Question

Draw the energy level diagrams and indicate the occupancy of the orbitals in the following complexes.

(a) $d^{6}$ , octahedral, low spin
(b) $d^{9}$ , octahedral with tetragonal elongation
(c) $d^{8}$ , square planar
(d) $d^{6}$ , tetrahedral

Calculate the difference in crystal field stabilization energy (in unit of $Δ_{o}$ ) between the complexes (a) and (d), assuming that the ligands are strong field ligands.

(Answer: octahedral $- 2.4 Δ_{o}$ , tetrahedral $- 0.27 Δ_{o}$ , difference $- 2.13 Δ_{o}$ )

Accepted Answer

(a) For $d^{6}$ , octahedral, strong field ligand:

Due to the strong field condition, metal forms a low spin complex.

(b) For $d^{8}$ , octahedral with tetragonal elongation:

In a tetragonal system, the $z$ axis get elongated, hence the repulsion with ligands gets decreased for this axis and feels the higher repulsion for the orbitals having $x$ axis.

(c) For $d^{8}$ , square planar:

The splitting in square planar and octahedral splitting can relate by an equation $∆_{sp} = 1.3 ∆_{o}$

(d) For $d^{6}$ , tetrahedral:

The splitting in the tetrahedral field can relate to the octahedral field by an equation, $∆_{t} = \frac{4}{9} ∆_{o}$

Octahedral:

CFSE $= [- 0.4 \times 6 + 0.6 \times 0] ∆_{o} = - 2.4 ∆_{o}$

Tetrahedral:

CFSE $= [- 0.6 \times 3 + 0.4 \times 3] ∆_{t}$

CFSE $d^{6}$

Difference $= (- 2.4) - (- 0.27) = - 2.13 ∆_{o}$

Important Questions on Coordination Compounds

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