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

Write names for the following coordination compounds: (a) \(\left[\mathrm{Cd}(\mathrm{en}) \mathrm{Cl}_{2}\right]\) (b) \(\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]\) (c) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{CO}_{3}\right] \mathrm{Cl}\) (d) \(\left[\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{3}\)

A Cu electrode is immersed in a solution that is \(1.00 \mathrm{M}\) in \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and \(1.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). When the cathode is a standard hydrogen electrode, the emf of the cell is found to be \(+0.08 \mathrm{~V}\). What is the formation constant for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\)

Write balanced chemical equations to represent the following observations. (In some instances the complex involved has been discussed previously in the text.) (a) Solid silver chloride dissolves in an excess of aqueous ammonia. (b) The green complex \(\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl},\) on treatment with water over a long time, converts to a brown-orange complex. Reaction of \(\mathrm{AgNO}_{3}\) with a solution of the product precipitates \(3 \mathrm{~mol}\) of AgCl per mole of Cr present. (Write two chemical equations.) (c) When an NaOH solution is added to a solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},\) a precipitate forms. Addition of excess \(\mathrm{NaOH}\) solution causes the precipitate to dissolve. (Write two chemical equations.) (d) A pink solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) turns deep blue on addition of concentrated hydrochloric acid.

Sketch the structure of the complex in each of the following compounds and give the full compound name: (a) cis- \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{Ru}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{5}\right]\) (c) trans- \(\mathrm{NH}_{4}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) (d) cis- \(\left[\mathrm{Ru}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\)

Explain why the \(d_{x y}, d_{x z}\), and \(d_{y z}\) orbitals lie lower in energy than the \(d_{z}^{2}\) and \(d_{x^{2}-y^{2}}\) orbitals in the presence of an octahedral arrangement of ligands about the central metal ion.

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