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…

Match List I with List II

  List I Complex   List II CFSE ∆0
A. CuNH362+ I. -0.6
B. TiH2O63+ II. -2.0
C. FeCN63- III. -1.2
D. NiF64- IV. -0.4

Choose the correct answer from the options given below:

  1. A(III), B(IV), C(I), D(II)
  2. A(I), B(IV), C(II), D(III)
  3. A(I), B(II), C(IV), D(III)
  4. A(II), B(III), C(I), D(IV)

Solution

CFSE=(-0.4 nt2g+0.6neg)∆0

nt2g= Number of electrons in t2g orbital

neg=number of electrons in eg orbital.

Crystal Field Stabilisation Energy for the given complexes is as follows:

(A)

CuNH362+

Cu2+.. 3d6, t2g6 eg3

CFSE

=-6×0.4+3×0.6∆0

=-0.6∆0
(B)

TiH2O63+

Ti3+: 3d1, t2g1 eg0

CFSE=-1×0.4∆0

=-0.4∆0
(C)

FeCN63-

Fe3+:3d5, t2g5 eg0

CFSE=-5×0.4∆0

=-2.0 ∆0
(D)

NiF64-

Ni2+:3d8, t2g6 eg2

CFSE

=-6×0.4+2×0.6∆0

=-1.2 ∆0

So the correct option is B.

Asked in: JEE Main 2023 (12 Apr Shift 1)

Practice more Coordination Compounds questions on Aicharya