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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