Question

What are the oxidation numbers of the underlined elements in the following and how do you rationalise your results?

H₂S₄O₆

Answer 1

H₂S₄O₆

+1	x	-2
H2	SO4	O6

Now 2(+1) + 4(x) + 6(-2) = 0

⇒ 2 + 4x - 12 = 0

⇒ 4x = 10

⇒ `x = +2 1/2`

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

\[\begin{array}{cc}
\ce{O}\phantom{..........}\ce{O}\\
||\phantom{...........}||\\
\ce{H - O - S - S - S - S - O - H}\\
||\phantom{...........}||\\
\ce{O}\phantom{..........}\ce{O}
\end{array}\]

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.