Step 1:

In ${\left[Ni\right(H}^{}$₂O)₆]²⁺, the oxidation state of Ni is +2. It is a 3d⁸ system. 

The orbital diagram is as follows:

Ni²⁺ = 3d⁸ ;  

Here, H₂O is a weak field ligand. Hence, a pairing of 3d electron does not takes place. Therefore, there are unpaired electrons in $N{i}^{2+}$.

In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d transition is present. Hence, $Ni{(H}_{}$₂O)₆]²⁺ is coloured. 

Step 2:

In [Ni(CN)₄]^2-, the oxidation state of Ni is +2. The electronic configuration is 3d⁸. 

Ni²⁺ = 3d⁸ ;  

But CN^- is a strong field ligand hence, the pairing of electrons takes place present in 3d orbital. 


Thus, no possibility of d-d transition i$[Ni{\left(CN\right)}_4{]}^{2-}$. Hence, it is colourless.