| Oxidation State |
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The increase in oxidation state of an atom is known as an oxidation: a decrease in oxidation state is known as a Reduction . Such reactions involve the transfer of electrons, a net gain in electrons being a reduction and a net loss of electrons being an oxidation. Calculation of oxidation states There are two common ways of computing the oxidation state of an atom in a compound. The first one is used for molecules when one has a Lewis Structure , as is often the case for organic molecules, while the second one is used for simple compounds (molecular or not) and does not require a Lewis structure. It should be remembered that the oxidation state of an atom does not represent the "real" charge on that atom: this is particularly true of high oxidation states, where the Ionization Energy required to produce a multiply positive ion are far greater than the energies available in chemical reactions. The assignment of electrons between atoms in calculating an oxidation state is purely a formalism, albeit a useful one for the understanding of many chemical reactions. For more about issues with calculating atomic charges, see Partial Charge . From a Lewis structure When a Lewis Structure of a molecule is available, the oxidation states may be assigned unambiguously by computing the difference between the number of valence electrons that a neutral atom of that element would have and the number of electrons that "belong" to it in the Lewis structure. For purposes of computing oxidation states, electrons in a bond between atoms of different elements belong to the most electronegative atom; electrons in a bond between atoms of the same element are split equally, and electrons in lone pair belong only to the atom with the lone pair. For example, let's consider acetic acid: The carbon atom on the left has 6 valence electrons from its bonds to the hydrogen atoms, because carbon is more electronegative than hydrogen, and 1 electron from its bond with the other carbon atom, because the electron pair in the C–C bond is split equally, for a total of 7 electrons. A neutral carbon atom would have 4 valence electrons, because carbon is in group 14 of the periodic table. The difference, 4 – 7 = –3, is the oxidation state—that is, the charge that the atom would have if all the bonds were 100% ionic. Following the same rules, the carbon on the right has an oxidation state of +3 (it only gets one valence electron from the C–C bond; the oxygen atoms get all the other electrons). The oxygen atoms both have an oxidation number of –2; they get 8 electrons each (4 from the lone pairs and 4 from the bonds), while a neutral oxygen atom would have 6. The hydrogen atoms all have oxidation state +1, because none of them get any electrons and a neutral hydrogen atom would have one electron. Without a Lewis structure The algebraic sum of oxidation states of all atoms in a neutral Molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. This fact, combined with the fact that some elements almost always have certain oxidation states, allows one to compute the oxidation states for atoms in simple compounds. For example, in most compounds
Exceptions:
Other elements with common oxidation states include
Example With the example, Cr(OH)3, oxygen has an oxidation state of −2 (no fluorine, O-O bonds present), and hydrogen
As the compound is neutral, Cr has to have an oxidation state of +3. Elements with multiple oxidation states Most elements have more than one possible oxidation state, with carbon having nine: –4: CH4 –3: C2H6 –2: CH3F –1: C2H2 0: CH2F2 +1: C2H2F4 +2: CHF3 +3: C2F6 +4: CF4 For an exhaustive list of the possible oxidation states of each element, see http://en.wikipedia.org/wiki/Periodic_Table#Standard_periodic_table See also External links
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