Electron Configuration

Electron Configuration - Electron Configuration: Self-Study...

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Unformatted text preview: Electron Configuration: Self-Study Assignment You will have a QUIZ on the attached pages on _____________________ . Your assignment is: READ the pages attached. WORK the examples in the lesson. Complete the pages as homework. To work the examples, • use a sheet of paper to cover below the * * * * * line, • try the problem on your paper, • then check your answer below the * * * * * line. Start early. This assignment will require 2-4 hours of work outside of class. Module 23 — Electron Configuration Module 23 — Electron Configuration Pretests: If you believe that you know the material in a lesson, try two problems at the end of the lesson. If you can do those calculations, you may skip the lesson. ***** Lesson 23A: The Multi-Electron Atom Orbitals For the Other Atoms Schrődinger’s wave equation predicts mathematically the observed behaviors of the hydrogen atom. However, for atoms with more than one electron, the wave equation provides less exact predictions for how the atom will behave. Why? In the case of hydrogen, one proton and one electron attract. Mathematics is able to precisely model the forces in this “two body” problem, but if a second electron is added to the atom, the situation is more complex. Because the protons are tightly packed into the nucleus, nuclei with more than one proton behave as a single point of positive charge. Multiple electrons will be attracted to those protons, but unlike the case of hydrogen with one electron, two or more electrons also repel each other. How much will they repel? It depends on the types of orbitals that the electrons occupy. At n = 2, an electron in the 2s orbital will on average be closer to the nucleus than an electron in a 2p orbital. This closer 2s electron, by repelling the 2p electron, will act to shield the 2p electron slightly from the attraction of the protons in the nucleus. This means that the 2p orbital will be slightly higher in energy than the 2s orbital. The result of these factors is • the wave equation predicts qualitatively, but not exactly, how electrons in atoms other than hydrogen will behave, and • the orbital diagram for multi-electron atoms will be different from that of hydrogen. In neutral atoms other than hydrogen, the orbitals have different energy values in each atom, but the orbitals are nearly always arranged in the same order. This means that for all of the elements in the periodic table, there will be only two types of energy level diagrams to learn: one for hydrogen, and one that works in most cases for all of the other atoms. The Orbital Diagram for Multi-Electron Atoms For elements other than hydrogen, these rules apply. 1. The orbital energy level diagram has “clusters” of energy levels. Large energy gaps separate these clusters, but within the clusters, energy levels are close. 2. The lowest energy level in a cluster is always an s orbital. The energy of this s orbital is relatively high compared to the energy of the orbitals below it, but the s orbital is relatively close in energy to the orbitals above it in the cluster. © 2008 ChemReview.Net v.2a Page 569 Module 23 — Electron Configuration 3. The d orbitals are shielded by both the s and p orbitals, so much so that the 3d orbital rises to between the energy of the 4s and 4p orbitals. All d orbitals have energies just below those of the p orbitals with one higher principal quantum number. For the five lowest energy clusters, the sublevels in order of increasing energy are 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 4. The f orbitals are more shielded than the d orbitals, so much so that the 4f orbitals for most elements have energy slightly above the 6s orbital, but below the 5d orbital. The sixth and seventh clusters have sublevels in the order 6s 4f 5d 6p 7s 5f 6d 7p . These rules result in the Energy-Level Diagram for a Multi-Electron Atom 7p 6d 6p 5d 7s 6s 5s 4s 3s 2s 5p 5f 32 4f 32 4d 3d 4p 18 18 3p 8 2p 8 2 1s The numbers at the right side of the diagram show the number of electrons that the cluster will hold. © 2008 ChemReview.Net v.2a Page 570 Module 23 — Electron Configuration When a series of orbitals is at the same energy, as in the three p orbitals or five d orbitals, the orbitals are said to be in a sublevel, and the orbitals are termed degenerate. At this point, you need to be able to draw the bottom four clusters of the above diagram from memory. At a later point, the complete diagram will need to be memorized, but as you will see, patterns that relate the orbital diagram to the periodic table will make this task much easier. The Orbital Electron Configuration A key step in understanding the chemical behavior of atoms is to write the electron configuration of the element. This electron configuration can be determined by filling the orbital-energy-level diagram with the element’s electrons. Rules for Filling the Orbital Energy Level Diagram To write the orbital electron configuration for an element, use these steps. 1. Find the number of electrons in the element. The atomic number is the number of protons in an atom. In a neutral atom, this is also the number of electrons. 2. Put the electrons into the orbitals one at a time. The electron will fall to the unfilled orbital with the lowest energy. The rule that the orbitals fill from the lowest energy up is called the aufbau principle. 3. An orbital becomes filled when it has two electrons; it cannot have more than two. In an orbital with two electrons, the electrons must have opposite spins. (This rule is required by the Pauli exclusion principle, which says that no two electrons in an atom can have the same four quantum numbers.) In an orbital with two electrons, the electrons are said to be paired and the orbital is filled. An orbital with only one electron is half-filled and is said to have an unpaired electron. 4. If a series of orbitals is at the same energy, fill each orbital with one electron, and give all the electrons in this sublevel the same spin (by convention, positive spins are assigned first), before you start to pair electrons. (This is Hund’s rule.) Example: The neutral element carbon has 6 electrons. Fill the orbital diagram with the six electrons from the bottom up. Each orbital is filled with two electrons that have opposite spins. For the 2p orbitals, since the three p orbitals are at the same energy, put one electron in each orbital before you start to pair electrons. The resulting carbon electron configuration is 2s ↑↓ 2p ↑ ↑ 1s ↑↓ Carbon has two filled orbitals, two half-filled orbitals, and two unpaired electrons. ***** © 2008 ChemReview.Net v.2a Page 571 Module 23 — Electron Configuration Practice Memorize the orbital energy-level diagram for the bottom 4 clusters (from the 1s to the 4p orbitals), then do the problems below. Use a periodic table. Check your answers at the end of the lesson after each part. 1. The following are electron configurations for neutral atoms. Name the atoms. a. 2s ↑↓ 2p ↑ b. 2s ↑↓ 1s ↑↓ 2p ↑ ↓ ↑ ↓ ↑ 1s ↑↓ 2. Draw the orbital electron configuration for a. Nitrogen b. Neon c. Phosphorous d. Copper 3. How many unpaired electrons are found in neutral atoms of a. Nitrogen (2a above) b. Copper (2d above) ANSWERS 1. a. 5 electrons = Boron (B) b. 9 electrons = Fluorine (F) 2. a. 2s ↑↓ 2p ↑ ↑ ↑ b. d. 3s ↑↓ 2s ↑↓ 2p ↑↓ ↑↓ ↑↓ 1s ↑↓ 1s ↑↓ c. 2s ↑↓ 3p ↑ ↑ ↑ 2p ↑↓ ↑↓ ↑↓ 1s ↑↓ 4s ↑↓ 3s ↑ ↓ 2s ↑ ↓ 4p 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑ 3p ↑↓ ↑↓ ↑↓ 2p ↑↓ ↑↓ ↑↓ 1s ↑↓ 3. a. Nitrogen: 3 unpaired electrons b. Copper: One unpaired electron ***** © 2008 ChemReview.Net v.2a Page 572 Module 23 — Electron Configuration Lesson 23B: Abbreviated Electron Configurations Shorthand Electron Configurations The shorthand electron configuration is a way of quickly determining the electron configuration of an element without drawing the orbital diagram. To write the shorthand electron configuration for an element, use these Steps: 1. Before writing the shorthand configuration for individual atoms, write the orbitals for the multi-electron atom in order of increasing energy (from the bottom up). This list is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p … The order of the orbitals in the list above does not seem regular, but it can 7s 7p be remembered with the help of the 6s 6p 6d 6f memory device at the right. 5s 5p 5d 5f 5g 4s 4p 4d 4f First draw the order for the orbitals of the 3s 3p 3d hydrogen atom (as shown), and then add 2s 2p the diagonal arrows pointed left and up. 1s As you write the orbital numbers and letters, leave a gap in front of each of the s orbitals in the series. This gap indicates the large gap in energy between the clusters in the orbital diagram, a factor that is important in determining electron behavior. In listing the orbitals and clusters in order, it may also help to note that each of the clusters begins with the number of the cluster followed by an s, and ends (except for the first cluster) with the number of the cluster followed by a p. 2. To write the shorthand electron configuration for an individual element, find the number of electrons in the element (the atomic number). 3. Rewrite the orbitals in the order listed above. As you go, fill each orbital full of electrons. Add superscripts to indicate the number of electrons that fill each orbital. • An s orbital is filled when it has 2 electrons. • The p orbital sublevel is full when they have 6 electrons. • The d orbital sublevel can hold 10 electrons. • The f orbital sublevel can hold 14 electrons. A full 1s orbital is written as 1s2 (read as “one s two”). A full 3d sublevel is written: 3d10 (read as “three d ten”). © 2008 ChemReview.Net v.2a Page 573 Module 23 — Electron Configuration 4. When not enough electrons remain to fill an orbital, write the number of remaining electrons as a superscript, then stop. Examples: a. Sodium is atomic number 11; the element Na has 11 electrons. Its shorthand electron configuration is written as: 1s2 2s2 2p6 3s1 The total for the superscripts must equal the number of electrons: 2 + 2 + 6 +1 = 11 Note that all of the sublevels before the last sublevel are filled. b. Iridium (Ir) has 77 electrons. Its shorthand configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d7 To check your answer, add up the superscripts. The total must be the number of electrons in the element. 4. As another check, note that for each atom, its highest cluster starts with the number of the row in which the element is found in the periodic table, followed by an s. Examples: a. Sodium is in the third row of the periodic table (rows go across). Its shorthand electron configuration is: 1s2 2s2 2p6 3s1 The highest cluster in which sodium has electrons begins with 3s . b. Iridium is in the sixth row of the table. Its shorthand configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d7 5. To find the number of unpaired electrons in an element, draw the orbital diagram, but do so only for the sublevel that is unfilled in the highest cluster. Examples: In sodium above, all of the levels below 3s1 are filled. They will contain no unpaired electrons. The final term is 3s1, which is shorthand for the orbital configuration 3s ↑ . Sodium therefore has one unpaired electron. In iridium above, all of the orbitals below 5d7 are filled. In the highest occupied cluster, the one sublevel with unfilled orbitals has the electron configuration 5d7, which is shorthand for the orbital configuration 5d ↑↓ ↑↓ ↑ ↑ ↑ . Iridium has 3 unpaired electrons. ***** © 2008 ChemReview.Net v.2a Page 574 Module 23 — Electron Configuration Practice A Use a periodic table. Check your answers after each part. 1. Write the memory device that lists the orbitals in order of increasing energy, then write the orbitals on one line, in order of increasing energy, from 1s to 7p. 2. Write the shorthand electron configuration for a. Oxygen b. Sulfur c. Fe d. Br e. Kr d. Br e. Kr 3. Write the number of unpaired electrons in a. Oxygen b. Sulfur c. Fe 4. a. What is the shorthand electron configuration for strontium? b. Sr is in which row of the periodic table? c. Into which orbital did strontium’s last (highest energy) electron go? Abbreviated Electron Configurations Shorthand electron configurations can be abbreviated by using the symbol for the appropriate noble gas to represent totally filled lower-energy clusters. Example: Lead, atomic number 82, has a shorthand electron configuration Pb: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p2 but this configuration can be abbreviated as Pb: [Xe] 6s2 4f14 5d10 6p2 The symbol [Xe] represents the five filled lower clusters. Because the abbreviated configuration is the fastest of the three types of configurations to write, it is the one written most often. Abbreviated configurations are also helpful because they separate the electrons that do not change in chemical reactions (those in the lower filled clusters) from those that do. In chemistry, the focus of our attention is the electrons written to the right of the noble gas symbol, since those are the electrons that participate in bonding and in chemical reactions. The electrons that change in chemical reactions are the electrons in the highest unfilled cluster. © 2008 ChemReview.Net v.2a Page 575 Module 23 — Electron Configuration Writing Abbreviated Electron Configurations For Elements 1. Use the symbol for the appropriate noble gas to represent totally filled lower-energy clusters. [He] = 1s2 = 2 electrons [Ne] = 1s2 2s2 2p6 = 10 e─ [Ar] = 1s2 2s2 2p6 3s2 3p6 = 18 e─ [Kr] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 = 36 e─ [Xe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 = 54 e─ [Rn] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 = 86 e─ Note that except for helium, all of the noble gas electron configurations end in p6, and the number in front of the p6 is the row of the periodic table that the noble gas completes. Example: Aluminum is atomic number 13. The shorthand electron configuration for a neutral Al atom is Al: 1s2 2s2 2p6 3s2 3p1 To write the abbreviated configuration, note that Al has 13 electrons. The noble gas in the chart above that can be used to represent the filled lower clusters is neon, which has 10 electrons. The abbreviated electron configuration for Al is Al: [Ne] 3s2 3p1 The symbol [Ne] represents two filled lower clusters containing 10 electrons. 2. For a simplified way to write abbreviated configurations for elements, use these steps. a. Find the element in the periodic table. Note its number of electrons. b. Write in brackets [ ] the symbol of the noble gas at the end of the row above the element. c. After the [noble gas], write the number of the row in the periodic table that the element is in, followed by an s. This will get you started on writing the orbitals in the highest unfilled cluster. Example: Find vanadium (V) in the 4th row of the table. Ar is at the end of the row above V. Start vanadium’s abbreviated configuration by writing V: [Ar] 4s d. Write the remaining orbitals in the cluster. Starting from the number of electrons in the noble gas, add electrons to fill the sublevels until you run out. e. To check your configuration, add the number of electrons in the noble gas (equal to its atomic number) plus the superscripts to the right of the noble gas symbol. This total must equal the number of electrons in the element (its atomic number). © 2008 ChemReview.Net v.2a Page 576 Module 23 — Electron Configuration Try the steps in Rule 2 on this question. Q. Write the abbreviated electron configuration for tungsten (W), atomic number 74. ***** Tungsten is in the 6th row of the periodic table. The noble gas at the end of the 5th row is xenon (Xe). Start the electron configuration by writing W: [Xe] 6s Complete the list of orbitals in the 6th cluster. W: [Xe] 6s 4f 5d 6p Xe has 54 electrons and W has 74. Fill the unfilled cluster until you run out of electrons. W: [Xe] 6s2 4f14 5d4 is the abbreviated configuration. Check: 54 + 2 + 14 + 4 = 74 ***** 3. The valence electrons establish many of the properties of an element, because they are the electrons most likely to be involved in chemical reactions and covalent bonding. Compared to the electrons in lower clusters, the valence electrons are “loosely bound.” For the main group elements (in the tall columns of the periodic table), the valence electrons are defined as all of the s and p electrons in the highest unfilled cluster (the s’s and p’s to the right of the noble gas symbol). Examples: Al: [Ne] 3s2 3p1 has two s and one p electron to the right of the noble gas symbol. Aluminum therefore has 3 valence electrons. I: [Kr] 5s2 4d10 5p5 has two s and five p electron to the right of the noble gas symbol, so iodine has 7 valence electrons. For main group elements, the d and f electrons are not considered to be valence electrons. For elements in the transition metals and rare earths, the valence electrons are those in the highest cluster in the s orbital, but in some cases may also include the d or f electrons. In reactions of the transition metals, the d and f electrons are not as loosely bound as the s electrons, but the d and f electrons may be lost in some chemical reactions. Try the following question. Q. How many valence electrons are in lead (Pb)? ***** Pb: [Xe] 6s2 4f14 5d10 6p2 has two s and two p electron to the right of the noble gas symbol; lead has 4 valence electrons. For main group elements, the valence electrons are limited to the s and p electrons in the highest cluster that contains electrons. ***** © 2008 ChemReview.Net v.2a Page 577 Module 23 — Electron Configuration Practice B: Use a periodic table. Check your answers after each part. 1. List in order the sublevels in the 4th cluster of the orbital energy-level diagram. 2. What noble gas symbol would be used to represent five full orbital clusters? 3. When writing an abbreviated electron configuration, what would be the first sublevel that would be written after these noble gas symbols? a. [He] b. [Kr] c. [Rn] 4. Write the abbreviated electron configuration for these elements. a. Li (3) b. Cl (17) c. Iodine (53) d. Yttrium (39) e. Polonium (84) 5. Write the number of valence electrons for these Problem 4 elements. a. Li (3) b. Cl (17) c. Iodine (53) e. Polonium (84) 6. What is the highest number of valence electrons that a main group element can have? 7. For the 2nd element in the 4th row of the periodic table, a. what is its abbreviated electron configuration? b. What ion does this element tend to form? c. What would be the electron configuration for the ion that this element tends to form? ANSWERS Practice A 1. 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p. 2. a. Oxygen: 1s2 2s2 2p4 b. Sulfur: 1s2 Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 e. Kr: 1s2 2s2 2p6 3s2 3p6 2s2 2p6 3s2 3p4 4s2 3d10 4p6 c. d. Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 3. To find the number of unpaired electrons, look only at the highest sublevel with electrons. a. Oxygen: 2p4 = 2p ↑↓ ↑ ↑ = 2 unpaired e─ b. Sulfur: 3p4 = 3p ↑↓ ↑ ↑ = 2 unpaired e─ c. Fe: 3d6 = 3d ↑↓ ↑ ↑ ↑ ↑ = 4 unpaired e─ d. Br: 4p5 = 4p ↑↓ ↑↓ ↑ e. Kr: 4p6 = 4p ↑↓ ↑↓ ↑↓ 4. a. Sr: 1s2 2s2 2p6 © 2008 ChemReview.Net v.2a 3s2 3p6 = 1 unpaired e─ = 0 unpaired e─ 4s2 3d10 4p6 5s2 b. Row 5 c. 5s Page 578 Module 23 — Electron Configuration Practice B 1. 4s 3d 4p 2. [Xe] at the end of the 5th row. 3. a. [He] 2s The number of the s orbital is the row number that the atom is in. b. [Kr] 5s 4. a. Li (3): [He] 2s1 b. Cl (17): [Ne] 3s2 3p5 d. Yttrium (39): [Kr] 5s2 4d1 c. [Rn] 7s c. Iodine (53): [Kr] 5s2 4d10 5p5 e. Polonium (84): [Xe] 6s2 4f14 5d10 6p4 5. Valence electrons: a. Li (3) one b. Cl (17) seven c. Iodine (53) seven e. Polonium (84) six 6. Eight. The highest number of s and p electrons possible in the highest cluster is 2 in the s and 6 in the p. 7. a. Ca: [Ar] 4s2 b. Ca2+ c. [Ar] . To form the 2+ ion, calcium loses its two valence electrons. ***** Lesson 23C: The Periodic Table and Electron Configuration The Shape of the Table Elements in the families of the periodic table have similar behavior because those elements have similar configurations in their highest energy electrons. The correlations between the periodic table and electron configuration include: 1. The periodic table starts with hydrogen, which has proton and one electron. One proton and one electron are then added for each element moving to the right across the rows of the periodic table. When a cluster in the orbital energy level diagram is filled, the result is the electron configuration of a noble gas. After a noble gas, a new row of the table is started. This convention places all of the noble gases in the last column. All noble gases have totally filled clusters. 2. The number of electrons a cluster can hold equals the number of elements in each row of the periodic table. • The lowest cluster of the orbital diagram can hold two electrons. The first row of the periodic table has two elements: H and He. • The second and third clusters in the orbital energy diagram each hold 8 electrons. The second and third rows of the periodic table both have 8 elements. Across these rows, 8 electrons are added to the elements, one electron at a time. • The fourth cluster has the orbitals 4s 3d 4p . These three sublevels can hold 18 electrons. The fourth row of the table has 18 elements. • The sixth cluster has the orbitals 6s 4f 5d 6p . These four sublevels can hold 32 electrons. The sixth row of the table has 32 elements. © 2008 ChemReview.Net v.2a Page 579 Module 23 — Electron Configuration 3. Elements in the same column of the table have similar electron configurations. • All of the elements in column one have an electron configuration ending in s1. H ends in 1s1, Li in the 2nd row ends in 2s1, Fr in the 7th row ends in 7s1. • All halogens (Group 7A) have electron configurations ending in p5. All halogens have seven valence electrons: two s and five p electrons in their highest cluster. • All of the elements in the carbon family (Group 4A) have their electron configuration ending in p2. All have four valence electrons: two s and two p electrons. • The number of valence electrons for an atom is the number of the main (A) group (using the A-B group notation) for its column of the periodic table. All elements in Group 7A (the halogens) have seven valence electrons. 4. The shape of the periodic table is determined by the order in which the orbitals fill in the orbital energy-level diagram. • All elements in the first two columns of the table have their highest energy electron going into an s orbital. For elements in the first column, all electron configurations end in s1. In the second column, all end in s2. • All of the elements in the six tall columns at the right in the table have their highest energy electron going into a p orbital (except helium). In Group 3A, all electron configurations end in p1. In the last column, all end in p6 (except for helium, which ends in s2 to fill the first cluster). • All transition metals have their highest energy electron going into a d orbital. The transition metals are placed between the s tall columns and the p tall columns because the d orbitals fill after the s, but before the p orbitals. The first transition metal, scandium (Sc), has an electron configuration that ends in 3d1. The second, titanium (Ti) ends in 3d2. The last transition metal in that row, zinc (Zn) has an electron configuration that ends in 3d10. Mercury (Hg), two rows below zinc in the table, has an electron configuration that ends in 5d10. As you might expect, cadmium (Cd), in the row between zinc and mercury, has an electron configuration that ends in ….? 4d10. • The rare-earth elements, in the two rows below the table, have their highest energy electrons going into f orbitals. Since there is room for 14 electrons in the seven f orbitals, there are 14 elements in each of the two rows of the rare earth elements. In the first row of the rare earths, called the lanthanides because the row begins with lanthanum, the 4f orbitals are filling. In the second row, the actinides, the 5f orbitals are filling. © 2008 ChemReview.Net v.2a Page 580 Module 23 — Electron Configuration The following diagram help in remembering the above. 1s2 2p6 2s2 3s2 4s2 5s2 (d filling) ““ 3d10 4d10 6s2 7s2 ““ 3p6 4p6 5d10 5p6 6p6 4f14 5f14 This diagram is easy to remember if you have access to a periodic table. Simply memorize the position of the six circled orbitals above, and then fill in the rest of that column based on the patterns. To write shorthand electron configurations, start with the top row (1s2), go across each row, and write the orbitals that are in bold. Leave a small gap when you start a new row. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6…. Knowing the above chart, all seven clusters of the orbital diagram are easy to write in order. This diagram will also quickly identify the configuration of the highest energy orbital. Example: Tellurium (Te) has 52 electrons. Find Te in the periodic table. Mark where it will be in the above table. Decide from the position of the box the configuration of the highest energy orbital. Based on its position, Te’s highest energy orbital will have the configuration: _______ ***** 5p4 . For the shorthand configuration, simply write all of the bold orbitals above, in order from the top and across the rows, until you get to Te’s 5p4 , then stop. Try it. ***** Te: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4 To write the abbreviated configuration, write the [noble gas] at the end of the column above Te, then write the bold filled orbitals in the row that Te is in, with the last orbitals being the highest energy sublevel for Te. Try it. ***** © 2008 ChemReview.Net v.2a Page 581 Module 23 — Electron Configuration Te: [Kr] 5s2 4d10 5p4 As a check for both types of configurations, count the electrons. You should get 52. ***** In 1872, the Russian chemist Dmitri Mendeleev proposed the first periodic table as a way to predict chemical behavior. Electrons would not be discovered until 1897, and the electron configurations predicted by Schrődinger’s wave equation were 50 years in the future. In fact, at the time, many of the elements currently in the periodic table had not yet been discovered, yet Mendeleev’s table is remarkably similar to the modern periodic table that organizes the patterns of chemical behavior. We know today what Mendeleev did not: why the elements are organized the way they are in his table. The shape of the table is determined by the order of the electron orbitals that are predicted to exist by Schrődinger’s wave equation. Elements in the columns have similar behavior because they have similar electron configurations. Practice: Use a periodic table. If needed, check your answers after each part. 1. What is the abbreviated electron configuration (using the noble gas symbol) for the first element in the 7th row of the periodic table? 2. Which orbitals are being filled in the 6th row of the periodic table, in order? 3. Where in the periodic table are of all the elements that have electron configuration ending in a. s2 b. p6 c. p2 4. Name the four elements whose electron configuration ends in d2 . 5. Which p orbitals are filling in the 5th row of the periodic table? 6. Which d orbitals are filling in the 4th row of the table? 7. Which f orbitals are filling in the 7th row of the table? 8. What do the rare earth elements have in common with regard to their electron configuration? 9. Why do the rare earth elements fit into the periodic table before the transition metals begin, rather than after? 10. Write the orbital configuration of just the highest unfilled sublevel for these elements. a. Iodine (53) b. Cobalt (27) c. Rubidium (37) 11. Write the shorthand electron notation (1s2 2s2…) for these elements. a. Gallium (31) b. Molybdenum (42) 12. Write the abbreviated electron configuration (using noble gas symbols) for a. Barium (56) © 2008 ChemReview.Net b. Osmium (76) v.2a c. Rutherfordium (104) Page 582 Module 23 — Electron Configuration ANSWERS 1. [Rn] 7s1 2. 6s 4f 5d 6p 3a. In Group 2A (the 2nd tall column). 3b . Noble gases below He. 3c. In the 2nd tall column to the right of the transition metals (the carbon family). 4. Titanium (Ti), Zirconium (Zr) Hafnium (Hf), and Rutherfordium (Rf). 5. 5p . The principal quantum number of the filling p orbitals is the row number in the table. 6. 3d . The principal quantum number of the d orbitals filling is one less than the row number. 7. 5f . The principal quantum number of the f orbitals filling is two less than the row number. 8. In most cases, the highest energy electron is going into an f orbital. 9. In most cases, the f orbitals are at lower energy than the d orbital with one higher principal quantum number; the f orbitals fill before the d’s. 10. a. Iodine 5p5 b. Cobalt 1s2 2s2 2p6 b. Mo: 1s2 2s2 2p6 11. a. Ga: 12. a. Ba: [Xe] 6s2 3d7 3s2 3p6 3s2 3p6 c. Rubidium 5s1 4s2 3d10 4p1 4s2 3d10 4p6 b. Os: [Xe] 6s2 4f14 5d6 5s2 4d4 c. Rf: [Rn] 7s2 5f14 6d2 ***** Lesson 23D: Electron Configurations: Exceptions and Ions Exceptions to the Standard Predictions Not all of the actual electron configurations follow the rules above, but most of the exceptions can be predicted. In predicting exceptions, the rules are 1. A series of orbitals at a sublevel (those with the same energy) often have lower potential energy if they are empty, half-filled or totally filled. The expected electron configurations predicted by the standard rules often rearrange to form these lower energy configurations that gain them special stability. Example: Write the standard predicted electron configuration for chromium (24): [ How many unpaired electrons would chromium have? ***** By the standard rules, Cr = [Ar] 4s2 3d4 4p ↑ 3d ↑ ↑ ↑ 4s ↑↓ As seen in the orbital diagram, chromium would have four unpaired electrons. The highest unfilled cluster would be However, if one of the 4s electrons is promoted slightly, up to the 3d orbitals, both the 4s and the 3d orbitals become half-filled. This configuration is © 2008 ChemReview.Net v.2a 4s ↑ 4p 3d ↑ ↑ ↑ ↑ ↑ = [Ar] 4s1 3d5 Page 583 Module 23 — Electron Configuration If a system can go to lower potential energy, it strongly tends to do so. Measurements of the behavior of neutral chromium atoms find that they have the six unpaired electrons predicted by this exception rule. Other elements below chromium in the same column may have similar behavior, but the possible exceptions do not always occur. Experimental measurement of the number of unpaired electrons is often required to determine the actual electron configuration of an element or ion when exceptions are possible. 2. The electron configurations of the rare -earth elements contain many exceptions, especially in the early columns. The reason is that the d and the f orbitals are very close in energy. Most of the exceptions have a non-predicted d1 electron, such as Ce (58): [Rn] 6s2 4f1 5d1 and U (92): [Rn] 7s2 5f3 6d1 But, by a slim majority, most rare earth elements have the s2 f* d0 configuration that is predicted by the standard rules for filling the orbital energy-level diagram. Some periodic tables list La and Ac in the 14 columns below the table. Others place La and Ac in the transition metals, and instead list Lu and Lr at the end of the 14 lower columns, though both Lu and Lr have an s2 f14 d1 configuration which make them similar to the first column of the transition metals. Arguments can be made for both conventions. Practice A: Use a periodic table. If needed, check your answers after each part. 1. Write the abbreviated electron configuration (using noble gas symbols) that would be predicted, using the standard rules, for a. Nickel (28) b. Palladium (46) After each configuration above, write the predicted number of unpaired electrons. 2. By experiment, palladium is found to have no unpaired electrons. Write what this exceptional electron configuration is likely to be, and explain why it would be stable. 3. Find the three coinage metals (copper, silver, and gold) in the periodic table. These elements have been treated as a group since ancient times. All are shiny metals with high stability. Their exceptional stability as metals arises in large measure from their exceptional electron configurations. a. Write the standard prediction of the abbreviated configuration for gold (79). b. What would be the exceptional configuration for gold, and why would it be stable? c. In its exceptional electron configuration, how many unpaired electrons does gold have? How many valence electrons? © 2008 ChemReview.Net v.2a Page 584 Module 23 — Electron Configuration Electron Configurations For Ions Elements tend to form ions that have the number of same number of valence electrons as a noble gas. The noble gases have eight valence electrons (except for helium that has 2). To write electron configurations for monatomic ions, follow these steps. 1. First write the neutral electron configuration. 2. To form negative ions, add electrons in the standard order. 3. To form positive ions, take away electrons from the highest cluster. Take the electrons out in this order: take first the p’s then the s’s, then the d’s, then the f’s. Note that in making positive ions, the electrons are taken out of the neutral atom in a different order than they go in. Electrons are lost first from the orbitals with the highest principal quantum number. For main group elements, the valence electrons are taken away first. Why? When an electron is taken away, the repulsion of all the electrons is reduced, and the shielding of the d and f orbitals is reduced. This drops the d and f orbitals in positive ions to a lower energy than they have in elements. For example, the 3d orbitals, which in a neutral atom are higher in energy than the 4s orbitals, in positive ions drop to an energy lower than the 4s orbitals. Apply the three rules above to these two examples, then check your answers below. Q1. Write the abbreviated electron configuration for the chloride ion, Cl─. Q2. Write the abbreviated electron configuration for the tin (IV) ion, Sn4+. ***** A1. Neutral chlorine is [Ne] 3s2 3p5, chloride ion has one more electron: [Ne] 3s2 3p6 This gives chloride the same electron configuration as argon, the nearest noble gas. When two atoms have the same electron configuration, they are said to be isoelectronic. Iso- is a prefix from ancient Greek meaning equal. A2. Neutral tin is [Kr] 5s2 4d10 5p2, and Sn4+ is [Kr] 4d10 . The 4+ ion has lost four electrons, first the two 5p electrons, then the two 5s electrons. This gives the tin ion zero valence electrons. Ions tend to have either zero or 8 valence electrons, giving them the same stable valence configuration as the nearest noble gas. It is difficult to add an electron to, or to remove an electron from, a noble gas electron configuration. Atoms with electron configurations close to that of a noble gas are very reactive, usually undergoing reactions that allow them to assume a noble gas electron configuration. That tendency is a driving factor in a wide range of chemical reactions. ***** © 2008 ChemReview.Net v.2a Page 585 Module 23 — Electron Configuration Practice B: Use a periodic table. If needed, check your answers after each part. 1. Add a charge to these symbol to show the monatomic ion that the following elements tend to form (review Lesson 7B if needed). a. Cs b. S c. At d. Al e. Mg f. F 2. Write the electron configurations for each ion in Problem 1. 3. Which ion in Problem 1 is isoelectronic with the noble gas xenon? 4. Silver (atomic number 47) forms a 1+ ion. a. Write the abbreviated electron configuration for Ag+. b. How many unpaired electrons are in Ag+? 5. Iron (26) forms two cations: Fe2+ and Fe3+. a. Write the abbreviated electron configuration for both ions. b. What is the number of unpaired electrons in each ion? c. Which ion lost more than its valence electrons? d. Why might that ion lose more than its valence electrons? 6. Write the electron configuration for the two ions formed by copper. Which ion has a configuration that is not predicted by the standard rules for filling the energy level diagram, and why might this ion form? ANSWERS Practice A 1a. Ni: [Ar] 4s2 3d8 ; 2 unpaired electrons: b. Pd: [Kr] 5s2 4d8 2 unpaired electrons: 4s ↑↓ 5s ↑↓ 4p 5p 3d ↑↓ ↑↓ ↑↓ ↑ ↑ 4d ↑↓ ↑↓ ↑↓ ↑ ↑ 2. Pd: is actually [Kr] 5s0 4d10 This configuration has empty s and full d orbitals. Orbitals at the same energy that are empty, half-filled, or totally filled have lowered potential energy. 3. a. Au: [Xe] 6s2 4f14 5d9 b. Au: [Xe] 6s1 4f14 5d10 This gives half-filled s , filled f , and filled d orbitals. Orbitals at the same energy that are empty, half-filled, or totally filled have special stability. c. The single 6s electron is the one unpaired and the one valence electron. © 2008 ChemReview.Net v.2a Page 586 Module 23 — Electron Configuration Practice B 1. a. Cs+ b. S2─ 2. a. [Xe] b. [Ne] 3s2 3p6 or [Ar] d. [Ne] e. [Ne] c. At─ d. Al3+ e. Mg2+ f. F─ c. [Xe] 6s2 4f14 5d10 6p6 or [Rn] f. [Ne] 3. Only 1a: Cs+ . The electron configuration for At─ can be written using Xe at the start, but At─ is isoelectronic with radon (Rn) (see 2c answer). 4. a. Neutral Ag would have a standard predicted configuration of [Kr] 5s2 4d9 , but if a coinage metal has a neutral configuration of s1 d10 , this results in half-filled s and totally filled d orbitals. The actual configuration for neutral silver is Ag: [Kr] 5s1 4d10 a configuration that has only once valence electron. Since silver’s most common ion is one plus, this fits the prediction that it only has one valence electron as an element. To make Ag+, take away the one valence electron. Ag+ = [Kr] 4d10 b. Ag+ = [Kr] 4d10 has no unpaired electrons. All of the orbital sublevels are filled. 5. a. Write the neutral atom configuration first. Fe: [Ar] 4s2 3d6 Fe2+ has lost two its two valence electrons. Fe2+: [Ar] 3d6 Fe3+ has lost one more electron than Fe2+: Fe3+: [Ar] 3d5 b. Fe2+ has four unpaired electrons. Fe2+: [Ar] 3d6 = 3d ↑↓ ↑ Fe3+ has five unpaired electrons. Fe3+: [Ar] 3d5 c. = 3d ↑ ↑↑ ↑ ↑↑↑ ↑ Fe3+ has lost its valence electrons, plus one more. d. By losing one extra electron, Fe3+ has gained a 3d5 configuration, which has degenerate d orbitals that are half-filled and therefore have special stability. 6. Copper’s two ions are copper (I) and copper (II). The neutral copper electron configuration predicted by the standard rules is [Ar] 4s2 3d9 . However, the actual, measured copper electron configuration is [Ar] 4s1 3d10 . In this exceptional configuration, by promoting one s electron to the d orbitals, Cu gains a totally filled series of d orbitals which have extra stability. In some cases, copper behaves as if it has one valence electron, and in other cases as if it has two. The Cu2+ configuration is [Ar] 3d9 . In this configuration, the copper forms an ion by losing two electrons. The Cu+ configuration is [Ar] 3d10 . In this configuration, Cu+ has a totally filled series of d orbitals which have extra stability. ##### © 2008 ChemReview.Net v.2a Page 587 ...
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This note was uploaded on 09/25/2010 for the course BIOLOGY Biology 10 taught by Professor D'arville during the Spring '10 term at Rutgers.

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