Question

The Co (II) has preference for a geometry T_d before one O_h. The Ni (II), however, prefers a Oh geometry. Explain why (compare EECC).

Answer 1

Cobalt (II)

Case (i): Tetrahedral geometry

The electronic configuration = e_g⁴ t_2g³

The crystal field stabilization energy (CFSE) = 4(0.6 ∆_t) - 3(0.4 ∆_t), i.e. 1.2 ∆_t

The tetrahedral CFSE can be converted to an octahedral CFSE using the relation: 4∆_o = 9∆_t

Therefore, 1.2 ∆_t = 1.2*(4/9 ∆_o), i.e. 0.533 ∆_o

Case (ii): Octahedral geometry

The electronic configuration = t_2g⁵ e_g²

The crystal field stabilization energy (CFSE) = 5(0.4 ∆_o) - 2(0.6 ∆_o), i.e. 0.8 ∆_o

From case(i) and case(ii), the tetrahedral CFSE (= 0.533 ∆_o) ~ the octahedral CFSE (= 0.8 ∆_o)

Since the stabilization energy is almost similar for Co(II) both in the tetrahedral and octahedral fields, it can adopt both the tetrahedral (Td) and octahedral (Oh) geometries almost with equal preference.

Nickel (II)

Case (i): Tetrahedral geometry

The electronic configuration = e_g⁴ t_2g⁴

The crystal field stabilization energy (CFSE) = 4(0.6 ∆_t) - 4(0.4 ∆_t), i.e. 0.8 ∆_t

The tetrahedral CFSE can be converted to an octahedral CFSE using the relation: 4∆_o = 9∆_t

Therefore, 0.8 ∆_t = 0.8*(4/9 ∆_o), i.e. 0.355 ∆_o

Case (ii): Octahedral geometry

The electronic configuration = t_2g⁶ e_g²

The crystal field stabilization energy (CFSE) = 6(0.4 ∆_o) - 2(0.6 ∆_o), i.e. 1.2 ∆_o

From case(i) and case(ii), the octahedral CFSE (= 1.2 ∆_o) >> the tetrahedral CFSE (= 0.355 ∆_o)

Since the stabilization energy in the octahedral field is much higher for Ni(II) compared to that in tetrahedral field, it prefers octahedral (Oh) geometry.