JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The oxidation state of Ni in all three complexes is +2. The electronic configuration of Ni^ 2+ is [Ar] 3d^ 8 . In [Ni(en)_3]^ 2+ (A), the coordination number is 6 (octahedral geometry). The d^ 8 configuration in an octahedral field is t_ 2g ^ 6 e_ g ^ 2 , which has 2 unpaired electrons. In [NiCl_4]^ 2- (B), the coordination number is 4 with weak field Cl^ - ligands (tetrahedral geometry). The d^ 8 configuration in a tetrahedral field is e^ 4 t_ 2 ^ 4 , which has 2 unpaired electrons. In [Ni(NH_3)_6]^ 2+ (C), the coordination number is 6 (octahedral geometry). The d^ 8 configuration in an octahedral field is t_ 2g ^ 6 e_ g ^ 2 , which has 2 unpaired electrons. Thus, the number of unpaired electrons in (A), (B), and (C) are 2, 2, and 2 respectively. The frequency of absorption of light is directly proportional to the crystal field splitting energy ( ). According to the spectrochemical series, the field strength of the ligands follows the order: en > NH_3 > Cl^ - . Furthermore, the crystal field splitting energy for octahedral complexes is significantly greater than that for tetrahedral complexes ( _o = 9 4 _t). Therefore, the order of crystal field splitting energy, and hence the frequency of absorption, is [Ni(en)_3]^ 2+ > [Ni(NH_3)_6]^ 2+ > [NiCl_4]^ 2- , which corresponds to (A) > (C) > (B). Answer: 2, 2, 2 and (A)>(C)>(B)