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The ionic compound \(A B\) is formed. The charges on the ions may be +1,\(-1 ;+2,-2 ;+3,-3 ;\) or even larger. What are the factors that determine the charge for an ion in an ionic compound?

Short Answer

Expert verified
The factors determining the charge of an ion in an ionic compound include the number of valence electrons, fulfillment of the octet rule, stability based on ionization energies and electron affinities, and electrostatic neutrality. These factors help in predicting the charges of the ions by determining how many electrons each atom needs to lose or gain to achieve a stable electron configuration. The charges of the cations and anions in the compound must balance each other out, maintaining electrostatic neutrality.

Step by step solution

01

Determine the valence electrons of the atoms involved

To predict the charges of the ions involved, find the number of valence electrons for each atom. Valence electrons are electrons in the outermost energy level of an atom and determine the chemical reactivity of atoms. Use the periodic table to identify the number of valence electrons for each atom.
02

Predict the charges based on octet rule

The octet rule states that atoms tend to lose, gain, or share electrons in order to achieve a stable electron configuration, usually having eight electrons in their outermost shell (except helium, which has two). Predict the charges of the ions by determining how many electrons each atom needs to lose or gain to achieve a stable electron configuration. For example: 1. An atom with one valence electron will lose that electron to achieve a full outer shell of electrons. This atom will form a +1 charged ion. 2. An atom with 6 valence electrons will gain 2 electrons to achieve a full outer shell. This atom will form a -2 charged ion.
03

Recognize charge stability based on ionization energy and electron affinity

Ionization energy is the energy needed to remove an electron from an atom, whereas electron affinity is the energy change when an electron is added to an atom. Some ions are more stable because they require less energy to form due to their ionization energies and electron affinities. The more stable an ion, the more likely it is to form. For instance, alkali metals (group 1) readily lose an electron due to their low ionization energy, forming +1 ions. Halogens (group 17) have high electron affinity and readily gain an electron, forming -1 ions.
04

Respecting the electrostatic neutrality

Ionic compounds must maintain electrostatic neutrality, meaning the overall charge of the compound must be equal to zero. The charges of the cations and anions in the compound must balance each other out. The formula of the ionic compound reflects this balance. For example, when magnesium (Mg) bonds with oxygen (O), magnesium loses two electrons and forms a +2 ion (Mg^2+), while oxygen gains two electrons and forms a -2 ion (O^2-). In this case, the charges balance out, and the formula for the ionic compound is MgO. In conclusion, the factors determining the charge of an ion in an ionic compound include the number of valence electrons, fulfillment of the octet rule, stability based on ionization energies and electron affinities, and electrostatic neutrality.

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Key Concepts

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

Valence Electrons
Valence electrons play a key role in the formation of ionic compounds. These are the electrons located in the outermost shell of an atom. They determine how an atom will react chemically with other atoms. By observing the periodic table, you can find out the number of valence electrons of an element.
  • Elements in Group 1 have 1 valence electron.
  • Group 2 elements possess 2 valence electrons.
  • Group 17 elements have 7 valence electrons.
The number of valence electrons influences whether an atom will lose or gain electrons.
For atoms with a small number of valence electrons, losing them to reach a stable electron configuration is often easier. For example, sodium (Na) with 1 valence electron will tend to lose it to achieve stability.
Octet Rule
The octet rule is a guiding concept in understanding how and why atoms form certain ions. It states that atoms tend to seek a stable electron configuration, often resembling that of noble gases, which usually means having eight electrons in their outermost shell. This rule is crucial for predicting the charge of ions in ionic compounds.
  • Atoms with a few valence electrons will tend to lose them, forming positive ions or cations.
  • Conversely, atoms that are close to having a full outer shell prefer to gain electrons, leading to the formation of negative ions or anions.
For example, chlorine (Cl) with 7 valence electrons will likely gain one electron to have a full octet, forming Cl\(^-\) as a result.
Ionization Energy
Ionization energy is a key factor when determining how atoms form ions. It is the energy required to remove an electron from a neutral atom. Elements with low ionization energy tend to lose electrons easily, forming positive ions.
  • Alkali metals, such as lithium (Li) and potassium (K), have low ionization energies and readily form +1 ions by losing electrons.
  • Noble gases have very high ionization energies, making them unlikely to form positive ions.
A low ionization energy means an atom can give away its electrons easily, whereas a high ionization energy means stronger attraction between the atom and its electrons.
Electron Affinity
Electron affinity is the measure of energy released when an electron is added to a neutral atom. Atoms with high electron affinity have a strong tendency to gain electrons to form stable ions.
  • Halogens, such as fluorine (F) and chlorine (Cl), demonstrate high electron affinity, which means they release more energy when they gain electrons, forming negative ions.
  • Elements with lower electron affinity release less energy and are less likely to gain electrons.
Understanding electron affinity can predict an element's inclination to gain electrons and form anions in ionic compounds.
Electrostatic Neutrality
Electrostatic neutrality is the principle that ionic compounds must adhere to. This principle dictates that the overall charge of an ionic compound must be zero. The positive charges of cations and the negative charges of anions must balance each other out.
  • If an element forms a +2 ion and pairs with another forming a -2 ion, the charges balance (e.g., MgO).
  • In cases where charges are not directly balanced (e.g., Na\(^+\) with O\(^{2-}\)), a stoichiometric adjustment is necessary, like two Na for every one O in Na\(_2\)O.
This balance ensures that the resulting compound remains electrically neutral, conforming to natural laws of charged interactions.

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Most popular questions from this chapter

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$\therefore N=N=O^{\cdot} \leftrightarrow: N \equiv N-\vec{O}: \longleftrightarrow: N-N \equiv 0$$ Given the following bond lengths, $$\begin{aligned} &\mathrm{N}-\mathrm{N} \quad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\\\ &\mathrm{N}=\mathrm{N} \quad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\\\ &\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm} \end{aligned}$$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

Given the following information: Energy of sublimation of \(\mathrm{Li}(s)=166 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{HCl}=427 \mathrm{kJ} / \mathrm{mol}\) Ionization energy of \(\mathrm{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}\) Electron affinity of \(\mathrm{Cl}(g)=-349 \mathrm{kJ} / \mathrm{mol}\) Lattice energy of \(\mathrm{LiCl}(s)=-829 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{H}_{2}=432 \mathrm{kJ} / \mathrm{mol}\) Calculate the net change in energy for the following reaction: $$ 2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g) $$

Use the following data formagnesium fluoride to estimate \(\Delta E\) for the reaction: $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s) \quad \Delta E=?$$ Lattice energy First ionization energy of \(\mathrm{Mg}\) Second ionization energy of \(\mathbf{M g}\) Electron affinity of \(\mathbf{F}\) Bond energy of \(\mathrm{F}_{2}\) Energy of sublimation for \(\mathrm{Mg}\) \(-2913 \mathrm{kJ} / \mathrm{mol}\) \(735 \mathrm{kJ} / \mathrm{mol}\) \(1445 \mathrm{kJ} / \mathrm{mol}\) \(-328 \mathrm{kJ} / \mathrm{mol}\) \(154 \mathrm{kJ} / \mathrm{mol}\) 150\. kJ/mol

Write Lewis structures that obey the octet rule (duet rule for H) for each of the following molecules. a. \(\mathrm{H}_{2} \mathrm{CO}\) b. \(\mathrm{CO}_{2}\) c. HCN Carbon is the central atom in all of these molecules.

The designations \(1 \mathrm{A}\) through \(8 \mathrm{A}\) used for certain families of the periodic table are helpful for predicting the charges on ions in binary ionic compounds. In these compounds, the metals generally take on a positive charge equal to the family number, while the nonmetals take on a negative charge equal to the family number minus \(8 .\) Thus the compound between sodium and chlorine contains \(\mathrm{Na}^{+}\) ions and \(\mathrm{Cl}^{-}\) ions and has the formula NaCl. Predict the formula and the name of the binary compound formed from the following pairs of elements. a. Ca and N b. \(K\) and 0 c. \(\mathrm{Rb}\) and \(\mathrm{F}\) d. \(\mathrm{Mg}\) and \(\mathrm{S}\) e. Ba and I f. Al and Se g. Cs and \(P\) h. In and Br

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