Question

The value of the crystal field stabilisation energies for a high spin d⁶ metal ion in octahedral and tetrahedral fields, respectively, are ______.

Options

• −2.4 Δ_o and −0.6 Δ_t

• −1.6 Δ_o and −0.4 Δ_t

• −0.4 Δ_o and −0.27 Δ_t

• −0.4 Δ_o and −0.6 Δ_t

Answer 1

The value of the crystal field stabilisation energies for a high spin d⁶ metal ion in octahedral and tetrahedral fields, respectively, are `bbunderline(-0.4 Δ_o and -0.6 Δ_t)`.

Explanation:

In an octahedral field, the CFSE is calculated by considering the splitting of the d-orbitals into t_2g​ and e_g​ orbitals. For high-spin d⁶, the electrons fill the orbitals with 2 electrons in the higher e_g​ orbitals and 4 electrons in the lower t_2g​ orbitals.

CFSE = [−0.4 nt_2g + 0.6ne_g] Δ_o + n(P)

= [−0.4 × 4 + 0.6 × 2] Δ_o + 0

= [−1.6 + 1.2] Δ_o

= [−0.4] Δ_o 

In a tetrahedral field, the splitting of the d-orbitals is smaller, and the CFSE is given by −0.6 Δ_t for a high-spin configuration. For high-spin d⁶, the electrons fill the orbitals with 3 electrons in the higher e_g​ orbitals and 3 electrons in the lower t_2g​ orbitals.

CFSE = [−0.6 ne_g + 0.4nt_2g] Δ_t 

= [−0.6 × 3 + 0.4 × 3] Δ_t

= [−1.8 + 1.2] Δ_t

= [−0.6] Δ_t 

Thus, the energy gap Δ_t​ is smaller than Δ_o​, leading to this value. The correct values are −0.4 Δ_o for the octahedral field and −0.6 Δ_t for the tetrahedral field.