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In general, the higher the charge on the ions in an ionic compound, the more favorable the lattice energy. Why do some stable ionic compounds have \(+1\) charged ions even though \(+4,+5,\) and \(+6\) charged ions would have a more favorable lattice energy?

Short Answer

Expert verified
In conclusion, some stable ionic compounds contain +1 charged ions because forming ions with higher charges like +4, +5, and +6 is not always energetically favorable. The increased ionization energies required to remove additional electrons, and the difficulty in forming stable ionic bonds with large-sized ions counterbalance the benefits of higher lattice energies. Therefore, ions with lower charges can still form sufficiently favorable lattice energies to create stable ionic compounds.

Step by step solution

01

Understand lattice energy

Lattice energy is the energy required to separate one mole of an ionic compound into its constituent ions in the gaseous state. The magnitude of the lattice energy depends on the charges of the ions and the size of the ions. In general, higher charges on the ions and smaller ionic radii will result in larger lattice energy.
02

Know the relation between lattice energy and the stability of an ionic compound

A more favorable (greater) lattice energy indicates higher stability of an ionic compound. This is because it takes more energy to separate the ions, meaning they are held more strongly within the crystal. As the charge on the ions in the compound increases, the electrostatic attraction between the ions also increases, resulting in a higher lattice energy.
03

Consider factors affecting ion formation and charges

Despite higher lattice energy being more favorable for the stability of the ionic compounds, there are other factors that also influence the formation of ions and their charges. These factors are: 1. Ionization energy: The energy required to remove an electron from an atom. As the charge on the ion increases, the ionization energy required to remove additional electrons also increases, making it less favorable to form ions with higher charges. 2. Electron affinity: The energy released when an electron is added to an atom. More energy is released as an atom gains an electron, making it more favorable to form negatively charged ions. 3. Atomic size: Smaller ions have stronger electrostatic interactions with their oppositely charged counterparts in the lattice. When the charges increase for large-sized ions, it is difficult for them to form stable ionic bonds due to the repulsive forces between the electrons in their electron cloud.
04

Conclude with the explanation

In conclusion, while ionic compounds with higher charges on their ions have more favorable lattice energies, forming +4, +5, and +6 charged ions is not always energetically favorable due to the increased ionization energies required to remove additional electrons and the inherent difficulty in forming stable ionic bonds with large-sized ions. Therefore, some stable ionic compounds have +1 charged ions, as they can be formed with lower ionization energies and still have sufficiently favorable lattice energies to form stable compounds.

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Key Concepts

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

Ionic Compounds
Ionic compounds are made up of positive and negative ions that are held together by electrostatic forces. These forces are usually quite strong, resulting in compounds with high melting and boiling points.
Ionic compounds are generally formed between metals (which lose electrons easily) and non-metals (which gain electrons easily). This exchange of electrons creates ions that are oppositely charged, leading to attraction and the formation of a lattice structure.
The stability of the ionic compound is often linked to the concept of lattice energy. The lattice energy is the amount of energy required to separate a mole of an ionic compound into its individual ions in the gaseous phase. The stronger the attraction between the ions, the higher will be the lattice energy. This means that compounds with higher lattice energies are usually more stable.
However, forming ions with very high charges can be energetically unfavorable due to other influential factors, such as ionization energy and electron affinity. This is why you sometimes find stable ionic compounds with ions having relatively low charges, like +1 instead of +4 or more.
Ionization Energy
Ionization energy is the energy needed to remove an electron from an atom or ion. It is a critical factor in determining how and why ions form in the first place.
When an atom loses an electron and forms a positive ion, this process requires overcoming the attraction between the negatively charged electron and the positively charged nucleus.
  • The greater the charge of the ion you are trying to form, the more energy is required to remove additional electrons.
  • Consequently, if a +1 charged ion is easier to form than a +5 ion because the latter involves significantly higher ionization energy.
This makes it energetically unfavorable to form ions with very high positive charges. Such high charges would require an exorbitant amount of energy that offsets the benefits of the resultant high lattice energy. Thus, compounds with lower charge ions, like +1, are often more favored energetically.
Electron Affinity
Electron affinity is the amount of energy released when an electron is added to a neutral atom to form a negative ion. This concept is particularly significant when looking at why and how negatively charged ions form.
As an atom gains an electron, energy is released, making negative ions quite stable and energetically favorable. This is why non-metals readily gain electrons and become negatively charged ions in ionic compounds.
  • The more negative the electron affinity, the more favorable it is for an atom to gain additional electrons and form highly charged negative ions.
  • However, reciprocating this into forming highly charged positive ions is not as energetically simple. It would require higher ionization energies, countering such stability advantages.
Therefore, when we are talking about forming stable ionic compounds, the balance of ionization energy for positive ions and electron affinity for negative ions is crucial. This is a significant reason why even +1 charge ions can be quite prevalent despite lower lattice energy symptoms.
Atomic Size
Atomic size plays a crucial role in the formation and stability of ionic compounds. Smaller atoms tend to form ions that can compact closely in a crystal lattice, leading to stronger electrostatic forces and a higher lattice energy.
However, if an atom is large, this close packing becomes challenging, especially when the ion has a high charge because of the repulsive forces among the inner electrons.
  • Increased atomic size may introduce additional shielding that decreases effective nuclear charge, thus requiring less energy to remove electrons.
  • However, larger atoms forming highly charged ions can face repulsion that makes high-charged ions less stable and less favorable energetically.
Thus, while atomic size can enable certain ions to form easily, it also sets limitations on how effectively high-charge ions can contribute to stability in ionic compounds. This explains why lower charged ions like +1 can form stable compounds despite seemingly lower lattice energy potential.

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Most popular questions from this chapter

Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{F}, \mathrm{Si}-\mathrm{F}, \mathrm{Ge}-\mathrm{F} \quad\) c. \(\mathrm{S}-\mathrm{F}, \mathrm{S}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}\) b. \(\mathrm{P}-\mathrm{Cl}\) or \(\mathrm{S}-\mathrm{Cl} \quad\) d. \(\mathrm{Ti}-\mathrm{Cl}, \mathrm{Si}-\mathrm{Cl}, \mathrm{Ge}-\mathrm{Cl}\)

When comparing the size of different ions, the general radii trend discussed in Chapter 7 is usually not very useful. What do you concentrate on when comparing sizes of ions to each other or when comparing the size of an ion to its neutral atom?

Predict the molecular structure (including bond angles) for each of the following. a. \(\mathrm{PCl}_{3}\) b. \(\mathrm{SCl}_{2}\) c. \(\mathrm{SiF}_{4}\)

Identify the five compounds of \(\mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) described as follows. For each compound, write a Lewis structure that is consistent with the information given. a. All the compounds are electrolytes, although not all of them are strong electrolytes. Compounds \(\mathrm{C}\) and \(\mathrm{D}\) are ionic and compound \(\mathrm{B}\) is covalent. b. Nitrogen occurs in its highest possible oxidation state in compounds \(\mathrm{A}\) and \(\mathrm{C}\); nitrogen occurs in its lowest possible oxidation state in compounds \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E}\). The formal charge on both nitrogens in compound \(\mathrm{C}\) is \(+1\); the formal charge on the only nitrogen in compound \(\mathrm{B}\) is 0. c. Compounds \(\mathrm{A}\) and \(\mathrm{E}\) exist in solution. Both solutions give off gases. Commercially available concentrated solutions of compound A are normally 16\(M .\) The commercial, concentrated solution of compound \(E\) is 15\(M .\) d. Commercial solutions of compound E are labeled with a misnomer that implies that a binary, gaseous compound of nitrogen and hydrogen has reacted with water to produce ammonium ions and hydroxide ions. Actually, this reaction occurs to only a slight extent. e. Compound \(D\) is 43.7\(\% \mathrm{N}\) and 50.0\(\%\) O by mass. If compound D were a gas at STP, it would have a density of 2.86 \(\mathrm{g} / \mathrm{L} .\) f. A formula unit of compound \(\mathrm{C}\) has one more oxygen than a formula unit of compound \(\mathrm{D}\). Compounds \(\mathrm{C}\) and \(\mathrm{A}\) have one ion in common when compound \(\mathrm{A}\) is acting as a strong electrolyte. g. Solutions of compound \(\mathrm{C}\) are weakly acidic; solutions of compound \(\mathrm{A}\) are strongly acidic; solutions of compounds \(\mathrm{B}\) and \(\mathrm{E}\) are basic. The titration of 0.726 g compound \(\mathrm{B}\) requires 21.98 \(\mathrm{mL}\) of 1.000 \(\mathrm{M} \mathrm{HCl}\) for complete neutralization.

Which of the following molecules have net dipole moments? For the molecules that are polar, indicate the polarity of each bond and the direction of the net dipole moment of the molecule. a. \(\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}, \mathrm{CCl}_{4}\) b. \(\mathrm{CO}_{2}, \mathrm{N}_{2} \mathrm{O}\) c. \(\mathrm{PH}_{3}, \mathrm{NH}_{3}\)

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