Rank ability of ligands to increase or decrease Δ in a square planar compound

In summary, constructing a spectrochemical series involves ranking ligands based on their ability to split d-orbitals in a coordination complex. Comparisons can be made by analyzing the absorption wavelengths and considering the relative strength of ligands as sigma donors and pi acceptors. [NiBr(Mes)(PPh3)2] cannot be directly compared to the other complexes given the data, but it can be said that it will have a larger Δ compared to [NiBr(Mes)(dppe)] due to the presence of multiple phosphorus atoms in the ligands.
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Homework Statement


Homework Equations



I am working on a formal lab report for an inorganic chem course and forget how to construct a spectrochemical series. I have 4 complexes, each has a single absorption band in the UV-Vis spectrum.

[NiBr(Mes)(PPh3)2] 450nm
[Ni(Mes)Br(bipy)] 487nm
[NiBr(Mes)(dppe)] 425nm
[NiBr2(dppe)] 477nm

I am told to rank the ability of the ligands (Br-, dppe, bipy, PPh3, Mes-) to increase or decrease Δ in a square planar compound based only on the experimental data (given above).

The Attempt at a Solution



I have decided that the each of the above complexes is square planar in geometry, so they can be compared directly to one another in terms of d-d electronic transitions.

Comparing [NiBr(Mes)(bipy)] and [NiBr(Mes)(dppe)], which differ by only one ligand, it can be deduced that dppe will cause an increase in Δ relative to bipy. Similarly, [NiBr2(dppe)] and [NiBr(Mes)(dppe)] can be compared to show that Mes- will cause an increase in Δ relative to Br-.

I am now unsure if I am able to make any further comparisons. Concerning [NiBr(Mes)(PPh3)2] in particular, I do not recall if it is possible to compare this complex to any of the others, given the above data. At best, it differs from [Ni(Mes)Br(bipy)] by bipy and (PPh3)2, and a similar comparison can be made to [NiBr(Mes)(dppe)]. However this only allows the comparison of (PPh3)2 and not a single PPh3. I believe that the first PPh3 will have a significantly different influence on the complex compared to the second PPh3, and so the ligand cannot be ranked using this data.

I haven't had to do this for quite a while so I just wanted to make sure I remember how to do it correctly. I tried to reference it but had trouble finding information on ranking ligands relative to one another, and in particular when they differ buy 1L_1 and 2L_2 as I detailed above. Any help is greatly appreciated. Thanks!
 
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Dear student,

Thank you for your question. Constructing a spectrochemical series involves ranking ligands based on their ability to split d-orbitals in a coordination complex. This series is usually constructed by comparing the absorption wavelengths of different complexes and determining the ligand that causes the largest splitting (Δ) in the d-orbitals.

In your case, you have correctly compared the complexes [NiBr(Mes)(bipy)] and [NiBr(Mes)(dppe)] and determined that dppe causes a larger Δ compared to bipy. This is because dppe is a stronger sigma donor and pi acceptor compared to bipy, which leads to a larger splitting of the d-orbitals. Similarly, you have also correctly compared [NiBr2(dppe)] and [NiBr(Mes)(dppe)] and determined that Mes- causes a larger Δ compared to Br-. This is because Mes- is a stronger pi donor compared to Br-, which leads to a larger splitting of the d-orbitals.

For the complex [NiBr(Mes)(PPh3)2], it is not possible to make a direct comparison with the other complexes given the data provided. This is because the ligands PPh3 and (PPh3)2 may have different effects on the splitting of d-orbitals. However, we can still make a comparison by considering the ligands as a group. In general, ligands that have multiple phosphorus atoms (such as PPh3 and (PPh3)2) are considered stronger sigma donors compared to ligands with a single phosphorus atom (such as dppe). Therefore, we can say that [NiBr(Mes)(PPh3)2] will have a larger Δ compared to [NiBr(Mes)(dppe)].

I hope this helps. Good luck with your lab report!
 

1. How does the size of a ligand affect its ability to increase or decrease Δ in a square planar compound?

The size of a ligand can greatly impact its ability to increase or decrease Δ in a square planar compound. Generally, smaller ligands are better at increasing Δ, while larger ligands are better at decreasing Δ. This is because smaller ligands can form stronger bonds with the central metal ion and create a greater difference in energy levels, leading to a larger Δ.

2. What is the relationship between the electron-donating or withdrawing ability of a ligand and its effect on Δ in a square planar compound?

The electron-donating or withdrawing ability of a ligand can also play a role in its ability to increase or decrease Δ in a square planar compound. Electron-donating ligands, such as amines, can increase Δ by destabilizing the d orbitals of the central metal ion. On the other hand, electron-withdrawing ligands, such as halides, can decrease Δ by stabilizing the d orbitals and reducing the energy difference between them.

3. Can the coordination number of a central metal ion affect the rank ability of ligands to increase or decrease Δ in a square planar compound?

Yes, the coordination number of the central metal ion can have an impact on the rank ability of ligands to increase or decrease Δ in a square planar compound. In general, ligands with a greater number of bonding atoms, such as bidentate ligands, are better at increasing Δ compared to ligands with fewer bonding atoms, such as monodentate ligands. This is because bidentate ligands can form more stable bonds with the central metal ion and create a larger energy difference between the d orbitals.

4. How does the electronegativity of a ligand affect its ability to increase or decrease Δ in a square planar compound?

The electronegativity of a ligand also plays a role in its ability to increase or decrease Δ in a square planar compound. Generally, ligands with higher electronegativity, such as oxygen or nitrogen, are better at increasing Δ because they can form stronger bonds with the central metal ion. Ligands with lower electronegativity, such as sulfur or phosphorus, are better at decreasing Δ because they form weaker bonds with the central metal ion.

5. Can the geometry of a ligand affect its ability to increase or decrease Δ in a square planar compound?

Yes, the geometry of a ligand can have an impact on its ability to increase or decrease Δ in a square planar compound. For example, ligands with a tetrahedral geometry, such as NH3, are better at increasing Δ compared to ligands with a square planar geometry, such as PPh3. This is because tetrahedral ligands can form stronger bonds with the central metal ion and create a larger energy difference between the d orbitals.

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