A) \[S\]in \[{{H}_{2}}S\]has - 2 oxidation state
B) \[S\]in \[S{{O}_{2}}\]has oxidation state + 4
C) Hydrogen in \[{{H}_{2}}S\]more +ve than oxygen
D) Oxygen is more - ve in \[S{{O}_{2}}\]
Correct Answer: A
Solution :
In \[{{H}_{2}}S\] sulphur shows -2 oxidation state and in \[S{{O}_{2}}\] shows +4 oxidation state. Hence \[S{{O}_{2}}\] shows both oxidising and reducing properties.
You need to login to perform this action.
You will be redirected in
3 sec
