4
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ I get it. But until the ligands approach, don't the $d$ orbitals exist in specific orientations? $\endgroup$ Aug 27, 2016 at 5:54
  • 4
    $\begingroup$ No, they don't. $\endgroup$ Aug 27, 2016 at 9:52
  • $\begingroup$ @IvanNeretin Could you please explain why not? Are the dx2-y2 orbitals not uniquely identifiable from, say, the dxy orbitals, until a ligands approach? $\endgroup$ Jun 29, 2021 at 6:15
  • $\begingroup$ @OneEyedMushroom This is already not the case for p orbitals. In fact, what we often think of as $\mathrm p_x$ and $\mathrm p_y$ would be a pair of complex-number orbitals $\mathrm p_{-1}$ and $\mathrm p_{+1}$ in isolation. Maybe there is a chance of identifying $\mathrm{p_0}$ (also known as $\mathrm p_z$) because of its quantum state but this would be outside of my knowledge of maths and quantum mechanics. $\endgroup$
    – Jan
    Jun 30, 2021 at 9:12
  • 1
    $\begingroup$ More fundamentally, the choice of p-orbital basis is arbitrary, because all three p-orbitals are eigenvalues of the Hamiltonian with the same energy. Consequently, any linear combination of p-orbitals, $c_x\mathrm{p}_x + c_y\mathrm{p}_y + c_z\mathrm{p}_z$, is also a perfectly valid 'orbital' with $n = 1$ and $\ell = 1$. The p-orbitals $\mathrm{p_x}$ (i.e. the choice of coefficients $(c_x, c_y, c_z) = (1, 0, 0)$), etc. are entirely arbitrary choices, and the fact that we typically choose them to be along interatomic axes is purely out of convenience. $\endgroup$
    – orthocresol
    Jul 2, 2021 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.