What is the average oxidation state of $\mathrm{S}$ atoms in ${\mathrm{H}}_2{\mathrm{S}}_4{\mathrm{O}}_6$?

Step 1: Given information

• The given molecule is ${\mathrm{H}}_2{\mathrm{S}}_4{\mathrm{O}}_6$ (Tetrathionic acid).

Step 2: Oxidation state or number

• The absolute number of electrons that a particle gains or loses to form a chemical bond with another molecule are known as the oxidation number.
• The oxidation number of any free element is zero.
• In neutral compounds, the total of all the constituent atoms' oxidation numbers equals zero.
• The polyatomic ion's net charge equals the total sum of all the oxidation numbers of the atoms that make up it.

Step 3: Calculate the average oxidation state of Sulfur atoms

• Let the oxidation state of the sulfur $\left(\mathrm{S}\right)$ atom in ${\mathrm{H}}_2{\mathrm{S}}_4{\mathrm{O}}_6$ be “$\mathrm{x}$.”
• The oxidation state of hydrogen $\left(\mathrm{H}\right)$ is $1$.
• The oxidation state of oxygen $\left(\mathrm{O}\right)$ is $-2$.
• The average oxidation state of sulfur atoms is as follows:

$\begin{array}{rcl}2\times \left(1\right)+4\times \left(\mathrm{x}\right)+6\times (-2)& =& 0\\ 2+4\mathrm{x}-12& =& 0\\ 4\mathrm{x}-10& =& 0\\ \mathrm{x}& =& \frac{10}{4}\\ & =& 2.5\end{array}$

Therefore, the oxidation state for sulfur atoms in ${\mathbf{H}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{6}}$ is $\mathbf{2}\mathbf{.5}$.