In quantitative analysis which of the following groups of metal ions are precipitated as insoluble sulphides when`H_(2)S(g)` is passed through an ammonical solution of the salt mixture?

Options

A. `III B`
B. `IV`
C. `IIA` and `IIB`
D. `V`

Answer

1

Explanation

The formation of insoluble metal sulphides is quite common in inorganic qualitative analysis. Hydrogen sulphide is bubbled through an acid solution containing unknown metal ions. The presence of the high hydrogen ion concentration reduces the sulphide ion concentration to extremely low levels:
`H_(2)S(aq.)+2H_(2)O(l) hArr 2H_(3)O^(+)(aq.)+S^(2-) (aq.)`
This very low level of sulphide ion is still enough to precipitate the most insoluble metal sulphides, those with a solubility product `K_(sp)` smaller than`10^(-30)`. These metal sulphides are separated by filtration or centrifugation. The pH of the filtrate is increased by adding base. This increase shifts the sulphide equilibrium to the right, thereby raising the concentretion of sulphide ions to the point where those metal sulphide with a solubility product between `10^(-20)` and `10^(-30)` (mainly the transition metals of period 4) precipitate `IIA` and `IIB`. groups of metal ions are precipitated as insoluble sulphides when `H_(2)S` is passed through a weakly acidified solution of the salt mixture whilst `III` B group of metal ions are separated as insoluble sulphides when `H_(2)S` is passed through an ammoniacal solution of the salt mixture.
Specific tests could then be used to identify which metal ions are present. More recently, thio- acetamide, a reagent that hydrolyzes to hydrogen sulphide, has been used for such tests. In addition to forming conventional sulphides, some elements form disulphides,`S_(2)^(2-)` ions analogous to dioxide`(2-)`. Thus, `FeS_(2)` does not contain iron in a high oxidation state, but the disulphide ion and `Fe^(2+)`. Also the alkali and alkaline earth metals form polysulphides, which contain the `S_(n)^(2-)` ion, where n has value between 2 and 6.