A) 0 to 2
B) -2 to 0
C) 2 to 0
D) -2 to -1
Correct Answer: B
Solution :
\[C{{l}_{2}}+{{H}_{2}}S2HCl+S\] Let the oxidation number of S in \[{{H}_{2}}S=x\] \[2\,(+1)+x=0\] \[+2+x=0\] \[x=-\,2\] oxidation number of S in s = 0 Thus, in reaction, oxidation number of sulphur changes from -2 to 0.You need to login to perform this action.
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