Sulphur has the highest oxidation state in

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(B) To find the oxidation state of sulphur $(S)$ in each compound:$1$. In $SO_2$: $x + 2(-2) = 0 \implies x = +4$.$2$. In $H_2SO_4$: $2(+1) + x + 4(-2

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$H_2SO_4$

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(B) To find the oxidation state of sulphur $(S)$ in each compound:$1$. In $SO_2$: $x + 2(-2) = 0 \implies x = +4$.$2$. In $H_2SO_4$: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$.$3$. In $Na_2S_2O_3$: $2(+1) + 2x + 3(-2) = 0 \implies 2 + 2x - 6 = 0 \implies 2x = 4 \implies x = +2$.$4$. In $Na_2S_4O_6$: $2(+1) + 4x + 6(-2) = 0 \implies 2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5$.Comparing the values $(+4, +6, +2, +2.5)$,the highest oxidation state is $+6$ in $H_2SO_4$.

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