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One of the following nitrogen compounds or ions is not capable of serving as a ligand: \(\mathrm{NH}_{4}^{+}, \mathrm{NH}_{3}, \mathrm{NH}_{2}^{-}\) Identify this species and explain your answer.

Short Answer

Expert verified
\( \mathrm{NH}_{4}^{+} \) cannot serve as a ligand because it lacks lone pairs.

Step by step solution

01

Definition of a Ligand

A ligand is a molecule or ion that donates a pair of electrons to a central atom, usually a metal, to form a coordination complex. To serve as a ligand, the species must have a lone pair of electrons that it can donate to form a bond.
02

Analyze \( \mathrm{NH}_{3} \)

Ammonia, \( \mathrm{NH}_{3} \), has a nitrogen atom with a lone pair of electrons available. This lone pair can be donated to form a coordinate bond with a metal atom, making \( \mathrm{NH}_{3} \) capable of acting as a ligand.
03

Analyze \( \mathrm{NH}_{2}^{-} \)

The amide ion, \( \mathrm{NH}_{2}^{-} \), has a nitrogen atom with two lone pairs of electrons. One of these lone pairs can be donated to a central atom, thus \( \mathrm{NH}_{2}^{-} \) can act as a ligand.
04

Analyze \( \mathrm{NH}_{4}^{+} \)

The ammonium ion, \( \mathrm{NH}_{4}^{+} \), does not have any lone pairs of electrons on the nitrogen atom since it is bonded to four hydrogen atoms and carries a positive charge. Thus, it cannot donate a pair of electrons to form a coordinate bond and cannot serve as a ligand.
05

Identify the Species

Based on the analysis, \( \mathrm{NH}_{4}^{+} \) is the nitrogen species that cannot act as a ligand because it lacks a lone pair of electrons to donate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Pair Donor
In chemistry, certain molecules or ions are known as electron pair donors. These are able to provide a pair of electrons to another atom. This process is key in forming bonds in coordination complexes. For a molecule to act as an electron pair donor, it needs to have a lone pair of electrons ready for donation. This concept is crucial in understanding why some molecules can serve as ligands while others cannot. For instance, ammonia (\(\mathrm{NH}_3\)) has a lone pair of electrons on the nitrogen atom, making it an electron pair donor. Similarly, the amide ion (\(\mathrm{NH}_2^-\)) has two lone pairs, one of which can be donated to a central metal atom.

However, not all molecules containing nitrogen can act as electron pair donors. The ammonium ion (\(\mathrm{NH}_4^+\)), unlike ammonia or the amide ion, does not have any lone pairs of electrons because its nitrogen atom is fully occupied by bonds with four hydrogen atoms. Therefore, it cannot donate an electron pair and cannot function as a ligand.
Coordination Complex
A coordination complex is a special type of chemical structure composed of a central atom and surrounding molecules or ions called ligands. These ligands attach to the central atom, typically a metal, via coordinate bonds. This type of bond is formed when a ligand donates a lone pair of electrons to the central metal atom. The interactions in these complexes are essential in many chemical processes, such as catalysis, biological systems, and materials science.

Coordination complexes often display remarkable properties due to the unique arrangements of ligands around the central atom. An example of a ligand forming a coordination complex is ammonia (\(\mathrm{NH}_3\)), which can engage its lone pair of electrons to bond with metal atoms. On the other hand, the ammonium ion (\(\mathrm{NH}_4^+\)) cannot form coordination complexes since it lacks the crucial lone pair to initiate a bond.
  • Examples of coordination complexes include metal-acetylacetonate and metal-ethylenediamine.
  • Various factors like the type of metal and the number and type of ligands can influence the properties of these complexes.
Ammonium Ion
The ammonium ion, represented as \(\mathrm{NH}_4^+\), is an interesting species due to its positive charge and bonding nature. This ion forms when ammonia (\(\mathrm{NH}_3\)) gains an additional hydrogen ion (\(\mathrm{H}^+\)), resulting in a positively charged nitrogen atom bonded to four hydrogen atoms.

The presence of the positive charge implies that the nitrogen atom in ammonium has shared its lone pair of electrons to bond with the extra hydrogen, meaning no lone pairs remain for further donation. This makes ammonium unique compared to species like ammonia (\(\mathrm{NH}_3\)), which can act as a ligand. Because it cannot donate an electron pair, ammonium cannot form the necessary coordinate bonds required to join a coordination complex.

The ammonium ion is often encountered in various chemical contexts, such as fertilizers and acid-base reactions, showcasing its versatility and importance beyond its inability to act as a ligand.

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Most popular questions from this chapter

Give the formula of a complex constructed from one \(\mathrm{Cr}^{3+}\) ion, two ethylenediamine ligands, and two ammonia molecules. Is the complex neutral or is it charged? If charged, give the charge.

In water, the chromium(II) ion, \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\) absorbs light with a wavelength of about \(700 \mathrm{nm}\). What color is the solution?

Four isomers are possible for \(\left[\mathrm{Co (\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{+}\) Draw the structures of all four. (Two of the isomers are chiral, meaning that each has a non-superimposable mirror image.)

An this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{f},\) for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: $$\begin{aligned} \mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) & \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq}) & & K_{f}=10^{8} \\ \mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) & \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq}) & & K_{f}=10^{18} \end{aligned}$$ The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta G^{\circ}=-R T \ln K\) and \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) We know from experiment that \(\Delta H^{\circ}\) for the \(\mathrm{NH}_{3}\) reaction is \(-109 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}\). Is the difference in \(\Delta H^{\circ}\) sufficient to account for the \(10^{10}\) difference in \(K_{f}\) ? Comment on the role of entropy in the second reaction.

An aqueous solution of iron(II) sulfate is paramagnetic. If \(\mathrm{NH}_{3}\) is added, the solution becomes diamagnetic. Why does the magnetism change?

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