Pentacoordination of Zinc: A Possibility or a Myth?

  • Thread starter Thread starter chem_tr
  • Start date Start date
  • Tags Tags
    Zinc
Click For Summary

Discussion Overview

The discussion revolves around the possibility of pentacoordination of zinc(II) ions, specifically examining the coordination of five ligands in a tetragonal monopyramidal environment. Participants explore theoretical implications, stability energy calculations, and the geometrical configurations involved in such coordination.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that zinc(II) ions have a d10 electronic configuration, leading to questions about the stability energy in pentacoordination.
  • There is a suggestion to calculate the stability energy for five ligands, with inquiries about whether the ligands are identical.
  • One participant clarifies that in certain configurations, four donors are in a planar arrangement with a fifth ligand approaching perpendicularly.
  • Another participant discusses the degeneracy split pattern for square pyramidal geometry and its implications for energy levels in coordination complexes.
  • There are speculations about the energy calculations required to determine the stability of pentacoordinated zinc, with references to complex formulas and the need for perturbation theory.
  • Some participants express uncertainty about how to apply the provided data to determine the feasibility of pentacoordination for zinc.

Areas of Agreement / Disagreement

Participants express varying levels of uncertainty regarding the calculations and implications of pentacoordination. There is no consensus on whether zinc can effectively bind a fifth ligand to achieve a pentacoordinated state, and multiple viewpoints on the theoretical aspects remain unresolved.

Contextual Notes

Limitations include the complexity of ligand interactions, the dependence on specific ligand types, and the unresolved nature of the mathematical steps required for accurate energy calculations.

chem_tr
Science Advisor
Gold Member
Messages
608
Reaction score
3
Hello,

I've heard rumors (frankly, not rumors, evidenced by some scientific papers), that zinc(II) ion has coordinated five ligands, four at a plane, and the fifth from top, to form a tetragonal monopyramidal environment.

I have some doubts about this, because zinc (II) ion has d10 electronic configuration in normal circumstances, which means, according to crystal field theory, that octahedral stability energy is zero.

What are your comments?

chem_tr
 
Chemistry news on Phys.org
Hi chem_tr,

Five ligands? Mmm... strange. It would be necessary to calculate the stability energy. Are the five ligands identical? I´ve found the degeneracy split pattern for a square piramidal geomety:
http://www.unine.ch/chim/chw/html%20files/CFSE%20geom.html

Can you calculate the stability energy from this data? I´m not sure about it.
 
Last edited by a moderator:
Hello altered-gravity,

Let me correct something first. There are four donors in a planar macrocyle (e.g., phthalocyanine), and the fifth one approaches from the perpendicular axis to the plane. There are also some examples in which zinc coordinates a pentadentate ligand, two thiolates, one pyridine nitrogen, and two azomethine nitrogens.

I will look through your table in detail, thank you for your interest.
 
Biochemical complexes are too much for my POOR "ab initio power" :biggrin:

Anyway the case you are talkig about (phtalocyanine ring) is a square based piramidal geometry (C4v point group). In the table I post before, that geometry is supposed for five identical and simple ligands, and says that the degeneracy splits into four different energy levels, three monodegenerate and one doubly degenerate.

In your case, although much more complicated, the point group is the same. So we may think that the energy scheme is the same (four levels). The task is to calculate the system energy provided that there are 10 electrons and the spin is 0. I think that the energy would be like this:
[tex]E=\sum_{i} e_i -\sum_i \sum_{j>i} J_{ij}[/tex]

i: each electron
ei: energy of the level where "i" electron is
Jij: Coulomb integral of electrons "i" and "j"

I´m just speculating so don´t pay too much attention to me if you´re working seriously on this.
 
Last edited:
Hello,

C4v symmetry is valid for phthalocyanines, you are right. I have no idea about how to use these data on the table I downloaded from your post. Can you say that, by using the values on the table, it is possible/known for Zn to bind another ligand to reach pentacoordinated state?

My mathematics has left me a long time ago, so I cannot make any comments about your double-sigma-containing huge formula :smile:

Thank you for your interest
 
chem_tr said:
Can you say that, by using the values on the table, it is possible/known for Zn to bind another ligand to reach pentacoordinated state?

Not exactly. That table gives you energies of the "d" orbitals for different geometries, but with ONLY ONE ligand type, for example Zn(OH)4 2-. From here you can calculate the square piramid geometry system energy, and compare it with others as planar square geom, in other words, the relative stability of the pentacoordinated. Then you can "imagine" or qualitatively approach the perturbation that the fifth ligand creates when it aproaches the square planar system.

The table gives energyes in terms of [tex]\Delta_0[/tex], this parameter is the energy perturbation of an octahedral geometry with the same ligand type, look at this:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/crystal.html
this parameter must be calculated for each ligand type.

For the phtalocyanine system you can only know that the energy pattern wold be similar due to the simetry, but not equal. If you want to calculate it exactly you must classically construct the whole field that all ligands create and start with parturbation theory calculus. This task colud be large with such complex ligands.

I´m sorry if this doesn´t help you. Good luck.
 
Last edited by a moderator:

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
7K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 48 ·
2
Replies
48
Views
14K
Replies
4
Views
11K
  • · Replies 237 ·
8
Replies
237
Views
20K
Replies
6
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K