Question

ELECTRON CONFIGURATIONS andEFFECTIVE NUCLEAR CHARGE (Zeff) INTRODUCTION This assignment is designed to accompany lecture and text regarding how properties of the elements depend on their electron configurations. The questions are intended to help you reach a higher level of mastery in using the quantum mechanics model. This assignment will count in the miscellaneous category of your grade for 25 points. We will work on it together as a class during lecture on Wednesday, March 1, then you will have time to work on it at home. It will be due Monday, March 6. The principles and results of quantum mechanics are generally far removed from our sensory experience and thus present a challenge to our understanding. However, we can gain confidence in the usefulness of the theory when we recognize that it allows us to predict or explain many properties of atoms that can be confirmed by measurement. Among these properties are charges of common ions, ratios in which elements combine to form compounds, atomic radius, ionization energy, electron affinity, paramagnetism, and atomic and molecular spectra. We can give reasonable explanations for many of these properties by making use of a small number of fundamental principles that come from the quantum mechanics model. The first is that we can use the aufbau principle, Hund’s rule, and predictions of the energy level ranking of orbitals to create an electron configuration of an atom. The electron configuration essentially summarizes a model of the atom in terms of the distribution of the mass and charge of electrons about the nucleus. The second is that we can analyze the net attraction of any electron for the nucleus (hence its motivation to stay in the atom) as the net effect of four factors: a) the electrostatic attraction between the electron and the nucleus, b) the average distance of the electron from the nucleus, c) The electrostatic repulsion caused by the presence of other electrons in the atomic system, and d) the magnetic interaction lowering electron energy when other electrons have like spin and raising the energy when others have opposite spin. One way to combine these factors is to start with the charge of the protons in the nucleus (a) and think of the other factors (b-d) as modifying it to give an effective nuclear charge. The third principle to consider is that most of the properties of an atom depend primarily on only the electrons at greatest average distance from the nucleus, i.e., those electrons sharing the highest principle quantum number present. These are referred to as the valence or outer-shell electrons. If we determine which electrons these are from an electron configuration and consider the strength of the effective nuclear charge at their average distance, we can make meaningful comments about how those electrons might behave in determining atomic properties. The text (pp. 306-307) and perhaps your instructor have referred to the concept of effective nuclear charge. It is usually discussed qualitatively, without attempting to assign a numerical value to it. In advanced texts, various methods have been proposed to give approximate values from calculations. Strictly for the sake of this exercise, let us adopt an approximating system. It may appear rather clumsy at first, and it is not accurate for elements in general. However, it does give relatively good estimates for elements of low atomic number and only requires simple arithmetic. The estimate of the effective nuclear charge, Zeff, as felt by the highest energy valence electron in an atom will be calculated as follows. HOW TO DETERMINE Zeff Determine which is the highest energy valence electron. It should have or share the highest principle quantum number, n, and of those at level n, it should have the highest azimuthal quantum number, l. If several electrons fit that description (i.e., there are several in the highest-energy valence subshell), choose an electron whose spin quantum number is in the minority. This will be the electron whose point of view we will take. Since an electron does not attract or repel itself, do not include this electron in the accounting below. i) Start with the number of protons in the nucleus, this is the atomic number, Z. ii) Subtract 0.95 times the number of electrons at lower principle quantum numbers than the one we are focusing on. These are inner-shell electrons. They are the most efficient at screening outer electrons from the nuclear attraction. iii) Subtract 0.85 times the number of electrons at the same principle quantum number, but with lower azimuthal quantum number, l (i.e., those at a lower energy subshell). iv) Subtract 0.75 times the number of electrons in the same subshell. Remember not to count the electron whose point of view we are taking. v) Add 0.1 for each other electron in the same subshell with the same spin as the one we are considering. vi) Subtract 0.1 for each other electron with the opposite spin. vii) Add 0.05 is this last subshell is full. In the two examples below, note the double arrow indicates a highest energy electron and that the steps are applied with regard to the other electrons. Example 1. Beryllium, Be 1s  2s  Z = 4 (atomic no.) - 0.95 × 2 (the 1s electrons are “inner shell” relative to n = 2) - 0.85 × 0 (no lower subshell than the 2s at n = 2) - 0.75 × 1 (the other 2s electron, same subshell) + 0.10 × 0 (no other 2s electron with same spin) - 0.10 × 1 (the other 2s electron has opposite spin) + 0.05 (last subshell, 2s, is full) Zeff = 1.3 Example 2. Fluorine, F 1s  2s  2p    Z = 9 (atomic no.) - 0.95 × 2 (the 1s electrons are “inner shell” relative to n = 2) - 0.85 × 2 (the 2s is a lower subshell than the 2p at n = 2) - 0.75 × 4 (the other 2p electrons, same subshell) + 0.10 × 1 (one other 2p electron with same spin) - 0.10 × 3 (the other 2p electrons have opposite spin) + 0.00 (last subshell, 2p, is not full) Zeff = 2.2 What can this comparison tell us? Both Be and F have electrons at n = 2. If all else was equal, that would suggest the same radius. But since the outer electrons of F experience a greater net attraction to the nucleus, F is actually smaller in radius. The smaller average distance and greater net attractive charge makes it require more energy to remove an electron from F, i.e., F has higher first ionization energy. If we add a new electron to each atom and perform the Zeff calculation again we can compare the attraction to the electron of highest energy in each ion, Be- and F-. The values are Zeff = 0.4 for Be- and Zeff = 1.6 for F-. This suggests that the added electron in F-would be more strongly attracted, so F has a more negative electron affinity (more energy is released upon adding an electron to a neutral F atom). Try adding another electron to F-. The eleventh electron would have to go to the 3s orbital making the other ten electrons inner shell. The value of Zeff for the outer electron would be -0.5. This explains why fluorine always takes on a -1 charge in its ionic compounds, not -2. This brief analysis should suggest how much information can be explained by analyzing electron configurations. The report for this experiment asks you to consider some comparisons and give reasonable explanations for the observations based on electron configurations. The exercise should help you to become more comfortable with the terms and symbols of quantum mechanics as well as clarifying the trends in properties observed for elements in rows and columns of the periodic table. In explaining the comparisons, you may sometimes find that there is conflicting information. If two elements or ions have the same Zeff, a higher principle quantum number would indicate greater average distance from the nucleus and thus a weaker force of attraction. If two elements have the same principle quantum number for the outer shell, then a greater Zeff would indicate a stronger attraction of electrons for the nucleus. But what if you are comparing two elements where one has a greater value for n and a higher Zeff? Based on the information we have available, we can not decide, from theory, which factor would be more influential. For this reason, the report questions will not ask you to predict an unknown comparison of properties. Instead, we merely need to explain a comparison whose outcome we already know from experimental information. ANSWERING ASSIGNMENT QUESTIONS In answering the questions, bring all of the tools we have discussed to bear on the issue. Show electron configurations, use Zeff calculations from class, compare energy levels and principle quantum numbers, etc. Decide what factors support the comparison and what (if any) factors work against it. Since we know the conclusions, you can state which must have been the more significant factors. You may refer to the text for assistance but do not merely repeat what is written there. Summarize the rationale for each item to prepare for writing the assignment. The assignment should be done on separate paper; it can be typed or handwritten. If possible, please type at least the written portion of your answers, at 1.5 spacing. Use complete sentences in paragraph form. When you include diagrams or show calculations, organize them separately so that they do not break up the continuity of the written arguments. You should label your diagrams and calculations so that they can be referenced in your written arguments. Be sure to label each of the ten responses. For species that appear in more than one question, you do not need to repeat diagrams and Zeff calculations, but do re-reference the prior figures and calculations.

ASSIGNMENT QUESTIONS

1. ATOMIC RADIUS. There are two trends in radius associated with the A-group elements of the periodic table: increase in radius for elements lower in columns and decrease in radius from left to right across a row. As examples, explain why: a) Sodium has a larger radius than lithium. b) Magnesium has a smaller radius than sodium. In questions 2-4, you may use radius comparisons as given information.

2. IONIZATION ENERGY. Ionization energy refers to energy required to remove an outer electron from an isolated atom. The energy tends to be less for elements with weaker attractions to the nucleus or higher initial energy levels. The general trend is for lower energy for elements lower in columns and higher energy from left to right across rows of the periodic table. There are some notable exceptions. Explain why: a) Lithium has higher ionization energy than sodium. b) Fluorine has higher ionization energy than boron. c) Oxygen has lower ionization energy than nitrogen.

3. ELECTRON AFFINITY. The electron affinity is the energy change occurring when an electron is added to an isolated atom. Values generally become more negative (exothermic) from left to right across rows and less negative for elements lower in columns of the table. Both trends have many exceptions. Explain why: a) The electron affinity is more negative for fluorine than for oxygen. b) The electron affinity is less negative for sodium than for lithium. c) The electron affinity is less negative for nitrogen than for carbon.

4. COMBINING RATIOS. Use an analysis of the electron configurations to explain in detail why magnesium and fluorine would make a compound with the formula MgF2. Explain first the signs and then the values of the oxidation numbers we would assign. 3 / 5

Answer 1

Q.2: (a): Due to the addition of extra shells down the group, the distance between the nucleus and valance shell electron increases. Hence force of attraction beteen the nucleus and valance shell electron decreases and less energy is required to remove electron as we move down the group. Hence Li (2nd group) has higher energy than Na (3rd group).

(b): Boron is prsent on left side in 2nd period and fluorine in present on extreme right in 2nd period. As we move from left to right in a period, the distance between the nucleus and the valance shell electron decreases. This increases the force of attraction between nucleus and valance shell electron and hence more elecgy is required to remove electron from valance shell.

Hence fluorine has higher ionization energy than boron.

(c): Although oxygen is present on extreme right in 2nd period incomparison to nitrogen, the later has more stable half-filled p-orbital configuration. Hence more energy is required to remove an electron by breaking the half-filled configuration.

N(7) --> 1s2, 2s2, 2p_x¹2p_y¹2p_z¹  (stable half-filled p-orbital configuration)

Hence Nitrogen has higher ionization energy than Oxygen.

Q.3: (a): Because after receiving an electron fluorine acquires the nearest stable inert gas (Ne) configuration, where as oxygen doesn't. Hence electron afinity is more negative for fluorine than oxygen.

(b): When we add an electron to sodium, it goes to a higher energy shell in comparison to lithium. Hence the added electron in sodium is less strongly bound to nucleus. Hence the electron affinity of sodium is less negative than that of lithium.

(c): Nitrogen atom has already stable half-filled p-orbital configuration. When we add an electron to nitrogen atom we disturb the stable half-filled p-orbital configuration. Hence the electron gets a strong repulsion. Hence the electron affinity of Nitrogen is less negative.