/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 In Table 7.8 , the bonding atomi... [FREE SOLUTION] | 91影视

91影视

In Table 7.8 , the bonding atomic radius of neon is listed as 0.58 A, whereas that for xenon is listed as 1.40 A. A classmate of yours states that the value for Xe is more realistic than the one for Ne. Is she correct? If so, what is the basis for her statement?

Short Answer

Expert verified
The classmate's statement that the value for Xe is more realistic than the one for Ne can be considered accurate, as xenon has more electron shells and stronger van der Waals forces compared to neon. These factors explain why xenon has a larger bonding atomic radius (1.40 脜) than neon (0.58 脜). However, it is essential to note that both values are relative and could be valid in the appropriate contexts, as they are usually experimentally determined or theoretically estimated using computational methods.

Step by step solution

01

Understand what bonding atomic radius is

Bonding atomic radius represents half the distance between two atoms that are chemically bonded together. It's a measure of the size of an atom in a chemical bond. For noble gases like neon and xenon, the bonding atomic radius mainly depends on their electron configurations and the van der Waals forces between gas atoms.
02

Analyze the electron configurations of Ne and Xe

Noble gases have full electron shells, which makes them stable and less reactive. We can analyze the electron configurations of neon (Ne) and xenon (Xe) to understand their atomic sizes. The electron configurations are: Neon (Ne): 1s虏 2s虏 2p鈦 Xenon (Xe): 1s虏 2s虏 2p鈦 3s虏 3p鈦 3d鹿鈦 4s虏 4p鈦 4d鹿鈦 5s虏 5p鈦 As we go down the noble gas group in the periodic table, the atomic number increases, and more electron shells are present with filled orbits. Xenon has more electron shells than neon, which would mean Xe is expected to have a larger atomic size compared to Ne.
03

Examine van der Waals forces

Van der Waals forces are weak intermolecular forces present between noble gas atoms due to temporary dipoles created by the movement of electrons. As the size of noble gas atoms increases down the group (from Ne to Xe), the van der Waals forces generally get stronger. This would result in potentially larger distances between larger gas atoms when bonded.
04

Evaluate the statement based on properties of Ne and Xe

Considering the electron configurations and van der Waals forces, it's reasonable to expect that xenon (Xe) would have a larger bonding atomic radius than neon (Ne). The given values (Xe 1.40 脜 and Ne 0.58 脜) are consistent with this expectation. Therefore, the classmate's statement that the value for Xe is more realistic than the one for Ne could be considered accurate. However, it is essential to note that both values are relative and could be valid in the appropriate contexts, as they are usually experimentally determined or theoretically estimated using computational methods.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Noble Gases
Noble gases, the last group in the Periodic Table, hold a unique place in chemistry due to their complete outer electron shells. These elements include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). As we move down the group, each noble gas has an additional electron shell compared to the one above it.

Because their outermost shell is full, noble gases are chemically inert, meaning they don鈥檛 react easily with other elements. This lack of reactivity gives them their 'noble' status, since they don鈥檛 readily form compounds. The completed shells also contribute to the atomic radius. As new shells are added with each step down the group, the atomic radius naturally increases. This is visible when comparing the bonding atomic radius of neon and xenon; neon has fewer electron shells than xenon, resulting in a smaller atomic radius.
Electron Configurations
Electron configurations are a shorthand way of illustrating the locations of electrons within an atom's electron shells and subshells. The configuration reflects the atom鈥檚 quantum mechanical properties and helps predict chemical behavior. For instance, a neon atom has the electron configuration of 1s虏 2s虏 2p鈦, signifying that its first and second electron shells are completely filled. This configuration confers stability to the atom.

Xenon, on the other hand, has a much more complex electron configuration: 1s虏 2s虏 2p鈦 3s虏 3p鈦 3d鹿鈦 4s虏 4p鈦 4d鹿鈦 5s虏 5p鈦. Each 'p鈦' sequence indicates a full p subshell, contributing to the element鈥檚 stability. Xenon's larger number of electron shells means that its electrons are further from the nucleus, compared to neon, increasing its atomic size. Understanding the electron configurations of noble gases is essential when considering properties like atomic radius and reactivity.
Van der Waals Forces
Van der Waals forces are the weak, attractive forces that occur between molecules or parts of molecules. These forces come into play due to fluctuations in electron density, which create temporary dipoles within atoms or molecules. Even though noble gases are nonreactive, van der Waals forces do exist between them and can influence their physical properties.

In noble gases, larger atoms like xenon have more electrons that can move around, creating stronger temporary dipoles compared to smaller atoms like neon. This results in stronger van der Waals forces in xenon, potentially increasing the distance between atoms when they are close to one another. While these forces are weak compared to chemical bonds, they are significant enough to affect states of matter and boiling points. A solid understanding of van der Waals forces is key for interpreting the behavior of non-polar substances and the noble gases in various conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Mercury in the environment can exist in oxidation states 0, +1, and +2. One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury鈥檚 4\(f\) orbitals at 105 eV, from an X-ray source that provided 1253.6 \(\mathrm{eV}\) of energy \(\left(1 \mathrm{ev}=1.602 \times 10^{-19} \mathrm{J}\right)\) The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV},\) corresponding to the 1 \(\mathrm{s}\) orbital of oxygen. Overall the researchers concluded that oxidation states were \(+2\) for \(\mathrm{Hg}\) and \(-2\) for \(\mathrm{O}\) (a) Calculate the wavelength of the X rays used in this experiment. (b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-} ;\) which electrons are the valence electrons in each case?

Detailed calculations show that the value of \(Z_{\text { eff }}\) for the outermost electrons in Si and Cl atoms is \(4.29+\) and \(6.12+\) , respectively.(a) What value do you estimate for \(Z\) eff experienced by the outermost electron in both Si and Cl by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text { eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text { eff? }} ?(\mathbf{d})\) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text { eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text { eff }}\) for a valence electron in \(\mathrm{P}\) , phosphorus, based on the calculations for Si and Cl.

Using only the periodic table, arrange each set of atoms in order from largest to smallest: \((\mathbf{a}) \mathrm{K},\) Li, \(\mathrm{Cs} ;(\mathbf{b}) \mathrm{Pb}, \mathrm{Sn}, \mathrm{Si} ;(\mathbf{c}) \mathrm{F},\) \(\mathrm{O}, \mathrm{N} .\)

The As\(-\)As bond length in elemental arsenic is 2.48 A. The \(\mathrm{Cl}-\mathrm{Cl}\) bond length in \(\mathrm{Cl}_{2}\) is 1.99 A. (a) Based on these data, what is the predicted \(\mathrm{As}-\mathrm{Cl}\) bond length in arsenic trichloride, \(A s C l_{3},\) in which each of the three Cl atoms is bonded to the As atom? (b) What bond length is predicted for \(A s C l_{3},\) , using the atomic radii in Figure 7.7?

(a) One of the alkali metals reacts with oxygen to form a solid white substance. When this substance is dissolved in water, the solution gives a positive test for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}.\) When the solution is tested in a burner flame, a lilac-purple flame is produced. What is the likely identity of the metal? (b) Write a balanced chemical equation for the reaction of the white substance with water.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.