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Topic: Why is MgF2 octahedrally coordinated? (with no d orbitals)  (Read 4117 times)

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Offline apexaviour

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Why is MgF2 octahedrally coordinated? (with no d orbitals)
« on: August 19, 2011, 05:33:11 PM »
Mg(II)F2 has a rutile structure like TiO2. It forms the same structure as MnF2, NiF2, CoF2 etc.. which are all transition metal cations and have d-orbitals forming your octahedral crystal field (ie surrounded by 6 anions). The d-orbitals split into eg and t2g in this field and the bonding is fairly straight forward.

Anyway... magnesium is element 12, no d orbitals. Okay thinking about it now the p-orbital lobes could line up pointing in 6 directions... hmmm is that the case or am I way off? Is there some hybrisation with the Mg 3s? Or am I barking up the wrong tree here...

Jebus I really need to find out more about the bonding in this compound. Any recommended reading or papers would be appreciated... I've exhausted my searching ability on web of science.

Cheers  ;)

-experimental physics grad student desperately trying to write up thesis.


Offline cheese (MSW)

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Re: Why is MgF2 octahedrally coordinated? (with no d orbitals)
« Reply #1 on: August 21, 2011, 12:51:09 PM »
If you had taken your BSc in chemistry (;)) you would have learned early on that an important factor in the determination of the structure of solids that are mainly ionic is the radius ratio [google] r+/r- where r+ is the radius of the cation and r- is the radius of the anion.   Electrostatic attraction is nondirectional (unlike covalent bonding): the more anions that can be packed around a cation the more energetically favorable it is; opposing this is the electrostatic repulsion between the anions.  It is predicted that if r+/r- is between 0.4 - 0.7, 6:6 (NaCl lattice) 6:3 (TiO2) will be preferred, whereas if r+/r- is >0.7,  8:8 (CsCl) 8:4 (CaF2) is predicted to be more stable.   There is only a rough correlation between r+/r- predictions and what is observed; the exceptions are rationalized that covalent contributions tip the structure in favor of 6:6 (AX) or 6:3 (AX2) coordination.  Note that Na^+ in NaCl is six-coordinate and doesn’t have any d AOs to talk about.
So calculate r+/r- for your cmpds [the data at http://www.webelements.com/  appears reliable] and confirm that radius ratio goes a long way in determining the structures of MF2 cmpds (note that CaF2 with the larger Ca^2+ ion has 8:8 coordination).  The r+ for Mg^2+ and say Mn^2+ are similar.
Lattice Energies (+ve values, kJ mol^-1) MgF2 2926 (calc) 2978(exp);  MnF2 2644(calc) –(exp)
(CRC Handbook).
For NiF2, CoF2 there is added CFSE that you are familiar with (F^- is however weak field ligand).  For  MnF2 (high spin d^5) and ZnF2 (d^10) there is no CFSE, but the cmpds are still stable.
The following is somewhat  outside my area of expertise.)  There is some interaction of the empty 3p AOs on Mg^2+ with the filled 2p AOs of F^- to give the valence band, but the interaction is small because there is a large energy difference between interacting AOs.  MgF2 is an insulator with a large band gap.
Further reading at your level: J. E. Huheey, E. A. Keiter, R. L. Keiter Inorganic Chemistry 4th ed. Chpt 4.

Offline apexaviour

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Re: Why is MgF2 octahedrally coordinated? (with no d orbitals)
« Reply #2 on: August 27, 2011, 05:59:23 PM »
I'm not sure I could have hoped for a better answer. Everything you wrote either directly or as a nucleation point, filled in pretty much all the gaps and questions I had. It also helped me remember some of the two years of undergrad chemistry I thought I'd forgotten.

You helped me plough through a section of my thesis I was really stumped and frustrated with. My sincerest thanks!

 

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