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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

If a transition-metal complex has an even number of valence \(\bar{d}\) electrons, does it necessarily mean that the complex is diamagnetic? Explain.

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.

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A manganese complex formed from a solution containing potassium bromide and oxalate ion is purified and analyzed. It contains \(10.0 \% \mathrm{Mn}, 28.6 \%\) potassium, \(8.8 \%\) carbon, and \(29.2 \%\) bromine by mass. The remainder of the compound is oxygen. An aqueous solution of the complex has about the same electrical conductivity as an equimolar solution of \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] .\) Write the formula of the compound, using brackets to denote the manganese and its coordination sphere.

The \(E^{\circ}\) values for two low-spin iron complexes in acidic solution are as follows: $$ \begin{aligned} \left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{3+}(a q)+\mathrm{e}^{-} \rightleftharpoons\left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{2+}(a q) & E^{\circ}=1.12 \mathrm{~V} \\\ \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}(a q)+\mathrm{e}^{-} \rightleftharpoons\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}(a q) & E^{\circ}=0.36 \mathrm{~V} \end{aligned} $$ (a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which complex, \(\left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{3+}\) or \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},\) is more difficult to reduce? (c) Suggest an explanation for your answer to (b).
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