CBSE Class 11-science Answered
how to calculate oxidation number
Asked by | 03 Dec, 2012, 10:04: PM
Expert Answer
First, there are some basic (general) rules that should be followed when calculating oxidation numbers:
1) For the atoms in a neutral species (i.e. an isolated atom, a molecule, or a formula unit), the sum of all the oxidation numbers is O.
Ex: Fe(s) atom has an oxidation number of 0; the sume of the oxidation numbers of all the atoms in Cl2 and C6H12O6 is 0; The sum of the oxidation numbers of the ions in MgBr2 is 0.
2) For the atoms in an ion, the sums of the oxidation numbers is equal to the charge on the ion.
Ex: The oxidation number of Cr in the Cr+3 ion is +3.
3) In compounds, the group 1 metals all have an oxidation number of +1 and the group 2 metals all have an oxidation number of +2.
Ex: The oxidation number of Na in Na2SO4 is +1.
4) In compounds, the oxidation number of fluorine is -1.
5) In compounds, hydrogen has an oxidation number of +1.
6) In most compounds, oxygen has an oxidation number of -2.
7) In binary compounds with metals, group 7 elements have an oxidation number of -1, group 6 have an oxidation number of -2, and group 5 elements have an oxidation number of -3.
Ex: The oxidation number of Br is -1 in CaBr2.
The Oxidation number of S here is to be calculated using the oxidation numbers of H and O, when the goal is that the complete compound will have an oxidation number of zero.
Oxidation numbers -
H = +1
O = -2
2x(+1)+(-2)x4= 2-8=(-6)
Since the combined oxidation numbers equal -6, to reach zero, S must be +6.
1) For the atoms in a neutral species (i.e. an isolated atom, a molecule, or a formula unit), the sum of all the oxidation numbers is O.
Ex: Fe(s) atom has an oxidation number of 0; the sume of the oxidation numbers of all the atoms in Cl2 and C6H12O6 is 0; The sum of the oxidation numbers of the ions in MgBr2 is 0.
2) For the atoms in an ion, the sums of the oxidation numbers is equal to the charge on the ion.
Ex: The oxidation number of Cr in the Cr+3 ion is +3.
3) In compounds, the group 1 metals all have an oxidation number of +1 and the group 2 metals all have an oxidation number of +2.
Ex: The oxidation number of Na in Na2SO4 is +1.
4) In compounds, the oxidation number of fluorine is -1.
5) In compounds, hydrogen has an oxidation number of +1.
6) In most compounds, oxygen has an oxidation number of -2.
7) In binary compounds with metals, group 7 elements have an oxidation number of -1, group 6 have an oxidation number of -2, and group 5 elements have an oxidation number of -3.
Ex: The oxidation number of Br is -1 in CaBr2.
The Oxidation number of S here is to be calculated using the oxidation numbers of H and O, when the goal is that the complete compound will have an oxidation number of zero.
Oxidation numbers -
H = +1
O = -2
2x(+1)+(-2)x4= 2-8=(-6)
Since the combined oxidation numbers equal -6, to reach zero, S must be +6.
Answered by | 04 Dec, 2012, 10:04: AM
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