The oxidation number of sulphur in H2S2O8 is

Question

The oxidation number of sulphur in H2S2O8 is (A) +2 (B) +4 (C) +6 (D) +7

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f1392bce96a9265bafe54f716cc673c2-.png" /> ∵ 2 O -atoms are in peroxide linkage ∴ Their oxidation state =-1 and, oxidation state of remaining 6O -atoms =-2 Oxidation state of H -atoms =+1 Let oxidation state of S atom =x S 2 H 2 O 6 O 2 ∴ 2 x+2(+1)+6(-2)+2(-1) =0 2 x+2-12-2 =0 x =+6

Q. The oxidation number of sulphur in $H_{2} S_{2} O_{8}$ is

$+ 2$

$+ 4$

$+ 6$

$+ 7$

Q. The oxidation number of sulphur in H2​S2​O8​ is

+2

+4

+6

+7

∵2O -atoms are in peroxide linkage ∴ Their oxidation state =−1 and, oxidation state of remaining 6O -atoms =−2 Oxidation state of H -atoms =+1 Let oxidation state of S atom =x S2​H2​O6​O2​ ∴2x+2(+1)+6(−2)+2(−1)=0 2x+2−12−2=0 x=+6

Q. The oxidation number of sulphur in $H_{2} S_{2} O_{8}$ is

$+ 2$

$+ 4$

$+ 6$

$+ 7$