The oxidation number of sulphur is -4 in | vedclass

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Question

(D) The correct option is $D$.Let us calculate the oxidation number of $S$ in the given compounds:$(i)$ In $H_2S$:Let the oxidation state of $S$ be $x

Vedclass · Accepted Answer

None of the above

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(D) The correct option is $D$.Let us calculate the oxidation number of $S$ in the given compounds:$(i)$ In $H_2S$:Let the oxidation state of $S$ be $x$.$2(+1) + x = 0$$x = -2$$(ii)$ In $CS_2$:Let the oxidation state of $S$ be $x$.$C$ is $+4$ (as $C$ is more electronegative than $S$ is not true,but $C$ is $+4$ in $CS_2$ based on standard rules).$+4 + 2(x) = 0$$2x = -4$$x = -2$$(iii)$ In $Na_2SO_4$:$2(+1) + x + 4(-2) = 0$$2 + x - 8 = 0$$x = +6$Since none of the compounds have an oxidation state of $-4$ for sulphur,the correct option is $D$.

The oxidation number of sulphur is $-4$ in

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