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Chapter 23: Problem 87

Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the +3 rather than in the +2 oxidation state (for metals that form stable +3 ions in the first place). Suggest an explanation, keeping in mind the Lewis acid-base nature of the metal-ligand bond.

Short Answer

Expert verified
In summary, the stability of a coordination compound is generally greater for a metal in the +3 oxidation state than in the +2 oxidation state due to the stronger positive charge and charge density of the metal in the +3 state. This leads to a stronger Lewis acid-base interaction between the metal ion and the ligand, contributing to a more stable coordination compound.

Step by step solution

01

Understanding Lewis acid-base interaction

In a metal-ligand bond, the metal acts as a Lewis acid, and the ligand acts as a Lewis base. A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor. The formation of a coordination compound involves the interaction of metal ions with ligands, and the strength of this interaction determines the stability of the compound.
02

Comparing the +2 and +3 oxidation states

A metal in the +3 oxidation state has one more positive charge compared to a metal in the +2 oxidation state. This means that a metal in the +3 oxidation state has a stronger positive charge, which can attract electron pairs more effectively from a ligand. As a result, the metal-ligand interaction is stronger in the +3 oxidation state, contributing to greater stability for the coordination compound.
03

Relating the oxidation state to the stability of the metal-ligand bond

When a metal is in a higher oxidation state, it has a greater charge density and an increased ability to accept electron pairs from ligands, resulting in stronger metal-ligand bonds and higher stability of the coordination compound. Since the +3 oxidation state has a higher positive charge and charge density compared to the +2 oxidation state, the Lewis acid-base interaction is generally stronger for metals in the +3 oxidation state.
04

Conclusion

The stability of a coordination compound with a metal in the +3 oxidation state is generally greater compared to the same metal in the +2 oxidation state. This is because a metal in the +3 oxidation state has a stronger positive charge and charge density, which results in a stronger Lewis acid-base interaction between the metal ion and the ligand. This stronger interaction leads to a more stable coordination compound.

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Most popular questions from this chapter

Polydentate ligands can vary in the number of coordination positions they occupy. In each of the following, identify the polydentate ligand present and indicate the probable number of coordination positions it occupies: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}(o\) -phen \()\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] \mathrm{Br}\) (c) \(\left[\mathrm{Cr}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}\) (d) \(\left[\mathrm{Zn}(\mathrm{en})_{2}\right]\left(\mathrm{ClO}_{4}\right)_{2}\)

Although the cis configuration is known for \(\left[\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2}\right], \mathrm{no}\) trans form is known. (a) Explain why the trans compound is not possible. (b) Would \(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) be more likely than en \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)\) to form the trans compound? Explain.

(a) Draw the structure for \(\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2} .\) (b) What is the coordination number for platinum in this complex, and what is the coordination geometry? (c) What is the oxidation state of the platinum? [Section 23.2]

Write out the ground-state electron configurations of (a) \(\mathrm{Ti}^{3+}\) (b) \(\mathrm{Ru}^{2+},(\mathrm{c}) \mathrm{Au}^{3+}\) (d) \(\mathrm{Mn}^{4+}\).

Suppose that a transition-metal ion was in a lattice in which it was in contact with just two nearby anions, located on opposite sides of the metal. Diagram the splitting of the metal \(\bar{d}\) orbitals that would result from such a crystal field. Assuming a strong field, how many unpaired electrons would you expect for a metal ion with six \(d\) electrons? (Hint: Consider the linear axis to be the \(z\) -axis)
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