Question
What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
H2S4O6
Solution 1
H2S4O6
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
