19.3 Spectroscopic and magnetic properties of coordination compounds  (Page 3/17)

In [Fe(CN) ₆ ] ⁴⁻ , the strong field of six cyanide ligands produces a large Δ _oct . Under these conditions, the electrons require less energy to pair than they require to be excited to the e _g orbitals (Δ _oct >P). The six 3 d electrons of the Fe ²⁺ ion pair in the three t _{2 g} orbitals ( [link] ). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.

In [Fe(H ₂ O) ₆ ] ²⁺ , on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δ _oct <P). Because it requires less energy for the electrons to occupy the e _g orbitals than to pair together, there will be an electron in each of the five 3 d orbitals before pairing occurs. For the six d electrons on the iron(II) center in [Fe(H ₂ O) ₆ ] ²⁺ , there will be one pair of electrons and four unpaired electrons ( [link] ). Complexes such as the [Fe(H ₂ O) ₆ ] ²⁺ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.

A similar line of reasoning shows why the [Fe(CN) ₆ ] ³⁻ ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H ₂ O) ₆ ] ³⁺ and [FeF ₆ ] ³⁻ ions are high-spin complexes with five unpaired electrons.

Cft for other geometries

CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the e _g set point directly at the ligands. For tetrahedral complexes, the d orbitals remain in place, but now we have only four ligands located between the axes ( [link] ). None of the orbitals points directly at the tetrahedral ligands. However, the e _g set (along the Cartesian axes) overlaps with the ligands less than does the t _{2 g} set. By analogy with the octahedral case, predict the energy diagram for the d orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral e _g set becomes a tetrahedral e set, and the octahedral t _{2 g} set becomes a t ₂ set.

Solution

Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so Δ _tet is usually small $\left({\text{Δ}}_{\text{tet}}=\frac{4}{9}{\text{Δ}}_{\text{oct}}\right):$

Check your learning

Explain how many unpaired electrons a tetrahedral d ⁴ ion will have.

Answer:

4; because Δ _tet is small, all tetrahedral complexes are high spin and the electrons go into the t ₂ orbitals before pairing

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