Question

A substance on treatment with $dil.H_{2}S{O}_4$ liberates a colorless gas which produces (i) turbidity with ba-ryta water and (ii) turns acidified dichromate solution green. The reaction indicates the presence of :

• A. $C{O}_3^{2-}$
• B. $S^{2-}$
• C. $S{O}_3^{2-}$
• D. $N{O}^{-}$

Answer 1

Answer: (C)

$S{O}_3^{2-}$

Option (C) is correct.
The reaction indicates the presence of $S{O}_3^{2-}$.
$BaS{O}_3+2HCl(dil.)⟶2NaCl+H_{2}O+S{O}_2$;$\begin{array}{ccc}{K}_{2}C{r}_2{O}_7+H_{2}S{O}_4+3S{O}_2⟶K_{2}S{O}_4+& C{r}_2{\left(S{O}_4\right)}_3& +H_{2}O\\ & Green& \end{array}$.