Test Prep Workbook for AP Chemistry The Central Science 13th Edition
Test Prep Workbook for AP Chemistry The Central Science 13th Edition
13th Edition
ISBN: 9780133598025
Author: Edward L. Waterman
Publisher: Pearson Education
bartleby

Videos

Textbook Question
Chapter 16, Problem 1DE

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 coreelectrons 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's 4f orbitals at 105 eV, from an X-ray source that provided 1253.6 eV of energy (1 ev = 1.602 x 10 19J). The oxygen on the mineral surface gave emitted electron energies at 531 eV, corresponding to the 1s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for Hg and -2 for 0. (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 He2+ and 02- ; which electrons are the valence electrons in each case?

(a)

Expert Solution
Check Mark
Interpretation Introduction

To determine: The wavelength of the X rays used in the experiment.

Answer to Problem 1DE

Solution: The wavelength of the X rays used in the experiment is 9.89A .

Explanation of Solution

Given

The energy of the X rays is 1253.6eV .

The conversion of eV into joules (J) is done as,

1eV=1.602×1019J

Hence, the conversion of 1253.6eV into joules (J) is,

1253.6eV=1.602×1019J1eV×1253.6eV=2.01×1016J

The wavelength of the X rays is calculated by the formula,

λ=hcE

Where,

  • E is the energy of theX rays.
  • h is the planck’s constant (6.626×1034Js) .
  • c is the speed of light (3.0×108m/s) .
  • λ is the wavelength of the photon.

Substitute the value of h , c , and E in the above equation,

λ=6.626×1034Js×3.0×108m/s2.01×1016J=9.89×1010m(1  A   10 10m)=9.89A

Conclusion

The wavelength of the X rays used in the experiment is 9.89A .

(b)

Expert Solution
Check Mark
Interpretation Introduction

To determine: The difference found in the first ionization energies of mercury and oxygen to that of XPS energy for the removal of 4f electron of mercury and 1s electron of oxygen.

Answer to Problem 1DE

Solution: For mercury, 10 times more energy is required to remove electron from 4f orbital as compared to the 6s orbital.

For oxygen, 40 times more energy is required to remove electron from 1s orbital as compared to the 2p orbital of oxygen.

Explanation of Solution

The XPS energy for the removal of 4f electron of mercury is 105eV .

The conversion of eV into joules (J) is done as,

1eV=1.602×10-19J

Hence, the conversion of 105eV into joules (J) is,

105eV=1.602×1019J1eV×105eV=1.68×1017J

Energy required for 1 mole of mercury is calculated as,

Energyreguiredby1Hgatom=1.68×10-17JEnergyreguiredby6.02×1023Hgatom=1.68×10-17J×6.02×1023=10,113.6kJ/mol

The first ionization energy of mercury is 1007kJ/mol and the XPS energy to remove electron from 4f orbital is 10,113.6kJ/mol which is 10 times greater than the first ionization energy. Hence, 10 times more energy is required to remove electron from 4f orbital as compared to the 6s orbital.

The XPS energy for the removal of 1s electron of oxygen is 531eV .

The conversion of eV into joules (J) is done as,

1eV=1.602×10-19J

Hence, the conversion of 531eV into joules (J) is,

531eV=1.602×1019J1eV×531eV=8.51×1017J

Energy required for 1 mole of oxygen is calculated as,

Energyreguiredby1Oatom=8.51×1017JEnergyreguiredby6.02×1023Oatom=8.51×1017J×6.02×1023=51,230kJ/mol

The first ionization energy of oxygen is 1314kJ/mol and the XPS energy to remove electron from 1s orbital is 51,230kJ/mol which is 40 times greater than the first ionization energy. Hence, 40 times more energy is required to remove electron from 1s orbital as compared to the 2p orbital of oxygen.

Conclusion

For mercury, 10 times more energy is required to remove electron from 4f orbital as compared to the 6s orbital.

For oxygen, 40 times more energy is required to remove electron from 1s orbital as compared to the 2p orbital of oxygen.

(c)

Expert Solution
Check Mark
Interpretation Introduction

To determine: The ground state electronic configuration of Hg2+ and O2 ion and the valence electron present in them.

Answer to Problem 1DE

Solution: The ground state electronic configuration of Hg2+ is [Xe]4f145d10 and the valence electrons in Hg2+ are 5d orbital electrons.

The ground state electronic configuration of O2 is [He]2s22p6 and the valence electrons in O2 are 2p orbital electrons.

Explanation of Solution

The atomic number of Hg is 80 and the electronic configuration is [Xe]4f145d106s2 . If Hg undergoes ionization to Hg2+ , the removal of 2 electrons will take place from outermost shell which is 6s in mercury. So, the ground state electronic configuration of Hg2+ is [Xe]4f145d10 . The valence electrons are those that present in outermost orbital of the atom or ion. Hence, valence electrons in Hg2+ are 5d orbital electrons.

The atomic number of O is 8 and the electronic configuration is [He]2s22p4 . If O gains electrons to form O2 , the 2 electrons will add to outermost shell which is 2p in oxygen. So, the ground state electronic configuration of O2 is [He]2s22p6 . The valence electrons are those that present in outermost orbital of the atom or ion. Hence, valence electrons in O2 are 2p orbital electrons.

Conclusion

The ground state electronic configuration of Hg2+ is [Xe]4f145d10 and the valence electrons in Hg2+ are 5d orbital electrons.

The ground state electronic configuration of O2 is [He]2s22p6 and the valence electrons in O2 are 2p orbital electrons.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Explain, in terms of Coulombic energy and Exchange energy, why the ground-state electron configuration of Cr is [Ar]4s13d5 instead of [Ar]4s23d4.
Describe the process of Sublevel Energy Splitting in Multi-Electron-Atoms?
Atoms of Group 1 elements have the lowest first ionization energy of all the elements on the periodic table but subsequently have the largest increase between the first and second ionization energy that any other elements have. Explain why this is the case, based on the electronic structure of the atoms/ions.

Chapter 16 Solutions

Test Prep Workbook for AP Chemistry The Central Science 13th Edition

Ch. 16.4 - Prob. 16.6.1PECh. 16.4 - Prob. 16.6.2PECh. 16.4 - Prob. 16.7.1PECh. 16.4 - Prob. 16.7.2PECh. 16.5 - Write a balanced equation for the reaction that...Ch. 16.5 - (a) As described in Section 7.7 , the alkali...Ch. 16.5 - Prob. 16.9.1PECh. 16.5 - Prob. 16.9.2PECh. 16.6 - Arrange each of the following sets of atoms and...Ch. 16.6 - Prob. 16.10.2PECh. 16.6 - In the ionic compoundsLiF,NaCI,KBr, andRbl, the...Ch. 16.6 - Prob. 16.11.2PECh. 16.6 - 7.38 Write equations that show the process for...Ch. 16.6 - Prob. 16.12.2PECh. 16.6 - Prob. 16.13.1PECh. 16.6 - Prob. 16.13.2PECh. 16.6 - (a) What is the trend in first ionization energies...Ch. 16.6 - Prob. 16.14.2PECh. 16.7 - Prob. 16.15.1PECh. 16.7 - Prob. 16.15.2PECh. 16.7 - Prob. 16.16.1PECh. 16.7 - Prob. 16.16.2PECh. 16.8 - Prob. 16.17.1PECh. 16.8 - Write an equation for the second electron affinity...Ch. 16.9 - If the electron affinity for an element is a...Ch. 16.9 - Prob. 16.18.2PECh. 16.9 - 7.52 What is the relationship between the...Ch. 16.9 - Prob. 16.19.2PECh. 16.10 - Prob. 16.20.1PECh. 16.10 - Prob. 16.20.2PECh. 16 - Mercury in the environment can exist in oxidation...Ch. 16 - When magnesium metal is burned in air (Figure 3.6...Ch. 16 - The dipole moment of chlorine monofluoride,...Ch. 16 - Prob. 3ECh. 16 - Consider the element silicon, Si. Write its...Ch. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Which of the these elements is most likely to from...Ch. 16 - Prob. 22ECh. 16 - Which of the following bond is the most polar? H-F...Ch. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Which of the following bonds is the most polar? a....Ch. 16 - Which of the following bonds is most polar: S-Cl,...Ch. 16 - Prob. 29ECh. 16 - How many valence electrons should appear in the...Ch. 16 - Compare the lewis symbol for neon the structure...Ch. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Which of the statements about resonance is true?...Ch. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - A portion of a two-dimensional "slab" of NaCl(s)...Ch. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Incomplete Lewis structures for the nitrous acid...Ch. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - True or false: The hydrogen atom is most stable...Ch. 16 - Prob. 50ECh. 16 - What is the Lewis symbol for each of the following...Ch. 16 - Using Lewis symbols, diagram the reaction between...Ch. 16 - Use Lewis symbols to represent the reaction that...Ch. 16 - Predict the chemical formula of the ionic compound...Ch. 16 - Prob. 55ECh. 16 - Prob. 56ECh. 16 - Prob. 57ECh. 16 - Is lattice energy usually endothermic or...Ch. 16 - NaCI and KF have the same crystal structure. The...Ch. 16 - Consider the ionic compounds KF, NaCl, NaBr, and...Ch. 16 - Which of the following trends in lattice energy is...Ch. 16 - Energy is required to remove two electrons from Ca...Ch. 16 - Prob. 63ECh. 16 - Use data from Appendix C, Figure 7.10, and Figure...Ch. 16 - Prob. 65ECh. 16 - Prob. 66ECh. 16 - Prob. 67ECh. 16 - Using Lewis symbols and Lewis structures, diagram...Ch. 16 - Use Lewis symbols and Lewis structures to diagram...Ch. 16 - Prob. 70ECh. 16 - What is the trend in electronegativity going from...Ch. 16 - Prob. 72ECh. 16 - By referring only to the periodic table, select...Ch. 16 - which of the following bonds are polar? B-F,...Ch. 16 - Prob. 75ECh. 16 - Prob. 76ECh. 16 - Prob. 77ECh. 16 - In the following pairs of binary compounds,...Ch. 16 - Prob. 79ECh. 16 - Prob. 80ECh. 16 - Draw the dominant Lewis structure for the...Ch. 16 - Prob. 82ECh. 16 - Prob. 83ECh. 16 - Prob. 84ECh. 16 - Prob. 85ECh. 16 - Prob. 86ECh. 16 - Prob. 87ECh. 16 - Prob. 88ECh. 16 - Prob. 89ECh. 16 - Prob. 90ECh. 16 - 8.62 For Group 3A-7A elements in the third row of...Ch. 16 - Draw the Lewis structures for each of the...Ch. 16 - Prob. 93ECh. 16 - Prob. 94ECh. 16 - 8.66 Describe the molecule xenon trioxide, XeO3,...Ch. 16 - 8.67 There are many Lewis structures you could...Ch. 16 - Prob. 97ECh. 16 - Using Table 8.3, estimate H for each of the...Ch. 16 - Using Table 8.3, estimate H for the following...Ch. 16 - Prob. 100AECh. 16 - Prob. 101AECh. 16 - Prob. 102AECh. 16 - Prob. 103AECh. 16 - Consider the stable elements through lead (Z =...Ch. 16 - 17.80]Figure 7.4 shows the radial probability...Ch. 16 - (a) If the core electrons were totally effective...Ch. 16 - Prob. 107AECh. 16 - Prob. 108AECh. 16 - Prob. 109AECh. 16 - The following observations are made about two...Ch. 16 - Prob. 111AECh. 16 - Prob. 112AECh. 16 - Prob. 113AECh. 16 - Prob. 114AECh. 16 - Prob. 115AECh. 16 - Prob. 116IECh. 16 - Prob. 117IECh. 16 - Prob. 118IECh. 16 - Prob. 119IECh. 16 - Prob. 120IECh. 16 - The electron affinities. in kJ/mol, for the group...Ch. 16 - 7.99 Hydrogen is an unusual element because it...Ch. 16 - Prob. 123IECh. 16 - Prob. 124IECh. 16 - Which of the following is the expected product of...Ch. 16 - Elemental cesium reacts more violently with water...
Knowledge Booster
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
    • SEE MORE QUESTIONS
    Recommended textbooks for you
  • Chemistry: The Molecular Science
    Chemistry
    ISBN:9781285199047
    Author:John W. Moore, Conrad L. Stanitski
    Publisher:Cengage Learning
    Introductory Chemistry: A Foundation
    Chemistry
    ISBN:9781337399425
    Author:Steven S. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    General Chemistry - Standalone book (MindTap Cour...
    Chemistry
    ISBN:9781305580343
    Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
    Publisher:Cengage Learning
  • Chemistry
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Chemistry
    Chemistry
    ISBN:9781133611097
    Author:Steven S. Zumdahl
    Publisher:Cengage Learning
    Chemistry: An Atoms First Approach
    Chemistry
    ISBN:9781305079243
    Author:Steven S. Zumdahl, Susan A. Zumdahl
    Publisher:Cengage Learning
  • Chemistry: The Molecular Science
    Chemistry
    ISBN:9781285199047
    Author:John W. Moore, Conrad L. Stanitski
    Publisher:Cengage Learning
    Introductory Chemistry: A Foundation
    Chemistry
    ISBN:9781337399425
    Author:Steven S. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    General Chemistry - Standalone book (MindTap Cour...
    Chemistry
    ISBN:9781305580343
    Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
    Publisher:Cengage Learning
    Chemistry
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Chemistry
    Chemistry
    ISBN:9781133611097
    Author:Steven S. Zumdahl
    Publisher:Cengage Learning
    Chemistry: An Atoms First Approach
    Chemistry
    ISBN:9781305079243
    Author:Steven S. Zumdahl, Susan A. Zumdahl
    Publisher:Cengage Learning
    Lanthanoids and its Position in Periodic Table - D and F Block Elements - Chemistry Class 12; Author: Ekeeda;https://www.youtube.com/watch?v=ZM04kRxm6tY;License: Standard YouTube License, CC-BY