[zeoblade]

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 d_xz and d_yz in e_g level then completely fill them before filling d_z² in a_1g level then half filling d_xy in b_2g level then filling d_x²_-y² in b_1g level?

B) Half filling all d-orbitals in e_g to b_2g then completely filling them?

Thanks in advance

[Francopper]

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 d_z², d_xz and d_yz 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 d_xy orbital is immediately higher in energy, so it's the next one to be filled, and finally the d_x²-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.

[zeoblade]

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 d_z² and d_x²_-y² are relatively close in energy to square planar d_xz d_yz and d_z². 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 d_xz d_yz to complete pairing. but d_xz d_yz is not far away from d_z² in energy being relatively close far below the barycentre than tetrahedral d_z² d_x²_-y². i have a feeling that in square planar, the 3rd electron might go to d_z² before a d_xz or d_yz orbital is paired, and pairing of d_xz d_yz before the 4th electron fills d_xy because of the difference in energy between d_z² and d_xy. is this plausible?

[Francopper]

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 e_g 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 e_g orbitals, needing at least 2.

[zeoblade]

thank you so much, your explanation is very appreciated, you just filled in the blank