When Ethane-1,2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be :

When ethane-1,2-diamine (ethylenediamine, abbreviated as "en") is added progressively to an aqueous solution of nickel(II) chloride ($$NiCl_2$$), the following stepwise complexation occurs:

$$NiCl_2$$ dissolves in water to form $$[Ni(H_2O)_6]^{2+}$$, which is green in colour. This is an octahedral complex where the d-d transitions in $$Ni^{2+}$$ ($$d^8$$) with weak-field water ligands produce a green colour.

$$[Ni(H_2O)_6]^{2+} + en \rightarrow [Ni(en)(H_2O)_4]^{2+} + 2H_2O$$

As the stronger field ligand "en" replaces two water molecules, the crystal field splitting increases slightly, shifting the absorption and producing a pale blue colour.

$$[Ni(en)(H_2O)_4]^{2+} + en \rightarrow [Ni(en)_2(H_2O)_2]^{2+} + 2H_2O$$

With further increase in field strength, the complex turns blue.

$$[Ni(en)_2(H_2O)_2]^{2+} + en \rightarrow [Ni(en)_3]^{2+} + 2H_2O$$

The tris(ethylenediamine)nickel(II) complex has the highest crystal field splitting in this series, producing a violet colour.

The observed colour sequence is: Green $$\rightarrow$$ Pale Blue $$\rightarrow$$ Blue $$\rightarrow$$ Violet.

The correct answer is Option 3.