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`        How do we calculate the n factor of a noon stoichiometric compound? `
one year ago

dolly bhatia
200 Points
```							Hypo solution (sodium thiosulfate, Na2S2O3.5H2O) is a good reductant.With iodine, l2, it gives (S4O6)2-, tetra-thionate. I2 + 2Na2S2O3 = Na2S4O6 + 2NaIHere, oxidation number per sulphur atom changes from +2 to +2.5.But, one thiosulfate ion has two sulphur atoms, hence, n-factor i.e., total change in oxidation number per molecule is 0.5 x 2 = 1.Sodium thiosulfate, Na2S2O3.5H2O, when reacts with stronger oxidant Cl2 or Br2, it is oxidised to greater extent and gives (SO4)2-, sulfate.Na2S2O3 + 4 Cl2 + 5 H2O → 2 NaHSO4 + 8 HCl Here, oxidation number per sulphur atom changes from +2 to +6.But, one thiosulfate ion has two sulphur atoms, hence, n-factor i.e., total change in oxidation number per molecule/ion is 4 x 2 = 8.For any reaction, n = ½ax – ay½To calculate n-factor of a salt, we take one mole of the reactant and find the number of mole of the element whose oxidation state is changing. This is multiplied with the oxidation state of the element in the reactant, which gives us the total oxidation state of the element in the reactant. Now, we calculate the total oxidation state of the same element in the product for the same number of mole of atoms of that element in the reactant. Remember that the total oxidation state of the same element in the product is not calculated for the number of mole of atoms of that element in the product.Also, keeping in mind the basic definition of n factor helps in finding out the n factor of any reaction logically.N factor is defined as the total moles of catioinic/anionic charge replaced in 1 mole of the salt.
```
one year ago
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