Question

Precipitate of group IV cations takes place when ${{H}_{2}}S$ is
A. Highly ionised
B. Less ionised
C. Not ionised
D. None of these

Answer 1

Hint: The solubility product of sulphides of group iv is very high as compared to group ii. A very small amount ${{H}_{2}}S$ in the form of ionised ${{S}^{2-}}$is enough for the precipitation of group ii cations. In the case of group iv cations, $N{{H}_{4}}OH$is required for the ionisation of hydrogen sulphide.

Complete step by step solution:
Group iv cations can not be precipitated in the presence of less ionised ${{H}_{2}}S$. Because their solubility product is very high. The term solubility product is defined by the product of concentration of the ions which exist in equilibrium with the solid compound in a saturated solution.

For their precipitation as their sulphides, a very high amount of sulphide ions are required. It is achieved by passing hydrogen sulphide $({{H}_{2}}S)$ gas through the group iv cations in the presence of ammonium hydroxide $(N{{H}_{4}}OH)$. Then ${{H}^{+}}$ions are reacting with $H{{O}^{-}}$ ions to form water. As a result, the concentration of sulphide ions$({{S}^{2-}})$ increases, as equilibrium favours the forward direction.
${{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}$
Therefore group iv cations are precipitated as their sulphides, as the ionic products are now greater than their solubility products.
So, precipitation of group IV cations takes place when ${{H}_{2}}S$is highly ionised.
Thus, Option (A) is correct.

Note: Like group iv cations, group ii cations are precipitated as their sulphides. Due to the low solubility product of group ii cations, a little amount of hydrochloric acid is required. Due to the presence of a common ion i.e, ${{H}^{+}}$, the concentration of sulphide ion(${{S}^{2-}}$) decreases. As a result precipitation occurs in the group ii cations as their sulphides.