Q. 2 B3.8( 31 Votes )
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
Let the O.N. of S be ‘x’.
Now,2(+ 1) + 4(x) + 6(-2) = 0
2 + 4x-12 = 0
X = + 2
However, the 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.
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