A) \[P{{b}^{2+}}\]
B) \[F{{e}^{3+}}\]
C) \[Z{{n}^{2+}}\]
D) \[C{{u}^{2+}}\]
Correct Answer: A
Solution :
Pb2+ as it?s precipitated as chloride and sulphide in Ist and IInd group respectively
You need to login to perform this action.
You will be redirected in
3 sec
