Question

Acidified $K_2{Cr}_2{O}_7$ solution turns green when ${Na}_{2}S{O}_3$ is added to it. This is due to the formation of:

• A. $Cr{O}_4^{2-}$
• B. ${Cr}_2{\left(S{O}_3\right)}_3$
• C. $CrS{O}_4$
• D. ${Cr}_2{\left(S{O}_4\right)}_3$

Answer 1

Answer: (D)

${Cr}_2{\left(S{O}_4\right)}_3$

Acidified $K_{2}C{r}_2{O}_7$ solution turns green when $N{a}_{2}S{O}_3$ is added to it. This is due to the formation of
${Cr}_2{\left(S{O}_4\right)}_3$

$K_{2}C{r}_2{O}_7+3N{a}_{2}S{O}_3+4H_{2}S{O}_4\to {Cr}_2{\left(S{O}_4\right)}_3+K_{2}S{O}_4+3N{a}_{2}S{O}_4+4H_{2}O$

Hence option D is correct.