Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The molecule methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) can act as a monodentate ligand. The following are equilibrium reactions and the thermochemical data at \(298 \mathrm{~K}\) for reactions of methylamine and en with \(\mathrm{Cd}^{2+}(a q):\) $$ \begin{array}{c} \mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) \\ \Delta H^{\circ}=-57.3 \mathrm{~kJ} ; \quad \Delta S^{\circ}=-67.3 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-37.2 \mathrm{~kJ} \\\ \mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \\ \Delta H^{\circ}=-56.5 \mathrm{~kJ} ; \quad \Delta S^{\circ}=+14.1 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-60.7 \mathrm{~kJ} \end{array} $$ (a) Calculate \(\Delta G^{\circ}\) and the equilibrium constant \(K\) for the following ligand exchange reaction: \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+2 \operatorname{en}(a q) \rightleftharpoons\) $$ \left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) $$ Based on the value of \(K\) in part (a), what would you conclude about this reaction? What concept is demonstrated? (b) Determine the magnitudes of the enthalpic \(\left(\Delta H^{\circ}\right)\) and the entropic \(\left(-T \Delta S^{\circ}\right)\) contributions to \(\Delta G^{\circ}\) for the ligand exchange reaction. Explain the relative magnitudes. (c) Based on information in this exercise and in the "A Closer Look" box on the chelate effect, predict the sign of \(\Delta H^{\circ}\) for the following hypothetical reaction: $$ \begin{aligned} \left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) &+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \\ \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \end{aligned} $$

Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains \(\mathrm{Zn}^{2+}\) in its active site, is responsible for rapidly interconverting dissolved \(\mathrm{CO}_{2}\) and bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\). The zinc in carbonic anhydrase is tetrahedrally coordinated by three neutral nitrogen- containing groups and a water molecule. The coordinated water molecule has a \(\mathrm{p} K_{a}\) of \(7.5,\) which is crucial for the enzyme's activity. (a) Draw the active site geometry for the \(\mathrm{Zn}(\mathrm{II})\) center in carbonic anhydrase, just writing \({ }^{4} \mathrm{~N}^{n}\) for the three neutral nitrogen ligands from the protein. (b) Compare the \(\mathrm{p} K_{a}\) of carbonic anhydrase's active site with that of pure water; which species is more acidic? (c) When the coordinated water to the \(\mathrm{Zn}(\mathrm{II})\) center in carbonic anhydrase is deprotonated, what ligands are bound to the \(\mathrm{Zn}(\mathrm{II})\) center? Assume the three nitrogen ligands are unaffected. (d) The \(\mathrm{p} K_{a}\) of \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is \(10 .\) Suggest an explanation for the difference between this \(\mathrm{p} K_{a}\) and that of carbonic anhydrase. (e) Would you expect carbonic anhydrase to have a deep color, like hemoglobin and other metal-ion containing proteins do? Explain.

Explain the difference between a diamagnetic substance and a paramagnetic substance.

\(\begin{array}{lllll}\text { Sketch all the possible } & \text { stereoisomers } & \text { of }\end{array}\) (a) \(\left[\mathrm{Rh}(\text { bipy })(o \text { -phen })_{2}\right]^{3+}\), (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\text { bipy }) \mathrm{Br}\right]^{2+},\) (c) square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right]\).

By writing formulas or drawing structures related to any one of these three complexes, \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Cl}\) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{ONO})_{2}\right]\) cis-[ \(\left.\mathrm{V}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) illustrate (a) geometric isomerism, (b) linkage isomerism, (c) optical isomerism, (d) coordination-sphere isomerism.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free