Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: qcan375 on February 22, 2006, 03:33:40 PM
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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.
???
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PtBr4 has square planer geometry, like most of Pt2+ complexes.
I am guessing it is square planer since most Pt complexes are.
Most of Pt4+ complexes are, on the contrary, octahedral: i.e. K2PtCl6.
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yep
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d8 and 4 ligand always show square planer geometry. Totally have 16 electrons.
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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"?
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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.