Maharashtra BoardEnglish MediumSTD 11 ScienceChemistryRedox Reactions2 Marks
Question
Calculate the oxidation number of underlined atoms. H2S4O6
✓
Answer
$H _2 S _4 O _6$ Oxidation number of $H =+1$ Oxidation number of $O =-2$ $H _2 S _4 O _6$ is a neutral molecule. $\therefore$ Sum of the oxidation numbers of all atoms $=0$ $\therefore 2 \times($ Oxidation number of $H )+4 \times$ (Oxidation number of $S )+6 \times$ (Oxidation number of $O )=0$ $\therefore 2 \times(+1)+4 \times($ Oxidation number of S) $+6 \times(-2)=0$ $\therefore 4 \times$ (Oxidation number of S) $+2-12=0$ $\therefore 4 \times($ Oxidation number of $S)=+10$ $\therefore$ Oxidation number of $S =+\frac{10}{4}$ $\therefore$ Oxidation number of $S$ in $H _2 S _4 O _6=+2.5$
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