Q.A solution of $[Ni(H_2O)_6]^{2+}$ is green but a solution of $[Ni(CN)_4]^{2-}$ is colourless. Explain.

The key idea is that the colour of a transition metal complex depends on the energy gap ($\Delta$) between its d-orbitals, which determines the wavelength of light absorbed during d-d transitions.

Reasoning:

1. $[Ni(H_2O)_6]^{2+}$ has a weak-field ligand ($H_2O$), resulting in a small crystal field splitting energy ($\Delta_o$). This allows d-d transitions in the visible region, absorbing red light and transmitting green — hence the green colour.

2. $[Ni(CN)_4]^{2-}$ is square planar with a strong-field ligand ($CN^-$), which causes a very large splitting ($\Delta$). For $Ni^{2+}$ ($d^8$), the eight electrons pair up in the four lower orbitals, leaving the highest orbital ($d_{x^2-y^2}$) vacant — a d-d transition is therefore still possible, but the gap up to that orbital is so large that the absorption falls in the ultraviolet, not the visible, region. …

The key is d-orbital splitting in a transition metal complex. Colour in such compounds comes from electrons jumping from lower-energy d-orbitals to higher-energy d-orbitals when they absorb visible light. The energy gap between these orbitals — the crystal field splitting energy, $\Delta$ — determines which colour (wavelength) is absorbed, and thus which colour we see (the complementary colour).

For nickel(II), the electron configuration is $3d^8$. In an octahedral field like $[Ni(H_2O)_6]^{2+}$, the d-orbitals split into two sets: the lower-energy $t_{2g}$ (three orbitals) and the higher-energy $e_g$ (two orbitals). With 8 electrons, the $t_{2g}$ set is fully filled (6 electrons), and the remaining 2 electrons go into the $e_g$ set. This leaves room for an electron to be excited from $t_{2g}$ to $e_g$ — a d-d transition.

Now, water is a weak-field ligand. It produces a small $\Delta$. That small energy gap falls right in the visible region — specifically, the complex absorbs light in the red-orange part of the spectrum. The complementary colour is green, which is why the solution looks green.

In $[Ni(CN)_4]^{2-}$, the situation is completely different. Cyanide ($CN^-$) is a strong-field ligand, and here the geometry is square planar, not octahedral. For a $d^8$ ion in a square planar field, the d-orbital splitting is very large — much larger than in the octahedral case. The energy gap $\Delta$ is so big that the d-d transition now requires ultraviolet (UV) light, not visible light. Since no visible light is absorbed, the complex appears colourless. …

Concept: Crystal Field Theory (CFT)

This question is about d-orbital splitting and electronic transitions in coordination complexes.

Method: Crystal Field Theory Analysis

Step 1: Identify the geometry and ligand field strength

• $[Ni(H_2O)_6]^{2+}$ — octahedral geometry, weak field ligand (H₂O)
• $[Ni(CN)_4]^{2-}$ — square planar geometry, strong field ligand (CN⁻)

Step 2: Determine the d-electron configuration of Ni²⁺

• Ni atomic number = 28, Ni²⁺ = 28 − 2 = 26 electrons
• Electronic configuration: $[Ar]\,3d^8$

Step 3: Analyze d-orbital splitting in each case

For $[Ni(H_2O)_6]^{2+}$ (octahedral, weak field):

• In octahedral field, d-orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy)
• For $d^8$ in an octahedral field the filling is fixed (there is no high/low-spin choice for $d^8$): $t_{2g}^6 e_g^2$ — two unpaired electrons in $e_g$
• d-d transitions are possible (electrons can jump from $t_{2g}$ to $e_g$)
• These transitions absorb light in the visible region → green colour

For $[Ni(CN)_4]^{2-}$ (square planar, strong field):

• Square planar has a large splitting, especially with strong field CN⁻
• For $d^8$ with strong field: all electrons pair up …

Here are the common mistakes students make on this question and how to avoid each.

Mistake 1: Forgetting to check the oxidation state and d-electron count

• The mistake: Students often assume both complexes have the same number of d-electrons without verifying. They might say both are $Ni^{2+}$ and leave it at that, missing the key difference.
• Why it’s wrong: The geometry and ligand field strength depend on the metal ion’s exact electronic configuration. For $[Ni(H_2O)_6]^{2+}$, Ni is in the +2 state with a $3d^8$ configuration. For $[Ni(CN)_4]^{2-}$, Ni is also +2, but the strong field cyanide ligand causes pairing.
• How to avoid: Always write the electronic configuration of the metal ion first. For $Ni^{2+}$: $[Ar] 3d^8$. Then note that in a tetrahedral or square planar field, the splitting pattern changes.

Mistake 2: Confusing geometry — octahedral vs square planar

• The mistake: Assuming $[Ni(CN)_4]^{2-}$ is tetrahedral (like many $[MCl_4]^{2-}$ complexes) and then incorrectly predicting d-orbital splitting.
• Why it’s wrong: $[Ni(CN)_4]^{2-}$ is square planar, not tetrahedral. Cyanide is a strong field ligand, and for $d^8$ systems, square planar geometry is favoured. In square planar, the d-orbital splitting is large, and all electrons are paired.
• How to avoid: Memorise common geometries: $[Ni(CN)_4]^{2-}$ is square planar; $[Ni(H_2O)_6]^{2+}$ is octahedral. For $d^8$ with strong field ligands, expect square planar.

Mistake 3: Ignoring the role of ligand field strength

• The mistake: Saying both complexes absorb visible light but giving no reason why one is coloured and the other is not.
• Why it’s wrong: Colour arises from d-d transitions. The energy gap ($\Delta$) determines which wavelength is absorbed. If $\Delta$ is very large (strong field), the transition may shift to UV, making the complex appear colourless.
• How to avoid: Compare $\Delta$ values: $H_2O$ is a weak field ligand (small $\Delta$), so $[Ni(H_2O)_6]^{2+}$ absorbs in the visible range (green colour). $CN^-$ is a strong field ligand (large $\Delta$), so $[Ni(CN)_4]^{2-}$ absorbs in the UV — no visible light absorbed, hence colourless.

Mistake 4: Forgetting that colourless means no d-d transition in visible range

• The mistake: Stating that $[Ni(CN)_4]^{2-}$ has no unpaired electrons and therefore is colourless — this is incomplete.
• Why it’s wrong: Colourless does not mean no d-d transition; it means the transition energy is outside the visible range (400–700 nm). Even paired electrons can undergo d-d transitions if the gap is small enough.
• How to avoid: Always mention that the energy gap is so large that absorption occurs in the UV region, not the visible. Use the formula: $\Delta E = h\nu$. If $\nu$ corresponds to UV, the complex appears colourless. …