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A classmate says, "A weak-field ligand usually means the complex is high spin." Is your classmate correct? Explain.

Short Answer

Expert verified
Yes, your classmate is correct. A weak-field ligand results in a smaller crystal field splitting energy (Δ), which is more likely to lead to a high-spin complex. This is because when Δ is smaller than the pairing energy (P), electrons prefer to occupy higher-energy orbitals with parallel spins rather than pairing up in lower-energy orbitals.

Step by step solution

01

Understanding Weak-Field Ligands and High-Spin Complexes

Weak-field ligands are ligands that do not produce a large crystal field splitting energy (Δ). They result in smaller differences between the energy levels of the d orbitals in a transition metal complex. This means that d electrons will not be strongly stabilized by the ligand, and the ligand has a weaker ability to create a field around the metal ion. On the other hand, high-spin complexes are those in which electrons prefer to occupy higher-energy orbitals with parallel spins, rather than pairing up in lower-energy orbitals. This occurs when the crystal field splitting energy (Δ) is smaller than the pairing energy (P), which is the energy required for an electron to pair up in a lower-energy orbital (Δ < P).
02

Examining the Correlation between Weak-Field Ligands and High-Spin Complexes

A weak-field ligand, as mentioned earlier, results in a smaller crystal field splitting energy (Δ). When the splitting energy is small, it is more likely that the complex will be high-spin because the electrons would prefer occupying the higher-energy orbitals with parallel spins rather than pairing up in the lower-energy orbitals. In other words, if the ligands are weak-field, the electrons will not be stabilized by their interaction with the ligands, so they will require less energy to move to the next energy level (as compared to moving to a higher orbit in a strong-field complex). Thus, it's more probable that the complex will be high-spin with weak-field ligands.
03

Conclusion

Based on this explanation, it can be said that the classmate's statement is indeed correct. A weak-field ligand usually means the complex will be high-spin, given that the crystal field splitting energy is smaller than the pairing energy, making it more favorable for electrons to occupy higher-energy orbitals with parallel spins.

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Most popular questions from this chapter

For each of the following metals, write the electronic configu- ration of the atom and its \(2+\) ion: \((\) a) \(M n,\) (b) \(R u,\) (c) \(R h\). Draw the crystal-field energy-level diagram for the \(d\) orbitals of an octahedral complex, and show the placement of the \(d\) electrons for each \(2+\) ion, assuming a strong-field complex. How many unpaired electrons are there in each case?

(a) What is meant by the term chelate effect? (b) What thermodynamic factor is generally responsible for the chelate effect? (c) Why are polydentate ligands often called sequestering agents?

A four-coordinate complex \(\mathrm{MA}_{2} \mathrm{~B}_{2}\) is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it?

In crystal-field theory, ligands are modeled as if they are point negative charges. What is the basis of this assumption, and how does it relate to the nature of metal-ligand bonds?

Indicate the coordination number of the metal and the oxidation number of the metal as well as the number and type of each donor atom of the ligands for each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CdBr}_{4}\right]\) (c) \(\left[\mathrm{Pt}(\mathrm{en})_{3}\right]\left(\mathrm{ClO}_{4}\right)_{4}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right]^{+}\) (e) \(\mathrm{NH}_{4}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{NCS})_{4}\right]\) (f) \(\left[\mathrm{Cu}(\text { bipy })_{2} \mathrm{I}\right] \mathrm{I}\)

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