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# How would you define oxidation states?

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#### Explanation

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#### Explanation:

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anor277 Share
Jun 3, 2017

This the charge left on the central atom when all the bonding pairs of electrons are broken, with the charge, the electron, assigned to the most electronegative atom. The resultant charge is the oxidation state/number.

#### Explanation:

From the earliest days of studying chemistry we are taught that ionic bonding results from the transfer of electrons to give ions, whereas covalent bonding results from the sharing of electrons between (bound) atoms. Note that this is a conceptual exercise that DOES NOT really have much physical or chemical significance. However, assignment of oxidation states allows us to balance equations effectively (it also gives us something to teach to students of chemistry).

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

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

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

$4.$ $\text{The sum of the oxidation numbers in a polyatomic ion is}$ $\text{equal to the charge of the ion.}$

So we look at a simple oxidation reaction, that of carbon to give carbon dioxide..........electron loss $\text{(oxidation)}$ is FORMALLY equal to electron gain $\text{(reduction)}$.

$\stackrel{0}{C} \left(s\right) + {\stackrel{0}{O}}_{2} \left(g\right) \rightarrow \stackrel{{\text{-II)O=stackrel(""^+IV)C=stackrel(}}^{-} I I}{O} \left(g\right)$

In inorganic chemistry, electron-transfer reactions, redox reactions, become more explicit because we include include electrons as virtual particles..........

2"Cr"^(3+) +"7H"_2"O" rarr stackrel(VI+)"Cr"_2"O"_7^(2-)+14H^(+) +"6e"^-

So is charge balanced, so is mass balanced in this equation? If the answer is no we cannot accept it as a descriptor of physical reality.

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$

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