrel{- I}{O}$ and $\stackrel{+ I}{H}$. The hydrogen peroxide molecule is NEUTRAL. Claro?

#### Explanation:

By definition the oxidation number is the charge left on the central atom, when all the bonding pairs of electrons are broken, with the charge, the extra electron, going to the more electronegative atom. Do this for water, $H - O - H$, we gets ${O}^{2 -}$, and $2 \times {H}^{+}$, i.e. $\stackrel{- I I}{O}$ and $\stackrel{+ I}{H}$. Break the homoelement bond in peroxide, the electrons are conceived to be shared, because the oxygens clearly have equal electronegativity: $H O - O H \rightarrow 2 \times \dot{O} H$; the oxidation state of the oxygen in the hydroperoxyl radical is CLEARLY $- I$.

When we do this for other element-element bonds, say for ${H}_{3} C - C {H}_{2} - C {H}_{3}$, we get $2 \times \stackrel{- I I I}{C} + \stackrel{- I I}{C} + 8 \times \stackrel{+ I}{H}$. As always the sum of the oxidation numbers equals the charge on the ion or species, and here it is ZERO as required.

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

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

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

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

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

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

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

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

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

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