Q. When sulphur dioxide is passed in an acidified $K_{2}{Cr}_{2}O_{7}$ solution, the oxidation state of sulphur is changed from
A) 4 to 0
B) 4 to 2
C) 4 to 6
D) 6 to 4
Explanation
(c)
Acidified $K_{2}{Cr}_{2}O_{7}$ solution oxidises ${SO}_{2}\$into ${Cr}_{2}\left( {SO}_{4} \right)_{3}.$
_{+4 +6}
$3{SO}_{2} + K_{2}{Cr}_{2}O_{7} + H_{2}{SO}_{4} \longrightarrow K_{2}{SO}_{4} + {Cr}_{2}\left( {SO}_{4} \right)_{3} + H_{2}O$
Hence, oxidation state of sulphur changes from +4 to +6.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.