Question

In an octahedral crystal field, the energies of which d-orbitals will be raised when ligands approach the central metal atom/ion? Give a reason in support of your answer.

Answer 1

Concept: In crystal field theory the ligands are treated as point negative charges. In an octahedral arrangement the six ligands come along the x, y and z axes. The d-orbitals that point directly along these axes feel more repulsion from the ligand electrons, so their energy rises more. Answer: The energies of the e_g set of d-orbitals, namely d_{x²–y²} and d_z², are raised. The reason is that these two orbitals have their lobes pointing straight along the x, y and z axes, which is exactly where the six ligands approach in an octahedral field. The negatively charged ligands therefore repel the electrons in these orbitals strongly, so the e_g orbitals are pushed to higher energy. The other three orbitals (d_xy, d_yz, d_zx, the t_2g set) lie between the axes, feel less repulsion, and are lowered in energy. This unequal repulsion is what splits the d-orbitals in an octahedral field.

Answer 2

Concept: In both complexes cobalt is in the +3 state, which is a d⁶ ion. How the six d-electrons fill the lower t_2g and upper e_g sets depends on the ligand. A weak-field ligand gives a small splitting, so electrons spread out (high spin); a strong-field ligand gives a large splitting, so electrons pair up in the lower set (low spin). Answer: Co is atomic number 27, so Co³⁺ is [Ar] 3d⁶. (i) In [CoF₆]^3–, fluoride is a weak-field ligand, so the splitting is small and the complex is high spin. The configuration is t_2g⁴ e_g², which gives four unpaired electrons (paramagnetic). (ii) In [Co(NH₃)₆]³⁺, ammonia is a strong-field ligand, so the splitting is large and the electrons pair up in the lower set, making it low spin. The configuration is t_2g⁶ e_g⁰, which has no unpaired electrons (diamagnetic). The difference comes only from the strength of the ligand field.

Answer 3

Concept: Magnetic behaviour depends on the oxidation state of nickel and on whether the ligand is weak-field or strong-field, because a strong-field ligand like CO can pair up the metal electrons. Answer: In [NiCl₄]^2–, nickel is in the +2 state, so it is Ni²⁺ (3d⁸). Chloride is a weak-field ligand and does not force pairing, so the 3d⁸ ion keeps two unpaired electrons. The complex uses sp³ hybridisation, is tetrahedral, and is paramagnetic. In [Ni(CO)₄], nickel is in the zero oxidation state, Ni (3d⁸ 4s²). Carbon monoxide is a very strong-field ligand, so it forces the 4s electrons into the 3d orbitals; the 3d electrons pair up completely to give 3d¹⁰ with no unpaired electrons. The empty 4s and 4p orbitals then form sp³ hybrids, giving a tetrahedral, diamagnetic complex. So although both are tetrahedral, the difference in oxidation state and ligand strength makes one paramagnetic and the other diamagnetic.

Answer 4

Concept: First find the oxidation state and d-electron count of iron, then decide the type of hybridisation from the ligand strength. Cyanide is a strong-field ligand, so it causes pairing and inner d-orbitals are used. Answer: In [Fe(CN)₆]^3–, each cyanide carries a –1 charge, so iron is in the +3 state, Fe³⁺. Iron is atomic number 26, so Fe³⁺ is 3d⁵. Cyanide is a strong-field ligand, so it forces the five d-electrons to pair up as far as possible, leaving only one unpaired electron in the 3d orbitals and emptying two inner 3d orbitals. These two 3d orbitals, together with the 4s and three 4p orbitals, form d²sp³ hybridisation, giving an octahedral, low-spin (inner orbital) complex. Because one unpaired electron remains, the complex is paramagnetic. So the hybridisation is d²sp³ and the complex is paramagnetic with one unpaired electron.