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Chapter 9: Problem 11

Explain why ions with charges greater than 3 are seldom found in ionic compounds.

Short Answer

Expert verified
Ions with charges greater than 3 are seldom found in ionic compounds because forming such ions by removing more than 3 valence electrons would require a substantial amount of energy, which is generally unavailable in ordinary chemical reactions and thus is energetically unfavorable.

Step by step solution

01

Understanding Atomic Structure

Each atom consists of a nucleus surrounded by electrons. These electrons reside in different energy levels. The first two electrons are in the first energy level, the next eight in the second, the next eight in the third, and so on. The outermost electrons, called valence electrons, determine an atom’s chemical properties, including its tendency to form ions.
02

Understanding Ionic Bonding

When an atom donates one or more of its valence electrons to another atom, it forms an ionic bond. The atom that loses electrons becomes a positive ion (or cation), and the atom that gains electrons becomes a negative ion (or anion). The resulting compound, due to the attraction between oppositely charged ions, is known as an ionic compound.
03

Reasoning Energy Consumption

The process of removing an electron from an atom requires energy to overcome the attraction between the electron and the nucleus. The more electrons you remove, the more the energy required because the positive charge of the nucleus holds the remaining electrons stronger.
04

Explaining Possible Charge for Ions

For ions with charges greater than 3, it implies that more than 3 electrons were removed. This removal process requires a substantial amount of energy that is usually not available in ordinary chemical reactions. Therefore, ions with charges greater than 3 are rarely encountered in ionic compounds because forming such ions is energetically unfavorable.

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Most popular questions from this chapter

Which of these molecules or ions are isoelectronic: \(\mathrm{NH}_{4}^{+}, \mathrm{C}_{6} \mathrm{H}_{6}, \mathrm{CO}, \mathrm{CH}_{4}, \mathrm{~N}_{2}, \mathrm{~B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6} ?\)

List these bonds in order of increasing ionic character: cesium to fluorine, chlorine to chlorine, bromine to chlorine, silicon to carbon.

Write Lewis dot symbols for the following atoms and ions: (a) I, (b) I \(^{-},\) (c) S, (d) \(\mathrm{S}^{2-},\) (e) \(\mathrm{P},\) (f) \(\mathrm{P}^{3-}\), (l) \(\mathrm{Al}^{3+}\) (g) \(\mathrm{Na},\) (h) \(\mathrm{Na}^{+},\) (i) \(\mathrm{Mg},\) (j) \(\mathrm{Mg}^{2+},\) (k) \(\mathrm{Al}\), \((\mathrm{m}) \mathrm{Pb},(\mathrm{n}) \mathrm{Pb}^{2+}\).

An ionic bond is formed between a cation \(\mathrm{A}^{+}\) and an anion \(\mathrm{B}^{-}\). How would the energy of the ionic bond [see Equation (9.2)\(]\) be affected by the following changes? (a) doubling the radius of \(\mathrm{A}^{+},\) (b) tripling the charge on \(A^{+},(\mathrm{c})\) doubling the charges on \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-},\) (d) decreasing the radii of \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-}\) to half their original values.

The species \(\mathrm{H}_{3}^{+}\) is the simplest polyatomic ion. The geometry of the ion is that of an equilateral triangle. (a) Draw three resonance structures to represent the ion. (b) Given the following information and $$ \begin{aligned} 2 \mathrm{H}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_{3}^{+} & \Delta H^{\circ}=-849 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \longrightarrow 2 \mathrm{H} & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}^{+}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3}^{+} $$
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