Question

The oxidation number of sulfur in ${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ is:
A) ${ +2 }$
B) ${ +4 }$
C) ${ +6 }$
D) ${ +7 }$

Answer 1

Hint: An oxidation number is a number assigned to an element in a chemical combination which represents the number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound.

Complete step-by-step answer:
The structure of ${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ is

Let, the oxidation number of sulfur is = x
In peroxydisulfuric acid, two oxygen atoms are involved in peroxide linkage, so the oxidation number will be ${ -1 }$ each. The remaining oxygen atoms have as usual a ${ -2 }$ charge.
So, the oxidation number of sulfur in ${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ can be calculated as:
${ 2 + 2x+ 2(-1)+ 6(-2) = 0 }$
$\Rightarrow { 2+2x-2-12 = 0 }$
$\Rightarrow { 2x-12 = 0 }$
$\Rightarrow { 2x = 12 }$
$\Rightarrow { x = 12/2 }$
$\Rightarrow x = { +6 }$
Hence, the oxidation number of sulfur in ${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$ is ${ +6 }$.
The correct option is C.

Additional Information:
>There are some rules for assigning oxidation numbers to an atom: The rules have been formulated on the basis of the assumption that electrons in a covalent bond belong entirely to the more electronegative atom.
>Oxidation number (O.N) of:
atoms in free elemental state (like ${ H }_{ 2 }$, Na, ${ O }_{ 2 }$, Ag etc) = 0
>Oxidation number of simple monatomic ions = Charge on them ( For example : Halogens (like Fluorine, chlorine) = -1, ${ Na }^{ + }$= +1, ${ Ca }^{ +2 }$= +2 etc)
>Oxygen = -2; in peroxides(-1); ${ F }_{ 2 }{ O }$ (+2); ${ F }_{ 2 }{ O }_{ 2 }$ (+1)
>Hydrogen = +1; however in metal hydrides it is (-1)
>Sum of O.N. of all the atoms in molecules = 0
>Sum of O.N. of atoms in polyatomic ions = overall charge on them

Note: The possibility to make a mistake is that you may choose option D. As in this compound, peroxide is present and the oxidation number of these are ${ -1 }$, not ${ -2 }$, so instead of 8 oxides, there are 6 oxides and 2 peroxides.