But there is an alternative formalism, which I will outline.....We gots #"sulfuric acid"#, #H_2SO_4#, and here CLEARLY we got #S(VI)#. And as is typical the sum of the oxidation numbers of each constituent element equals the charge on the molecule, i.e. here #0#...
And so for sulfuric acid.....
#S(VI+) + 4xxO(-II)+2xxH(+I)=0#
....but when we use #"thiosulfuric acid"# I like to think that we have replaced ONE of the oxygen atoms with a SULFUR atom, and this sulfur atom assumes the SAME oxidation state as the oxygen atom it replaces...i.e. we gots #S(-II)#...and so here....
#S(VI+) +S(-II)+ 3xxO(-II)+2xxH(+I)=0#....as before....
And, clearly, both sulfur, and oxygen are Group 16 atoms, so this formalism is not totally whack.
#S_"average oxidation number"=(""^+VI-II)/2=+II#....but the designation of individual oxidation numbers here can be useful as a formalism.
