So for #"sulfate anion"# we have, FORMALLY, #""^(-)O-S(=O)_2O^-#. The oxidation number of oxygen is usually #-II# and it is here.
Thus #S_"ON"+4xx(-2)=-2#; clearly, #S_"ON"=+VI#. Oxygen gets both electrons from a #S-O# bond because it is more electronegative than sulfur.
For arsenic sulfide, #As_2S_3#, which is think is used by artists as the vivid yellow pigment, #"King's yellow"#, sulfur is now bound to an element whose electronegativity is LESS than its own and thus sulfur is here an analogue of oxygen, and assumes a #S^(-2)# i.e. #S(-II)# oxidation state; i.e. here it is an oxide analogue; and we necessarily have #As(+III)# centres.
What is the oxidation state of elemental sulfur, #S_8#?
