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प्रश्न
What are the oxidation numbers of the underlined elements in the following and how do you rationalise your results?
H2S4O6
उत्तर
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.
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