Question

The oxidation state of $S$ in $H_{2}S{O}_5$ is

• A. +8
• B. +6
• C. -2
• D. +2

Answer 1

The correct option is B

+6

Type I. In molecules containing peroxide linkage in addition to element -oxygen bonds. Example,

(i) Oxidation number of $S$ in $H_{2}S{O}_5$ (Peroxymonosulphuric acid or Caro's acid).

By usual method; $H_{2}S{O}_5$

$2\times 1+x+5\times (-2)=0orx=+8$

But this cannot be true as maximum oxidation number for $S$ cannot exceed +6 since $S$ has only 6 electrons in its valence shell. This erroneous, exceptional value for oxidation state is due to the fact that two oxygen atoms in $H_{2}S{O}_5$ form a peroxide linkage as shown below,

Therefore the evolution of O.N. of Sulphur here should be made as follows,

$2\times (+\stackrel{H}{1})+\stackrel{S}{x}+3\times (-\stackrel{O}{2})+2\times (-\stackrel{O-O}{1})2x-6-2=0orx=+6$