The potassium ferrocyanide solution gives a Prussian blue colour, when added to:

First, we recall that the qualitative test for ferric ions $$\text{Fe}^{3+}$$ involves adding potassium ferrocyanide, whose formula is $$\text{K}_4[\,\text{Fe(CN)}_6\,]$$.

Potassium ferrocyanide contains the complex anion $$[\,\text{Fe(CN)}_6\,]^{4-}$$. When this anion meets ferric ions $$\text{Fe}^{3+}$$, a double-exchange (double displacement) reaction occurs and an intensely coloured, insoluble coordination compound is produced.

Writing the reaction explicitly, we have three moles of potassium ferrocyanide reacting with four moles of ferric chloride:

$$ 4\,\text{FeCl}_3 \;+\; 3\,\text{K}_4[\,\text{Fe(CN)}_6\,] \;\longrightarrow\; \text{Fe}_4[\,\text{Fe(CN)}_6\,]_3 \;\downarrow\; +\; 12\,\text{KCl}. $$

The solid $$\text{Fe}_4[\,\text{Fe(CN)}_6\,]_3$$ is known as Prussian blue. Its deep blue colour is characteristic and is used as a confirmatory test for the presence of $$\text{Fe}^{3+}$$ ions.

Now we consider each chloride salt in the options:

• $$\text{CoCl}_3$$ contains $$\text{Co}^{3+}$$ ions, which do not produce Prussian blue with potassium ferrocyanide.

• $$\text{CoCl}_2$$ contains $$\text{Co}^{2+}$$ ions, which also do not give the blue precipitate.

• $$\text{FeCl}_2$$ provides $$\text{Fe}^{2+}$$ ions. With potassium ferrocyanide this gives a white precipitate of $$\text{K}_2\text{Fe}[\,\text{Fe(CN)}_6\,]$$ that slowly turns blue on oxidation by air; the immediate Prussian blue colour is therefore absent.

• $$\text{FeCl}_3$$ supplies the required $$\text{Fe}^{3+}$$ ions, and the reaction shown above takes place instantly, giving the characteristic deep Prussian blue precipitate.

Hence, the correct answer is Option D.