What is the Oxidation no. of Sulphur in #(CH_"3")_"2"SO# ?

1 Answer
May 31, 2017

Answer:

Well, sulfur is slightly more electronegative than carbon......so it is formally zerovalent......

Explanation:

And oxidation state is the charge assigned to the central atom, when all the bonds are (conceptually) broken, with the electronic charge (i.e. the two bonding electrons) assigned to the most electronegative atom.

We do this for #H_3C-S(=O)-CH_3#, and get...........

#H_3C-S(-O^+)-CH_3rarr2xx""^(+)CH_3+""^(-2)(S=O)#

And breaking the #""^-2(S=O)# bond........

#""^(-2)(S=O)rarrstackrel(0)S+stackrel(-II)O#.

And thus we have #stackrel(0)S#, #stackrel(-II)O#, #stackrel(+I)H#, and #stackrel(-II)C#.

For more on the assignment of oxidation states, see here.