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

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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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Most popular questions from this chapter

Name the compounds in parts a-d and write the formulas for the compounds in parts e-h. a. \(\mathrm{NaBr}\) b. \(\mathrm{Rb}_{2} \mathrm{O}\) c. CaS d. \(\mathrm{AlI}_{3}\) e. strontium fluoride f. aluminum selenide g. potassium nitride h. magnesium phosphide

The most common exceptions to the octet rule are compounds or ions with central atoms having more than eight electrons around them. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3},\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions. a. \(\mathrm{Na}^{+}\) b. \(\mathrm{Ca}^{2+}\) \(\mathbf{c} . \mathrm{Al}^{3+}\) d. \(\mathbf{R} \mathbf{b}^{+}\)

An alternative definition of electronegativity is Electronegativity \(=\) constant (I.E. \(-\) E.A.) where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 2 to calculate the (I.E. \(-\) E.A.) term for \(\mathbf{F}, \mathbf{C l}, \mathbf{B r}\) and I. Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are \(1678,1255,1138,\) and \(1007 \mathrm{kJ} / \mathrm{mol},\) respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals \(4.0 .\) Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)

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