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What is the Oxidation State of Sulphur in H2S2O8 ??
What is the Oxidation State of Sulphur in
H2S2O8 ??
+6. This is known as peroxy di sulphuric acid(marshals acid) where peroxylinkage is there ,so six oxygen normal with -2 charge while two with -1 chrge due to peroxy linkage so sulphur oxidation number is +6 .HO-S-O-O-S-OH both sulphur attach with two cordinate oxygen atom each.
0=2*1+2x+ (-2)*8 where x is the oxidation state therefore the oxidation state is 14 correct me i i am wrong somewhere Regards
0=2*1+2x+ (-2)*8
where x is the oxidation state
therefore the oxidation state is 14
correct me i i am wrong somewhere
Regards
O.S for H is +1 O.S for O is -2 let O.S for S be x , then 2+2x-16=0 x=+7 is O.S of sulphur
O.S for H is +1
O.S for O is -2
let O.S for S be x , then
2+2x-16=0
x=+7 is O.S of sulphur
Hi, in this the two S atoms peroxy linkage between them as shown; (-S-O-O-S-) & each S has 2 double bonded O atoms & 1 -OH group. Hence, O.N. of S =+6.
OXIDATION STATE OF SULPHUR IS 7
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