Question

When $S{O}_2$ is passed in acidified potassium dichromate solution, the oxidation number of $S$ is changed from:

• A. $+4$ to zero
• B. $+4$ to $+2$
• C. $+4$ to $+6$
• D. $+6$ to $+4$

Answer 1

Answer: (C)

$+4$ to $+6$

The reaction of sulphur dioxide with acidified $K_{2}C{r}_2{O}_7$ in acidic medium is

$3S{O}_2+C{r}_2{O}_7^{2-}+2H^{+}\to 3S{O}_4^{2-}+2C{r}^{3+}+H_{2}O$

So Oxidation state of Sulphur in $S{O}_2$ is $+4$ which is changed to $+6$ in $S{O}_4^{2-}$ .

Hence option $C$ is correct.