Find the n factor of `Na_(2) S_(2) O_(3)` in the following reactions.
`Na_(2) S_(2) O_(3) + I_(2) rarr NaI + Na_(2) S_(4) O_(6)`

Explanation

Change in oxidation number of sulphur
`=|2-6| = 4`
Change in oxidation number per molecule
`:. `n factor of `Na_(2) S_(2) O_(3) = 8`
`rArr` consider the salt `A_(x) B_(y)` to undergo a reaction so that the element A undergo a change in O.S. but is present in more than one product with the same O.S. i.e.
`A_(x) ^(+e) By rarr A_(a) ^(+f) +A_(x-a)^(+f) +yB^(-(xc)/(y))` ( the superscipts denote the oxidation state of the respective elements ) .
`:. n = |x c - cf|`
`rArr` Salts that react in such a way that more than one type of atom in the salt undergoes O.S. change.
`E + A_(x) ^(+c) By rarr A_(a)^(+f) E + J_(x) B^(-1)`
In thic case both and A and B are changing their O.S.'s and both of them are either getting oxidised or reduced . In such a case n factor of the compound is the sum of the individual n factors of A and B. i.e., `|xc-xf| + |-xC - ( yi) |`. Then n factor of A can be understood which is `|x c -xf|` . Then n factor of B is `| -xc-(-yi)|`