How to calculate oxidation Number..???

Question

how to calculate oxidation Number..???

Answer 1

Actually

First, there are some basic (general) rules that should be followed when assigning 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.

So lets do an example:

KClO4

According to rule 3, the oxidation number of K is +1. According to rule 6, the oxidation number of O is -2 and the total for 4 O atoms is -8. For these two elements , nsthe total is +1 - 8 = -7. The oxidation number of Cl must be +7 to give a total of zero for all atoms in this formula unit.

Answer 2

First, there are some basic (general) rules that should be followed when assigning 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.

So lets do an example:

KClO4

According to rule 3, the oxidation number of K is +1. According to rule 6, the oxidation number of O is -2 and the total for 4 O atoms is -8. For these two elements , nsthe total is +1 - 8 = -7. The oxidation number of Cl must be +7 to give a total of zero for all atoms in this formula unit.

Make sense?

Answer 3

calculation of o.n.is a very relative method,it only holds meaning for substnces that can be oxiidisedorreduced.eg.KMno4,now we know that k with atomic no. 19 needs to lose one electron to attain noble gas configuration  n thus acquires a charge of +1.similarly O needs to gain 2 more electrons to fill up its valence shell,thus is in -2 state.now electrons lost by Mn vary due to its configuration.now take charges of various atoms into considertn n equatr it to total charge on the molecule thus O.N. of Mn=>Mn+(-2)*4+1=0=>+7.similarly in k2Cr207,o.number of cr is +6{+1*2+(-2)*7+2*Cr=0)