(Solved) - Consider a d8 compound (like Pt(II)). What is more likely to form... (1 Answer) | Transtutors (2024)

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Subhash P

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Valence Bond Theory (VBT) Crystal Field Theory (CFT) Considers formation of covalent bonds between metal and ligands via intermixing of their orbitals. Based on hybridization of orbitals. Considers effect of atomic orbitals of both metal and ligands. Covalent...

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(Solved) - Consider a d8 compound (like Pt(II)). What is more likely to
form...  (1 Answer) | Transtutors (2024)


Why is square planar geometry preferred for the second and third row transition metals with D8 configuration over octahedral and tetrahedral? ›

For this reason, most first row transition metals prefer tetrahedral while the second and third-row transition metals prefer square planar. Square planar is preferred because of the higher splitting of the d orbitals which is advantageous when the metal ion does not have filled d orbitals.

Why is NiCl4 tetrahedral and ptcl4 square planar? ›

The molecule [PdCl4]2− is diamagnetic, which indicates a square planar geometry as all eight d-electrons are paired in the lower-energy orbitals. However, [NiCl4]2− is also d8 but has two unpaired electrons, indicating a tetrahedral geometry.

Is d8 always square planar? ›

d-electron effects

imparting a strong preference for non-Kepert coordination geometries. This is why, for example, 2nd and 3rd row complexes in which the metal has a d8 electron configuration are almost always square planar.

Which is more stable square planar or octahedral? ›

c) Square planar complexes are more stable than octahedral complexes. d) The [Fe(CN). 14 ion is paramagnetic but [Fe(CN).] is diamagnetic.

Is PT tetrahedral or square planar? ›

Coordination number = 4

Most of the four-coordinated complexes of Zn(II), Cd(II), and Hg(II) are tetrahedral, while the square planar arrangement is preferred by Pd(II), Pt(II), and Cu(II) complexes. These two isomers are called geometrical isomers.

Why is PT 2 square planar? ›

That means its electrons generate little repulsion while chloride interacts with nickel, and the ligand-field splitting energy is small. That allows for square planar complexes to form, even though those are higher in energy. Pt(II) has a similar electron configuration as Ni(II) , also acting as a d8 transition metal.

How to know if tetrahedral or square planar? ›

A square planar structure has 2 lone pairs on the central atom (AX4E2) while a tetrahedral structure has no lone pairs on the central atom (AX4).

Why do d8 metals show strong preference for a square planar geometry? ›

D8 metals like Au(III) and Pt(II) prefer square planar geometry because they have eight d-electrons in their outermost shell. In square planar complexes, the metal ion is surrounded by four ligands arranged in a square plane.

Why do 4d and 5d metals form square planar complexes but not tetrahedral complexes? ›

Coordination number 4 allows for either square planar or tetrahedral shapes. However, when metals from the 4d and 5d series are present, all ligands act as strong field ligands. Strong field strength usually results in a square planar geometry.

Why is square planar Favoured over tetrahedral? ›

Additionally, if the central metal ion has unpaired electrons, it may favor a tetrahedral geometry to maximize the spatial separation of the unpaired electrons and if it has few or no unpaired electrons, it may be more likely to adopt a square-planar geometry.

Which metal ions with d3 and d8 electronic configuration prefers octahedral geometry? ›

Chromium (+3) (d3) and nickel (+2) (d8) do indeed show an exceptional preference for octahedral geometries. So if you consider the CFSE then predictions can be made that d3 and d8 ions strongly prefer an octahedral geometry over a tetrahedral one; for other configurations the preference is less pronounced.

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