Question

A white substance $A$ reacts with dilute $H_{2}S{O}_4$ to produce a colourless gas $B$ and a colourless solution $C$. The reaction between $B$ and acidified $K_{2}C{r}_2{O}_7$ solution produces a green solution and a slightly coloured precipitate $D$. The substance $D$ burns in air to produce a gas $E$ which reacts with $B$ to yield $D$ and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous $N{H}_3$ or $NaOH$ to $C$ produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify $A,B,C,D$ and $E$. Write the equations of the reactions involved

Answer 1

Since, the white substance $A$ gives a colourless gas $B$ with dil, $H_{2}S{O}_4$, such gas may be $H_{2}S$ So, the substance $A$ may be a metal sulphide ($Na/K/Zn$, etc.) When $H_{2}S$ gas reacts with acidified $K_{2}C{r}_2{O}_7$, it gives green coloured solution of $C{r}_2(S{O}_4{)}_3$ along with slightly yellow ppt of $D$ as sulphur. $K_{2}C{r}_2{O}_7+4H_{2}S{O}_4+3H_{2}S⟶K_{2}S{O}_4+\underset{\text{Green}}{C{r}_2(S{O}_4{)}_3}$ $+\underset{D}{3S}+7H_2$ $S$ on burning in air gives $S{O}_2(E)$. Substance $E$ on reaction with $B(H_{2}S)$ produces $D(s)$ : $\underset{B}{2H_{2}S}+\underset{E}{S{O}_2}⟶2H_{2}O(l)+\underset{D}{3S\downarrow }$ Anhydrous $CuS{O}_4$ produces blue colour in water. Solution $C$ produces ppt first with $N{H}_3/NaOH$ which dissolve in excess $N{H}_3/NaOH.$ Hence, A must be $ZnS$. $\underset{A}{\mathrm{ZnS}}+dil.H_{2}S{O}_4⟶\underset{C}{ZnS{O}_4(aq)}+H_{2}S(g)$ $\underset{C}{ZnS{O}_4}+2NaOH⟶\underset{\text{White}}{Zn(OH{)}_2\downarrow }+N{a}_{2}S{O}_4$ $Zn(OH{)}_2+2NaOH⟶[Zn(OH{)}_4{]}^{2-}+2N{a}^{+}$