[qcan375]

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.
 

[Albert]

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.

[AgG]

yep

[organometallic]

d8 and 4 ligand always show square planer geometry. Totally have 16 electrons.

[plu]

d8 and 4 ligand always show square planer geometry. Totally have 16 electrons.

I don't believe this is true.  NiCl₄^2- 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"?

[letk0]

I don't believe this is true. NiCl₄^2- 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 d_x^2-y^2 is not split by the coordinating ligands, therefore raising its energy.  On the other hand, the ligands will split the d_xy, d_xz, d_yz and d_z^2 orbitals, which will place them at a lower energy.  Filling these four orbitals with the 8 electrons increases their gap from d_x^2-y^2, resulting in a stable complex.  Google should bring up some more in depth intros to crystal field theory involving various geometries.