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Chapter 3: Problem 54

Give an example of an ionic compound where both the anion and the cation are isoelectronic with each of the following noble gases. a. Ne b. Arr c. Kr d. Xe

Short Answer

Expert verified
a. Sodium fluoride (NaF) b. Potassium chloride (KCl) c. Rubidium bromide (RbBr) d. Cesium iodide (CsI)

Step by step solution

01

a. Ne (Neon)

The electron configuration of neon is 1s虏 2s虏 2p鈦. An ion is isoelectronic with Ne if it has the same number of electrons (10 electrons). To find possible cations and anions, we look for elements which lose or gain electrons to achieve this configuration. Possible ions that are isoelectronic with Ne: Sodium cation (Na鈦) and Fluorine anion (F鈦). Na鈦 has 1s虏 2s虏 2p鈦 configuration and F鈦 has 1s虏 2s虏 2p鈦 configuration. Ionic compound: Sodium fluoride (NaF).
02

b. Ar (Argon)

The electron configuration of argon is 1s虏 2s虏 2p鈦 3s虏 3p鈦. An ion is isoelectronic with Ar if it has the same number of electrons (18 electrons). Possible ions that are isoelectronic with Ar: Potassium cation (K鈦) and Chlorine anion (Cl鈦). K鈦 has 1s虏 2s虏 2p鈦 3s虏 3p鈦 configuration and Cl鈦 has 1s虏 2s虏 2p鈦 3s虏 3p鈦 configuration. Ionic compound: Potassium chloride (KCl).
03

c. Kr (Krypton)

The electron configuration of krypton is 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦. An ion is isoelectronic with Kr if it has the same number of electrons (36 electrons). Possible ions that are isoelectronic with Kr: Rubidium cation (Rb鈦) and Bromine anion (Br鈦). Rb鈦 has 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 configuration and Br鈦 has 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 configuration. Ionic compound: Rubidium bromide (RbBr).
04

d. Xe (Xenon)

The electron configuration of xenon is 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 5p鈦. An ion is isoelectronic with Xe if it has the same number of electrons (54 electrons). Possible ions that are isoelectronic with Xe: Cesium cation (Cs鈦) and Iodine anion (I鈦). Cs鈦 has 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 5p鈦 configuration and I鈦 has 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 5p鈦 configuration. Ionic compound: Cesium iodide (CsI).

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Key Concepts

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

Isoelectronic
When ions or atoms have the same number of electrons, they are known as isoelectronic. This means they share an identical electron configuration. Imagine two different elements, such as sodium and fluorine, that achieve the same configuration by gaining or losing electrons. They become isoelectronic. Isoelectronic ions are crucial in forming ionic compounds where the ions are stable and energy-efficient due to having filled electron shells. This concept helps us understand why elements tend to form certain ions, often aspiring for the stable configuration of noble gases.
Electron Configuration
The electron configuration of an atom describes the arrangement of its electrons in orbitals. This configuration is vital in determining chemical properties and reactivity. For example, neon, with an electron configuration of 1s虏 2s虏 2p鈦, is stable and non-reactive. Elements like sodium or fluorine adjust their electron configuration through ionic bonding, often achieving a noble gas configuration. This adaptation results in stability. Understanding electron configurations helps us predict how and why elements form specific ions.
Noble Gases
Noble gases are a group of elements in the far right column of the periodic table. These include helium, neon, argon, krypton, xenon, and radon. What makes these elements unique is their full valence electron shells, rendering them highly stable and mostly inert. This stability is due to having a complete outer shell, meaning they rarely react with other elements. Many elements and ions aim to achieve the same configuration as these gases, deeming them a role model in chemistry for stability and peace.
Cations and Anions
Cations and anions are charged ions formed when atoms lose or gain electrons. A cation is a positively charged ion, achieved when an atom loses electrons. For instance, sodium (Na) becomes a cation (Na鈦) by losing one electron. An anion, conversely, is a negatively charged ion formed by gaining electrons, like fluorine (F) gaining an electron to become fluoride (F鈦). Understanding the formation of cations and anions is crucial in ionic bonding, where oppositely charged ions attract to form stable compounds.

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Most popular questions from this chapter

A polyatomic ion is composed of \(\mathrm{C}, \mathrm{N},\) and an unknown element \(X\). The skeletal Lewis structure of this polyatomic ion is \([\mathrm{X}-\mathrm{C}-\mathrm{N}]^{-} .\) The ion \(\mathrm{X}^{2-}\) has an electron configuration of [Ar]4s^{2} 3 d ^ { 1 0 } 4 p ^ { 6 } . \text { What is element X? Knowing the identity of } X, complete the Lewis structure of the polyatomic ion, including all important resonance structures.

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part \(c,\) why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl}\) ? \(\mathrm{NaCl}_{2}\) ? What about \(\mathrm{MgO}\) compared to \(\mathrm{MgO}_{2} ?\) \(\mathrm{Mg}_{2} \mathrm{O} ?\)

Name each of the following compounds. Assume the acids are dissolved in water. a. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) b. \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) c. \(\mathrm{Co}_{2} \mathrm{S}_{3}\) d. ICl e. \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) f. \(\mathrm{KClO}_{3}\) g. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(\mathbf{h} . \mathrm{Sr}_{3} \mathrm{N}_{2}\) i. \(\quad \mathrm{Al}_{2}\left(\mathrm{SO}_{3}\right)_{3}\) J. \(\mathrm{SnO}_{2}\) \(\mathbf{k} . \mathrm{Na}_{2} \mathrm{CrO}_{4}\) I. \(\mathrm{HClO}\)

Use data in this chapter (and Chapter 2 ) to discuss why \(\mathrm{MgO}\) is an ionic compound but CO is not an ionic compound.

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.)
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