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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: zeoblade on April 05, 2010, 07:40:19 AM

Title: Square Planar Crystal Field Theory
Post by: zeoblade on April 05, 2010, 07:40:19 AM
In square planar complexes, how is the order of electron filling supposed to proceed? Specifically for d4 and d8 1st row transition metals

A) Half fill the dxz and dyz in eg level then completely fill them before filling dz2 in a1g level then half filling dxy in b2g level then filling dx2-y2 in b1g level?

B) Half filling all d-orbitals in eg to b2g then completely filling them?

Thanks in advance
Title: Re: Square Planar Crystal Field Theory
Post by: Francopper on April 06, 2010, 08:46:05 PM
Greetings.
In square planar complexes the orbitals that are involved with the z axe are extremely stabilized because of the null interaction of the z axe with the ligands. Thus, the dz2, dxz and dyz have very low energy, so these are the ones to fill first (they have about the same energy so you can half fill the three of them, then completely fill them). The dxy orbital is immediately higher in energy, so it's the next one to be filled, and finally the dx2-y2 orbital, the most unstable (highest in energy) one is the last one to be fill since it's directly poiting to the ligands in the square plane complex, having maximum electrostatic repulsion.
Hope this helps.
Title: Re: Square Planar Crystal Field Theory
Post by: zeoblade on April 06, 2010, 08:58:32 PM
it really does help, thanks francopper.

apparently d4 d8 favour square planar rather than tetrahedral. i feel maybe square planar for strong field ligands and tetrahedral for weak field ligands? i have a feeling there's more occurrence in tetrahedral <---> square planar conversion for d4 because the tetrahedral dz2 and dx2-y2 are relatively close in energy to square planar dxz dyz and dz2. but d8 just likes being square planar as seen in Ni, Pd, Pt, Au.

i'm curious what is the filling order of electrons for square planar. i'm also assuming hund's rule for for the 3rd and 4th electrons fill the dxz dyz to complete pairing. but dxz dyz is not far away from dz2 in energy being relatively close far below the barycentre than tetrahedral dz2 dx2-y2. i have a feeling that in square planar, the 3rd electron might go to dz2 before a dxz or dyz orbital is paired, and pairing of dxz dyz before the 4th electron fills dxy because of the difference in energy between dz2 and dxy. is this plausible?
Title: Re: Square Planar Crystal Field Theory
Post by: Francopper on April 06, 2010, 09:51:24 PM
Yes, d8 transition metal ion always form square plane with exception of Ni2+ which with strong field ligands forms square plane and weak field ligands tetrahedral. The reason why all other d8 ions form tetrahedral lies in the fact that as they are for the second and third row of transition metals (think of Pd, Pt, Au, Rh, Ir) form complexes with bigger Δ than the first row transition metals (like Ni) because of having larger orbitals (3d<4d<5d), causing more repulsion, increasing Δ, just like strong field ligands do.

d4 I think have more tendency to form tetrahedral complexes because in order to form square plane they first need to form an octahedral arrangement and distort in order to stabilize electrons formerly occupying degenerated eg orbitals. When one makes the diagram it's easy to see that d4 high spin octahedral complexes can distort, according to Jahn-Teller's theorem, in order to stabilize the z axe, increasing the length of the metal-ligand bond in that axe, and hence forming a tetragonal structure. But it can hardly derive on square plane complexes because it only has one electron (an odd number of electrons) in eg orbitals, needing at least 2.
Title: Re: Square Planar Crystal Field Theory
Post by: zeoblade on April 06, 2010, 09:58:45 PM
thank you so much, your explanation is very appreciated, you just filled in the blank