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Oxidation Number


The concept of oxidation number is very useful in the study of chemical reactions. By definition, the oxidation number of an element in a compound is the charge it is supposed to carry if all compounds are considered to be ionic. In case of truly ionic compounds such as NaCl and KCl, the assignment of oxidation numbers presents no problem as the atoms in them actually carry charges. For example, in NaCl the oxidation numbers of Na and Cl are +1 and -1 respectively. For covalent molecules such as CCl4 and CO2, the bonding electrons are assigned to the more electronegative atom constituting the bond. For example in CCl4, each chlorine atom is assigned -1 as the oxidation number. Since there are four chlorine atoms attached to carbon, the latter will have an oxidation number of +4.

In general, the oxidation number of an element in a compound can be determined by following certain arbitrary rules that are given below.
  1. A free element (regardless of whether it exists in monatomic or polyatomic form, e.g. Hg, H2, P4 and S8) is assigned an oxidation number of zero.
  2. A free monoatomic ion is assigned an oxidation number equal to the charge it carries. For example, the oxidation numbers of Al3+, S2- and Cl- are +3, -2 and -1 respectively.
  3. In their compounds, the alkali and alkaline earth metals are assigned oxidation numbers of +1 and +2, respectively.
  4. The oxidation number of hydrogen in its compounds is generally +1 except in the ionic hydrides such as LiH, LiAIH4 where its oxidation number is -1.
  5. The oxidation number of fluorine in all its compounds is -1, The oxidation number of all other halogens is -1 in all compounds except those with oxygen (i.e.) and halogens having a lower atomic number (e.g. ICl3). Their oxidation numbers are determined via oxygen and halogens of lower atomic number.
  6. The oxidation number of both, oxygen and sulphur in their normal oxides (e.g. Na2O) and sulphides (e.g. CS2) is -2. The exceptions are the peroxide (e.g. H2O2 and Na2O2), superoxides (e.g. KO2).
  7. The algebraic sum of oxidation numbers of atoms in a chemical species (compound or ion) is equal to the net charge on the species or equal to zero if the molecule is neutral.
A few examples of computing the oxidation number of S in various compounds are described below.
  1. S8; rule 1, oxidation number = 0
  2. S2-; rule 2, oxidation number = -2
  3. ; rules 6 and 7, 2x + 3(-2) = -2, which gives x = +2
  4. H2SO4; rules 4, 6 and 7, 2(+1) + 1(x) + 4(-2) = 0, which gives x = +6
  5. S2Cl2; rules 5 and 7, 2 (x) + 2(-1) = 0, which gives x = +1
  6. Na2S4O6; rules 3, 6 and 7, 2(+1) + 4(x) + 6(-2) = 0, which gives x = 5
Problem
Calculate the oxidation number of carbon in the compounds CH4, CH3Cl, CH2Cl2, CHCl3, CCl4, C2H2, C2H4 and C2H6.

Solution
The oxidation numbers of H and Cl are +1 and -1, respectively (rules 4 and 5). If x is the oxidation number of carbon, then we will have
  1. CH4: x + 4(+1) = 0 which gives x = -4
  2. CH3Cl: x + 3(+1) + (-1) = 0 which gives x = -2
  3. CH2Cl2: x + 2(+1) + 2(-1) = 0 which gives x = 0
  4. CHCl3: x + 1(+1) +3(-1) = 0 which gives x = +2
  5. CCl4: x + 4(-1) = 0 which gives x = +4
  6. C2H2: 2x + 2(+1) = 0 which gives x = -1
  7. C2H4: 2x + 4(+1) = 0 which gives x = -2
  8. C2H6: 2x + 6(+1) = 0 which gives x = -3




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