Oxidation number of sulphur in H2S2O8 is:

Question

Oxidation number of sulphur in H2S2O8 is: (A) 6 (B) 8 (C) 21 (D) 7

Tardigrade Admin · Accepted Answer

In H2S2O8 (morshall?s acid), two O-atom have-1 oxidation state each (peroxide Imkage) and rest six O-atom have-2 oxidation state. Hence 2(+1)+2x+2(-1)+6(-2)=0 2+2x-2-12=0 2x=12 x=+6

Q. Oxidation number of sulphur in $H_{2} S_{2} O_{8}$ is:

6

8

21

7

In $H_{2} S_{2} O_{8}$ (morshall?s acid), two O-atom have-1 oxidation state each (peroxide Imkage) and rest six O-atom have-2 oxidation state. Hence $2 (+ 1) + 2 x + 2 (- 1) + 6 (- 2) = 0$ $2 + 2 x - 2 - 12 = 0$ $2 x = 12$ $x = + 6$

Q. Oxidation number of sulphur in H2​S2​O8​ is:

6

8

21

7

In H2​S2​O8​ (morshall?s acid), two O-atom have-1 oxidation state each (peroxide Imkage) and rest six O-atom have-2 oxidation state. Hence 2(+1)+2x+2(−1)+6(−2)=0 2+2x−2−12=0 2x=12 x=+6

Q. Oxidation number of sulphur in $H_{2} S_{2} O_{8}$ is:

In $H_{2} S_{2} O_{8}$ (morshall?s acid), two O-atom have-1 oxidation state each (peroxide Imkage) and rest six O-atom have-2 oxidation state. Hence $2 (+ 1) + 2 x + 2 (- 1) + 6 (- 2) = 0$ $2 + 2 x - 2 - 12 = 0$ $2 x = 12$ $x = + 6$