charco
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#21
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#21
(Original post by -Kav-)
I appreciate that simplified models have to be used in teaching and in prediction, of course I do, but I don't buy it in this case. A hybridisation model of coordinate complexes is all but useless, as it explains nothing but the geometry. Degenerate orbitals would make coloured complexes impossible, for one thing, and I contest that A-level students time would be much better spent learning the fundamentals of Crystal Field Theory (for which they already have most of the 'tools') rather than learning an enormous list of complex colours without any explanation as to how they arise. Hybridisation is a very useful model for organic chemistry and, to some extent, for aspects of main group chemistry, but I can see no excuse for applying it to co-ordinate chemistry.
Hybridisation offers a model that does not exclude the principles of transition metal chemistry. It can explain colour, magnetism etc if the 3d orbitals remain intact, but 'split' by a ligand field. (nobody suggested that hybridisation be used in isolation)
The hybridisation used for lone pair acceptance would be between 4s, 4p and 4d. This leaves transitions between 3d orbitals, paramagnetism etc perfectly explainable to pre-university students, and it fits observable geometry.
You can easily argue that this leaves inner orbitals either partially, or wholly, unoccupied. That's fine as it has a precedent in, for example, the configuration of chromium and copper, if you consider the 4s orbital to be of lower energy, or any of the transition metal atoms if you consider 3d to be of lower energy.
These models are useful and with a little simplified ligand field theory lead to a greater understanding of concepts that can be developed at a later date.
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-Kav-
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#22
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#22
(Original post by EierVonSatan)
I'm afraid that i have to agree with charco, crystal field theory has never been an A-level topic and frankly shouldn't be. Hybrid models sp2d (square planar), sp3d (triginal bipyramidal), sp3d2 (octohedral) hybrids are reasonable approximations, even though yes i know there are many exceptions - this is why MO theory has largely replaced it. Its too much to ask A-level students to cope with non-bonding and antibonding orbitals, symmetry mixing and the variation principle etc
Rudimentary CFT doesn't require any of that, though. If you know your orbital shapes and your electron-electron repulsion you'll be fine. I suppose it's just a personal thing. I've been too indoctrinated by our Inorganic faculty, some of whom would take up torches and pitchforks at the notion of using hybridisation in coordinate chemistry at any level.

(Original post by charco)
Hybridisation offers a model that does not exclude the principles of transition metal chemistry. It can explain colour, magnetism etc if the 3d orbitals remain intact, but 'split' by a ligand field. (nobody suggested that hybridisation be used in isolation)
The hybridisation used for lone pair acceptance would be between 4s, 4p and 4d. This leaves transitions between 3d orbitals, paramagnetism etc perfectly explainable to pre-university students, and it fits observable geometry.
You can easily argue that this leaves inner orbitals either partially, or wholly, unoccupied. That's fine as it has a precedent in, for example, the configuration of chromium and copper, if you consider the 4s orbital to be of lower energy, or any of the transition metal atoms if you consider 3d to be of lower energy.
These models are useful and with a little simplified ligand field theory lead to a greater understanding of concepts that can be developed at a later date.
Aye, your right, I've gone off on one without considering things properly. That makes sense, but it strikes me as a good deal more complicated than CFT. It's actually a better model than pure CFT in many ways, but there's no way I'd be happy presenting it to A-level students, which is presumably why it's not examined.
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aster100
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#23
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#23
Now that's what you call a discussion.
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TheRandomer
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#24
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Woaaaah. *intimidation*

Erm, why does Cu onlys bond to 4 ammonia ligands, instead of the full 6 like Co or Cr, though? Any particular reason? Is Copper smaller or something?
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EierVonSatan
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#25
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#25
It's because of something called jahn-teller distortions, which you won't need to know about for A-level...but it essentially means that ligand substitution occurs at only 4 positions for Cu2+
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TheRandomer
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#26
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(Original post by EierVonSatan)
It's because of something called jahn-teller distortions, which you won't need to know about for A-level...but it essentially means that ligand substitution occurs at only 4 positions for Cu2+
Ok thankyou
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zzzzzoe
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#27
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coppers just an oddball with excess NH3. the rest (if they do react with excess) have 6 surrounding the metal ion without any OH groups.
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cpchem
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#28
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#28
(Original post by EierVonSatan)
It's because of something called jahn-teller distortions, which you won't need to know about for A-level...but it essentially means that ligand substitution occurs at only 4 positions for Cu2+
How I love Jahn-Teller distortions. The way they just get rid of those nasty degenerate ground states...

It's becoming obvious that my exams are over... I'm getting nostalgic for chemistry. Worrying.
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