Arrange the following in order of increasing number of unpaired electrons in them i. [ Fe ( CN )₆ ]³⁻ ii. [ MnCl ₆ ]³⁻ iii. [ FeF ₆ ]³⁻ iv. [ Co ( NH ₃ )₆ ]³⁺

Step-by-step solution

(i) [ Fe ( CN )₆ ]³⁻:- Fe is in +3 state and CN ⁻ is a strong-field ligand. Thus, Fe ³⁺=[ Ar ] 3 ~d ^5 or Thus, there is only one unpaired electron. (ii) [ MnCl ₆ ]³⁻ :- Mn is in +3 state and Cl ⁻ is a weak-field ligand. Thus, Mn ³⁺=[ Ar ] 3 ~d ^4 or Thus, there are four unpaired electrons. (iii) [ FeF ₆ ]³⁻:- Fe is in +3 state and F ⁻ is a weak-field ligand. Thus, Fe will be [ Ar ] 3 ~d ^5 with five unpaired electron. (iv) [ Co ( NH ₃ )₆ ]³⁺ :- Co is in +3 state here and NH ₃ is a strong-field ligand. Thus, Co wil