Answer:
Sb3+ is present.
Ions that could be absent include copper II ions(Cu2+) and lead II ions(Pb2+)
The procedure for finding ions in doubt is explained below under explanation.
Explanation:
Sb3+ is the only ion that's definitely present. This group 2B sulfide is the only one that will be orange in colour when re - precipitated from the hydroxide ion(OH-) extract. Nevertheless, the orange color could serve as a masking of the yellow Arsenic trisulphide and the Tin sulphide precipitates, but not the black mercury sulphide precipitate.
Now, back to the black original precipitate. For it to be so, a group IIA cation must be present. The only group IIB cation with a black sulfide is mercury cation Hg2+ which we said could not be masked earlier.
The group IIA cations that must be present are Pb2+, Cu2+, Cd2+, Bi3+. When the acid extract of any of these 4 cations are present are made alkaline with NH3 then Copper ion (Cu2+) and Cadmium ion (Cd+) will form soluble amine complexes, which is very possible.
Bismuth Cation(Bi3+) would form the white precipitate of Bi(OH)3 and Lead II cation(Pb2+). However, if they are not removed in Group I, they will form solid Pb(OH)2, which is also white.
To find out the ions that are still in doubts. For example, when adding Ammonia(NH3) noted earlier to the Group IA extract, if the solution turns blue, then copper II ion(Cu2+) is confirmed otherwise, it is absent or present in very low concentration. However, if we first treated the precipitate with Sodium Hydroxide(NaOH), and it dissolves, then it confirms that Pb2+. Bi(OH)3 will not dissolve in Sodium Hydroxide(NaOH) solution.
