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JEE Advanced Chemistry Coordination Compounds 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Chemistry Question (2023) — Solution

Question

Match the electronic configurations in List-I with appropriate metal complex ions in List-II and choose the correct option. [Atomic Number: Fe   =   26 ,   Mn   =   25 ,   Co   =   27 ]   List I   List II P t 2 g 6 e g 0 1 Fe H 2 O 6 2 + Q t 2 g 3 e g 2 2 Mn H 2 O 6 2 + R e 2 t 2 3 3 Co NH 3 6 3 + S t 2 g 4 e g 2 4 FeCl 4 -     5 CoCl 4 2 -  

Options

  1. A.   P   →   1 ;   Q   →   4 ;   R   →   2 ;   S   →   3
  2. B. P   →   1 ;   Q   →   2 ;   R   →   4 ;   S   →   5
  3. C. P   →   3 ;   Q   →   2 ;   R   →   5 ;   S   →   1  
  4. D. P   →   3 ;   Q   →   2 ;   R   →   4 ;   S   →   1

Answer

D. P   →   3 ;   Q   →   2 ;   R   →   4 ;   S   →   1

Step-by-step solution

In the complex  Fe H 2 O 6 2 +  the electronic configuration of  Fe 2 +  is  Ar 3 d 6 .  H 2 O  is a weak ligand. Hence, the crystal filed splitting can be shown as follows, In the complex  Mn H 2 O 6 2 +  the electronic configuration of  Mn 2 +  is  Ar 3 d 5 .  H 2 O  is a weak ligand. Hence, the crystal filed splitting can be shown as follows, In the complex  Co NH 3 6 3 +  the electronic configuration of  Fe 3 +  is  Ar 3 d 6 .  NH 3  is a strong filed ligand. Hence, the crystal filed splitting can be shown as follows,     In the complex  FeCl 4 -  the electronic configuration of  Fe 3 +  is  Ar 3 d 5 .  Cl -  is a weak ligand. Hence, the crystal filed splitting can be shown as follows,       In the complex  CoCl 4 2 -  the electronic configuration of  Co 2 +  is  Ar 3 d 7 .  Cl -  is a weak ligand. Hence, the crystal filed splitting can be shown as follows,    

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