Question

Given below are two statements:
Statement (I): A solution of \([Ni(H_2O)_6]^{2+}\) is green in color.
Statement (II): A solution of \([Ni(CN)_4]^{2-}\) is colorless.  
In the light of the above statements, choose the most appropriate answer from the options given below.

• A. Both Statement I and Statement II are incorrect
• B. Both Statement I and Statement II are correct
• C. Statement I is incorrect but Statement II is correct
• D. Statement I is correct but Statement II is incorrect

Answer 1

The Correct Option is B

This question requires an analysis of the color of two nickel coordination complexes, \([Ni(H_2O)_6]^{2+}\) and \([Ni(CN)_4]^{2-}\), based on the principles of coordination chemistry and crystal field theory.

Concept Used:

The color of transition metal complexes is attributed to d-d electronic transitions. When light passes through a solution of the complex, it absorbs photons of a specific energy (and color) from the visible spectrum to promote an electron from a lower-energy d-orbital to a higher-energy d-orbital. The color we perceive is the complementary color of the light absorbed.

The possibility and energy of these d-d transitions depend on:

1. The electronic configuration of the central metal ion (presence of partially filled d-orbitals).
2. The nature of the ligand (strong-field or weak-field), which determines the magnitude of the crystal field splitting energy (\( \Delta \)).
3. The geometry of the complex (e.g., octahedral, tetrahedral, square planar), which determines the pattern of d-orbital splitting.

If all d-orbitals are empty (\(d^0\)), completely filled (\(d^{10}\)), or if the energy gap (\(\Delta\)) is too large (requiring UV light), no d-d transition occurs in the visible range, and the complex appears colorless.

Step-by-Step Solution:

Step 1: Analysis of Statement (I) - A solution of \([Ni(H_2O)_6]^{2+}\) is green in color.

First, we determine the properties of the complex \([Ni(H_2O)_6]^{2+}\).

• Central Metal Ion: Nickel (Ni).
• Oxidation State: Let the oxidation state of Ni be \(x\). Water (\(H_2O\)) is a neutral ligand (charge 0). The overall charge is +2. So, \( x + 6(0) = +2 \Rightarrow x = +2 \). The ion is \(Ni^{2+}\).
• Electronic Configuration: The atomic number of Ni is 28 (\([Ar] 3d^8 4s^2\)). For \(Ni^{2+}\), the configuration is \([Ar] 3d^8\).
• Ligand and Geometry: The ligand is \(H_2O\), which is a weak-field ligand. With six ligands, the geometry is octahedral.
• Crystal Field Splitting: In an octahedral field, the five d-orbitals split into two sets: a lower energy \(t_{2g}\) set and a higher energy \(e_g\) set. Since \(H_2O\) is a weak-field ligand, the crystal field splitting energy (\(\Delta_o\)) is small, and electrons will fill the orbitals according to Hund's rule before pairing.
• Electron Filling: For a \(d^8\) configuration in a weak octahedral field, the electrons fill as \(t_{2g}^6 e_g^2\).

\[ \begin{array}{c} \underline{\uparrow\phantom{\downarrow}} \quad \underline{\uparrow\phantom{\downarrow}} \quad e_g \\ \text{Energy} \uparrow \\ \underline{\uparrow\downarrow} \quad \underline{\uparrow\downarrow} \quad \underline{\uparrow\downarrow} \quad t_{2g} \end{array} \]

The \(e_g\) orbitals contain two unpaired electrons. The presence of these unpaired electrons in the partially filled d-orbitals allows for d-d electronic transitions. The complex absorbs light in the red region of the visible spectrum, and it transmits the complementary color, which is green. Thus, a solution of \([Ni(H_2O)_6]^{2+}\) is green.

Conclusion for Statement (I): Statement (I) is correct.

Step 2: Analysis of Statement (II) - A solution of \([Ni(CN)_4]^{2-}\) is colorless.

Now, we determine the properties of the complex \([Ni(CN)_4]^{2-}\).

• Central Metal Ion: Nickel (Ni).
• Oxidation State: Let the oxidation state of Ni be \(y\). The cyanide ligand (\(CN^-\)) has a charge of -1. The overall charge is -2. So, \( y + 4(-1) = -2 \Rightarrow y = +2 \). The ion is again \(Ni^{2+}\).
• Electronic Configuration: The configuration of \(Ni^{2+}\) is \([Ar] 3d^8\).
• Ligand and Geometry: The ligand is \(CN^-\), which is a strong-field ligand. For a \(d^8\) metal ion with strong-field ligands and a coordination number of 4, the complex adopts a square planar geometry. This involves \(dsp^2\) hybridization.
• Crystal Field Splitting: In a square planar field, the d-orbitals split into four different energy levels. The strong-field \(CN^-\) ligand causes a large splitting energy, forcing all eight d-electrons to pair up in the lower energy orbitals.
• Electron Filling: The 8 electrons fill the lower four d-orbitals completely.

\[ \begin{array}{c} \underline{\phantom{\uparrow\downarrow}} \quad d_{x^2-y^2} \\ \text{Energy} \uparrow \\ \underline{\uparrow\downarrow} \quad d_{xy} \\ \underline{\uparrow\downarrow} \quad d_{z^2} \\ \underline{\uparrow\downarrow} \quad \underline{\uparrow\downarrow} \quad d_{xz}, d_{yz} \end{array} \]

As shown, there are no unpaired electrons in the d-orbitals. All electrons are paired. The energy gap between the highest occupied d-orbital (\(d_{xy}\)) and the lowest unoccupied d-orbital (\(d_{x^2-y^2}\)) is very large. An electronic transition would require a large amount of energy, corresponding to light in the ultraviolet (UV) region. Since the complex does not absorb any light from the visible region, it appears colorless.

Conclusion for Statement (II): Statement (II) is correct.

Final Result:

Both Statement (I) and Statement (II) are correct. \([Ni(H_2O)_6]^{2+}\) is green due to d-d transitions allowed by its unpaired electrons in a weak octahedral field. \([Ni(CN)_4]^{2-}\) is colorless because it is a square planar complex with a strong-field ligand, resulting in all d-electrons being paired and a large energy gap for any electronic transition, which prevents absorption of visible light.

Therefore, both Statement (I) and Statement (II) are true.

Answer 2

The compound \([ \text{Ni}(\text{H}_2\text{O})_6 ]^{2+}\) appears green due to d-d transitions in the visible spectrum.

\([ \text{Ni}(\text{CN})_4 ]^{2-}\) is diamagnetic and does not exhibit any d-d transitions, rendering it colourless.
Thus, the correct answer is option 2.