According to the Crystal Field Theory (CFT), it is given that ligands (their lone pairs) approach the central metal along the axes. When they approach, the degeneracy of the $\mathrm{d}$ orbitals is lost. The explanation:

The $\mathrm{d}_{x^2-y^2}$ and the $\mathrm{d}_{z^2}$ orbitals(the $\mathrm{e_g}$ orbitals) are oriented along the axes while others are oriented in between the axes. In octahedral geometry having complexes, when the ligands approach, they approach along the axes and thus the $\mathrm{e_g}$ orbitals which are along the axes experience more repulsion. Hence, these orbitals increase in energy more than the $\mathrm{t_{2g}}$ orbitals. Thus, the orbitals split in energy due to the approach of ligands.

My question is why do the ligands (their lone pairs) have to approach along the axes. They can approach from anywhere.

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    $\begingroup$ Isotropy of space. The axes aren't uniquely defined in space, you can choose them to point anywhere you like. It doesn't matter where you define them (it doesn't affect the final result) and it turns out it is just easier to do the maths that way. $\endgroup$ – orthocresol Aug 25 '16 at 15:30
  • $\begingroup$ There are no axes in space. The ligands approach from wherever they like. And then along come quantum chemists and draw their axes. $\endgroup$ – Ivan Neretin Aug 25 '16 at 15:31
  • $\begingroup$ That's exactly what I'm saying. The orbitals are not uniquely defined in space. If I approach the ligands along 3 t_2g orbitals, the energy changes or splitting should be different. $\endgroup$ – Apoorv Potnis Aug 25 '16 at 15:36
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    $\begingroup$ The splitting isn't different regardless of what direction you approach along. Hard to believe but it's true. $\endgroup$ – orthocresol Aug 25 '16 at 23:59
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    $\begingroup$ Okay, I get it. Even if I approach the ligands along the t2g orbitals, the total energy change and the splitting is the same. I also thank you for editing(formatting my question) $\endgroup$ – Apoorv Potnis Aug 26 '16 at 3:39

You basically said it.

They can approach from anywhere.

Well, I can define the axes to be anywhere.

An isolated atom in vacuum with degenerate d-orbitals does not know where $x,y$ or $z$ are. Hence, the $\mathrm{d}_{z^2}$ orbital does not have the ‘orientation’ you know from pictures — it only gets that once in some way axes are defined. You could now go ahead and define that one of your octahedral ligand approaches the central metal (placed at $(0,0,0)$) from the coordinates $(1,1,1)$. Get out a large sheet of paper and a new pencil and do the maths, calculating what the final orbital distribution will be. The maths will be a lot more complicated, but the result will be the same. Hence, chemists decide to not do complicated maths and just have octahedral ligands approach along the coordinate axes.

Note that in tetrahedral complexes, the ligands would indeed approach from $(1,1,1)$ and the symmetry-equivalent positions of a tetrahedron.

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    $\begingroup$ I get it. But until the ligands approach, don't the $d$ orbitals exist in specific orientations? $\endgroup$ – Apoorv Potnis Aug 27 '16 at 5:54
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    $\begingroup$ No, they don't. $\endgroup$ – Ivan Neretin Aug 27 '16 at 9:52

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