The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet colour in non-luminous flame under hot condition in borax bead test is

Electronic configuration of nickel atom is $$[Ar]\,3d^8\,4s^2$$, hence for $$Ni^{2+}$$ it becomes $$[Ar]\,3d^8$$. Thus the d-electron count is 8.

In an octahedral ligand field, the five d-orbitals split into a lower energy set $$t_{2g}$$ and a higher energy set $$e_g$$. The way electrons occupy these levels depends on the relative magnitudes of the octahedral crystal field splitting energy $$\Delta_o$$ and the pairing energy $$P$$.

Case 1: Weak field ligand
Here $$\Delta_o < P$$, so electrons occupy orbitals singly before pairing (high spin).
For d⁸: first fill all three $$t_{2g}$$ orbitals with six electrons (each pair in one orbital), then place the remaining two electrons one in each $$e_g$$ orbital. The configuration is $$t_{2g}^6\,e_g^2\,. $$

Case 2: Strong field ligand
Here $$\Delta_o > P$$, so electrons pair in the lower set before occupying the higher set (low spin).
For d⁸: all six electrons pair in $$t_{2g}$$, and the last two go into $$e_g$$, again giving $$t_{2g}^6\,e_g^2\,. $$

In both cases the electronic configuration remains $$t_{2g}^6\,e_g^2$$. Hence the distribution of d-electrons in $$Ni^{2+}$$ is not affected by ligand strength.

In the borax bead test, $$Ni^{2+}$$ imparts a violet colour to the non-luminous flame under hot conditions.

Therefore, the correct answer is Option C: $$Ni^{2+}$$.