The incorrect statement is:

Let us examine each statement one by one.

Option A: $$Cu^{2+}$$ ion gives a chocolate coloured precipitate with potassium ferrocyanide solution.

The reaction is: $$2Cu^{2+} + [Fe(CN)_6]^{4-} \rightarrow Cu_2[Fe(CN)_6] \downarrow$$

Copper(II) ferrocyanide is indeed a chocolate (reddish-brown) coloured precipitate. This is a standard confirmatory test for $$Cu^{2+}$$ ions. So this statement is correct.

Option B: $$Cu^{2+}$$ and $$Ni^{2+}$$ ions give black precipitate with $$H_2S$$ in presence of $$HCl$$ solution.

In the qualitative analysis scheme, $$H_2S$$ is passed in acidic medium (dilute $$HCl$$) to precipitate Group II cations. $$Cu^{2+}$$ belongs to Group II and precipitates as $$CuS$$ (black) in acidic medium:

$$Cu^{2+} + H_2S \rightarrow CuS \downarrow + 2H^+$$

However, $$Ni^{2+}$$ belongs to Group IV (ammonium sulphide group). It does NOT precipitate with $$H_2S$$ in the presence of $$HCl$$ because the solubility product of $$NiS$$ is not exceeded in acidic conditions. $$Ni^{2+}$$ requires alkaline or ammoniacal conditions ($$NH_4OH + H_2S$$ or $$(NH_4)_2S$$) to precipitate as $$NiS$$ (black).

Therefore, while $$Cu^{2+}$$ gives a black precipitate with $$H_2S$$ in $$HCl$$, $$Ni^{2+}$$ does not. This statement is incorrect.

Option C: Ferric ion gives blood red colour with potassium thiocyanate.

The reaction is: $$Fe^{3+} + 3SCN^{-} \rightarrow Fe(SCN)_3$$

Iron(III) thiocyanate is blood red in colour. This is a well-known confirmatory test for $$Fe^{3+}$$. The statement is correct.

Option D: $$Cu^{2+}$$ salts give red coloured borax bead test in reducing flame.

In the borax bead test, borax ($$Na_2B_4O_7 \cdot 10H_2O$$) on heating first loses water and then decomposes:

$$Na_2B_4O_7 \rightarrow 2NaBO_2 + B_2O_3$$

The glassy bead of $$B_2O_3$$ (boron trioxide) and $$NaBO_2$$ dissolves metal oxides to form coloured metaborates. For copper:

In the oxidising flame: $$CuO + B_2O_3 \rightarrow Cu(BO_2)_2$$ which is green/blue.

In the reducing flame: $$Cu_2O$$ is formed, which gives $$Cu$$ metal or $$Cu_2O$$ metaborate, producing a red (opaque) bead. This statement is correct.

Therefore, the incorrect statement is Option B, since $$Ni^{2+}$$ does not precipitate with $$H_2S$$ in the presence of $$HCl$$.

The correct answer is Option B.