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Chapter 8: Problem 133

What do each of the following sets of compounds/ions have in common with each other? See your Lewis structures for Exercises 113 through 116 . a. \(\mathrm{XeCl}_{4}, \mathrm{XeCl}_{2}\) b. \(\mathrm{ICl}_{5}, \mathrm{TeF}_{4}, \mathrm{ICl}_{3}, \mathrm{PCl}_{3}, \mathrm{SCl}_{2}, \mathrm{SeO}_{2}\)

Short Answer

Expert verified
a. Both \(\mathrm{XeCl}_{4}\) and \(\mathrm{XeCl}_{2}\) are Xenon-chlorine compounds. b. All of the compounds in the second set (\(\mathrm{ICl}_{5}, \mathrm{TeF}_{4}, \mathrm{ICl}_{3}, \mathrm{PCl}_{3}, \mathrm{SCl}_{2}, \mathrm{SeO}_{2}\)) have central atoms with lone pairs of electrons and are polar molecules.

Step by step solution

01

Draw Lewis structures

First, we need to draw the Lewis structures for the compounds in each set. The Lewis structures are representations of molecules that show how the valence electrons are arranged around the atoms. a. For the first set, the compounds are \(\mathrm{XeCl}_{4}\) and \(\mathrm{XeCl}_{2}\). Their Lewis structures are: \(\rm{XeCl_4}\): The Xe atom is the central atom, surrounded by four Cl atoms with single bonds. The Xe atom has two lone pairs of electrons as well. \(\rm{XeCl_2}\): The Xe atom is the central atom, surrounded by two Cl atoms with single bonds. The Xe atom has three lone pairs of electrons. b. For the second set, the compounds are \(\mathrm{ICl}_{5}, \mathrm{TeF}_{4}, \mathrm{ICl}_{3}, \mathrm{PCl}_{3}, \mathrm{SCl}_{2},\) and \( \mathrm{SeO}_{2}\). Their Lewis structure is as follows: \(\mathrm{ICl}_{5}\): The I atom is the central atom, surrounded by five Cl atoms with single bonds. The I atom has one lone pair of electrons. \(\mathrm{TeF}_{4}\): The Te atom is the central atom, surrounded by four F atoms with single bonds. The Te atom has one lone pair of electrons. \(\mathrm{ICl}_{3}\): The I atom is the central atom, surrounded by three Cl atoms with single bonds. The I atom has two lone pairs of electrons. \(\mathrm{PCl}_{3}\): The P atom is the central atom, surrounded by three Cl atoms with single bonds. No lone pairs on the P atom. \(\mathrm{SCl}_{2}\): The S atom is the central atom, surrounded by two Cl atoms with single bonds. The S atom has two lone pairs of electrons. \(\mathrm{SeO}_{2}\): The Se atom is the central atom, surrounded by two O atoms with double bonds. The Se atom has two lone pairs of electrons.
02

Identify common properties

Now that we have the Lewis structures, we need to examine them and find any common properties between the compounds in each set. a. In the first set, both \(\mathrm{XeCl}_{4}\) and \(\mathrm{XeCl}_{2}\) have xenon (Xe) as their central atoms, and both have chlorine (Cl) atoms bonded to the central atom. The common property they have is that they are both examples of Xenon-chlorine compounds. b. For the second set, we can see that all of the compounds contain a central atom and bonded atoms around it. In addition, all of these central atoms have lone pairs of electrons. We can also observe that all these molecules have polar bonds, making them polar through dipole parameters.
03

Write down the common properties

Now that we have identified the common properties, we can write them down for each set: a. Both \(\mathrm{XeCl}_{4}\) and \(\mathrm{XeCl}_{2}\) are Xenon-chlorine compounds. b. All of the compounds in the second set (\(\mathrm{ICl}_{5}, \mathrm{TeF}_{4}, \mathrm{ICl}_{3}, \mathrm{PCl}_{3}, \mathrm{SCl}_{2}, \mathrm{SeO}_{2}\)) have central atoms with lone pairs of electrons and are polar molecules.

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Key Concepts

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

Valence Electrons
In chemistry, valence electrons are the electrons that are found in the outermost shell of an atom. These electrons play a crucial role because they are involved in chemical bonding. To illustrate, when drawing Lewis structures, we pay special attention to valence electrons as they determine how atoms will bond together. For example, in the case of xenon in both \( \text{XeCl}_4 \) and \( \text{XeCl}_2 \), xenon requires a specific number of valence electrons to form bonds with the surrounding chlorine atoms.
  • Valence electrons are crucial in forming bonds.
  • They are found in the outermost electron shell.
  • They help predict an element's chemical properties and its reactivity.
Understanding the configuration of valence electrons enables chemists to predict how atoms will interact, how they form molecules, and the types of bonds that are likely to occur.
Lone Pairs
Lone pairs refer to the pairs of valence electrons that are not involved in forming bonds. In a molecule, these electrons are positioned around an atom but do not share a bond with another atom. For example, in \( \text{XeCl}_4 \), xenon not only bonds with four chlorine atoms but also retains two additional pairs of electrons as lone pairs. Lone pairs are significant because they influence the shape and polarity of molecules.
  • Each lone pair on an atom affects the geometry of the molecule.
  • Lone pairs can cause repulsion, changing bond angles.
  • They may also play a role in a molecule's reactivity.
By considering lone pairs, we can better understand the spatial arrangement of molecules, which is essential for predicting molecular behavior.
Polar Molecules
Polar molecules are characterized by the presence of permanent dipoles due to the uneven distribution of electron density. This occurs when two atoms in a molecule have differing electronegativity, leading to unequal sharing of electrons. For example, in \( \text{ICl}_5 \) and \( \text{TeF}_4 \), the central atoms have uneven pull towards the bonded atoms due to differences in electronegativity, resulting in polar bonds.
  • Polarity is determined by electronegativity differences between bonded atoms.
  • Polar molecules can interact through dipole interactions.
  • This property affects the solubility and boiling/melting points of substances.
Understanding this concept is vital as polarity plays a key role in the solubility and interaction of molecules with others, which has many practical implications in fields such as chemistry and biology.
Central Atom
The central atom in a molecule is typically the atom with the lowest electronegativity (aside from hydrogen) or the one that forms the largest number of bonds. This atom typically bears the greater share of electrons involved in bonding. In our examples, compounds like \( \text{ICl}_5 \) and \( \text{TeF}_4 \) illustrate how central atoms such as iodine and tellurium accommodate several atoms around them, forming multiple bonds.
  • The central atom is crucial in determining a molecule's geometry.
  • It is usually surrounded by atoms or groups bonded to it.
  • The choice of central atom affects molecular stability.
Choosing the correct central atom is critical in the construction of accurate Lewis structures. These structures help predict molecular properties and reactivities.

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Most popular questions from this chapter

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 A and 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 A and 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 \(\mathrm{E}\) is \(15 \mathrm{M}\). d. Commercial solutions of compound \(\mathrm{E}\) 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 \(\mathrm{D}\) is \(43.7 \% \mathrm{~N}\) and \(50.0 \% \mathrm{O}\) 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 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 A are strongly acidic; solutions of compounds \(\mathrm{B}\) and \(\mathrm{E}\) are basic. The titration of \(0.726 \mathrm{~g}\) compound \(\mathrm{B}\) requires \(21.98 \mathrm{~mL}\) of \(1.000 M \mathrm{HCl}\) for complete neutralization.

Hydrogen has an electronegativity value between boron and carbon and identical to phosphorus. With this in mind, rank the following bonds in order of decreasing polarity: \(\mathrm{P}-\mathrm{H}\), \(\mathrm{O}-\mathrm{H}, \mathrm{N}-\mathrm{H}, \mathrm{F}-\mathrm{H}, \mathrm{C}-\mathrm{H}\).

The most common type of exception 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?

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 \(\mathrm{NaCl}\) stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about \(\mathrm{MgO}\) compared to \(\mathrm{MgO}_{2}\) ? \(\mathrm{Mg}_{2} \mathrm{O} ?\)

Compare the electron affinity of fluorine to the ionization energy of sodium. Is the process of an electron being "pulled" from the sodium atom to the fluorine atom exothermic or endothermic? Why is NaF a stable compound? Is the overall formation of NaF endothermic or exothermic? How can this be?
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