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Oxidation of Elements

When element is in free state - forms an elementary substance - the motion of the electrons about all the atoms of this substance occurs identically. This holds for all elementary substances regardless of their structure. For example, in a Hydrogen molecule, the electrons travel about both atoms to an equal extent - the molecule H2 is non-polar. For crystals with a covalent bond, the chemical bonds between the atoms are also symmetrical relative to the joined atoms. For metals, the distribution of both the bound and the free electrons is also uniform on an average.

Matters are different in compound substances. The chemical bonds between the atoms of different elements are not symmetrical; polar bonds are generally the rule in molecules of compounds. This nonuniformity in the distribution of the electrons is the greatest in ionic compounds - in the formation of substances with an ionic bond, the valence electrons pass virtually completely from the atom of one element to the atom of another one.

The element whose electrons pass to atoms of another element (com­pletely with an ionic bond or partly with a polar one) is said to be positively oxidized. An element to whose atoms electrons from atoms of another element pass is negatively oxidized (or reduced).

The number of electrons that have passed from one atom of a given positively oxidized element or to one atom of a given negatively oxidized or reduced element is called the oxidation number (or oxidation state) of the element.

In elementary substances, the oxidation number of an element is always zero. In compounds, some elements always display the same oxidation number, but for most elements it differs in different com­pounds.

The elements having a constant oxidation number are the alkali metals (+1), the alkaline-earth metals (+2), and Fluorine (- 1). Hydrogen in most compounds is characterized by an oxidation number of +1, while in metal hydrides and in some other compounds it is - 1. The oxidation number of Oxygen, as a rule, is - 2. The most important exceptions here are the peroxide compounds, where it is - 1, and Oxygen Fluoride OF2, in which the oxidation number of Oxygen is + 2. For elements with changing oxidation number, its value is always simple to find knowing the formula of a given compound and taking into consideration that the sum of the oxidation numbers of all the atoms in a molecule is zero.

Let us determine as an example the oxidation number of carbon in CO, CO2, CH4, C2H6, and C2H5OH. We shall denote it by x. Hence, remembering that the oxidation number of Hydrogen is +1 and that of Oxygen - 2, we get

CO x + (-2) = 0 x = + 2
CO2 x + 2·(-2) = 0 x = + 4
CH4 x + 4·(+1) = 0 x = - 4
C2H6 2x + 6·(+1) = 0 x = - 3
C2H5OH 2x + 6·(-1) + (-2) = 0 x = - 2

To find the oxidation number of elements in compounds, the table of electronegativity of elements (Appendix 7) can be used. It must be borne in mind here that when a chemical bond is formed, the electrons are displaced to the atom of the element with the higher electronegativity. For instance, the relative electronegativity of Phosphorus is 2,2, and that of Iodine 2,5. Consequently, in the compound PI3, the shared electrons are displaced to the Iodine atoms, and the oxidation numbers of phosphorus and Iodine are +3 and - 1, respectively. In compound NI3, however, the oxidation numbers of Nitrogen and Iodine are - 3 and +1 because the electronegativity of Nitrogen (3,0) is higher than that of Iodine.



 


Date: 2015-01-12; view: 815


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Experiment 2. Influence of temperature to hydrolysis | Oxidation-Reduction Reactions
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