Question

What is the oxidation number of 'S' bond with oxygen in $S_4O_6^{2-}$ ion? $4x - 2 \times 6 = -2$

• A. +10
• B. +5
• C. +4
• D. -4

Answer 1

Answer: (B)

+5

Answer 2

In the tetrathionate ion, $\mathrm{S_4O_6^{2-}}$, the overall oxidation numbers sum to $-2$. If we assume each sulfur had the same oxidation state $x$, then:

$$4x + 6(-2) = -2 \quad\Longrightarrow\quad 4x - 12 = -2 \quad\Longrightarrow\quad 4x = 10 \quad\Longrightarrow\quad x = +2.5$$

This is the average oxidation number of sulfur.

However, the tetrathionate ion actually contains two different types of sulfur atoms:

• Terminal S atoms (bonded to oxygen): These are part of sulfonate ($\mathrm{-SO_3}$) groups and hence have oxidation states similar to those in sulfate, i.e., $+5$.
• Central (bridging) S atoms: Their oxidation number adjusts so that the overall average becomes $+2.5$. (Here, each central S comes out as 0.)

Thus, the sulfur atoms that are bonded to oxygen (the terminal ones) have an oxidation number of $+5$.