What is the oxidation number of the sulfur atom in #Li_2SO_4#?

2 Answers
May 29, 2017

Answer:

We has #S^(VI+)#

Explanation:

Sulfur in sulfate expresses it maximum oxidation state, its Group Number, #+VI#.

How? Well, as you know the sum of the oxidation numbers MUST equal the charge on the ion..........

And so #S_"oxidation number"+4xxO_"oxidation number"=-2#

Now oxygen generally assumes an oxidation number of #-II# in its compounds, and certainly it does so here.........

#S_"oxidation number"+4xx(-2)=-2#

Add #8# to both sides of the equation..........

#S_"oxidation number"+cancel8+ cancel(4xx(-2))=-2+8#

#S_"oxidation number"=+6=+VI#..........

What about #"thiosulfate"#, #S_2O_3^(2-)#? What are the oxidation numbers of sulfur here?

May 29, 2017

Answer:

#+6#

Explanation:

Charge of #Li_2SO_4# is balanced

The charge of #Li# is #+1# (because it is a #1^(st)# group element)

There are #2# #Li# atoms therefore the charge of #Li_2# is #+2# and #SO_4# must have a total charge of #-2# (to balance the charge)

Oxygen "in this case (because it is not a peroxide)" has a charge of #-2# (if it was a peroxide "#(O_2)^(2-)#" it would have had a charge of #-1#). Because there is #4# oxygen atoms, the total charge of oxygen will be #4*(-2)=color(red)(-8)#

Sulfur must have a charge that will make( #color(red)(-8)+# its charge)#=-2# ("charge of #SO_4#")

Therefore the charge of sulfur #=-2+8=color(blue)(+6)#