# Square Planar Complex Ion Arrangements

#### lavalamp

Why is it that sometimes ligands that surround a transition metal ion form a square planar arrangement?

Surely the best thing to do (according to electron pair repulsion theory) would be to form a tetrahedral arrangement.

I have heard that complex ions with Nickel as the transition metal ion form square planar arrangements quite often, so I will use one of those as an example, Ni[Cl]2[NH3]2:

Code:
      Cl
|
NH[sub]3[/sub] - Ni - NH[sub]3[/sub]
|
Cl

or:

NH[sub]3[/sub]
|
Cl - Ni - NH[sub]3[/sub]
|
Cl
If it helps, we have just done d-orbital splitting.

#### Bystander

Homework Helper
Gold Member
Ugh. Topics to check: ligand field theory; crystal field theory. This is probably worse than no explanation, but given what you're paying --- .

You've been through "splitting," and you'll want to keep in mind that the orbitals you're splitting are "hydrogen-like." For us to expect that transition metals be "hydrogen-like" is a bit flaky --- however, it's the only quantum mechanical result available to us to explain the behaviors of atoms more complex than H. All H orbitals of a given principal quantum number, n, have identical energies; orbitals for real atoms more complex than H do not, and it is reasonable to regard/consider/analyze/explain the energy differences as being the result of "splitting" due to internal fields from other occupied orbitals, and from external fields (external fields can be used to split H orbitals --- Zeeman, etc.). The magnitudes of the splits observed are functions of which element we examine, its oxidation state, the external fields (ligands, complexing agents), and can be large enough that it becomes energetically favorable for electrons to pair up in a single orbital rather than half-filling all available orbitals. In the "square-planar" case, you CAN (don't have to --- it's better to buy into an instructor's explanation) regard the complex in question to be a special case of the octahedral complexes in which two of the ligands are electron pairs belonging to the complexed atom/ion occupying two opposite vertices of an octahedron, and the more loosely attracted ligands occupying the other four vertices.

Don't commit yourself too heavily to this version --- try to find a few comments in standard texts on ligand field theory and crystal field theory and synthesize your own picture of what's going on.

#### lavalamp

Why is it that the "simple" questions always have comlicated answers?

I'll look into ligand and crystal field theory on google or yahoo.

It has been pointed out to us by our teacher that there is a square planar arrangement of ligands in an octahedral molecule. At the time I did wonder whether it was a lone pair of electrons either side of the planar molecule that caused it to stay planar, but if this were the case then there would also be another shape similar to pyramidal but with four bonds and two lone paris at the "top".

The reason that I want to know all of this is because I find it realy annoying when my teachers tell us that something happens but don't tell you why it happens. My physics teacher is doing this to me right now, "this happens because they say so".

#### Bystander

Homework Helper
Gold Member
"Simple." Uh-huh --- and, depending on how things split, unoccupied high energy orbitals.

A "simple" answer would be, "Quantum mechanics does NOT work for anything more complicated than the hydrogen atom, hydrogen like ions, and positronium; all the things that go on in the rest of the periodic table and in real chemical compounds, ions, complexes, and the like are the 'way they are because they are the way they are.' " This approach isn't satisfying to anyone --- the modifications to H-like orbitals leave a lot to be desired, but, that is the present state of the art.

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