A) \[HgS\]
B) \[PbS\]
C) \[CdS\]
D) \[CuS\]
Correct Answer: D
Solution :
Precipitation of II group radicals by \[{{H}_{2}}S\] takes place in presence of\[HCl\]. \[HCl\] Suppresses the ionisation of weakly dissociated \[{{H}_{2}}S\] \[{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}\] and thus, only the ionic product of the sulphide of group II radicals exceeds their corresponding Ky. So, \[CuS\] is only completely precipitated.
You need to login to perform this action.
You will be redirected in
3 sec
