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Topic: Geometry of K2PtBr4 Complex?  (Read 6566 times)

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qcan375

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Geometry of K2PtBr4 Complex?
« on: February 22, 2006, 03:33:40 PM »
Just curious if anyone has seen the geometry of this complex usually seen in the form K2PtBr4.

Is the complex PtBr4 with a negative 2 charge exhibit square planer geometry or tetrahedral geometry or octahedral geometry?

I am guessing it is square planer since most Pt complexes are. Is this correct?

Really appreciate it if anyone knows.
 ???

Offline Albert

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Re:Geometry of K2PtBr4 Complex?
« Reply #1 on: February 22, 2006, 04:21:03 PM »
PtBr4 has square planer geometry, like most of Pt2+ complexes.

Quote
I am guessing it is square planer since most Pt complexes are.

Most of Pt4+ complexes are, on the contrary, octahedral: i.e. K2PtCl6.
« Last Edit: February 22, 2006, 04:21:43 PM by Albert »

AgG

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Re:Geometry of K2PtBr4 Complex?
« Reply #2 on: February 27, 2006, 04:21:20 PM »
yep

organometallic

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Re:Geometry of K2PtBr4 Complex?
« Reply #3 on: April 06, 2006, 03:28:24 AM »
d8 and 4 ligand always show square planer geometry. Totally have 16 electrons.

Offline plu

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Re:Geometry of K2PtBr4 Complex?
« Reply #4 on: April 29, 2006, 08:27:18 PM »
d8 and 4 ligand always show square planer geometry. Totally have 16 electrons.

I don't believe this is true.  NiCl42- has a tetrahedral arrangement (Ni(CO)4 does as well, I believe).

On a side note, why is it that d8 transition metal ions tend to form 4-coordinate metal complexes that give total electron counts of 16, which seem to disobey the "18-electron rule"?

Offline letk0

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Re: Geometry of K2PtBr4 Complex?
« Reply #5 on: May 25, 2006, 10:37:31 PM »

I don't believe this is true. NiCl42- has a tetrahedral arrangement (Ni(CO)4 does as well, I believe).

On a side note, why is it that d8 transition metal ions tend to form 4-coordinate metal complexes that give total electron counts of 16, which seem to disobey the "18-electron rule"?

Crystal field theory explains this pattern.  In a square planar environment around a d8 metal's 5 d-orbitals, the dx^2-y^2 is not split by the coordinating ligands, therefore raising its energy.  On the other hand, the ligands will split the dxy, dxz, dyz and dz^2 orbitals, which will place them at a lower energy.  Filling these four orbitals with the 8 electrons increases their gap from dx^2-y^2, resulting in a stable complex.  Google should bring up some more in depth intros to crystal field theory involving various geometries.

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