ELECTRON CONFIGURATIONS andEFFECTIVE NUCLEAR CHARGE (Zeff) INTRODUCTION This ass
ID: 489503 • Letter: E
Question
ELECTRON CONFIGURATIONS andEFFECTIVE NUCLEAR CHARGE (Zeff)
INTRODUCTION
This assignment is designed to accompany lecture and text regarding how properties ofthe elements depend on their electron configurations. The questions are intended tohelp you reach a higher level of mastery in using the quantum mechanics model. Thisassignment will count in the miscellaneous category of your grade for 25 points. Wewill 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 oursensory experience and thus present a challenge to our understanding. However, wecan gain confidence in the usefulness of the theory when we recognize that it allows usto 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 combineto 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 asmall 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 theenergy level ranking of orbitals to create an electron configuration of an atom. Theelectron configuration essentially summarizes a model of the atom in terms of thedistribution 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) theelectrostatic attraction between the electron and the nucleus, b) the average distance ofthe electron from the nucleus, c) The electrostatic repulsion caused by the presence ofother electrons in the atomic system, and d) the magnetic interaction lowering electronenergy when other electrons have like spin and raising the energy when others haveopposite spin. One way to combine these factors is to start with the charge of theprotons in the nucleus (a) and think of the other factors (b-d) as modifying it to give aneffective nuclear charge. The third principle to consider is that most of the properties of an atom dependprimarily on only the electrons at greatest average distance from the nucleus, i.e., thoseelectrons 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 froman electron configuration and consider the strength of the effective nuclear charge attheir average distance, we can make meaningful comments about how those electronsmight behave in determining atomic properties.The text (pp. 306-307) and perhaps your instructor have referred to the concept ofeffective nuclear charge. It is usually discussed qualitatively, without attempting toassign a numerical value to it. In advanced texts, various methods have been proposedto give approximate values from calculations. Strictly for the sake of this exercise, let usadopt an approximating system. It may appear rather clumsy at first, and it is notaccurate for elements in general. However, it does give relatively good estimates forelements of low atomic number and only requires simple arithmetic. The estimate ofthe effective nuclear charge, Zeff, as felt by the highest energy valence electron in an atomwill be calculated as follows.
HOW TO DETERMINE Zeff
Determine which is the highest energy valence electron. It should have or share thehighest principle quantum number, n, and of those at level n, it should have the highestazimuthal quantum number, l. If several electrons fit that description (i.e., there areseveral in the highest-energy valence subshell), choose an electron whose spin quantumnumber 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 theaccounting 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 numbersthan the one we are focusing on. These are inner-shell electrons. They are the mostefficient at screening outer electrons from the nuclear attraction.iii) Subtract 0.85 times the number of electrons at the same principle quantumnumber, but with lower azimuthal quantum number, l (i.e., those at a lower energysubshell).iv) Subtract 0.75 times the number of electrons in the same subshell. Remember not tocount 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 onewe 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 electronand 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 wasequal, that would suggest the same radius. But since the outer electrons of F experiencea greater net attraction to the nucleus, F is actually smaller in radius. The smalleraverage distance and greater net attractive charge makes it require more energy toremove 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 cancompare the attraction to the electron of highest energy in each ion, Be- and F-. Thevalues 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 (moreenergy 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 3sorbital making the other ten electrons inner shell. The value of Zeff for the outer electronwould be -0.5. This explains why fluorine always takes on a -1 charge in its ioniccompounds, not -2.
This brief analysis should suggest how much information can be explained by analyzingelectron configurations. The report for this experiment asks you to consider somecomparisons and give reasonable explanations for the observations based on electronconfigurations. The exercise should help you to become more comfortable with theterms and symbols of quantum mechanics as well as clarifying the trends in propertiesobserved for elements in rows and columns of the periodic table. In explaining the comparisons, you may sometimes find that there is conflictinginformation. If two elements or ions have the same Zeff, a higher principle quantumnumber would indicate greater average distance from the nucleus and thus a weakerforce of attraction. If two elements have the same principle quantum number for theouter shell, then a greater Zeff would indicate a stronger attraction of electrons for thenucleus. But what if you are comparing two elements where one has a greater value forn and a higher Zeff? Based on the information we have available, we can not decide, fromtheory, which factor would be more influential. For this reason, the report questionswill not ask you to predict an unknown comparison of properties. Instead, we merelyneed to explain a comparison whose outcome we already know from experimentalinformation.
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 andprinciple 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 whichmust have been the more significant factors. You may refer to the text for assistance butdo not merely repeat what is written there. Summarize the rationale for each item toprepare for writing the assignment. The assignment should be done on separate paper; it can be typed or handwritten. Ifpossible, please type at least the written portion of your answers, at 1.5 spacing. Usecomplete sentences in paragraph form. When you include diagrams or showcalculations, organize them separately so that they do not break up the continuity of thewritten arguments. You should label your diagrams and calculations so that they can bereferenced in your written arguments. Be sure to label each of the ten responses. Forspecies that appear in more than one question, you do not need to repeat diagrams andZeff 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-groupelements of the periodic table: increase in radius for elements lower in columns anddecrease 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 anouter electron from an isolated atom. The energy tends to be less for elements withweaker attractions to the nucleus or higher initial energy levels. The general trend isfor lower energy for elements lower in columns and higher energy from left to rightacross 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 whenan 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 incolumns 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 indetail 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
Explanation / Answer
1. a) Sodium has a larger radius than lithium: Both lithium and sodium are alkali metals which belong to the same group. As we move down from Lithium to sodium, the number of electrons and consequently the energy levels increases. Each subsequent energy level is far away from the nucleus than the previous one. hence, the atomic radius of sodium is more than lithium.
b) Magnesium has a smaller radius than sodium: As we move from sodium to magnesium across the same period a valence electrons is added to the same energy level. Consequently, the number of proton also increases increasing the nuclear charge. Thus, electrons are pulled more strongly in magnesium resulting in a smaller radius as compared to sodium.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.