#### Question

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

H_{2}__S___{4}O_{6}

#### Solution 1

H_{2}__S___{4}O_{6}

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.

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

#### Solution 2

By conventional method O.N of S in `H_2S_4O_6` =

or 2(+1) + 4x + 6(-2) = 0 or x = 2.5(wrong)

But it is wrong because all the four S atoms cannot be in same oxidation state.

By chemical bonding method. The structure of `H_2S_4O_6` is shown below

The O.N. of each of the S-atoms linked with each other in the middle is zero while that of each of the reamining two S-atoms is +5