Sulphur has the lowest oxidation number in

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(D) To find the oxidation number of sulphur $(S)$ in each compound:$1$. In $H_2SO_3$: $2(+1) + x + 3(-2) = 0$ $\Rightarrow 2 + x - 6 = 0$ $\Rightarrow

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$H_2S$

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(D) To find the oxidation number of sulphur $(S)$ in each compound:$1$. In $H_2SO_3$: $2(+1) + x + 3(-2) = 0$ $\Rightarrow 2 + x - 6 = 0$ $\Rightarrow x = +4$.$2$. In $SO_2$: $x + 2(-2) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow x = +4$.$3$. In $H_2SO_4$: $2(+1) + x + 4(-2) = 0$ $\Rightarrow 2 + x - 8 = 0$ $\Rightarrow x = +6$.$4$. In $H_2S$: $2(+1) + x = 0$ $\Rightarrow 2 + x = 0$ $\Rightarrow x = -2$.Comparing the values $(+4, +4, +6, -2)$,the lowest oxidation number is $-2$,which corresponds to $H_2S$.

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