Question
Calculate the oxidation number of underlined atoms.
H2SO4
H2SO4
✓
Answer
$H _2 SO _4$
Oxidation number of $H =+1$
Oxidation number of $O =-2$
$H _2 SO _4$ is a neutral molecule.
$\therefore$ Sum of the oxidation numbers of all atoms of $H _2 SO _4=0$
$\therefore 2 \times($ Oxidation number of $H )+($ Oxidation number of $S )+4 \times($ Oxidation number of $O )$ $=0$
$\therefore 2 \times(+1)+$ (Oxidation number of $S)+4 \times(-2)=0$
$\therefore$ Oxidation number of $S+2-8=0$
$\therefore$ Oxidation number of $S$ in $H _2 SO _4=+6$
Oxidation number of $H =+1$
Oxidation number of $O =-2$
$H _2 SO _4$ is a neutral molecule.
$\therefore$ Sum of the oxidation numbers of all atoms of $H _2 SO _4=0$
$\therefore 2 \times($ Oxidation number of $H )+($ Oxidation number of $S )+4 \times($ Oxidation number of $O )$ $=0$
$\therefore 2 \times(+1)+$ (Oxidation number of $S)+4 \times(-2)=0$
$\therefore$ Oxidation number of $S+2-8=0$
$\therefore$ Oxidation number of $S$ in $H _2 SO _4=+6$
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