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
- A.   P   →   1 ;   Q   →   4 ;   R   →   2 ;   S   →   3
- B. P   →   1 ;   Q   →   2 ;   R   →   4 ;   S   →   5
- C. P   →   3 ;   Q   →   2 ;   R   →   5 ;   S   →   1
- 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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Related: Chemistry — Coordination Compounds · All PYQ Banks