Question

The table shows some reactions of a white compound, G.

Test	Observation
Silver nitrate is added to a solution of G followed by aqueous ammonia	A precipitate is formed which does not dissolve when the ammonia is added
Solid G is warmed with concentrated sulfuric acid	A mixture of gases is formed including hydrogen sulphide

What could be the identity of G?

Answer 1

Hint: The reaction of the halides with the silver nitrate results in the precipitate of silver halides. These precipitates when treated with ammonia do not dissolve in the solution. The halides are reducing agents .They reduce the sulphuric acid in the oxides of sulphur. The halide which has a strong reducing property can further reduce the oxides and liberates a $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ gas.


Complete step by step answer:
The ionic compounds can form the precipitate in the solution if the concentration of ions increases above a certain value. The solubility product expresses the solubility of the compound. If the value exceeds this value, the product cannot be soluble and it forms a precipitate.
The solubility product of $\text{ AgI }$ is equal to the$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }8.3\text{ }\times \text{ }{{10}^{-17}}\text{ }$. The silver iodide is pretty insoluble in the solution.
When the sodium iodide is treated with the silver nitrate, the silver nitrate combines with the iodide and forms a silver iodide ($\text{AgI}$) compound. The reaction is as shown below,
$\text{ NaI + AgN}{{\text{O}}_{\text{3}}}\text{ }\to \text{ AgI(s) }\downarrow \text{ + NaN}{{\text{O}}_{\text{3}}}\text{ }$

Ammonia $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$ is added to the $\text{NaI}$ and $\text{AgN}{{\text{O}}_{\text{3}}}$solution which also contains the precipitate of$\text{AgI}$. The ammonia interacts with the silver ions from the solution and forms a stable diamine silver (I) ion $\text{ }{{\left[ \text{Ag(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}} \right]}^{\text{+}}}\text{ }$ . This is a very stable complex. The equilibrium of the reaction always lies to the right.
$\text{ A}{{\text{g}}^{\text{+}}}\text{(aq) + 2N}{{\text{H}}_{\text{3}}}\text{(aq) }\rightleftharpoons \text{ }{{\left[ \text{Ag(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}} \right]}^{\text{+}}}\text{(aq) }$

Silver ions are already precipitated as the silver iodide. Thus the solution contains a very small concentration of silver ions. These silver ions react with the ammonia and lower the concentration of silver ions from the solution.
Thus, the yellow colour of the precipitate does not dissolve when ammonia is added to it.
The halides react differently with$\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$. Iodide or iodine ions are strong reducing agents these ions can oxidize to iodine in presence of$\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$.

$\text{ 2}{{\text{I}}^{-}}\text{ }\to \text{ }{{\text{I}}_{\text{2}}}\text{ + 2}{{\text{e}}^{-}}\text{ }$
The reduction of acid is more complicated. The iodide reduces the $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$ to the hydrogen sulphide. The half equation is as:
$\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }+\text{ 8}{{\text{H}}^{\text{+}}}\text{ + 8}{{\text{e}}^{-}}\text{ }\to \text{ }{{\text{H}}_{\text{2}}}\text{S + 4}{{\text{H}}_{\text{2}}}\text{O }$

The equation for the reaction of iodide (from sodium iodide) with the $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$is as follows,${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }+\text{ 8}{{\text{H}}^{\text{+}}}\text{ + 8}{{\text{I}}^{-}}\text{ }\to \text{ 4}{{\text{I}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{S + 4}{{\text{H}}_{\text{2}}}\text{O}$
Here, we know that the compound which reacts with the silver nitrate and precipitates followed by the treatment with the sulphuric acid is sodium iodide $\text{NaI}$.
Hence, $\text{ G = NaI }$

Note: Generally all halides give the precipitate of silver halide. However, the precipitate of chloride is white, for the bromide, it is a cream colour precipitate and the iodide forms a pale yellow colour precipitate. The $\text{ }{{\text{F}}^{-}}\text{ }$ and $\text{C}{{\text{l}}^{-}}$ are not oxidized by the acid, bromine reduces the acid to form $\text{ S}{{\text{O}}_{\text{2}}}\text{ }$ , but since iodide is a strong reducing agent it further reduces the $\text{ S}{{\text{O}}_{\text{2}}}\text{ }$to$\text{ }{{\text{H}}_{\text{2}}}\text{S }$.