TS EAMCET202314 May 2023Morning ShiftChemistryStructure of AtomActual
Match the following List-I (Complex) (A) [ CoF ₆ [³⁻ . . (B) [ Co ( C ₂ O ₄ )₃ ]³⁻ (C) [ FF ₆ ]³⁻ (D) [ Mn ( CN )₆ ]³⁻List-II (Spin only Magnetic moment) (I) 0 (II) 24 (III) 8 (IV) 35 (V) 15 The correct answer is:
Options
- AA-V, B-II, C-IV, D-I
- BA-II, B-I, C-IV, D-III
- CA-II, B-I, C-V, D-III
- DA-III, B-II, C-I, D-V
Correct answer
B. A-II, B-I, C-IV, D-III
Step-by-step solution
[ CoF ₆ ]³⁻: Co is present as Co ³⁺ and F ⁻ is a weakfield ligand. Thus, Co ³⁺=[ Ar ] 3 ~d ^6 with four unpaired electrons. = n ( n +2) B.M. = 4(4+2) = 24 [ Co ( C ₂ O ₄ )₃ ]³⁻: Co is present as Co ³⁺ and C ₂ O ₄ ²⁻ is strong-field ligand. Thus, Co ³⁺=[ Ar ] 3 ~d ^6 with all paired electrons. = n ( n +2) = 0(0+2) =0 [ FeF ₆ ]³⁻: Fe is present as Fe ³⁺ and F ⁻ is a weak-field ligand. Thus, Fe ³⁺=[ Ar ] 3 ~d ^5 with five unpaired electrons. =- n ( n +2) = 5(5+2) = 35 [ Mn ( CN )₆ ]³⁻: Mn is present as Mn ³⁺ and CN ⁻