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Topic: why [CoCl4] has regular tetrahedral but not [CuCl4]  (Read 19302 times)

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SS

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why [CoCl4] has regular tetrahedral but not [CuCl4]
« on: November 22, 2004, 09:26:39 PM »
Hi, does anyone know why [CuCl4] has a flattened tetrahedral structure, not regular tetrahedron? ???
« Last Edit: November 24, 2004, 12:11:39 PM by SS »

Offline Donaldson Tan

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Re:[CuCl4] form distorted tetrahedral
« Reply #1 on: November 23, 2004, 02:04:03 PM »
I think it's because the electron pair geometry of the central copper ion is octahedral. 4 vertices are taken up by Cl atoms, and the other 2 vertices are taken up by lone pairs. This should explain why CuCl42- is a distorted tetradedron.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

puru

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Re:why [CoCl4] has regular tetrahedral but not [CuCl4]
« Reply #2 on: February 04, 2005, 09:41:04 PM »
Guys Are Forgetting the crystal feild theory by Bethe and van Vlecke.This also has in connection the Jahn And Teller distortion of Octahedral Complexes.
the d orbitals split as you know into t2g and eg orbitals.
Due to octahedral feild splitting and weak ligand effect of Cl [CuCl4]2- is distorted octahedron or squashed tetrahedon.

Offline Donaldson Tan

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Re:why [CoCl4] has regular tetrahedral but not [CuCl4]
« Reply #3 on: May 21, 2005, 09:42:48 AM »
Al Basra: care to enlighten us on Crystal Field Theory?
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Grafter

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Re:why [CoCl4] has regular tetrahedral but not [CuCl4]
« Reply #4 on: May 30, 2005, 09:38:38 AM »
Crystal Field Theory concerns transition metal complexes, in which ligands (which are treated as point charges) approach at various coordinations (eg octahedral, tetrahedral, square planar), and as a result raise and lower the d orbitals in energy. This splitting of orbitals can be sufficient to pair electrons in some d orbitals before all are filled. This gives rise to 'low spin' and 'high spin' states (as observed in iron in haemoglobin).

The energy jumps betweeb the raised orbitals and lowered orbitals are in the region of visible light, giving the characteristic colours of transition metal complexes.

In answer to your question, copper is a d9 transition metal, which experiences the Jahn-Teller effect, which effectively distorts an octahedral complex so much that two ligands on the z axis are sp loosely held, that they depart, leaving a square planar complex.

Further reading: Greenwood and Earnshaw

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