Chemistry: A Guided Inquiry 5e

Chemistry: A Guided Inquiry 5e

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Printed in the United States of America 10 9 8 7 6 5 4 3 2

Table of Contents Chem Activity

Topic To the Instructor

1 3 4 5 8 9 10 11

The Nuclear Atom Coulombic Potential Energy The Shell Model (I) The Shell Model (II) Photoelectron Spectroscopy The Shell Model (III) Electron Configurations Electron Configurations and the Periodic Table

Appendix 1 Appendix 2

Answers to Critical Thinking Questions Answers to Exercises and Problems

To the Instructor We are very pleased to be able to provide these selected activities from Chemistry: A Guided Inquiry, 5th Edition for use in high school classrooms – particularly for the new AP Chemistry Curriculum. Our main goal is to provide materials that might be helpful in developing concepts related to photoelectron spectroscopy and the determination of electronic structure of atoms from photoelectron spectroscopy. Activity 3 develops the important concepts related to coulombic potential energy. Activities 4 and 5 develop a shell model of the atom that can be deduced from ionization energies. Activities 8 and 9 introduce photoelectron spectroscopy and show how it can be used to determine the distribution of electrons in the atom. Activities 10 and 11 provide another example of how photoelectron spectroscopy can be used to determine an electron configuration. These activities were designed for use in college-level general chemistry courses, and have been used successfully by tens of thousands of undergraduates at a variety of postsecondary institutions for more than 15 years. They are designed to be used by students working in groups of three or four, with the instructor serving as a facilitator, following the principles of cooperative learning. The POGIL Project supports the dissemination and implementation of these materials at both the high school and college levels. Information about professional development workshops to learn more about implementing and developing these types of activites can be found on The POGIL Project website at http://www.pogil.org. In addition, many useful tips about implementing POGIL activities and using groups in your classroom can be found at http://www.pogil.org/resources/implementation. Of particular value may be the High School POGIL Initiative (HSPI) Implementation Guide, available from that location. If you have further questions about POGIL or The POGIL Project, please contact The POGIL Project Office at [email protected].

2

ChemActivity

1 The Nuclear Atom (What Is an Atom?)

Model: Schematic Diagrams for Various Atoms.

1H

and 2H are isotopes of hydrogen.

12C

and 13C are isotopes of carbon.

ChemActivity 1

The Nuclear Atom

3

Critical Thinking Questions 1. How many protons are found in 12C? 13C? 13C–?

2. How many neutrons are found in 12C? 13C? 13C–?

3. How many electrons are found in 12C? 13C? 13C–?

4. Based on your answers to CTQs 1-3, what do all carbon atoms (and ions) have in common?

5. Based on the model, what do all hydrogen atoms (and ions) have in common?

6. Based on your answers to CTQs 4 and 5, what is the significance of the atomic number, Z, above each atomic symbol in the periodic table?

7. Based on your answer to CTQ 6, what do all nickel (Ni) atoms have in common?

8. In terms of the numbers of protons, neutrons and electrons: 13

C– have a negative sign in the upper right hand

a)

Why does the notation corner?

b)

What feature distinguishes a neutral atom from an ion?

c)

Provide an expression for calculating the charge on an ion.

4

ChemActivity 1

The Nuclear Atom

9. Determine the number of protons, neutrons, and electrons in one 1H+ ion. Explain how you found your answer.

10.

What structural feature is different in isotopes of a particular element?

11.

How is the mass number, A, (left-hand superscript next to the atomic symbol as shown in the Model) determined (from the structure of the atom)?

12.

Show that the mass number and charge given for Model 1.

13.

Based on the information in Model 1, where is most of the mass of an atom, within the nucleus or outside of the nucleus? Explain your reasoning using grammatically correct English sentences.

16

O2– and

23

Na+ are correct in

ChemActivity 1

The Nuclear Atom

5

Exercises 1. Complete the following table. Isotope Atomic Number Z 31P 15

Mass Number A

Number of Electrons

8

18O

19

39

18

58

58Ni2+

2. What is the mass (in grams) of a) one 1H atom? b) one 12C atom? 3. What is the mass (in grams) of 4.35 × 106 atoms of 12C? 4. What is the mass (in grams) of 6.022 × 1023 atoms of 12C? 5. What is the mass (in grams) of one molecule of methane which has one 12C atom and four 1H atoms, 12C1H4? 6. a) Define mass number. b) Define atomic number. 7. Indicate whether the following statement is true or false and explain your reasoning. An 18O atom contains the same number of protons, neutrons, and electrons. 8.

How many electrons, protons, and neutrons are found in each of the following? 24Mg

23Na+

35Cl

35Cl–

56Fe3+

15N

16O2–

27Al3+

6

ChemActivity 1

The Nuclear Atom

9. Complete the following table. Isotope

Atomic Number Z 27

Mass Number A 59

Number of Electrons

3

7

3

3

6

3

25

14N

58Zn2+ 19F–

10. Using grammatically correct English sentences, describe what the isotopes of an element have in common and how they are different.

Problems 1. Estimate the mass of one 14C atom (in amu) as precisely as you can (from the data in the model). Explain your reasoning. 2. Use the data in Model 1 to estimate the values (in amu) of a) the mass of an electron, b) the mass of a proton, and c) the mass of a neutron. 3. The mass values calculated in Problem 2 are only approximate because when atoms (up through iron) are made (mainly in stars) from protons, neutrons, and electrons, energy is released. Einstein’s equation E = mc2 enables us to relate the energy released to the mass loss in the formation of atoms. Use the known values for the mass of a proton, 1.0073 amu, the mass of a neutron, 1.0087, and the mass of an electron, 5.486 × 10–4 amu, to show that the mass of a 12C atom is less than the sum of the masses of the constituent particles.

8

ChemActivity

3

Coulombic Potential Energy (What Is Attractive about Chemistry?)

Model 1: Two Charged Particles Separated by a Distance "d". d particle 1

particle 2 charge on particle 1 = q1 charge on particle 2 = q2

V=

kq1q2 d

According to Coulomb, the potential energy (V) of two stationary charged particles is € given by the equation above, where q1 and q2 are the charges on the particles (for example: –1 for an electron), d is the separation of the particles (in pm), and k is a positive-valued proportionality constant. 1 pm = 10–12 m

Critical Thinking Questions 1. Assuming that q1 and q2 remain constant, what happens to the magnitude of V if the separation, d, is increased? 2. If the two particles are separated by an infinite distance (that is, d = ∞), what is the value of V? 3. If d is finite, and the particles have the same charge (that is, q1 = q2), is V > 0 or is V < 0? Explain your answer.

4. If q for an electron is –1, a) what is q for a proton? b) what is q for a neutron? c) what is q for the nucleus of a C atom?

ChemActivity 3

Coulombic Potential Energy

9

5. Recall that a 1H atom consists of a proton as the nucleus and an electron outside of the nucleus. Is the potential energy, V, of a hydrogen atom a positive or negative number? Explain your answer.

Model 2: Ionization Energy. The ionization energy (IE) is the amount of energy needed to remove an electron from an atom and move it infinitely far away. Ionization energies are commonly measured in joules, J. Figure 1.

Ionization of a hypothetical atom L with one proton and one stationary electron.

Atom L d = 1000 pm

Adding 0.231 × 10–18 J to Atom L results in a bare proton and a free electron. IE = 0.231 × 10–18 J Atom L + 0.231 × 10–18 J

Figure 2.

p+ + e –

Ionization energies of two hypothetical atoms, each with one proton and one stationary electron separated by distance "d".

10

ChemActivity 3


Table 1.

Ionization energies of several hypothetical atoms, each with one proton and one stationary electron separated by distance "d". Hypothetical d IE V –18 –18 Atom (pm) (10 J) (10 J) A ∞ 0 E 5000. 0.0462 L 1000. 0.231 Q 500.0 0.462 T 200.0 1.16 Z 100.0 2.31

Critical Thinking Questions 6. Do you expect the potential energy, V, of the hypothetical atoms in Table 1 to be positive or negative numbers? Explain your reasoning.

7. Without using a calculator, predict what trend (if any) you expect for the values of V for these hypothetical atoms.

8. Calculate the potential energies of the hypothetical atoms to complete Table 1. Use the value k = 2.31 × 10–16 J.pm.

9. What is the relationship between IE and V for these hypothetical atoms?

10. Which of the following systems will have the larger ionization energy? Explain your reasoning. a) an electron at a distance of 500 pm from a nucleus with charge +2 b) an electron at a distance of 700 pm from a nucleus with charge +2

ChemActivity 3

11.


11

Which of the following systems will have the larger ionization energy? Explain your reasoning. a) b)

an electron at a distance d1 from a nucleus with charge +2 an electron at a distance d1 from a nucleus with charge +1

12. How many times larger is the larger of the two ionization energies from CTQ 11? Show your work.

13. Consider a hydrogen atom and a helium ion, He+. Which of these do you expect to have the larger ionization energy? Explain your reasoning, including any assumptions you make.

12

ChemActivity 3


Exercises 1. For a hypothetical atom (as in Table 1) with V = –5.47 × 10–18 J, what would the IE be? 2. Which of the following systems will have the larger ionization energy? Show your work. a) an electron at a distance d1 from a nucleus with charge +2 b) an electron at a distance 2d1 from a nucleus with charge +1 3. Which of the following systems has the larger ionization energy? a) an electron at a distance 5d1 from a nucleus with a charge of +6 b) an electron at a distance 6d1 from a nucleus with a charge of +7

Problems 1. According to the Coulombic Potential Energy equation, if a particle with a charge of –1 is extremely close to a particle with a charge of +2, the potential energy is: a) large and positive b) large and negative c) small and negative d) small and positive 2. Two electrons and one helium nucleus are arranged in a straight line as shown below. The electron on the left is 300 nm from the nucleus; the electron on the right is 400 nm from the nucleus. Write the three Coulombic Potential Energy terms for this arrangement of charges.

electron –1

nucleus +2

electron –1

14

ChemActivity

4 The Shell Model (I) (How Are Electrons Arranged?)

Electrons in atoms are attracted to the nucleus by a Coulombic force. Thus, energy must be supplied (by some means) if the electron is to be pulled away from the nucleus, thereby creating a positively charged species, or cation, and a free electron. For real atoms, the ionization energy (IE) of an element is the minimum energy required to remove an electron from a gaseous atom of that element. Ionization energies are usually obtained experimentally. One method of measuring ionization energies is the electron impact method. Atoms are bombarded with fastmoving electrons. If these electrons have sufficient energy, they will, on colliding with an atom, eject one of the atom's electrons. The ionization energy described above (often called the first ionization energy) corresponds to the smallest amount of energy that a bombarding electron needs to be able to knock off one of the atom's electrons.

Model 1: First Ionization Energy (IE1). M(g)

M+ (g) + e–

For a H atom, IE = 2.178 × 10–18 J. The first ionization energy, IE1, for a single atom is a very small number of joules. For reasons of convenience, chemists have chosen to report the ionization energies of elements in terms of the minimum energy necessary to remove a single electron from each of a mole of atoms of a given element. This results in ionization energies for the elements that are in the range of MJ/mole. (Note that 1 MJ = 106 J.)

Critical Thinking Questions 1. How much total energy would it take to remove the electrons from a mole of H atoms? Write this energy in MJ/mole. 2. For atoms with many electrons, not all electrons are at the same distance from the nucleus. In this case, which electron would have the lowest ionization energy: the electron that is closest to the nucleus or the electron that is farthest from the nucleus? Explain.

ChemActivity 4

The Shell Model (I)

15

3. Predict the relationship between IE1 and atomic number by making a rough graph of IE1 vs. atomic number. DO NOT PROCEED TO THE NEXT PAGE UNTIL YOU HAVE COMPLETED THIS GRAPH.

IE1 (MJ/mole)

Atomic Number

16

ChemActivity 4

The Shell Model (I)

Information Based on our previous examination of ionization energies, it is expected that the ionization energy of an atom would increase as the nuclear charge, Z, increases. In addition, the ionization energy of an atom should decrease if the electron being removed is moved farther from the nucleus (that is, if d increases). Table 1 below presents the experimentally measured ionization energies of the first 20 elements. We will examine these results and attempt to propose a model for the structure of atoms based on these data. Table 1. Z 1 2 3 4 5 6 7 8 9 10

First Ionization energies of the first 20 elements. Z IE1 IE1 (MJ/mole) (MJ/mole) H 1.31 11 Na 0.50 He 2.37 12 Mg 0.74 Li 0.52 13 Al 0.58 Be 0.90 14 Si 0.79 B 0.80 15 P 1.01 C 1.09 16 S 1.00 N 1.40 17 Cl 1.25 O 1.31 18 Ar 1.52 F 1.68 19 K 0.42 Ne 2.08 20 Ca 0.59

Critical Thinking Questions 4. Compare your answer to CTQ 3 to the data in Table 1. similarities and differences.

Comment on any

5. Using grammatically correct English sentences: a) provide a possible explanation for why IE1 for He is greater than IE1 for H.

b) provide a possible explanation for why IE1 for Li is less than IE1 for He.

ChemActivity 4

The Shell Model (I)

17

Model 2: Simple Model Diagrams for Hydrogen and Helium Atoms. One simple model of the hydrogen atom pictures the H atom as a nucleus of charge +1 surrounded by an electron at some distance, as shown in Figure 1. Figure 1.

Model diagram of a hydrogen atom. electron the electron "sees" a +1 charge at the nucleus +1

Figure 2.

Model diagram of a helium atom. each electron "sees" a +2 charge at the nucleus +2

Examining the data in Table 1, we note that the ionization energy of He (Z=2) is larger than that of H (Z=1) by approximately a factor of 2. This is consistent with the two electrons in the He atom at a distance (from the nucleus) approximately the same as that in H.

Critical Thinking Question 6. The value of the ionization energy of He given in Table 1 is described as being consistent with two electrons in a "shell" approximately the same distance from the nucleus as the one electron in H. Use the Coulombic Potential Energy equation, kq q V = 1 2 to explain how this conclusion can be reached. Hint: recall the d relationship between V and IE1.

€

18

ChemActivity 4

The Shell Model (I)

Information Because the He nucleus has a charge of +2, we would expect that the ionization energy to remove an electron from (approximately) the same distance as in a H atom would be (approximately) twice that of the H atom. That is what we observe. We can say that there are two electrons in a shell around the He nucleus. Although we will present figures in which the shells appear to be circular (mostly because it is difficult to present three-dimensional representations on paper), we recognize that the model we develop is qualitatively consistent with spherical shells. Thus, within our Shell Model, He consists of a nucleus surrounded by 2 electrons in a single shell.

Critical Thinking Questions 7. Recall that the IE of H is 1.31 MJ/mole. If all three electrons in Li were in the first shell at a distance equal to that of hydrogen, which of the following values would be the better estimate of the IE1 of Li: 3.6 MJ/mole or 0.6 MJ/mole? Explain.

8. A student proposes a model for the Li atom in which all three electrons are in the first shell at a distance approximately the same as in H. Why is this model not consistent with the IE1 for Li in Table 1?

9. Another student proposes a model for the Li atom in which two electrons are in the first shell (as in He) and the third electron is much farther from the Li nucleus. Explain why this model is consistent with the IE1 for Li in Table 1.

ChemActivity 4

The Shell Model (I)

19

Model 3: The Shell Model for Lithium. For Li, there is a change in the trend of the ionization energy. The ionization energy of a Li atom is less than that of He. In fact, it is significantly smaller than that of the H atom! This is not consistent with a model of placing a third electron in the first shell, for doing so would result in an ionization energy which is larger than that of He. In order for Li to have a lower ionization energy than H, either the nuclear charge Z must be lower than that of H, or the distance of the easiest-to-remove electron from the nucleus must be greater than in H (and He), or both. We know that the nuclear charge is not lower than that of H; thus, the electron being removed must be farther from the nucleus than the first shell. Although the data we have does not require us to choose the following model, let us assume that the structure of Li involves two electrons in a first shell (as in He) with the third electron placed in a second shell, with a significantly larger radius, as shown in Figure 3.

Figure 3.

Diagram of a lithium atom using the shell model.

+3

Critical Thinking Question 10. Is the amount of energy required to remove one of the electrons from the first shell of Li greater than, less than, or equal to the IE1 for Li? Explain your reasoning.

20

ChemActivity 4

The Shell Model (I)

Exercises 1. A scientist proposes a model for the helium atom in which both electrons are in a "shell" which is half the distance from the nucleus as the electron in a hydrogen atom. Is this model consistent with the data in Table 1? Explain your reasoning. (Hint: according to the Coulombic Potential Energy equation, how much more strongly does a nuclear charge of +2, as in He, hold an electron than a nuclear charge of +1, as in H? According to the Coulombic Potential Energy equation, how much more strongly does a nuclear charge hold an electron if it is at d/2, rather than d?) 2. Propose an alternative model for the lithium atom which is consistent with the data in Table 1.

Problem 1. a) Write the three Coulombic Potential Energy terms for the helium atom model in Figure 2. Assume that the distance between each electron and the nucleus is d and that the distance between the two electrons is 2d.

b) Based on your answer to part a) explain why the IE of He is slightly less than twice the IE of H even though both atoms are about the same size.

22

ChemActivity

5 The Shell Model (II)

Model 1: Valence Electrons, Inner-Shell Electrons, and Core Charge. The electrons in the outermost shell of an atom are referred to as valence electrons. Electrons in shells closer to the nucleus are called inner-shell electrons. Thus, Li has one valence electron and two inner-shell electrons. H has one valence electron and no innershell electrons. The nucleus plus the inner shells of electrons constitute the core of the atom, and the net overall charge on the core is called the core charge. We can represent the Li atom in terms of core charge as shown in Figure 1.

Figure 1.

Diagram of a lithium atom using the shell model (a) and the core charge concept (b).

(a) valence electron

inner-shell or core electron

+3

core charge

(b) valence electron

+1

ChemActivity 5

The Shell Model (II)

23

Critical Thinking Questions 1. How many electrons are in the valence shell of a) H? b) Li? C) He? 2. How many inner-shell electrons are there in a) H? b) Li? C) He? 3. What is the core charge of Li?

Model 2: Core Charge and Electron-Electron Repulsion. Notice that within the model of the Li atom, shown in Figure 1, the valence electron is farther from the nucleus than the two inner-shell electrons. Although we have ignored it up to this point, we should remember that all of the electrons repel each other because they are negatively charged. Of particular interest is the repulsion of the valence electron by the two inner-shell electrons. This dramatically decreases the overall force of attraction pulling the valence electron toward the nucleus.

Critical Thinking Question 4. Two possible models for arrangement of electrons in Li are shown below: electron "a" electron "b"

ra

rb

3+

3+

ra = rb

Explain why the IE1 of electron “b” would be less than the IE1 of electron “a”.

24

ChemActivity 5


Model 3: The Beryllium Atom. The next element, Be, has an ionization energy which is larger than that for Li. This is consistent with the fourth electron in Be being added to the second shell. Thus, Be has 2 valence electrons and a core charge of +2. Two representations of the Be atom are given in Figure 2.

Figure 2.

Diagram of a Be atom using the shell model (a) and the core charge concept (b).

(a)

(b)

+4

+2

the valence electron "sees" a core charge of +2

Critical Thinking Questions 5. a) Why is the nuclear charge of Be "+4"? b) How many inner-shell electrons does Be have? c) How many valence electrons does Be have? d) Show how the core charge for Be was calculated. e) What is the relationship between the number of valence electrons and the core charge of a neutral atom? 6. Assuming that the valence shells of Li and Be are at approximately the same distance from their nuclei, explain how the core charges of Li and Be are consistent with the IE1 values for Li (0.52 MJ/mole) and Be (0.90 MJ/mole).

ChemActivity 5


25

Information As described above, the outer-shell valence electrons experience the charge of the core rather than the full charge of the nucleus. The inner electrons that surround the nucleus are said to shield the nucleus. In fact, because the valence electrons are all negatively charged, they repel each other also. Thus the net resulting charge acting on a valence electron to attract it toward the nucleus differs from the core charge. This overall resulting charge acting on a valence shell electron is known as the effective nuclear charge, and it is generally less than the core charge. Since there is no simple way to obtain values for the effective nuclear charge, we will use the core charge as a basis for our qualitative explanations. It is only an approximation, but it is adequate for our purposes.

Model 4: The Neon Atom. Although there are some slight variations, in general there is an increase in ionization energy as the atomic number further increases up to Z = 10 (Ne). This is qualitatively consistent with an increase in core charge. (The slight variations will be addressed later.) There is no large drop in ionization energy to a value less than that of H, as we observed in going from He to Li, to indicate that a third shell is needed. This suggests that as we move from Be up to Ne, the number of electrons in the second shell increases. Figure 3.

Diagram of a Ne atom using the shell model (a) and the core charge concept (b).

(a)

(b)

+10

+8

the valence electron "sees" a core charge of +8

Ne has 8 electrons in the second (valence) shell, and 2 electrons in the inner (first) shell. Notice that we can number the shells based on their distance from the nucleus. We can let the number "n" represent the number of the shell an electron is in. Thus, Ne has 2 electrons in the n = 1 shell and 8 electrons in the n = 2 shell.

Critical Thinking Questions 7. Show how the core charge for Ne was calculated.

26

ChemActivity 5


8. Make two diagrams, similar to Figures 3(a) and (b), for the nitrogen atom.

9.

10.

a)

Make two diagrams, similar to Figures 3(a) and (b), for the sodium atom, assuming that the 11th electron goes into the second shell.

b)

What is the core charge for the sodium atom in CTQ 9a?

c)

The IE1 of Ne is 2.08 MJ/mole. Predict whether the IE1 for the Na atom in CTQ 9a would be greater than, less than, or equal to 2.08 MJ/mole. Explain your reasoning.

a)

Make two diagrams, similar to Figures 3(a) and (b), for the sodium atom, assuming that the 11th electron goes into a new, third shell.

b)

What is the core charge for the sodium atom in CTQ 10a?

c)

Predict whether the IE1 for the Na atom in CTQ 10a would be greater than, less than, or equal to 2.08 MJ/mole. Explain your reasoning.

11. The experimental IE1 for Na is 0.50 MJ/mole. Use this datum to explain why the model for Na suggested in CTQ 10 is a better model than the one suggested in CTQ 9.

ChemActivity 5


27

Model 5: The Sodium Atom. Diagram of a Na atom using the shell model (a) and the core charge concept (b).

(a)

(b)

+11

+1

Critical Thinking Questions 12. How many electrons does Na have in shell n = 1? n = 2? n = 3?

13. How does the core charge for Na compare to the core charge for Li?

14. Based on your answer to CTQ 13 and the ionization energy data, Table 1 of ChemActivity 4: Shell Model (I), is the radius of the valence shell of Na larger, smaller or the same as the radius of the valence shell of Li?

15. Consider the models of Ne and Na shown in Models 4 and 5. Explain how the core charges of Na and Ne are qualitatively consistent with the IE1 data in Table 1 of ChemActivity 4: Shell Model (I).

28

ChemActivity 5


Information The IE1 for Na is 0.50 MJ/mole, much less than the IE1 for Ne. This decrease is analogous to (and similar in magnitude to) that observed in going from He to Li. Note that the ionization energy of Na is only 0.50 MJ/mole, even less (although only slightly so) than that of Li. Analogous to the conclusions we reached concerning the structure of the Li atom, these results suggest that the eleventh electron in Na should be placed in a third shell (n = 3), at a slightly greater distance from the nucleus than the second shell is for Li. Thus, it appears that the n = 2 shell can accommodate only eight electrons. (Recall that the n = 1 shell holds only two.) The pattern of ionization energies for the elements with Z = 11 to Z = 18 follows the trend we previously identified for Z = 3 to Z = 10: a general increase (with slight variations).

Critical Thinking Question 16. Use the core charge concept to propose an explanation for the increase of IE1 from Na through Ar. Clearly state any assumptions that you make.

Information As suggested by the data in Table 1 of ChemActivity 4: Shell Model (I), all of the atoms in Group 1A, the alkali metals, have a core charge of +1 and all of the atoms in Group 7A, the halogens, have a core charge of +7. In fact, for Groups 1A through 7A, the atoms in each group all have the same number of valence electrons, and that number is reflected by the group number. In all cases, the ionization energy decreases as we move down the group. This pattern is also observed in Group 8A, the Noble (or Inert) gases . However, not all of the atoms we have examined in Group 8A have eight valence electrons (and a core charge of +8). Helium has only 2 electrons, a seeming violation of the pattern we have uncovered. The resolution of this apparent inconsistency is that although He has only 2 valence electrons, its valence shell is completely filled. The same is true of Ne, although for Ne a filled valence shell has 8 electrons. Thus, we find that the structure of the elements using this shell model is reflected in the placement of the elements in the periodic table.

ChemActivity 5


29

Model 6: The Shell Model and Ionization Energies. Table 1. Atomic Properties of Various Atoms. Element Valence Shell Number of Core Charge (n) Valence Electrons H 1 1 +1 Li 2 1 +1 Na 3 1 +1 Rb F 2 7 +7 Cl 3 7 +7

IE1 (MJ/mole) 1.31 0.52 0.50 0.40 1.68 1.25

Critical Thinking Questions 17. Locate H, Li, and Na on the periodic table.

18.

a)

Describe any relationship between the core charge of these atoms, the number of valence electrons, and their positions in the periodic table.

b)

Describe any relationship between the valence shell of these atoms and their positions in the periodic table.

c)

Based on its position in the periodic table, predict the valence shell, core charge, and number of valence electrons for Rb and add these values to Table 1.

d)

Using the shell model and referring to the Coulombic Potential Energy relationship (equation in Model 1, CA3), explain clearly how the IE1 for Rb is consistent with your answer to part c.

Construct a shell model diagram of F that is consistent with the information in Table 1.

30

19.

ChemActivity 5


Locate F and Cl on the periodic table. a)

Describe any relationship between the core charge of these atoms, the number of valence electrons, and their position in the periodic table.

b)

Describe any relationship between the valence shell of these atoms and their position in the periodic table.

c)

Within our model and referring to the Coulombic Potential Energy expression, explain why the IE1 of Cl is less than that of F.

20.

Based on its position in the periodic table, what is the valence shell and what is the core charge for C? Explain your reasoning.

21.

How does the core charge on the neutral atom change as we move from left to right across a row (period) of the periodic table?

22. Within our model and referring to the Coulombic Potential Energy expression, explain why the IE increases from left to right across a row of the periodic table.

ChemActivity 5


31

Exercises 1. How many valence electrons are there in: a) C? b) O? c) N? d) Ne? 2. What is the core charge for: a) C? b) O? c) N? d) Ne? 3. Based on the information in Table 1 of ChemActivity 4: Shell Model (I), estimate the ionization energy for Br. Explain your reasoning. 4. If a single electron is removed from a Li atom, the resulting Li+ cation has only two electrons, both in the n = 1 shell. In this respect it is very similar to a He atom. How would you expect the ionization energy of a Li+ cation to compare to that of a He atom? Explain your reasoning. 5. If a single electron is somehow added to a F atom, the resulting F– anion has a total of 8 valence electrons in the n = 2 shell. In this respect it is very similar to a Ne atom. How would you expect the ionization energy of a F– anion to compare to that of a Ne atom? Explain your reasoning. 6. Predict the order of the ionization energies for the atoms Br, Kr, and Rb. Explain your reasoning. 7. The radius of the outer shell in Li is larger than the radius of the inner shell. Which electron is harder to remove—the valence electron or one of the inner shell electrons? Explain.

Problems 1. Indicate whether each of the following statements is true or false and explain your reasoning. a) The core charge of Br is +7. b) Helium has the largest 1st ionization energy. 2. Explain how the model of the structure of Be having the fourth electron in a third shell, further from the nucleus than any of the three electrons in Li, is not consistent with the experimentally obtained ionization energies.

32

ChemActivity

8

Photoelectron Spectroscopy (What Is Photoelectron Spectroscopy?) From our previous examination of the ionization energies of the atoms, we proposed a shell model of the atom, and noted that the number of valence electrons in the outermost shell is related to the position of the element in the periodic table, and therefore is an important factor in determining the physical and chemical properties of the element. Within this model, the electrons in an atom are arranged in shells about the nucleus, with the successive shells being farther and farther from the nucleus. The ionization energy described previously is the minimum energy needed to remove an electron from the atom. The most easily removed electron always resides in the valence shell, since that is the shell that is the farthest from the nucleus. For atoms with many electrons, we would expect that the energy needed to remove an electron from an inner shell would be greater than that needed to remove an electron from the valence shell, because an inner shell is closer to the nucleus and is not as fully shielded as the outer valence electrons. Thus, less energy is needed to remove an electron from an n = 2 shell than from an n = 1 shell, and even less is needed to remove an electron from an n = 3 shell. But do all electrons in a given shell require precisely the same energy to be removed? In order to answer this question, we must consider ionization energies in greater detail.

Model 1: Ionization Energies and Energy Levels From the Coulombic Potential Energy expression, we know that an electron in a given shell will require a certain energy to be separated from the atom. Thus, an electron can be said to occupy an energy level in an atom. Within our model, each electron must be in a shell at a particular distance from the nucleus, and the energy levels corresponding to these shells are quantized—that is, only certain discrete energy levels should be found. The electron at this energy level is easier to remove than electrons closer to the nucleus. γ α β

nucleus

Each of the two electrons at this energy level is harder to remove than the electron that is farther from the nucleus.

ChemActivity 8

Photoelectron Spectroscopy

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Critical Thinking Question 1. Suppose that the values for the two energy levels for the atom in Model 1 are –0.52 MJ/mole and –6.26 MJ/mole. a)

How much energy, in MJ, is required to remove electron "α" in Model 1 from one mole of neutral atoms?

b)

What is the potential energy of each of the three electrons in Model 1? (Hint: see CA3.)

c)

Determine the ionization energies of each of the three electrons in Model 1.

Information When comparing the energy level of two different electrons, the electron with the higher ionization energy is said to occupy the lower energy level.

Critical Thinking Question 2. Provide a statement, similar to the Information statement, that uses the potential energies of the electrons rather than the ionization energies.

Model 2: Photoelectron Spectroscopy. Ionization energies may be measured by the electron impact method, in which atoms in the gas phase are bombarded with fast-moving electrons. These experiments give a value for the ionization energy of the electron that is most easily removed from the atom—in other words, the ionization energy for an electron in the highest occupied energy level. An alternative, and generally more accurate, method that provides information on all the occupied energy levels of an atom (that is, the ionization energies of all electrons in the atom) is known as photoelectron spectroscopy; this method uses a photon (a packet of light energy) to knock an electron out of an atom. Electrons obtained in this way are called photoelectrons. Very high energy photons, such as very-short-wavelength ultraviolet radiation, or even x-rays, are used in this experiment. The gas phase atoms are irradiated with photons of a particular energy. If the energy of the photon is greater than the energy necessary to remove an electron from the atom, an electron is ejected with the excess energy 1 appearing as kinetic energy, mν 2 , where v is the velocity of the ejected electron. In 2 other words, the speed of the ejected electron depends on how much excess energy it has

€

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ChemActivity 8 Photoelectron Spectroscopy

received. So, if IE is the ionization energy of the electron and KE is the kinetic energy with which it leaves the atom, we have Ephoton = IE + KE or, upon rearranging the equation, IE = Ephoton – KE Thus, we can find the ionization energy, IE, if we know the energy of the photon and we can measure the kinetic energy of the photoelectron. The kinetic energy of the electrons is measured in a photoelectron spectrometer. Figure 1: Photoelectron spectroscopy of a hypothetical atom. Before Photon Interaction photon energy = 143.4 MJ/mole (for example)

Atom

After Photon Interaction

Atom+ e– kinetic energy of ionized electron = 114.8 MJ/mole

If photons of sufficient energy are used, an electron may be ejected from any of the energy levels of an atom. Each atom will eject only one electron, but every electron in each atom has an (approximately) equal chance of being ejected. Thus, for a large group of identical atoms, the electrons ejected will come from all possible energy levels of the atom. Also, because the photons used all have the same energy, electrons ejected from a given energy level will all have the same energy. Only a few different energies of ejected electrons will be obtained, corresponding to the number of energy levels in the atom. The results of a photoelectron spectroscopy experiment are conveniently presented in a photoelectron spectrum. This is essentially a plot of the number of ejected electrons (along the vertical axis) vs. the corresponding ionization energy for the ejected electrons (along the horizontal axis). It is actually the kinetic energy of the ejected electrons that is measured by the photoelectron spectrometer. However, as shown in the equation above, we can obtain the ionization energies of the electrons in the atom from the kinetic energies of the ejected electrons. Because these ionization energies are of most interest to us, a photoelectron spectrum uses the ionization energy as the horizontal axis.

ChemActivity 8

Figure 2:


35

A simulated photoelectron spectrum of the hypothetical atom in Figure 1.

Critical Thinking Questions 3. Use the data presented in Figure 1 to verify that the IE of the ejected electron is 28.6 MJ/mole.

4. What determines the position of each peak (where along the horizontal axis the peak is positioned) in a photoelectron spectrum?

5. What is the numerical value at the position of the hatch mark in the photoelectron spectrum of Figure 2?

6. What energy is associated with the energy level of the electron in Figure 2?

7. What determines the height (or intensity) of each peak in a photoelectron spectrum?

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8. Explain why it is not possible to determine the number of electrons in an individual hypothetical atom from the photoelectron spectrum in Figure 2.

Model 3: The Energy Level Diagram of Another Hypothetical Atom. A hypothetical atom in a galaxy far, far away has 2 electrons at one energy level and 3 electrons at another energy level as shown in the energy level diagram below:

E (energy level)

IE (ionization energy)

Critical Thinking Questions 9. How many peaks (1,2,3,4,5) will appear in a photoelectron spectrum of a sample of this hypothetical atom? Why?

10.

Describe the relative height of the peaks in the photoelectron spectrum of a sample of this hypothetical atom.

11.

Suppose that the two energy levels are –0.85 MJ/mole and –4.25 MJ/mole. On the axes below, make a sketch of the photoelectron spectrum of a sample of this hypothetical atom. Make sure to label the axes appropriately.

3 2 1

ChemActivity 8

Model 4:


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A Simulated Photoelectron Spectrum of an “Unknown” Atom.

Critical Thinking Questions 12.

Based on the number of peaks (one), the intensity of the peak, and your understanding of the shell model: a)

Explain why it is not possible to determine if the “unknown” atom is H or He?

b)

Explain why the “unknown” atom cannot be Li.

13. Based on the value of the IE given in Model 4 and on the values given in Table 1 of CA 4, identify the “unknown” atom.

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Model 5: The Neon Atom.

+10

Let us now predict what the photoelectron spectrum of Ne will look like, based on our current model of the Ne atom. In this model, there are 2 electrons in the n =1 shell, and 8 electrons in the n = 2 shell of a Ne atom. Assuming that all of the electrons in each of the shells has the same energy, we would expect two peaks in the photoelectron spectrum. One peak, from the electrons in the n = 2 shell, should appear at an energy of 2.08 MJ/mole, because that is the first ionization energy of Ne as determined previously. The second peak should be at a significantly higher energy, because it corresponds to the ejection of electrons from the n = 1 shell, which is significantly closer to the nucleus. At this point we do not have any good way of estimating what that energy is, but we know that it will be a lot higher than 2.08 MJ/mole. Finally, we also can predict the relative sizes of the two peaks—that is, the relative areas under the two curves on the spectrum. Recall that in photoelectron spectroscopy, the bombarding photon ejects an electron at random from each of the atoms in the sample. Thus, of the 10 electrons in Ne, we would expect that 2/10 of the time the electron is ejected from the n = 1 shell, and 8/10 of the time it is ejected from the n = 2 shell. The size of the peak in the spectrum is determined by the relative number of electrons with a given IE. Thus, the peak at 2.08 MJ/mole should be 4 times as large as the peak at a much higher energy, which corresponds to the ejection of electrons from the n = 1 shell. To summarize, our prediction is that the photoelectron spectrum of Ne should consist of two peaks, one at an energy of 2.08 MJ/mole and one at much higher energy, and the relative sizes of these two peaks should be 4:1.

Critical Thinking Questions 14. The peak due to the n = 1 shell is predicted to be at a much higher ionization energy than the n = 2 peak because the n = 1 shell is "significantly closer to the nucleus." Why is the distance of the shell from the nucleus important in determining the corresponding peak position in the photoelectron spectrum?

15.

Why is it expected that 2/10 of the ejected electrons will come from the n = 1 shell, and 8/10 of the electrons from the n = 2 shell?

ChemActivity 8

16.


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Make a sketch of the predicted photoelectron spectrum of Ne based on the description given above. Indicate the relative intensity (peak size) and positions of the two peaks.

Exercises 1. In a photoelectron spectrum, photons of 165.7 MJ/mole impinge on atoms of a certain element. If the kinetic energy of the ejected electrons is 25.4 MJ/mole, what is the ionization energy of the element? 2. The ionization energy of an electron from the first shell of lithium is 6.26 MJ/mole. The ionization energy of an electron from the second shell of lithium is 0.52 MJ/mole. a) Prepare an energy level diagram (similar to the one in Model 3) for lithium; include numerical values for the energy levels. b)

Sketch the photoelectron spectrum for lithium; include the values of the ionization energies.

3. An atom has the electrons in the energy levels as shown below:

E Make a sketch of the PES of this element.

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ChemActivity

9

The Shell Model (III) (How Many Peaks Are There in a Photoelectron Spectrum?)

Model 1. Simulated Experimental Photoelectron Spectrum of Neon.

Critical Thinking Questions 1.

a)

Is the photoelectron spectrum for Ne shown in Model 1 consistent with our model of the structure of atoms?

b)

Describe the ways (if any) in which our prediction was correct and the ways (if any) in which it was not correct.

2. Why is it not important to place numbers along the vertical axis of photoelectron spectra?

3. Based on the spectrum in Model 1, estimate the number of electrons at each of the three energy levels in Ne. Explain your reasoning clearly. (Hint: Recall that the total number of electrons should equal the number of electrons in a Ne atom.)

ChemActivity 9

The Shell Model (III)

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Information: The Neon Atom Revisited. Contrary to our predictions, there are three peaks in the spectrum, not two! It appears that we must now modify our model to remain consistent with these new experimental results. We note that there is a peak at 2.08 MJ/mole (as expected) and one at a much higher energy—in this case, 84.0 MJ/mole. We can safely assume that this higher energy peak corresponds to the electrons in the n = 1 shell, and that the electrons in that shell have an energy of –84.0 MJ/mole. Thus, the other two peaks must both arise from electrons in the n = 2 shell. This suggests that there are electrons with two different energies in the n = 2 shell, some with an energy of –4.68 MJ/mole, and others with the expected energy of –2.08 MJ/mole. In other words, there are two different energy levels associated with the n = 2 shell. To differentiate between them, we label them the 2s and 2p levels (or subshells), with the s designation corresponding to the lower energy level (and higher ionization energy) of the two (IE = 4.68 MJ/mole). (The labels s and p are used for historic reasons. They do not have any particular significance in the context of our model, but are used to be consistent with the designations used by contemporary practicing scientists.) The lowest energy level of a given shell is always designated as an s level, and so the electrons in the n = 1 shell are considered to be in a 1s energy level. Note that the 2s peak is approximately equal in size to that of the 1s peak, whereas the 2p peak is about three times as big. Thus, we can conclude that there are two electrons (of the eight total) in the 2s level of Ne, and six electrons in the 2p level. Our refined shell model of the Ne atom has the 10 electrons distributed in three different energy levels: 2 electrons in a 1s level, 2 electrons in a 2s level, and 6 electrons in a 2p level. At this point we will not be concerned about the details of the differences between the 2s and 2p levels. The important point is that the 2s level is slightly lower in energy than the 2p, but not by a large amount. This suggests that the electrons in both levels of the n = 2 shell are at nearly the same distance from the nucleus, and are clearly much farther from the nucleus than the electrons in the n = 1 shell. Also, we have found that there appears to be a limit of 2 on the number of electrons that can be placed in an s subshell.

Critical Thinking Question 4. What is the reasoning behind the assumption that the peak at 84.0 MJ/mole (for neon; Model 1) corresponds to the electrons in the n = 1 shell?

5. Based on the information provided by the photoelectron spectrum in Model 1, why are two of the three peaks in the spectrum of neon assigned to the n = 2 shell, rather than to the n = 1 shell?

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ChemActivity 9 The Shell Model (III)

Model 2. Energy Level Diagrams for Helium, Neon, and Argon.

Helium 1s

–2.37

Neon 2p 2s

Argon –2.08 –4.68

3p 3s

–24.1

2p

1s

–1.52 –2.83

2s

–31.5

1s

–309

–84.0

Energy levels are not to scale. Energies are given in MJ/mole.

Critical Thinking Questions 6. Based on the energy level diagram in Model 2, sketch a photoelectron spectrum for Ar. Make sure to indicate the relative intensities and positions of all peaks.

ChemActivity 9

The Shell Model (III)

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Exercises 1. a) Based on our revised shell model, how many peaks would be expected in a photoelectron spectrum of lithium? b) What would you expect the relative sizes of the peaks to be? c) The 1s ionization energies for H, He, and Li are 1.31, 2.37, and 6.26 MJ/mole, respectively. Explain this trend. d) The first ionization energies for H and Li are 1.31 and 0.52 MJ/mole, respectively. Explain why the Li first ionization energy is lower. 2. Answer Ex. 1a and 1b for beryllium and for carbon. 3. Sketch the energy level diagram (as in Model 2) for Be and for C. 4. What element do you think would give rise to the photoelectron spectrum shown below? Explain your reasoning.

Problems 1. Indicate whether each of the following statements is true or false and explain your reasoning: a) The photoelectron spectrum of Mg2+ is expected to be identical to the photoelectron spectrum of Ne. b) The photoelectron spectrum of 35Cl is identical to the photoelectron spectrum of 37Cl. 2. The energy required to remove a 1s electron from F is 67.2 MJ/mole. The energy required to remove a 1s electron from Cl is: a) 54 MJ/mole; b) 67.2 MJ/mole; c) 273 MJ/mole; d) a 1s electron cannot be removed from Cl. Explain.

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ChemActivity 10 Electron Configurations

ChemActivity

10

Electron Configurations (How Are Electrons Arranged?)

Model: Ionization Energies and Electron Configurations. Table 1.

Ionization energies (MJ/mole) for the first 18 elements. Element 1s 2s 2p 3s H 1.31 He 2.37 Li 6.26 0.52 Be 11.5 0.90 B 19.3 1.36 0.80 C 28.6 1.72 1.09 N 39.6 2.45 1.40 O 52.6 3.04 1.31 F 67.2 3.88 1.68 Ne 84.0 4.68 2.08 Na 104 6.84 3.67 0.50 Mg 126 9.07 5.31 0.74 Al 151 12.1 7.19 1.09 Si 178 15.1 10.3 1.46 P 208 18.7 13.5 1.95 S 239 22.7 16.5 2.05 Cl 273 26.8 20.2 2.44 Ar 309 31.5 24.1 2.82

Table 2.

Electron configurations of selected elements. Element Configuration 1 H 1s He 1s2 Be 1s2 2s2 C 1s2 2s22p2 Ne 1s2 2s22p6 Mg 1s2 2s22p6 3s2

3p

0.58 0.79 1.06 1.00 1.25 1.52


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Critical Thinking Questions 1. What is the first ionization energy of: a)

N?

b)

Ar?

2. Give an experimental method for obtaining the data in Table 1. 3. What information is provided by an electron configuration?

4. What is the relationship between the data in Tables 1 and 2?

5. Is it possible to deduce the electron configuration for an atom from its photoelectron spectrum? If so, describe how. If not, describe why not.

6. We will now construct two possibilities for the photoelectron spectrum of K. a)

First, consider the first 18 electrons of K, in shells n = 1, 2, and 3. For these 18 electrons, estimate the IEs [Hint: compare to Ar.] and indicate their relative intensities.

b)

If the 19th electron of K is found in the n = 4 shell, would the ionization energy be closest to 0.42, 1.4, or 2.0 MJ/mole? Explain. [Hint: compare to Na and Li.] Show a predicted photoelectron spectrum based on this assumption.

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c)

If the 19th electron of K is found in the third subshell of n = 3, would the ionization energy be closest to 0.42, 1.4, or 2.0 MJ/mole? Explain. [Hint: compare to other cases in which a new subshell appears.] Show a predicted photoelectron spectrum based on this assumption.

Exercises 1. Explain why more energy is required to remove an electron from the 1s orbital of Na (104 MJ/mole) than to remove an electron from the 1s orbital of Ne (84 MJ/mole). 2. According to the data in Table 1, would it require less than 0.50 MJ/mole, 0.50 MJ/mole, or more than 0.50 MJ/mole to remove a 3s electron from the Mg+ ion? Explain.

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ChemActivity

11

Electron Configurations and the Periodic Table (How Many Subshells Are There?)

Model 1. Simulated Photoelectron Spectrum of Potassium.

Table 1. Ionization energies (MJ/mole) for selected elements. Element 1s 2s 2p 3s 3p 3d 4s K 347 37.1 29.1 3.93 2.38 0.42 Ca 390 42.7 34.0 4.65 2.90 0.59 Sc 433 48.5 39.2 5.44 3.24 0.77 0.63

Critical Thinking Questions 1. Which of your predicted spectra from CTQ 6 of ChemActivity 10 provides the better match to the experimental spectrum, Model 1? Explain.

ChemActivity 11 Electron Configurations and the Periodic Table

49

2. Based on the analysis we have used to assign peaks in photoelectron spectra to shells and subshells in atoms, why is the peak at 0.42 MJ/mole in the K spectrum assigned to the n = 4 shell (as opposed to being another subshell of n = 3)? Refer to the data in Table 1 of ChemActivity 10: Electron Configurations.

Model 2. Simulated Photoelectron Spectrum of Scandium. (The 1s peak occurs at 433 MJ/mole and is not shown in this spectrum.)

Critical Thinking Question 3. In the photoelectron spectrum of Sc, the peak at 0.63 MJ/mole is assigned to the 4s subshell. Why is the peak at 0.77 MJ/mole in the Sc spectrum assigned as a third subshell of n = 3 (named 3d) as opposed to being a second subshell of n = 4 (that is, 4p)?

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Model 3: The Periodic Table. Note that the periodic table has an unusual form. The elements are arranged in "blocks" of columns—a block of two columns on the left, six columns on the right, and ten columns in the middle.

Critical Thinking Questions 4. What is the relationship between the form of the periodic table and the electron configurations of the elements?

5. Based on the form of the periodic table, how many electrons is the 3d subshell capable of holding?

6. Predict the electron configuration of Ga.

7. What is the common feature of the electron configurations of the elements in a given column of the periodic table?

ChemActivity 11 Electron Configurations and the Periodic Table

51

Exercises 1. Identify the element whose simulated photoelectron spectrum is shown below:

IMPORTANT NOTE: In the above spectrum, the peak which arises from the 1s electrons has been omitted. 2. Place the following in order of increasing energy to remove an electron from the 1s energy level: C

Pt

Ba

Ne

Zn

Gd

3. Make a rough sketch of the photoelectron spectrum of vanadium. Indicate the subshell that gives rise to each peak and the relative height of each peak. 4. Provide the electron configuration for: P, P3–, Ba, Ba2+, S, S2–, Ni, Zn. 5. How many valence electrons does Ga have?

Problems 1. As atomic orbitals are filled, the 6p orbitals are filled immediately after which of the following orbitals? 4f, 5d, 6s, 7s. 2+

3+

2. Provide the electron configuration for: Pd, Pd , Gd, Gd .

1

Answers to Critical Thinking Questions Please do not provide these answers to students or post them online. In some cases, the answers given here are less detailed than what might be expected for a detailed student’s answer. ChemActivity 1 1. 6, 6, 6 2. 6, 7, 7 3. 6, 6, 7 4. All carbon atoms and ions have six protons in the nucleus. 5. All hydrogen atoms and ions have one proton in the nucleus. 6. Z is the number of protons in the nucleus of that atom. 7. Twenty-eight protons in the nucleus. 8. a) Because there is a net charge on the species: there are six protons and 7 electrons. b) A neutral atom has equal numbers of protons and electrons; in an ion, the number of protons and electrons are not equal. c) The charge on an ion = # of protons – # of electrons. 9. Because every H atom must have one proton, the number of protons is 1. All of the 1H species in the model have no neutrons, so by analogy there will be 0 neutrons in this ion. Also, the number of electrons is zero because the charge is 1+ and there has to be one proton, which has a charge of +1. 10. Different isotopes on a particular element have a different number of neutrons in the nucleus (but the same number of protons). 11. The mass number, A, is the sum of the number of protons and the number of neutrons in the nucleus. 12. The O atom has 8 protons and 8 neutrons in its nucleus; hence, the mass number is 16. The O atom has 8 protons and 10 electrons; hence the charge on the ion is 2–. The Na atom has 11 protons and 12 neutrons; hence, the mass number is 23. The Na atom has 11 protons and 10 electrons; hence the charge on the ion is 1+. 13. Most of the mass is in the nucleus—where the protons and neutrons are. Note that the difference in mass between a 13C and a 13C– atom (which differ by only one electron) is only 0.0005 amu. ChemActivity 3 1. The magnitude of V decreases. 2. V = 0. 3. V must be positive. That is, V > 0 4. a) q = +1 for a proton. b) q = 0 for a neutron. c) q = +6 for the nucleus of a carbon atom. 5. V is a negative number because k(1)(–1)/d is negative. 6. V is negative because the electron is negative and the proton is positive. 7. I would expect V to be become more negative as d becomes smaller. 8. V (10–18 J) = 0, –.0462, –0.231, –0.462, –1.16, –2.31.

2 9. V = – IE 10. The electron that is closer to the nucleus would have the larger ionization energy because it is at a lower (more negative) potential energy according to Coulomb’s potential energy equation. 11. The electron that is at d1 from the +2 nucleus would have the larger ionization energy because it is at a lower (more negative) potential energy according to Coulomb’s potential energy equation. 12. The ionization energy is larger by a factor of 2. 13. He+ would have a larger ionization energy than H because q = +2 for He and it would have a lower (more negative) potential energy according to Coulomb’s potential energy equation. ChemActivity 4 1. 1.31 MJ/mole 2. The electron that is farthest from the nucleus will have the least negative potential energy (d is larger) and the lowest ionization energy. 3. Open-ended question. Students might predict that IE1 increases with atomic number (higher nuclear charge). Or, they might predict that IE1 decreases if they put additional electrons farther away from the nucleus. 4. The answer here depends on the students’ answer to CTQ 3. 5. a) The IE1 for He is greater than the IE1 for H because the nuclear charge on a He atom is

6.

7.

8.

9.

+2 whereas the nuclear charge on H is +1. b) The IE1 for Li is less than the IE1 for He because at least one of the electrons of Li must be farther away from the nucleus than any electron of He. Otherwise, the +3 nuclear charge of Li would hold the electron more tightly. Each electron is held by a +2 charge, rather than a +1 as in H. Therefore, if the electrons in He and H are at the same distance the Coulombic Potential Energy for He ought be twice the Coulombic Potential for He and IE1 for He ought to be about 2 × IE1 for H. If the 3rd electron was at the same distance as H, the IE1 would be about 3 × IE1 for H— 3.93 MJ/mole. However, it would be a bit less because of the repulsion between the electrons; the best answer is 3.6 MJ/mole. For this model the IE1 of Li would be about 3.93 MJ/mole (probably a bit less due to the repulsion between the electrons—see the previous CTQ). The IE1 of Li is 0.52 MJ/mole— much smaller than 3.93 MJ/mole (see the previous CTQ). This low value is inconsistent with the third electron being close to the nucleus. If the 3rd electron is farther away from the nucleus than the electron in H the IE1 would be

much lower that the IE1 for H. Because the IE1 of Li is 0.52 MJ/mole and the IE1 of H is 1.31 MJ/mole, this model is consistent with the experimental data. 10. The electrons in the first shell of Li are closer to the nucleus and should be harder to ionize (have a greater ionization energy) than the electron in the second shell.

3 ChemActivity 5 1. a) H, one electron b) Li, one electron c) He, two electrons 2. a) H, zero b) Li, two c) He, zero; He, zero 3. The core charge of Li is +1. 4. Electron “b” experiences more electron-electron repulsion than electron “a” because electron “a” is farther from the other electrons than electron “b”. Therefore the IE1 of 5. 6. 7. 8. 9.

10.

11. 12. 13. 14. 15.

electron “b” would be less than the IE1 of electron “a”. a) Be has 4 protons in its nucleus. b) two c) two d) 4 – 2 = 2 e) The core charge is equal to the number of valence electrons for a neutral atom. Be has a higher core charge (+2) than Li (+1) and a higher IE1 than Li (about twice as big). Ne: 10–2 = 8 Diagram (a) should have a nuclear charge of +7, two electrons in the first circle (shell), and five electrons in the second circle (shell). Diagram (b) should have a core charge of +5 and a circle with five electrons (at about the same distance as the second circle in diagram (a). a) Diagram (a) should have a nuclear charge of +11, two electrons in the first circle (shell), and nine electrons in the second circle (shell). Diagram (b) should have a core charge of +9 and a circle with nine electrons (at about the same distance as the second circle in diagram (a). b) +9 c) Because the core charge of Ne is +8 and the core charge of Na is +9 (and because the electrons are about the same distance from the nuclei) the IE1 for Na should be greater than 2.08 MJ/mole. a) Diagram (a) should have a nuclear charge of +11, two electrons in the first circle (shell), eight electrons in the second circle (shell), and one electron in the third circle (shell). Diagram (b) should have a core charge of +1 and a circle with one electron (at about the same distance as the third circle in diagram (a). b) +1 c) Because the core charge of Ne is +8 and the core charge of Na is +1 (and because the outermost Na electron is farther from the nucleus) the IE1 for Na should much less than 2.08 MJ/mole. The experimental IE1 for Na is 0.50 MJ/mole. This is consistent with the model in CTQ 10. 2, 8, 1 Both have a core charge of +1. The radius of the valence shell of Na is larger than the radius of the valence shell of Li because they both have a core charge of +1 and Na has the lower IE1. Na has a core charge of +1 and Ne has a core charge of +8. Na should have a much lower IE1. The is consistent with the experimental data.

16. IE1 increases as we move from left to right from Na to Ar because the core charge increases for +1 to +8 and the potential energy of the electron becomes more negative according to the Coulombic potential energy equation (the distance from the nucleus remaining fairly constant). 17. a) All three atoms have a core charge of +1 and all three atoms are in the first column (labeled I).

4 b) The valence shell of H is 1 and it is found in the first row of the table. The valence shell of Li is 2 and it is found in the second row. The valence shell of Na is 3 and it is found in the third row. c) For Rb, the valence shell is 5, the core charge is +1, and the number of valence electrons is 1. d) Rb has the lowest IE1 of these four elements. This is consistent with all four having one valence electron and a core charge of +1, but the valence electron of Rb is farther from the nucleus (higher numbered valence shell). 18. Diagram (a) should have a nuclear charge of +9, two electrons in the first shell, and seven electrons in the second shell. Diagram (b) should have a core charge of +7 and a shell of seven electrons. 19. a) Both atoms have a core charge of +7 and both are found in the column labeled VII. b) The valence shell of F is 2 and it is found in the second row. The valence shell of Cl is 3 and it is found in the third row. c) The IE1 of Cl is less than the IE1 of F because they both have a core charge of +7 but the valence electrons of Cl are farther from the nucleus than the valence electrons of F. 20. C is found in column IV, second row. Therefore, the core charge is +4 and the four valence electrons are found in the second shell. 21. The core charge increases as we move from left to right across a period. 22. IE1 increases as we move from left to right across a period because the core charge increases and the potential energy of the electron becomes more negative according to the Coulombic potential energy equation (the distance from the nucleus remaining fairly constant). ChemActivity 8 1. a) 6.26 MJ. b) α: -6.26 MJ/mole β: -6.26 MJ/mole γ: -0.52 MJ/mole c) α: 6.26 MJ/mole β: 6.26 MJ/mole γ: 0.52 MJ/mole 2. When comparing the energy level of two different electrons, the electron with the lower (more negative) potential energy is said to occupy the lower energy level. 3. 143.4 MJ/mole - 114.8 MJ/mole = 28.6 MJ/mole. 4. The position of a peak is determined by the ionization energy of the electron ejected. 5. 28.6 MJ/mole 6. -28.6 MJ/mole. 7. The number of electrons at that energy level. 8. Regardless of the number of electrons at this ionization energy (28.6 MJ/mole), one two, three,…) the spectrum would look the same. 9. There should be two peaks because there are electrons at two different energy levels. 10. The relative heights of the peaks will be 3 to 2. 11. The peak representing the two electrons will be at a higher ionization energy than the peak representing the three electrons because the two electrons are at a lower energy level (harder to remove). 12. a) The single peak indicates that all (one or more than one) electrons are at the same energy level.

5 b) Li has two different types of electrons (in two different shells) so there would be 2 peaks. 13. The unknown atom is He. The ionization energy matches that of He. 14. Electrons that are closer to the nucleus are harder to remove. 15. Because there are 2 electrons in the first shell and 8 electrons in the second shell. 16. The spectrum should have two peaks with a ratio of intensities of 8 (low ionization energy) to 2 (high ionization energy). ChemActivity 9 1. a) No. b) Not correct – there are three peaks rather than two. Correct – there is a peak at 2.08 MJ/mole and there is a peak at much higher energy – around 84 MJ/mole. 2. Only the relative peak height is important. For example, if the relative peak heights of two peaks are 4 to 1, the number of electrons at the two energy levels must have the same ratio: 4 to 1, or 8 to 2, or 16 to 4, … . 3. The relative peak heights are 3:1:1 (low IE to high IE). Because a Ne atom has ten electrons, the number of electrons must be 6:2:2 (low IE to high IE). 4. The 84 MJ/mole represents a high IE and the 2 electrons in the first shell of Ne would be hard to remove, much harder than for the electrons in the valence shell. 5. Because they have, relatively, low IEs. The 1s electrons have much higher ionization energies. 6. Should have 5 peaks. The relative intensities of 3p:3s;2p:2s:1s should be 3:1:3:1:1 and the peaks will appear with IEs equal to the negative of the energy levels given in the figure. ChemActivity 10 1. a) N, 1.40 MJ/mole. b) Ar, 1.52 MJ/mole. 2. Use photoelectron spectroscopy. 3. The number of electrons at each energy level. 4. Table 1 indicates how many different energy levels there are for the electrons in each element, but not how many electrons have each of those energies. Table 2 gives the number of electrons at each energy level (but not what the values of those energy levels are). 5. Yes (other than for H and He). The electrons with the highest IE will be 1s electrons. The IE of electrons in other shells will be grouped according to valence shell (the higher the n value, the lower the IE). Within the same shell, a p electron has a lower IE than an s electron. 6. a) 1s, 340 MJ/mole (higher than 309 of Ar 1s). 2s, 36 MJ/mole (higher than 31.5 of Ar 2s). 2p, 36 MJ/mole (higher than 31.5 of Ar 2p). 3s, 3.2 MJ/mole (higher than 2.8 of Ar 3s). 3p, 1.8 MJ/mole (higher than 1.52 of Ar 3p). Relative intensity, starting with 3p and working to lower IE— 6:2:6:2:2. b) Closest to 0.42 MJ/mole because the 4s electron in K should be easier to remove than the 3s electron in Na. c) Closest to 1.4 MJ/mole because the electron would be slightly easier to remove than a 3p electron in Ar. Note that Be 2s is 0.90 whereas B 2p is 0.80. Also note that Mg 3s is 0.74 whereas Al 3p is 0.58. ChemActivity 11

6 1. The experimental spectrum is similar to the predicted spectrum that has the last electron at the 4s energy level (the IE is closest to 0.42 MJ/mole). 2. The peak at 0.42 MJ/mole is assigned to the 4s energy level in K because it should be easier to remove than the 3s electron in Na (0.50 MJ/mole). If it were assigned to another subshell of n=3, the ionization energy would be expected to be about 1.4 MJ/mole, as described in CTQ 6c of CA 10. 3. If the electron was in a 4p energy level, it would be easier to remove than a 4s electron. That is, the peak would be at an IE < 0.63 MJ/mole. However it has a higher ionization energy than the 4s electron, but lower ionization energy than the 3p electron at 3.24 MJ/mole. 4. All of the elements in a group (column) have the same number of valence electrons and the same valence-shell electron configuration. Each new period (row) begins a new valence shell. 5. 10. 6. 1s2 2s2 2p63s2 3p6 4s2 3d10 4p1. 7. All of the elements in a group (column) have the same number of valence electrons and the same valence-shell electron configuration.

1

Answers to Exercises and Problems ChemActivity 1 1. A = 31, no. of e– = 15. Z = 8, A = 18. 39K+. Z = 28, no. of e– = 26. 2. 1.674 × 10–24 g. 1.993 × 10–23 g. 3. 8.67 × 10–17 g. 4. 12.00 g. 5. 7.305 × 10–23 g. 6. a) sum of protons and neutrons in the nucleus. b) number of protons in the nucleus. 7. False. 18O has 8 protons and 10 neutrons. 8. 12, 12, 12. 10, 11, 12. 17, 17, 18. 18, 17, 18. 23, 26, 30. 7, 7, 8. 10, 8, 8. 10, 13, 14. 9. 59Co2+ . Z = 7, A = 14, no. of e– = 7. 7Li . Z = 30, A = 58, no. of e– = 28. Z = 9, A = 19, no. of e– = 10. 10. All isotopes of an element have the same number of protons in the nucleus. One isotope of an element is differentiated from another isotope of the same element by the number of neutrons in the nucleus. Problems 1. Using carbon-13 and carbon-12, approx mass of neutron = 13.0034 – 12 = 1.0034 amu. approx mass of 14C = 13.0034 + 1.0034 = 14.0068 amu. – 2. a) Using 14C and 14C , mass of electron is approximately 13.0039 amu – 13.0034 amu = 0.0005 amu 1 b) Using H, mass of proton is approximately 1.0078 amu – 0.0005 amu = 1.0073 amu 1

3.

2

c) Using H and H, mass of neutron is approximately 2.0140 amu – 1.0078 amu = 1.0062 amu (Note that slightly different values for the masses of protons and neutrons will be obtained if different elements/isotopes are used to calculate these masses.) 12 The calculated mass of C based on the masses of the constituent particles is 12.099 amu; and the actual mass of

12

C is exactly 12 amu.

ChemActivity 3 1. 5.47 × 10–18 J. 2. a) IEa = –(2)(–1)/d1 = 2/d1 b) IEb = –(1)(–1)/2d1 = 1/2d1

IEa > IEb

3. The ionization energy of case (a) is larger, 1.20 k/d1, than that of case (b), 1.17 k/d1. Problems 1. large and negative (+2)(–1)k + 2. V = 300pm

(+2)(–1)k 400pm

+

(–1)(–1)k 700pm

2 ChemActivity 4 1. No. The ionization energy of He would be about 4× (twice the charge and half the distance) the ionization energy of H if this were the case. 2. Several possible answers. A typical one would have all three electrons at a much farther distance than in H from the nucleus – at least six or seven times as far. Problem 1.

a)

V=

(+2)(−1)k d

+

(+2)(−1)k d

+

(−1)(−1)k 2d

b) The IE of He is slightly less than twice the IE of H because the electron-electron repulsion makes the potential energy more positive. Note that the first two terms in 1a) are negative and the third term is positive. ChemActivity 5 1. a) 4. b) 6. c) 5. d) 8. 2. a) +4. b) +6. c) +5. d) +8. 3. The IE of Br should be less than the IE of Cl. There is about a 0.4 MJ/mole difference between the IEs of F and Cl. Prediction: Br, 0.8 MJ/mole. 4. The IE of Li+ should be larger than the IE of He because both atoms have 2 electrons in the 1st shell and Li+ has a core charge of +3 whereas He only has a core charge of +2. 5. The IE of F– should be less than the IE of Ne because both atoms have eight electrons in the 2nd shell and F– has a core charge of +7 whereas Ne has a core charge of +8. 6. IE of Kr > IE of Br because they are in the same valence shell and Kr has the higher core charge (+8 vs. +7). IE of Rb is the lowest because core charge is +1 and its valence shell (n = 5) is larger than the valence shell (n = 4) of Kr and Br. 7. One of the inner shell electrons is harder to remove because it is closer to the nucleus. Problems 1. a) True. Br is a group VII element. The number of valence electrons is seven. b) True. The first ionization energy increases as one moves from left to right across a period and as one moves up a group. 2. If the fourth electron in Be were added to a third shell, it would be easier to remove and the IE would be less than the IE of Li.

3 ChemActivity 8 1. 140.3 MJ/mole 2. a) –0.52 E (M/Jmole)

–6.26

(b)

6.26 0.52

8

4

0

Ionization Energy (MJ/mole)

3.

Ionization Energy (MJ/mole)

4 ChemActivity 9 1. a) Two. b) Lower energy peak (1s) is 2 x the intensity of the higher energy peak (2s). c) The nuclear charge for H, He, and Li is 1, 2, and 3, respectively. Therefore, the electrons in the first shell will be held most tightly by Li and least tightly by H. d) H and Li have the same core charge; the electron is farther away in Li. Therefore, Li will hold its valence electron less tightly than H. 2. Be. Two peaks. Both peaks have the same intensity. C. Three peaks. All three peaks have the same intensity. 3.

Be

C 2p

2s

2s

1s

1s

4. Mg. Two electrons in the 1s. Two electrons in the 2s. Six electrons in the 2p. Two electrons in the 3s. Problems 1. a) False. Both have 10 electrons. The number of peaks and the relative intensities will be the same, but the IEs of Mg2+ will be greater than the equivalent IEs of Ne. b) True. Both have 17 electrons and 17 protons. 2. 273 MJ/mole. The energy required to remove an electron from the 1s of Cl must be much higher than the energy required to remove an electron from the 1s of F because Cl has 17 protons in its nucleus and F only has 9 protons in its nucleus.

5 ChemActivity 10 1. 2.

Na has 11 protons in its nucleus and Ne only has 10 protons. Therefore, the 1s electrons will be held more tightly in Na. +1 It would require more than 0.50 MJ/mole because Mg and Na are isoelectronic and Mg has an additional electron in its nucleus.

ChemActivity 11

3.

Relative Number of Electrons

1. Kr 2. C
2s

3p 3s

4s 3d

500

60 Ionization Energy (MJ/mole)

20

0

4. P: [Ne] 3s2 3p3 P3–: [Ar] Ba: [Xe] 6s2 Ba2+: [Xe] S2–: [Ar] Ni: [Ar] 4s2 3d8 Zn: [Ar] 4s2 3d10 5. three

S: [Ne] 3s2 3p4

Problems 1. 5d 2 8 2+ 8 2 1 7 2 8 2. Pd [Kr] 5s 4d . Pd [Kr] 4d . Gd either [Xe] 6s 5d 4f or [Xe] 6s 4f is correct (depending on how you use the periodic table to determine electron configurations; 2 1 7 3+ 7 experimentally, we find the configuration to be [Xe] 6s 5d 4f ). Gd [Xe] 4f (regardless of your answer for Gd!).

Chemistry: A Guided Inquiry 5e

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