Choose the correct tests with respective observations.

(A) $$CuSO_4$$ (acidified with acetic acid) + $$K_4[Fe(CN)_6]$$ $$\rightarrow$$ Chocolate brown precipitate.

(B) $$FeCl_3 + K_4[Fe(CN)_6]$$ $$\rightarrow$$ Prussian blue precipitate.

(C) $$ZnCl_2 + K_4[Fe(CN)_6]$$, neutralised with $$NH_4OH$$ $$\rightarrow$$ White or bluish white precipitate.

(D) $$MgCl_2 + K_4[Fe(CN)_6]$$ $$\rightarrow$$ Blue precipitate.

(E) $$BaCl_2 + K_4[Fe(CN)_6]$$, neutralised with NaOH $$\rightarrow$$ White precipitate.

Choose the correct answer from the options given below:

The reagent $$K_4[Fe(CN)_6]$$ (potassium ferrocyanide) gives characteristic coloured precipitates with only certain metal ions. We shall examine tests (A) to (E) one by one.

$$Cu^{2+} + K_4[Fe(CN)_6] \xrightarrow{\text{acidic}} Cu_2[Fe(CN)_6] \downarrow$$
Copper(II) ferrocyanide is chocolate-brown in colour.
Observation stated in (A) - chocolate brown precipitate - is correct.

$$Fe^{3+} + K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow$$
Ferric ferrocyanide is the well-known Prussian blue.
Observation stated in (B) - Prussian blue precipitate - is correct.

$$Zn^{2+} + K_4[Fe(CN)_6] \xrightarrow{\text{neutral/alkaline}} Zn_2[Fe(CN)_6] \downarrow$$
Zinc ferrocyanide separates as a white or faint bluish-white precipitate; neutralisation with $$NH_4OH$$ furnishes the required medium.
Observation stated in (C) - white / bluish-white precipitate - is correct.

Magnesium ions do not give an insoluble ferrocyanide; $$Mg_2[Fe(CN)_6]$$ is soluble. Hence no blue precipitate is obtained.
Observation stated in (D) is incorrect.

Ferrocyanides of the alkaline-earth metals Ca, Sr and Ba are readily soluble in water; therefore $$Ba^{2+}$$ does not precipitate with $$K_4[Fe(CN)_6]$$ even after neutralisation with $$NaOH$$.
Observation stated in (E) is incorrect.

Thus only statements (A), (B) and (C) are correct.

The option containing A, B and C only is Option C.