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In my high school chemistry book, it is written that when ligands approach the central metal ion (transition metal ion) to form dative bonds, the $3d$ orbitals split into two: two which are in higher energy level and the other 3 in lower energy orbital...

One thing that doesn't make intuitive sense to me is that why 3 orbitals move to a lower energy level... if the ligands repel the electrons in the $3d$ orbitals, how can some of the orbitals possibly lose energy and be at a lower enrgy state than they were currently at before... enter image description here

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The picture you pose is only a part. The full picture is posted below.enter image description here

This solves your doubt. The splitting does not occur about the original level, i.e. not about the energy level the d-orbital has in complete absence of the ligand's electrostatic field. But, the splitting occurs about a hypothetical Barycentre. This is the energy level of the d-orbitals (still degenerate) in a hypothetical spherical electric field (with spherical symmetry). Due to the symmetry of the hypothetical barycentre field, the d-orbitals remain degenerate but at an energy higher than in the absence of the ligands. Now, the $t_{2g}$ group which is lowered in case of an octahedral field, is still above the original free ion energy level in the complete absence of the ligands.

EDIT
By conservation of energy, we mean that the energy is conserved during splitting. The very process of splitting occurs symmetrically about the barycentre, so that the net amount by which $t_{2g}$ is lowered ($3\times 0.4\Delta_o$) is equal to the amount by which $e_g$ is raised ($2\times 0.6\Delta_o$). But the net energy in absence of field and in presence of field need not be same for energy conservation to hold. Whatever the rise in the net energy (from free ionic to barycentre) will be taken from the kinetic energy (or potential) of the approaching ligands.

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  • $\begingroup$ That makes more sense, so all the orbitals have a higher energy level than the original.However, in my book, it says that: "from the point of view of conservation of energy, it is necessary that this split happens in such a way that the total energy of the 3d orbitals remains same "... doesn't that contradict the diagram you have given (i.e. how is energy conserved?) $\endgroup$ – Eliza Jan 26 '14 at 16:56
  • $\begingroup$ @Eliza Please see the edit in my answer $\endgroup$ – Satwik Pasani Jan 26 '14 at 17:12
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    $\begingroup$ so as the ligand approaches the 3d orbitals, the KE of the ligands is converted to the energy of the 3d orbitals... the "barycentre" just shows what would have happened if splitting was symmetrical... so the energy of the higher and lower energy 3d orbitals will just equal to the total energy of the barycentre.. is that it? I just want to check :) $\endgroup$ – Eliza Jan 26 '14 at 18:01
  • $\begingroup$ @Eliza To the best of my knowledge, yes. Just a minor clarification, saying that it is the kinetic energy of the ligands converting to the energy of the orbitals would not always be correct. The energy is a mixture of that released by the electrostatic attraction of the overall positive ion and the negative ligands, kinetic energy of either species, or any other energy involved in the interaction. $\endgroup$ – Satwik Pasani Jan 27 '14 at 2:58

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