Given below are two statements :
Statement (I) : A solution of [Ni(H₂ O)₆]²⁺ is green in colour.
Statement (II) : A solution of [Ni(CN)₄]²⁻ is colourless.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Same nickel(II), different ligands: water keeps the d–d gap in the visible, cyanide pushes it out.

1. Same metal ion. In both complexes nickel is Ni²⁺, a 3d⁸ ion. The colour therefore depends on the ligand, because the ligand sets the size of the crystal field splitting.
2. Water, weak field. H₂O is a weak-field ligand, so [Ni(H₂O)₆]²⁺ is octahedral with a small gap and two unpaired electrons (t_(2g)⁶ e_g²). A d–d transition absorbs red light, so you see the complementary colour, green.
3. Cyanide, strong field. CN⁻ is a strong-field ligand, so all eight d electrons pair up. This gives a square planar dsp² complex with no unpaired electrons. The only empty d orbital (d_(x²-y²)) lies very high, so the gap is too large for visible light. The solution is taken as colourless.

   Complex	Ligand	Shape, unpaired electrons	Colour
   [Ni(H₂O)₆]²⁺	weak field, small gap	octahedral, 2	green
   [Ni(CN)₄]²⁻	strong field, large gap	square planar, 0	colourless

4. So. Statement I is correct and Statement II is correct, which is option (b).

The idea to keep. Colour in a complex comes from d–d jumps across the splitting gap. A weak-field ligand gives a small gap and visible absorption. A strong-field ligand can push the absorption out of the visible range.

Answer from the NTA final answer key. Solution by PrepWiser.