Question

Consider the following statements and arrange in the order of true/false as given in the codes.
$S_1$: Like $C{O}_3^{-2},S{O}_3^{2-}$ ion also gives test with baryta water, $Ba(OH{)}_2$.
$S_2$: $A{g}_{2}S{O}_3$ is insolube in dilute $HN{O}_3$.
$S_3$: Triodide ions $\left(I_3^{2-}\right)$ produced by the reaction of $C{u}^{2+}$ and KI solution are not reduced by excess of sodium thiosulphate solution.
$S_4$: Black precipitate of $Cu(SCN{)}_2$ turns into white precipitate, when it reacts with saturated solution of $S{O}_2$ in water.

• A. FTTF
• B. TFFT
• C. FTFT
• D. TTFT

Answer 1

Answer: (B)

TFFT

$S_1:$ It is true, as $C{O}_3^{-2},S{O}_3^{2-}$ ions also give test with baryta water, $Ba(OH{)}_2$.

$S_2:$ It is false, as $A{g}_{2}S{O}_3$ is soluble in dilute $HN{O}_3$.

$S_3:$ It is also false. The reaction is as follows, and forms $I_2$

$2C{u}^{2+}+2I^{-}\to 2C{u}^{+}+I_2$

The copper (I) sulphate formed is a white insoluble salt (that slowly disproportionates in aqueous conditions $2C{u}^{+}\to C{u}^{2+}+C{u}^{-}$ but too slowly to affect this reaction scheme)

Reaction with sodium thiosulphate reduces the iodine formed back to iodide ions and the yellow colour (iodine in aqueous solution) disappears

$2S_2{O}_3^{2-}+I_2\to S_4{O}_6^{2-}+2I^{-}$

$S_4:$ It is also true, as black precipitate of $Cu(SCN{)}_2$ turns into white precipitate when it reacts with saturated solution of $S{O}_2$ in water.