JEE Main202312 Apr 2023Morning ShiftChemistryCoordination CompoundsActual
Match List I with List II List I Complex List II C F S E ∆ 0 A. C u N H 3 6 2 + I. - 0 . 6 B. T i H 2 O 6 3 + II. - 2 . 0 C. F e C N 6 3 - III. - 1 . 2 D. N i F 6 4 - IV. - 0 . 4 Choose the correct answer from the options given below:
Options
- AA(III), B(IV), C(I), D(II)
- BA(I), B(IV), C(II), D(III)
- CA(I), B(II), C(IV), D(III)
- DA(II), B(III), C(I), D(IV)
Correct answer
B. A(I), B(IV), C(II), D(III)
Step-by-step solution
CFSE = ( - 0 . 4   nt 2 g + 0 . 6 ne g ) ∆ 0 nt 2 g = Number of electrons in t 2 g orbital ne g = number of electrons in eg orbital. Crystal Field Stabilisation Energy for the given complexes is as follows: (A) Cu NH 3 6 2 + Cu 2 + . .   3 d 6 ,   t 2 g 6   e g 3 C F S E = - 6 × 0 . 4 + 3 × 0 . 6 ∆ 0 = - 0 . 6 ∆ 0 (B) Ti H 2 O 6 3 + Ti 3 + :   3 d 1 ,   t 2 g 1   e g 0 C F S E = - 1 × 0 . 4 ∆ 0 = - 0 . 4 ∆ 0 (C) Fe CN 6 3 - Fe 3 + : 3 d