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(D) To find the oxidation number of sulphur $(S)$ in $H_2SO_4$:Let the oxidation number of $S$ be $x$.$2 	imes (+1) + x + 4 	imes (-2) = 0$$2 + x -

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$+6$; $1s^2 2s^2 2p^6$

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(D) To find the oxidation number of sulphur $(S)$ in $H_2SO_4$:Let the oxidation number of $S$ be $x$.$2 	imes (+1) + x + 4 	imes (-2) = 0$$2 + x - 8 = 0$$x = +6$In $H_2SO_4$,sulphur is in its $+6$ oxidation state,meaning it has lost all its valence electrons $(3s^2 3p^4)$.The electronic configuration of neutral sulphur $(S)$ is $1s^2 2s^2 2p^6 3s^2 3p^4$.After losing $6$ electrons to reach the $+6$ state,the configuration becomes $1s^2 2s^2 2p^6$.

The oxidation number and the electronic configuration of sulphur in $H_2SO_4$ is

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