Which choice gives the correct oxidation numbers for all three elements in rb2so3 in the order that the elements are shown in the formula?

             Rb  =  + 1

              S    =  + 4

              O   =  - 2

                   Oxidation states of the elements were calculated keeping in mind the basic rules of assigning oxidation states which included assignment of +1 charge to first group elements i.e. Rubidium (Rb) and assignment of -2 charge to Oxygen atom. Then the oxidation state of Sulfur was calculated as follow,

Rb₂ + S + O₃  =  0

Above zero (0) means that the overall molecule is neutral.

Putting values of Rb and O,

                                           (+1)₂ + S + (-2)₃  =  0

                                           (+2) + S + (-6)  =  0

                                           +2 + S - 6  = 0

                                           S - 6  =  -2

                                          S  =  -2 + 6

                                           S  =  + 4

56 moles of   lieter in chemistry