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Chapter 14: Problem 141

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

Short Answer

Expert verified
In the given species, \(\mathrm{Fe}^{3+}\) has a higher positive charge and electron configuration of \([Ar]~3d^5\) compared to the \(\mathrm{Fe}^{2+}\) ion with an electron configuration of \([Ar]~3d^6\). Since both ions have empty 4s orbitals, the higher positive charge of \(\mathrm{Fe}^{3+}\) results in a stronger attraction to electron pairs. Therefore, \(\mathrm{Fe}^{3+}\) would be the stronger Lewis acid.

Step by step solution

01

Understanding Lewis acids

A Lewis acid is a chemical species that can accept an electron pair from a Lewis base in a chemical reaction. In other words, Lewis acids have an empty orbital that can accommodate a pair of electrons donated by a Lewis base.
02

Analyzing \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}^{2+}\) ions

To determine which species is the stronger Lewis acid, we need to compare their charge and electron configurations. The electron configuration for a neutral Fe atom is \([Ar] 4s^2~3d^6\). When Fe loses 2 electrons and becomes the \(\mathrm{Fe}^{2+}\) ion, its electron configuration becomes \([Ar]~3d^6\). When it loses 1 more electron, forming the \(\mathrm{Fe}^{3+}\) ion, its electron configuration is \([Ar]~3d^5\).
03

Comparing charge and electron configurations

The \(\mathrm{Fe}^{3+}\) ion has one more positive charge than the \(\mathrm{Fe}^{2+}\) ion. The increased positive charge attracts the negatively charged electron pairs of Lewis bases more strongly. Additionally, both \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}^{2+}\) have empty 4s orbitals that can accommodate electron pairs. Their 3d orbitals are partially filled, but the higher positive charge on the \(\mathrm{Fe}^{3+}\) ion results in a stronger attraction to electron pairs.
04

Conclusion

Based on the higher positive charge and similarity in electron configurations, we would expect the \(\mathrm{Fe}^{3+}\) ion to be the stronger Lewis acid, as it has a greater capacity to attract and accept electron pairs from Lewis bases.

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Most popular questions from this chapter

Calculate the percent dissociation for a \(0.22-M\) solution of chlorous acid \(\left(\mathrm{HClO}_{2}, K_{\mathrm{a}}=0.012\right)\)

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$

Calculate the pH of the following solutions. a. 0.10\(M\) NaOH b. \(1.0 \times 10^{-10} M \mathrm{NaOH}\) c. 2.0 \(\mathrm{M} \mathrm{NaOH}\)

Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is pH- dependent. The relevant equilibrium reaction is $$ \mathrm{HbH}_{4}^{4+}(a q)+4 O_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q) $$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, HbH \(_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4},\) is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygen-binding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise \(146 .\) ) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: CO, blood levels increase during cardiac arrest.)

Calculate the concentrations of all species present in a \(0.25-M\) solution of ethylammonium chloride $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)$
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