Answer Key ALE 7. Ionization Energies & Electron Affinities (Reference: Sections 8.4 - Silberberg 5 th edition) How can one use the Periodic Table to make predictions of atomic properties? The Model: First Ionization Energies The first ionization energy (IE 1 ) is the amount of energy that must be added to 1 mole of gas-phase atoms (X) to yield 1 mole of gas-phase cations and 1 mole of gas-phase electrons: X(g) + IE 1 X + (g) + e - (g) We can understand the periodic trends in ionization energy through the use of Coulomb s Law and what we already know (e.g., relative atomic radii) about the atoms we re comparing. Any deviations in the overall periodic trend in ionization energy can usually be explained in terms of the electron configurations of the atoms we re comparing. Key Questions 1. Suppose we re comparing two atoms, A and B. The force of attraction between the nucleus of A and A s outermost electron is stronger than the force of attraction between the nucleus of B and B s outermost electron. Which element, is going to have the greater first ionization energy (i.e., which element requires more energy to ionize 1 mole of gas-phase atoms): A or B? Circle the correct answer and then briefly justify your answer. The stronger the force of attraction between the outermost electron and the nucleus, the harder it will be for the electron to be removed. Therefore it will take more energy to remove the electron from the atom with the stronger force of attraction for its outmost electron. 2a. When comparing elements within the same group, we ve already said that the atoms will have basically the same effective nuclear charge. But the atomic radii of two elements in the same group will not be the same. Which element will have the smaller atomic radius: an element toward the top or one toward the bottom of a column of the Periodic Table? (Circle your answer.) b. According to Coulomb s Law, what happens to the force of attraction between the outermost electron and the nucleus when the distance between the nucleus and that electron decreases? Does the force decrease or increase in strength? (Circle your answer.) c. Which element will have the greater first ionization energy: an element toward the top or one toward the bottom of a column of the Periodic Table? (Circle your answer.) 3a. According to Coulomb s Law, does the force of attraction between the outermost electron and the nucleus decrease or increase when the effective nuclear charge increases? (Circle your answer.) b. Which element will have the smaller atomic radius: an element toward the left or one toward the right of a period of the Periodic Table? (Circle your answer.) c. Which element will have a greater effective nuclear charge: an element toward the left or one toward the right of a period of the Periodic Table? (Circle your answer.) d. Which element will have the greater first ionization energy: an element toward the left or one toward the right of a period of the Periodic Table? (Hint: Look at your answers to Questions 2b and 3a-c.) (Circle your answer.) Page 1 of 6
4a. Hund s Rule says that when there is more than one equal-energy orbital into which an electron may go, the electrons should be placed in their own orbital first before being paired up i.e. the most stable arrangement of electrons occurs when they're all unpaired within a sublevel. Therefore, the electron configurations of Carbon, Nitrogen, and Oxygen expressed as abbreviated orbital diagrams are shown to the right. Circle the highest-energy electron (i.e., the one that is removed when IE 1 is added) in each of the orbital diagrams. b. The 1 st ionization energies of Carbon, Nitrogen, and Oxygen are 1086 kj/mol, 1400 kj/mol, and 1314 kj/mol, respectively. We can think of the deviation in the periodic trend being that Nitrogen has too great (rather than Oxygen has too small ) an ionization energy. What does this say about the stability of an atom having a p subshell that is half-filled? (Such can also be said about atoms that have a half-filled d or f subshell.) Having a half-filled p subshell grants an atom an additional amount of stability. This is evidenced by Nitrogen having a half-filled 2p subshell and a larger ionization energy than Oxygen. 5a. The 1 st ionization energies of Copper, Zinc, and Gallium are 745 kj/mol, 906 kj/mol, and 579 kj/mol. Draw arrows to represent electrons ( for spin up, for spin down, for a pair of electrons in the same orbital) to complete the electron configurations (orbital diagrams) of the three neutral atoms. (Hint: Because a completely-filled d subshell grants an atom a significant amount of stability, the electron configuration of ground state Copper is not what you d expect it to be based on the Periodic Table.) Cu: [Ar] 4s 3d 4p Zn: [Ar] 4s 3d 4p Ga: [Ar] 4s 3d 4p b. Circle which electrons are removed when each of the atoms is ionized. c. Explain the deviation in the trend, bearing in mind that having a completely-filled subshell grants an atom an additional measure of stability. The general trend in first ionization energies is that the ionization energy increases going from the left to the right in a period of the Periodic Table. Therefore, we might expect Zinc to have a lower first ionization energy than Gallium. However, Zinc s IE 1 is greater than that of Gallium. The reason why is that both the 4s and the 3d subshells of Zinc are filled the more filled subshells that an atom has, the more stable it is. The electron that is to be removed from Gallium is the single electron in the 4p subshell. When this electron is removed, Gallium achieves having an electron configuration such that all subshells are completely filled, and thus its IE 1 is lower than what might be expected. Page 2 of 6
The Model: Electron Affinity The electron affinity (EA) is the amount of energy that is released when 1 mole of gas-phase atoms (X) accepts 1 mole of gas-phase electrons to become 1 mole of gas-phase anions: X(g) + e - (g) X - (g) + EA When comparing two elements on the Periodic Table, we will want to keep in mind Coulomb s Law as well as which atom will have the smaller ionic radius and which atom will have the greatest effective nuclear charge once the newly-added electron becomes part of the atom. Key Questions 6. It is an expectation that electron affinities are exothermic quantities. Explain why, most of the time, heat is released when a gas-phase atom accepts an electron. (Hint: The nucleus of the atom has protons. What kind of force, attractive or repulsive, exists between the nucleus of the atom and the newly-added electron? When that electron is added to the atom, will the species be stabilized or destabilized?) Opposite charges attract, so the protons in the nucleus are attracted to the newlyadded electron. When that electron is added to the atom, the system is stabilized. As a result of the final state of the system (combined nucleus with positively-charged protons and the bound electron) being more stable than the initial state (separated nucleus and free electron), heat energy is released. When heat energy is released from a system into its surroundings, we label the heat exchange exothermic. 7. Suppose we re comparing two atoms, A and B, which are both accepting electrons. The force of attraction between the nucleus of A and A s newly-added outermost electron is stronger than the force of attraction between the nucleus of B and B s newly-added outermost electron. Which element, is going to have the greater electron affinity (i.e., which element expels more energy when 1 mole of gas-phase atoms accepts 1 mole of gas-phase electrons): A or B? Circle the correct answer and then briefly justify your answer. The stronger the force of attraction between a nucleus and the newly-added electron, the more the system is stabilized and the more heat energy is released when the electron is added to the atom. Since the force of attraction between the nucleus of A and A s newly-added electron is stronger, A s EA will be greater. 8. We already know that an atom toward the bottom of a column on the Periodic Table is larger than an atom toward the top of that column. Suppose we re comparing two atoms, A and B, which are both accepting electrons. A is an atom toward the top of a column on the Periodic Table and B is an atom toward the bottom of that column. a. Which species, A - or B -, is going to have the smaller ionic radius? (Circle your answer.) b. According to Coulomb s Law, what happens to the force of attraction between the newly-added outermost electron and the nucleus when the distance between the nucleus and that electron decreases? Does the force decrease or increase in strength? (Circle your answer.) c. Which element will have the greater electron affinity: an element toward the top or one toward the bottom of a column of the Periodic Table? (Circle your answer.) Page 3 of 6
9a. According to Coulomb s Law, does the force of attraction between the newly-added outermost electron and the nucleus decrease or increase when the effective nuclear charge increases? (Circle your answer.) b. Which element will have the smaller atomic radius: an element toward the left or one toward the right of a period of the Periodic Table? (Circle your answer.) c. Which element will have a greater effective nuclear charge: an element toward the left or one toward the right of a period of the Periodic Table? (Circle your answer.) d. Which element will have the greater electron affinity: an element toward the left or one toward the right of a period of the Periodic Table? (Circle your answer.) 10. While most electron affinities are exothermic, all of the electron affinities for all of the nobel gases are endothermic. Explain why. The noble gases have electron configurations with completely-filled subshells and in some cases filled energy levels (e.g. He and Ne). In short, since noble gases have a stable electron configuration a newly added electron destabilizes the atom, thus the need to use energy to add another electron hence noble gases have an endothermic EA. In more detail: A newly-added electron is put into a new shell of considerably higher energy than the shells that are already occupied. The core electrons of the lower energy levels shield the nucleus charge from the newly-added electron so newly-added electron doesn t efficiently experience the positive charge in the nucleus. When an electron is added to a neutral noble gas atom, the system is destabilized, and therefore requires an input of energy. Heat energy can cause both the atoms of noble gas and the gas-phase electrons to move so chaotically that they are forced to collide and combine. But without this addition of heat energy the noble gas atom will not accept another electron. 11. A general trend in electron affinity is that as one goes from the left to the right on a period of the Periodic Table the electron affinity increases as illustrated in the table to the right. However, it is consistently seen in the 2 nd through the 5 th periods that as one goes from Group 3A to Group 4A the electron affinity increases dramatically, then electron affinity decreases in going from Group 4A to Group 5A. Use orbital diagrams to explain this apparent deviation in the general trend. Consider the elements of the 2 nd period of the Periodic Table. Boron and Carbon have the following electron configurations as neutral atoms and as anions with 1- charges: B: [He] C: [He] 2s 2p 2s 2p B - : [He] C - : [He] 2s 2p 2s 2p Electron affinities (in kj/mol) of the main group elements. Carbon achieves a half-filled 2p subshell and, thus, achieves an additional amount of stability. This additional stability is evidenced by the extra exothermic electron affinity of Carbon. On the other hand, neutral Nitrogen starts off with a half-filled 2p subshell, and it must be forced to accept a new electron (the electron affinity of Nitrogen is endothermic). Page 4 of 6
12. Another general trend in electron affinity is that as one goes from the top to the bottom of a column of the Periodic Table the electron affinity decreases. However, it is consistently seen in Group IIIA through Group VIIA that as one goes from the 2 nd to the 3 rd period the electron affinity increases (see the figure with question #11 on p. 4). Explain this apparent deviation in the general trend, taking into account that elements in the same group are isoelectronic and also that an element in the 3 rd period is considerably larger than the corresponding 2 nd -period element of the same group. (Hint: think of the electron electron repulsion and the relative sizes (volumes) of the atomic orbitals involved.) Both the 2 nd -period element and the 3 rd -period element have the same number of electrons in the valence shell. (For example, both F - and Cl - have ns 2 np 6 electron configurations.) But the 3 rd -period element is much larger than the 2 nd -period element in the same family. (For example, the ionic radii of F - and Cl - are 133 pm and 181 pm, respectively.) Electrons are all negatively charged, and they repel each other. So the valence shells of the 2 nd -period and 3 rd -period atoms have the same number of electrons, but the 2 nd -period elements have a smaller volume in which those electrons are found, making electron-electron repulsions greater. With more electron-electron repulsions comes less stabilization when the new electron is added to the 2 nd -period atom. And with less stabilization there will be less heat energy released when the 2 nd -period atom accepts a new electron (compared to the amount of heat energy released when the larger 3 rd -period atom accepts its new electron). Exercises 13. Problem 8.49 (modified): In a plot of IE 1 for the Period 2 and Period 3 elements (see the figure to the right), why do the values for elements in Groups 3A and 6A drop slightly below the generally increasing trend? Explain in terms of the electronic structure of the relevant elements. (i.e. For period 2: Be vs. B and O vs. N or for period 3: Mg vs. Al and P vs. S). Periodicity of first ionization energy, IE 1 The first drop occurs because the 3p sublevel is higher in energy than the 3s, so the 3p electron of Al is pulled off more easily than a 3s electron of Mg. The second drop occurs because the 3p 4 electron occupies the same orbital as another 3p electron. The resulting electron-electron repulsion raises the orbital energy and thus it is easier to remove an electron from S (3p 4 ) than P (3p 3 ). Page 5 of 6
14. Problem 8.51: a.) The EA 2 of an oxygen atom is positive, even though its EA 1 is negative. Explain why this change in sign occurs. This can be seen in the electron affinity values for oxygen. EA 1 for oxygen is negative because energy is released when an electron is added to the neutral atom due to its attraction to the atom's nuclear charge. The EA 2 for oxygen is positive. The second electron affinity is always positive (greater energy) because it requires energy to add a (negative) electron to a (negative) anion. b.) Which other elements exhibit a positive EA 2. Explain. The second electron affinity is always positive (greater energy) because it requires energy to add a (negative) electron to a (negative) anion. 15. Problem 8.60: Circle the element in each of the following sets would you expect to have the lowest third ionization energy, IE 3? Briefly explain your reasoning. a.) Na, Mg, Al Reasoning: Core electrons are removed from Na and Mg, while easily removed valence electrons from Al b.) K, Ca, Sc Reasoning: Core electrons are removed from K and Ca, while easily removed outer shell electrons from Sc c.) Li, Al, B Reasoning: Core electrons are removed from Li and B, while easily removed valence electrons from Al Page 6 of 6