JEE Main201910 Apr 2019Evening ShiftChemistryCoordination CompoundsActual
The crystal field stabilization energy (CFSE) of F e H 2 O 6 C l 2 and K 2 N i C l 4 , respectively, are:
Options
- A- 0.4 ∆ 0 and - 1.2 ∆ t
- B- 2.4 ∆ 0 and - 1.2 ∆ t
- C- 0.4 ∆ 0 and - 0.8 ∆ t
- D- 0.6 ∆ 0 and - 0.8 ∆ t
Correct answer
C. - 0.4 ∆ 0 and - 0.8 ∆ t
Step-by-step solution
F e H 2 O 6 2 + ; F e + 2 ⇒ 3 d 6 , Since H 2 O is Weak field ligands, CFSE = 4 × - 0.4 + 0.6 × 2 = - 0.4 ∆ 0 CFSE = 4 × 0.4 + - 0.6 × 4 = - 0.8 ∆ 0 and - 0.8 ∆ t