Question

What is the oxidation no. Of sulphur in Na2S2O6

Answer 1

Oxidation state of sulfur S: +5.

Step-by-step explanation

Known oxidation states:

• Na: +1. Sodium Na is a main group metal. The oxidation state of Na in compounds should be positive and the same as its group number. The new IUPAC group number (1 ~ 16) of Na is 1. The oxidation state of Na should thus be +1.
• O: -2. Oxygen is the second most electronegative element on the periodic table. The oxidation state of oxygen is mostly -2 with a few exceptions:
• 
  `\begin{array}{cc}\text{Compound} & \text{Oxidation state of Oxygen}\\\textbf{O}\text{F}_2&+2\\\text{Peroxides like H}_2\textbf{O}_2 & -1\end{array}\\`.

What the question is asking for:

• The oxidation state of S can vary from compound to compound. Let the oxidation state of S in
  `\text{Na}_2\text{S}_2\text{O}_6` be
  `x`.

The oxidation state of each atom in a compound should add up to zero.

• There are two Na atoms in this formula. The oxidation state on each is +1. Add
  `{\bf 2}*(+1) = +2`;
• There are six O atoms in this formula. The oxidation state on each is -2. Add
  `{\bf 6}* (-2) = -12`;
• There are two S atoms in this formula. The oxidation state of each is assumed to be
  `x`. Add
  `{\bf 2} \; x`.

Sum of oxidation states on each atom in
`\text{Na}_2\text{S}_2\text{O}_6`:

`{\bf 2}*(+1) + {\bf 6}* (-2) +{\bf 2} \; x`.

Again, this value should equals to zero since
`\text{Na}_2\text{S}_2\text{O}_6` is overall a neutral compound.

`{\bf 2}*(+1) + {\bf 6}* (-2) +{\bf 2} \; x = 0`;

`2 -12 + 2\;x = 0\\2\;x = 10\\x = 5`.

Thus the oxidation state of sulfur in
`\text{Na}_2\stackrel{+5}{\textbf{S}}_2\text{O}_6` should be +5.