The oxidation states of sulphur in S_8,S_2F_2 | vedclass

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(A) $1$. For $S_8$: Since it is an elemental form of sulphur,the oxidation state is $0$.$2$. For $S_2F_2$: Let the oxidation state of $S$ be $x$. The

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$0, +1$ and $-2$

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(A) $1$. For $S_8$: Since it is an elemental form of sulphur,the oxidation state is $0$.$2$. For $S_2F_2$: Let the oxidation state of $S$ be $x$. The oxidation state of $F$ is $-1$. Thus,$2x + 2(-1) = 0$,which gives $2x = 2$,so $x = +1$.$3$. For $H_2S$: Let the oxidation state of $S$ be $x$. The oxidation state of $H$ is $+1$. Thus,$2(+1) + x = 0$,which gives $2 + x = 0$,so $x = -2$.Therefore,the oxidation states are $0, +1, -2$.

Compound	Oxidation number
$1$. $C_6H_{12}O_6$ (Carbon)	$a$. $+1$
$2$. $CCl_4$ (Carbon)	$b$. $0$
$3$. $NH_4Cl$ (Nitrogen)	$c$. $+4$
$4$. $Ba(H_2PO_2)_2$ (Phosphorus)	$d$. $-3$

The oxidation states of sulphur in $S_8$,$S_2F_2$,and $H_2S$ respectively,are:

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