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Chapter 15: Problem 51

Explain why (a) oxalic acid removes rust stains. (b) there are no geometric isomers of tetrahedral complexes. (c) cations such as \(\mathrm{Co}^{2+}\) act as Lewis acids. (d) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is a chelating agent. (e) \(\mathrm{NH}_{3}\) can be a ligand but \(\mathrm{NH}_{4}{ }^{+}\) is not.

Short Answer

Expert verified
Answer: Ammonia (\(\mathrm{NH}_3\)) can function as a ligand because it has a lone pair of electrons on its nitrogen atom, which can be donated to a metal to form a coordination bond. In contrast, ammonium ion (\(\mathrm{NH}_{4}{ }^{+}\)) has a positive charge and no lone pairs available to donate to a metal ion, making it incapable of acting as a ligand.

Step by step solution

01

Understanding oxalic acid properties

Oxalic acid is a dicarboxylic acid with the formula \(\mathrm{C}_2\mathrm{H}_2\mathrm{O}_4\). It has strong reducing properties due to its ability to donate electrons or to accept electrons easily from other chemical species. Rust is essentially iron oxide (\(\mathrm{Fe}_2\mathrm{O}_3\)), formed due to the oxidation of iron. When oxalic acid comes into contact with rust, it can form a soluble complex with the iron ions, effectively dissolving the rust and removing the stain. The general reaction can be written as: \(\mathrm{C}_2\mathrm{H}_2\mathrm{O}_4 + \mathrm{Fe}_2\mathrm{O}_3 \rightarrow [\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-} + \mathrm{H}_2\mathrm{O}\).
02

Analyzing tetrahedral complex structures

In tetrahedral complexes, there are four ligands attached to the central metal atom. These ligands are located symmetrically around the metal atom, with each pair of ligands forming a bond angle of 109.5 degrees. The symmetry of the tetrahedral structure does not allow for any variations or different arrangements of the ligands, preventing the formation of different geometric isomers. Isomerism occurs when there are multiple arrangements of ligands or multiple bond angles between which ligands can be interchanged, which is not possible in tetrahedral complexes.
03

Understanding Lewis acids concept

According to the Lewis acid-base theory, a Lewis acid is a chemical species that can accept an electron pair. Metal cations, such as \(\mathrm{Co}^{2+}\), possess a positive charge, which means they are electron-deficient and are attracted to electron-rich species, known as Lewis bases. When a Lewis base donates its lone pair of electrons to a metal cation, it forms a coordination bond. In this case, the \(\mathrm{Co}^{2+}\) ion acts as a Lewis acid by accepting the electron pair and forming a bond with the Lewis base.
04

Recognizing \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) as a chelating agent

Chelating agents are ligands that can form multiple coordination bonds with a single metal ion, effectively surrounding the metal ion with their electron-rich donor atoms. Oxalate ion, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\), is a chelating agent because it has two oxygen donor atoms, each with a lone pair of electrons, which can form coordination bonds with a metal ion. This creates a stable ring structure, known as a chelate complex, with the metal ion at the center.
05

Comparing \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}{ }^{+}\) as ligands

A ligand is a molecule or ion that can donate a lone pair of electrons to a metal ion, forming a coordination bond. Ammonia (\(\mathrm{NH}_3\)) has a lone pair of electrons on its nitrogen atom, which can be donated to a metal to form a bond. This makes ammonia a suitable ligand. On the other hand, ammonium ion (\(\mathrm{NH}_{4}{ }^{+}\)) has a positive charge and no lone pairs available to donate to a metal ion, so it cannot function as a ligand.

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Most popular questions from this chapter

Consider three complexes of \(\mathrm{Ag}^{+}\) and their formation constants, \(K_{\mathrm{f}}\) $$\begin{array}{ll}\hline \text { Complex lon } & K_{\mathrm{f}} \\\\\hline \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+ & 1.6 \times 10^{7} \\ \mathrm{Ag}(\mathrm{CN})_{2}^{-} & 5.6 \times 10^{18} \\\\\mathrm{AgBr}_{2}^{-} & 1.3 \times 10^{7} \\ \hline\end{array}$$ Which statements are true? (a) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}\) is more stable than \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\). (b) Adding a strong acid \(\left(\mathrm{HNO}_{3}\right)\) to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) will tend to dissociate the complex ion into \(\mathrm{Ag}^{+}\) and \(\mathrm{NH}_{4}^{+} .\) (c) Adding a strong acid \(\left(\mathrm{HNO}_{3}\right)\) to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{AgBr}_{2}^{-}\) will tend to dissociate the complex ion into \(\mathrm{Ag}^{+}\) and \(\mathrm{Br}^{-} .\) (d) To dissolve AgI, one can add either \(\mathrm{NaCN}\) or \(\mathrm{HCN}\) as a source of the cyanide-complexing ligand. Fewer moles of NaCN would be required. (e) Solution \(A\) is \(0.10 M\) in \(B r^{-}\) and contains the complex ion \(\mathrm{AgBr}_{2}^{-}\). Solution B is \(0.10 M\) in \(\mathrm{CN}^{-}\) and contains the complex ion \(\mathrm{Ag}(\mathrm{CN})_{2}-\). Solution B will have more particles of complex ion per particle of \(\mathrm{Ag}^{+}\) than solution \(\mathrm{A}\).

\(\mathrm{MnF}_{6}{ }^{2-}\) has a crystal field splitting energy, \(\Delta_{0}\) of \(2.60 \times 10^{2} \mathrm{~kJ} / \mathrm{mol}\). What is the wavelength responsible for this energy?

\(\mathrm{Ti}\left(\mathrm{NH}_{3}\right)_{6}{ }^{3+}\) has a d-orbital electron transition at \(399 \mathrm{~nm}\). Find \(\Delta_{o}\) at this wavelength.

In the \(\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ion, the splitting between the \(\mathrm{d}\) levels, \(\Delta_{\mathrm{o}}\) is \(55 \mathrm{kcal} / \mathrm{mol}\). What is the color of this ion? Assume that the color results from a transition between upper and lower d levels.

Consider the complex ion \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+} .\) Its \(K_{\mathrm{f}}\) is \(2.1 \times 10^{18} .\) At what concentration of \(e n\) is \(67 \%\) of the \(\mathrm{Ni}^{2+}\) converted to \(\left.\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\right\\}\)
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