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What are the oxidation number of the underlined element and how do you rationalise your results?\mathrm{H}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}
CBSE | Chemistry | Class 11 | Exercise 1 | NCERT | Redox Reactions
What are the oxidation number of the underlined element and how do you rationalise your results?\mathrm{H}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}

Solution:

H2S4O6:

Let x be the oxidation number of S.

    \[\begin{array}{*{35}{l}} Oxidation\text{ }no.\text{ }of\text{ }H\text{ }=\text{ }+1 \\ Oxidation\text{ }no.\text{ }of\text{ }O\text{ }=\text{ }-2 \\ ~ \\ \end{array}\]

Then,

    \[\begin{array}{*{35}{l}} 2\left( +1 \right)\text{ }+\text{ }4\left( x \right)\text{ }+\text{ }6\left( -2 \right)\text{ }=\text{ }0 \\ 2\text{ }+\text{ }4x\text{ }-12\text{ }=\text{ }0 \\ 4x\text{ }=\text{ }10 \\ \end{array}\]

\mathrm{x}=+2 \frac{1}{2}​

Oxidation number cannot be fractional. Therefore, S would be present with different oxidation state in molecule.

 

Ncert Solutions Cbse Class 11-science Chemistry Chapter - Redox Reactions

The oxidation number  of two out of the four S atoms is +5 while that of other two atoms is 0.

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