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

Solution

Chemistry ma’am · 3 steps

Use two filters: a d-configuration that ignores the ligand, then a violet bead in the hot oxidising flame.

1. Ligand-proof configuration. In an octahedral field, d¹, d², d³, d⁸ and d⁹ ions fill t_(2g) and e_g the same way whether the ligand is weak or strong. Only d⁴ to d⁷ ions switch between high spin and low spin.
2. Flame meaning. The non-luminous flame is the oxidising flame. So you need the bead colour in the oxidising flame while the bead is still hot.

   Ion	d count	Weak vs strong ligand	Bead, oxidising flame, hot
   Cr³⁺	d³	t_(2g)³ in both, unaffected	green
   Ti³⁺	d¹	t_(2g)¹ in both, unaffected	colourless
   Ni²⁺	d⁸	t_(2g)⁶ e_g² in both, unaffected	violet (reddish brown on cooling)
   Mn²⁺	d⁵	t_(2g)³ e_g² or t_(2g)⁵, affected	violet (amethyst)

3. So. Only Ni²⁺ passes both tests. Its d⁸ configuration stays t_(2g)⁶ e_g² with any ligand, and it gives a violet bead in the hot oxidising flame. The answer is (c).

The idea to keep. Only d⁴ to d⁷ ions care whether the ligand is strong or weak. For bead tests, always note which flame is used (non-luminous means oxidising) and whether the bead is hot or cold.

Answer from the NTA final answer key. Solution by PrepWiser.