Question

calculate oxidation number of sulphur from Na2S4O6.

Answer 1

2+4x-12=0
by solving it we get 2.5 as an answer
the oxidation no of Na is 2 as it valency and no of Na ion the sulphur no let x and ion is 4 then it will be 4x then oxygen valency 2- and ion is6 it will become -12 by that we get 2.5 mark as briliant

Answer 2

Explanation:

It is known that oxidation state of sodium is +1, so oxidation state for two sodium atoms will be +2. Oxidation state of oxygen is -2 so, the oxidation state of six oxygen atoms will be -2 × 6 = -12.

Let the oxidation state of sulphur is x.

Calculate oxidation number of sulphur from $Na_{2}S_{4}O_{6}$ is as follows.

                          2 + 4x - 12 = 0

                                4x = +10

                           $x = \frac{10}{4}$

                                      = 2.5

Therefore, the oxidation number of sulphur in $Na_{2}S_{4}O_{6}$ is 2.5.