A) \[A{{s}_{2}}{{S}_{3}}\]and \[CdS\]
B) \[CdS,NiS\]and ZnS
C) NiS and ZnS
D) Sulphide of all ions
Correct Answer: D
Solution :
\[A{{s}^{3+}}\] and \[C{{d}^{2+}}\] are the radicals of group II, whereas \[N{{i}^{2+}}\And Z{{n}^{2+}}\] are the radicals of group IV. The solubility product of group IV radicals is higher as compared to group II. \[N{{H}_{4}}OH\] increases the ionisation of \[{{H}_{2}}S\] by removing \[{{H}^{+}}\] of \[{{H}_{2}}S\] as unionisable water. \[{{H}_{2}}S2{{H}^{+}}+{{S}^{2-}};{{H}^{+}}+O{{H}^{-}}\xrightarrow[{}]{{}}{{H}_{2}}O\] Thus, excess of sulphide ions are present which leads to the precipitation of all the four ions. Note: HCl decreases ionisation of \[{{H}_{2}}S\]whereas \[N{{H}_{4}}OH\]increases the ionisation of\[{{H}_{2}}S\]
You need to login to perform this action.
You will be redirected in
3 sec
