Respuesta :
Oxidation state of sulfur S: +5.
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:
- [tex]\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}\\[/tex].
What the question is asking for:
- The oxidation state of S can vary from compound to compound. Let the oxidation state of S in [tex]\text{Na}_2\text{S}_2\text{O}_6[/tex] be [tex]x[/tex].
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 [tex]{\bf 2}\times(+1) = +2[/tex];
- There are six O atoms in this formula. The oxidation state on each is -2. Add [tex]{\bf 6}\times (-2) = -12[/tex];
- There are two S atoms in this formula. The oxidation state of each is assumed to be [tex]x[/tex]. Add [tex]{\bf 2} \; x[/tex].
Sum of oxidation states on each atom in [tex]\text{Na}_2\text{S}_2\text{O}_6[/tex]:
[tex]{\bf 2}\times(+1) + {\bf 6}\times (-2) +{\bf 2} \; x[/tex].
Again, this value should equals to zero since [tex]\text{Na}_2\text{S}_2\text{O}_6[/tex] is overall a neutral compound.
[tex]{\bf 2}\times(+1) + {\bf 6}\times (-2) +{\bf 2} \; x = 0[/tex];
[tex]2 -12 + 2\;x = 0\\2\;x = 10\\x = 5[/tex].
Thus the oxidation state of sulfur in [tex]\text{Na}_2\stackrel{+5}{\textbf{S}}_2\text{O}_6[/tex] should be +5.