1. Identify the complexes and their geometries / electronic situations

All three complexes contain $\text{Ni}^{2+}$, so the metal ion is $3d^8$.

• $[Ni(H_2O)_6]^{2+}$ : octahedral, weak-field ligand
• $[NiCl_4]^{2-}$ : tetrahedral, weak-field ligand
• $[Ni(CN)_4]^{2-}$ : square planar, strong-field ligand

2. Color intensity depends on allowedness of $d$-$d$ transitions

The intensity of color is related to how strongly the complex absorbs light. For coordination compounds, this depends largely on whether the electronic transition is Laporte allowed or forbidden.

• In octahedral complexes with a center of symmetry, $d$-$d$ transitions are Laporte forbidden $\Rightarrow$ weak color.
• In tetrahedral complexes, there is no center of symmetry, so $d$-$d$ transitions are more allowed $\Rightarrow$ stronger color.
• In square planar complexes, especially with strong-field ligand $CN^-$, the splitting is large and the principal transitions are comparatively less intense in the visible region; such complexes are generally less intensely colored than tetrahedral $[NiCl_4]^{2-}$.

3. Compare the three complexes

(i) $[NiCl_4]^{2-}$

• Tetrahedral complex
• No inversion center
• $d$-$d$ transitions relatively more intense

So this should have the highest color intensity.

(ii) $[Ni(H_2O)_6]^{2+}$

• Octahedral complex
• Has inversion center
• $d$-$d$ transitions Laporte forbidden

So its color intensity is less than tetrahedral $[NiCl_4]^{2-}$.

(iii) $[Ni(CN)_4]^{2-}$

• Square planar, strong-field complex
• For $d^8$ nickel, this is low-spin and often shows comparatively weak visible absorption for the relevant transition pattern.
• Hence its observed color intensity is lower than the other two.

4. Therefore, the order of intensity of colors is

5. Match with the options

This corresponds to Option C.

6. Comparison with stored correct answer

Stored correct answer = C

Our derived answer = C

Hence, they agree.