Ch 8 Electron configurations and periodicity.pdf

Ch. 8 – Electron Configuration and Chemical Periodicity Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribu...

Ch. 8 – Electron Configuration and Chemical Periodicity Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. 8.1 Quantum Mechanics of Atoms other than Hydrogen The Schrödinger equation cannot be solved for the He atom or any other atom or molecule with 2 electrons or more. The electron-electron repulsion complicates the expression of the potential energy and introduces too many variables in the equation. Over the years, physicists have developed many approximation techniques in Quantum Mechanics to address multi-electron atoms and molecules. All approximations show that the atomic orbitals in elements heavier than Hydrogen have essentially the same shapes and quantum numbers as those in the Hydrogen atom. We call them hydrogen-like atomic orbitals. Main Theme Today: How do many electrons populate the sublevels and orbitals of atoms? ©McGraw-Hill Education. Detection of the Electron Spin Figure 8.1. Detection of the electron spin in a Stern – Gerlach experiment. ©McGraw-Hill Education. Electron Spin A famous experiments by Stern and Gerlach (1922) showed that a beam of silver atoms is split in two by a strong magnetic field. The unusual part is that the Ag beam is split in 2 directions, corresponding to the North and South of the external magnet. A beam of neutral Ag atoms does not form an electric current and should not be deflected by a magnetic field. Later, it was shown that the valence electron of Ag responds to the external magnetic field, in essence acting as a tiny magnet. The experiment reveals that the electrons have their own magnetic moment which interacts with the external magnetic field. The electron, as an electrically charged particle, was expected to have a magnetic moment. It was thought that the electron, as a particle, “spins” around an axis. This was the motion, it was thought, that generates the magnetic moment. Quantum Mechanics showed that electrons do not actually “spin”. Rather, the magnetic moment is an intrinsic property of the electron. The name “spin” stuck. © 2017 Pearson Education, Inc. The Property of Electron Spin Spin is a fundamental property of all electrons, whether free or bound. All electrons have the same amount of spin. The orientation of the electron spin in an external magnetic field is quantized: it can be in only one direction or its opposite: – Spin “up” or spin “down” The electron’s spin adds a fourth quantum number to the description of electrons in an atom, called the spin (magnetic moment) quantum number, ms. The Schrödinger equation cannot predict the existence of electron spin, therefore this additional quantum number is introduced after the fact (ad hoc). © 2017 Pearson Education, Inc. Summary of Quantum Numbers of Electrons in Atoms Table 8.1 ©McGraw-Hill Education. Pauli’s Exclusion Principle No two electrons in an atom may have the same set of four quantum numbers: n, l, ml and ms. Therefore, no quantum state (orbital) may have more than two electrons, and they must have opposite spins. For example, a 3s2 electron configuration has 2 electrons with the following sets of quantum numbers: ▪ n = 3, l = 0, ml = 0, ms = +1/2 ▪ n = 3, l = 0, ml = 0, ms = - 1/2 Knowing the number of orbitals in a sublevel allows us to determine the maximum number of electrons in the sublevel: – s sublevel has 1 orbital; therefore, it can hold 2 electrons. – p sublevel has 3 orbitals; therefore, it can hold 6 electrons. – d sublevel has 5 orbitals; therefore, it can hold 10 electrons. – f sublevel has 7 orbitals; therefore, it can hold 14 electrons. © 2017 Pearson Education, Inc. Coulomb’s Law: Electrostatic Field (Physics refresher) Coulomb’s law describes the attractions and repulsions between charged particles. For like charges (electron-electron), the potential energy (E) is positive and decreases as the particles get farther apart as r increases. For opposite charges (electron-nuleus), the potential energy is negative and becomes more negative as the particles get closer together. The strength of the interaction increases as the size of the charges increases. – Electrons are more strongly attracted to a nucleus with a 2+ charge than a nucleus with a 1+ charge. © 2017 Pearson Education, Inc. Shielding and Effective Nuclear Charge Each electron in a multi-electron atom experiences both the attraction by the nucleus and the repulsion by other electrons in the atom. These repulsions cause the electron to have a net reduced attraction to the nucleus; it is shielded from the nucleus by the inner electrons. – Example: 2s electrons are shielded by the 1s electrons. The total amount of attraction by the nucleus that an electron experiences is called the effective nuclear charge of the electron. Example: – Lithium (Z = 3) ground state electron configuration is 1s22s1 – The 2s1 electron does not experience a coulomb attraction of +3 protons in the nucleus because it is shielded by the 1s2 electrons. The actual or effective positive charge at the 2s1 electron is closer to +1. © 2017 Pearson Education, Inc. Shielding and Orbital Energy Electrons in the same energy level shield each other to some extent. Electrons in inner energy levels shield the outer electrons very effectively. The farther from the nucleus an electron is, the lower the Zeff for that particular electron. ©McGraw-Hill Education. Penetration and Sublevel Energy Orbital shape causes electrons in some orbitals to “penetrate” close to the nucleus. Penetration increases nuclear attraction and decreases shielding. Figure 8.4 © McGraw Hill Penetration of an Orbital at Nucleus The degree of penetration is related to the orbital’s radial distribution function. – s orbitals penetrate to the nucleus – p, d, f orbitals do not penetrate well to the nucleus because they have angular nodes at the nucleus. The radial distribution function shows that the 2s orbital penetrates more deeply into the 1s orbital than does the 2p. The weaker penetration of the 2p sublevel means that electrons in the 2p sublevel experience more repulsive force; they are more shielded from the attractive force of the nucleus. The deeper penetration of the 2s electrons means electrons in the 2s sublevel experience a greater attractive force to the nucleus and are not shielded as effectively. © 2017 Pearson Education, Inc. Order for Filling Energy Sublevels With Electrons All other atoms In a multielectron Hydrogen atom species, sublevels within the same level no longer have the same energy: 2s < 2p 3s < 3p < 3d 4s < 4p < 4d < 4f Sublevels from different levels may have energy inversions: 4s < 3d 5s < 4d 6s < 4f < 5d ©McGraw-Hill Education. The Aufbau Principle The aufbau principle: in the ground state of atoms, sublevels are populated with electrons in the order of increasing energy (to yield the lowest total energy) Aufbau (German) = building up The energy sequence of sublevels is: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p You have to memorize it! Or use a Periodic Table if available. © 2017 Pearson Education, Inc. Filling the Orbitals with Electrons 1) Energy levels and sublevels fill from lowest energy to highest: – – 2) 3) 4) Within a level: s → p → d → f Between levels (the aufbau principle): 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p Orbitals that are in the same sublevel have the same energy. They are degenerate. No single way of populating with electrons is preferred. Pauli exclusion principle: Not more than two electrons per orbital. The Hund’s rule of maximum spin multiplicity: When filling orbitals that have the same energy, place one electron in each orbital before completing pairs and align all the spins in the same direction. – Friedrich Hund (1925), based on atomic spectroscopy observations – Electrons in different degenerate orbitals (e.g., px, py, pz) will decrease the coulomb repulsion compared to when they reside in the same orbital with opposite spins. © 2017 Pearson Education, Inc. 8.2 Electron Configurations The electrons in atoms exist in orbitals because the orbitals are created by bound electrons: an orbital is a standing de Broglie wave of an electron in the electric field of the atomic nucleus and the other electrons. A description of the sublevels occupied by electrons is called an electron configuration. Indicate the value of the principal quantum number n, followed by the type of sublevel (s, p, d, or f), along with the number of electrons in that sublevel as a superscript. Electron configuration of H atom in the ground state: 1s1 © 2017 Pearson Education, Inc. Spin Quantum Number, ms, and Orbital Diagrams ms can have values of +½ or −½, corresponding to up and down in an external magnetic field. Orbital diagrams use a square to represent each orbital and an arrow to represent each electron in the orbital. By convention, an arrow pointing up is used to represent an electron in an orbital with spin up. or or If 2 electrons occupy the same orbital, spins must be paired, one up and one down, to align the attraction forces of their magnetic poles. – Paired © 2017 Pearson Education, Inc. H atom and ions: electron configurations 1) Ground state: 1s1 2) Excited states. Write the possible electron configurations for an excited H atom in the n = 3 level. Answer: there are 3 types of orbitals in the n = 3 level: 3s, 3p and 3d. Possible electron configurations: 3s1 or 3p1 or 3d1. 3) Ground state electron configuration for ions: a) Positive ion: H(g) → H+(g) + e1s0 b) Negative ion: H(g) + e- → H-(g) 1s2 © 2017 Pearson Education, Inc. Helium’s Electron Configuration Helium, Z = 2, has two electrons. Both electrons are in the first energy level, n = 1, to minimize the total energy of the atom. This is the ground state. Both electrons are in a 1s orbital of the first energy level because it is the only available orbital. Because both electrons are in the same orbital (3 identical quantum numbers n, l, and ml), they must have opposite spins. He ground state electron configuration: 1s2 He excited states: 1s12p1, 1s13d1, 2s2, etc. He ions: He+ ground state: 1s1 He2+: 1s0 © 2017 Pearson Education, Inc. Ground State Electron Configurations n = 2 © 2017 Pearson Education, Inc. Building Orbital Diagrams: Period 2 N (Z = 7) 1s22s22p3 O (Z = 8) 1s22s22p4 F (Z = 9) 1s22s22p5 Ne (Z = 10) 1s22s22p6 ©McGraw-Hill Education. Depicting Orbital Occupancy for the First 10 Elements Figure 8.8 ©McGraw-Hill Education. Sample Problem 8.1 (8th edition) Correlating Quantum Numbers and Orbital Diagrams PROBLEM: Use the orbital diagram for fluorine to write sets of quantum numbers for the third and eighth electrons of the F atom. PLAN: Referring to the orbital diagram, we identify the electron of interest and note its level (n), sublevel (l), orbital (ml), and spin (ms). ©McGraw-Hill Education. Sample Problem 8.1 - Solution SOLUTION: The orbital diagram with the electrons of interest in red: The third electron is in the 2s orbital. Recall that l = 0 for an s orbital. The upward arrow indicates a spin of + 1/2 : – n = 2 , l = 0 , ml = 0 , ms = + 1/2 The eighth electron is in the first 2p orbital (l = 1 for a p orbital), which is designated ml = −1, and is represented by a downward arrow because it was the second electron added to that 2p orbital: – n = 2 , l = 1 , ml = 0 or +1 or − 1 (degenerate orbitals), ms = − 1/2 ©McGraw-Hill Education. Partial Orbital Diagrams and Condensed Configurations A partial orbital diagram shows only the highest energy sublevels being filled. – Al (Z = 13) 1s22s22p63s23p1 A condensed electron configuration has the element symbol of the previous noble gas in square brackets. – Al has the condensed configuration [10Ne]3s23p1 ©McGraw-Hill Education. Period 3 ©McGraw-Hill Education. Building Up Period 4: The First Transition Series The 3d sublevel is filled in Period 4, but the 4s sublevel is filled first. The 4s orbital penetrates closer to the nucleus part of the time than the 3d orbital and is thus lower in energy than the 3d orbital. Figure 8.9 © McGraw Hill Period 4 Full Electron Configuration Condensed Electron Configuration K 1s22s22p63s23p64s1 [Ar] 4s1 20 Ca 1s22s22p63s23p64s2 [Ar] 4s2 21 Sc 1s22s22p63s23p64s23d1 [Ar] 4s23d1 22 Ti 1s22s22p63s23p64s23d2 [Ar] 4s23d2 23 V 1s22s22p63s23p64s23d3 [Ar] 4s23d3 24 Cr 1s22s22p63s23p64s13d5 [Ar] 4s13d5 25 Mn 1s22s22p63s23p64s23d5 [Ar] 4s23d5 26 Fe 1s22s22p63s23p64s23d6 [Ar] 4s23d6 27 Co 1s22s22p63s23p64s23d7 [Ar] 4s23d7 28 Ni 1s22s22p63s23p64s23d8 [Ar] 4s23d8 29 Cu 1s22s22p63s23p64s13d10 [Ar] 4s13d10 30 Zn 1s22s22p63s23p64s23d10 [Ar]4s23d10 31 Ga 1s22s22p63s23p64s23d104p1 [Ar] 4s23d104p1 32 Ge 1s22s22p63s23p64s23d104p2 [Ar] 4s23d104p2 33 As 1s22s22p63s23p64s23d104p3 [Ar] 4s23d104p3 34 Se 1s22s22p63s23p64s23d104p4 [Ar] 4s23d104p4 35 Br 1s22s22p63s23p64s23d104p5 [Ar] 4s23d104p5 36 Kr 1s22s22p63s23p64s23d104p6 [Ar] 4s23d104p6 Atomic Number Element 19 ©McGraw-Hill Education. Partial Orbital Diagram (4s, 3d, and 4p Sublevels Only) Stability of Half-filled and Filled Sublevels In Period 4, there are two exceptions to the expected electron configurations. © McGraw Hill Irregular Electron Configurations Predicted: – – – – – – Cr = [Ar]4s23d4 Cu = [Ar]4s23d9 Mo = [Kr]5s24d4 Ru = [Kr]5s24d6 Pd = [Kr]5s24d8 Ag = [Kr] 5s2 4d9 Found experimentally: – – – – – – Cr = [Ar]4s13d5 Cu = [Ar]4s13d10 Mo = [Kr]5s14d5 Ru = [Kr]5s14d7 Pd = [Kr]5s04d10 Ag = [Kr] 5s1 4d10 Half-filled and filled sublevels tend to have lower energy. Exceptions: W and other heavier elements © 2017 Pearson Education, Inc. A Periodic Table of Partial Ground-state Electron Configurations ©McGraw-Hill Education. Similar Reactivities in a Group Figure 8.12 © McGraw Hill Source: © McGraw-Hill Education/Stephen Frisch, photographer Electron Configuration and Group Elements in the same group of the periodic table have the same outer electron configuration. Elements in the same group of the periodic table exhibit similar chemical behavior. Similar outer electron configurations correlate with similar chemical behavior. ©McGraw-Hill Education. Orbital Filling and the Periodic Table The order in which the orbitals are filled can be obtained directly from the periodic table: The period number is the n value of the outermost s and p sublevels. The period number − 1 is the n value of the outermost d sublevel. The period number − 2 is the n value of the outermost f sublevel. ©McGraw-Hill Education. Categories of Electrons Inner (core) electrons are those an atom has in common with the previous noble gas and any completed transition series. The inner electrons form the closed electron shells of the atom and have a much lower energy than the outer electrons. The inner electrons do not participate in chemical bonding between atoms. P: [10Ne] 3s2 3p3 ; Mn: [18Ar] 4s2 3d5 ; Te: [36Kr] 5s2 4d10 5p4 10 inner e- 18 inner e- 46 inner e- Outer electrons are those in the highest energy level (highest n value). P: [10Ne] 3s2 3p3 ; Mn: [18Ar] 4s2 3d5 ; Te: [36Kr] 5s2 4d10 5p4 5 outer e- 2 outer e- 6 outer e- Valence electrons are those involved in forming chemical bonds. – For main group elements, the valence electrons are the outer electrons. – For transition elements, the valence electrons include the outer electrons and some, all or none of the (n -1)d electrons. ©McGraw-Hill Education. Sample Problem 8.1 (9th edition) Determining Electron Configurations PROBLEM: Using only the periodic table and assuming a regular filling pattern, give the full and condensed electron configurations, partial orbital diagram showing valence electrons only, and number of inner electrons for the following elements: a) Potassium (K; Z = 19) b) Technetium (Tc; Z = 43) c) Lead (Pb; Z = 82) PLAN: The atomic number tells us the number of electrons, and the periodic table shows the order for filling sublevels. In the partial orbital diagrams, we include all electrons added after the previous noble gas except those in filled inner sublevels. The number of inner electrons is the sum of those in the previous noble gas and in filled d and f sublevels. ©McGraw-Hill Education. The Modern Periodic Table Figure 2.10 ©McGraw-Hill Education. Sample Problem 8.1 - Solution (a) SOLUTION (a): For K (Z = 19), the full electron configuration is 1s22s22p63s23p64s1. The condensed configuration is [Ar] 4s1. The partial orbital diagram showing valence electrons is: The inner electrons in K are those in the [Ar] part of the condensed electron configuration. There are 18 inner electrons. ©McGraw-Hill Education. Sample Problem 8.1 - Solution (b) SOLUTION (b): For Tc (Z = 43), assuming the expected pattern, the full electron configuration is 1s22s22p63s23p64s23d104p65s24d5. The condensed configuration is [Kr] 5s24d5. The partial orbital diagram showing valence electrons is: The inner electrons in Tc are those in the [Kr] part of the condensed electron configuration. There are 36 inner electrons. ©McGraw-Hill Education. Sample Problem 8.1 - Solution (c) SOLUTION (c): For Pb (Z = 82), the full electron configuration is 1s22s22p63s23p64s23d104p65s2 4d105p66s24f145d106p2. The condensed configuration is [Xe] 6s24f145d106p2. The partial orbital diagram showing valence electrons is: The inner electrons in Pb are the 54 electrons represented by [Xe] plus the 14 electrons in the filled 4f sublevel plus the 10 electrons in the filled 5d sublevel. There are 78 inner electrons. ©McGraw-Hill Education. 8.3 Trends in Three Atomic Properties Three atomic properties directly affected by electron configuration and effective nuclear charge: – Atomic size (a.k.a. “atomic radius”) – Ionization energy → positive ions – Electron affinity → negative ions These properties are periodic. ©McGraw-Hill Education. Defining Atomic Size The metallic radius of a metal atom is half the distance between 2 nuclei, as measured in a single crystal of metal by Xray crystallography. The covalent radius of a non-metal atom in a diatomic molecule is half the equilibrium distance between the 2 nuclei, as measured by spectroscopic techniques (e.g., rotation spectroscopy, electron diffraction). Known covalent radii and distances between nuclei can be used to find unknown radii. ©McGraw-Hill Education. Atomic Radii of the Main-group and Transition Elements Figure 8.13 ©McGraw-Hill Education. Periodicity of Atomic Radius Figure 8.14 ©McGraw-Hill Education. Trends in Atomic Size Atomic size increases as the principal quantum number n increases. – As n increases, the size of the outer orbital increases. Atomic size decreases as the effective nuclear charge Zeff on the outer electrons increases. Prevalent across a period. – As Zeff increases, the outer electrons are pulled closer to the nucleus. For main group elements: atomic size increases down a group in the periodic table and decreases across a period. ©McGraw-Hill Education. Sample Problem 8.2 - Problem and Plan Ranking Elements by Atomic Size PROBLEM: Using only the periodic table, rank each set of main- group elements in order of decreasing atomic size: a) Ca, Mg, Sr b) K, Ga, Ca c) Br, Rb, Kr d) Sr, Ca, Rb PLAN: To rank the elements by atomic size, we find them in the periodic table. They are main-group elements, so size increases down a group and decreases across a period. ©McGraw-Hill Education. Sample Problem 8.2 ©McGraw-Hill Education. Sample Problem 8.2 - Solution SOLUTION: a) Sr > Ca > Mg. These three elements are in Group 2A(2), and size increases down the group. b) K > Ca > Ga. These three elements are in Period 4, and size decreases across a period. c) Rb > Br > Kr. Rb is largest because it has one more energy level (Period 5) and is farthest to the left. Kr is smaller than Br because Kr is farther to the right in Period 4. d) Rb > Sr > Ca. Ca is smallest because it has one fewer energy level. Sr is smaller than Rb because it is farther to the right. ©McGraw-Hill Education. Exercise on atomic size Choose the larger atom in each pair (if possible): a) Sn or I b) Cr or W c) F or Se d) Ge or Po © 2017 Pearson Education, Inc. Trends in Ionization Energy Ionization energy (IE) is the energy required for the removal of 1 mol of electrons from 1 mol of gaseous atoms, molecules, or ions (for subsequent ionization): 𝐴 𝑔 → 𝐴+ 𝑔 + 𝑒 − ∆𝐸 = 𝐼𝐸1 > 0 It is always an endothermic process, therefore the ionization energy is always a positive number. Many-electron atoms can lose more than one electron. We can define multiplestep ionization energies. First Ionization Energy, IE1: M(g) → M1+(g) + 1 e– Second Ionization Energy, IE2: M+1(g) → M2+(g) + 1 e– 𝑁𝑎 𝑔 → 𝑁𝑎+ 𝑔 + 𝑒− 𝑁𝑎+ 𝑔 → 𝑁𝑎2+ 𝑔 + 𝑒 − ©McGraw-Hill Education. 𝑘𝐽 𝐼𝐸1 = 496 𝑚𝑜𝑙 𝑘𝐽 𝐼𝐸2 = 4,560 𝑚𝑜𝑙 First Ionization Energies of the Main-group Elements Figure 8.16 ©McGraw-Hill Education. Periodicity of First Ionization Energy (IE1) N Zn Mg Ga ©McGraw-Hill Education. Exceptions in the IE1 Trends First ionization energy generally increases from left to right across a period. Except from 2A to 3A and 5A to 6A Removing the 4th p electron by decoupling the pair of spins in O costs less energy than breaking up the triplet of parallel spins in N: 𝑂( 𝐻𝑒 2𝑠 2 2𝑝4 ) → 𝑂+ ( 𝐻𝑒 2𝑠 2 2𝑝3 ) Removing the only 3p electron in Al costs less energy than breaking up a pair of electrons in 3s2 in Mg. © 2017 Pearson Education, Inc. Trends in the First Ionization Energies ©McGraw-Hill Education. Sample Problem 8.3 - Problem and Plan Ranking Elements by First Ionization Energy PROBLEM: Using the periodic table only, rank the elements in each set in order of decreasing IE1: a) Kr, He, Ar b) Sb, Te, Sn c) K, Ca, Rb d) I, Xe, Cs PLAN: We find the elements in the periodic table and then apply the general trends of decreasing IE1 down a group and increasing IE1across a period. ©McGraw-Hill Education. The Modern Periodic Table Figure 2.10 ©McGraw-Hill Education. Sample Problem 8.3 - Solution SOLUTION: a) He > Ar > Kr. These are in Group 8A(18), and IE1decreases down a group. b) Te > Sb > Sn. These are in Period 5, and IE1 increases across a period. c) Ca > K > Rb. IE1 of K is larger than IE1 of Rb because K is higher in Group 1A(1). IE1 of Ca is larger than IE1 of K because Ca is farther to the right in Period 4. d) Xe > I > Cs. IE1 of I is smaller than IE1 of Xe because I is farther to the left. IE1 of I is larger than IE1 of Cs because I is farther to the right and in the previous period. ©McGraw-Hill Education. Successive Ionization Energies Removal of each successive electron costs more energy: – Shrinkage in ion size due to having more protons than electrons leads to a 𝑞 𝑞 stronger Coulomb attraction: 𝐸 = 1 2 4𝜋𝜀0 𝑟 There is a much larger increase in the ionization energy when the first core electron is removed. ©McGraw-Hill Education. The First Three Ionization Energies of Beryllium Figure 8.17 ©McGraw-Hill Education. Sample Problem 8.4 - Problem and Plan Ranking Elements by Atomic Size PROBLEM: Name the Period 3 element with the following ionization energies (kJ/mol), and write its full electron configuration: IE1 IE2 IE3 IE4 IE5 IE6 1012 1903 2910 4956 6278 22,230 PLAN: We look for a large jump in the IE values, which occurs after all valence electrons have been removed. Then we refer to the periodic table to find the Period 3 element with this number of valence electrons and write its electron configuration. ©McGraw-Hill Education. Sample Problem 8.4 - Solution SOLUTION: The large jump occurs after IE5, indicating that the element has five valence electrons and, thus, is in Group 5A(15). This Period 3 element is phosphorus (P; Z = 15). Its full electron configuration is 1s22s22p63s23p3. ©McGraw-Hill Education. Trends in Electron Affinity: Thermochemistry of Ionization Electron Affinity (EA) is the energy change that occurs when 1 mol of electrons is added to 1 mol of gaseous atoms: 𝐴 𝑔 + 𝑒 − → 𝐴− 𝑔 ∆𝐸 = 𝐸𝐴1 The first electron added is attracted by the atom’s nucleus, so, in most cases, EA1 is negative (energy is released). Atoms with a low EA also have low IE1 and tend to form cations. Atoms with a high EA also have high IE1 and tend to form anions. The trends in electron affinity are not as regular and periodic as those for atomic size or IE. © McGraw Hill Electron Affinities of the Main-group Elements (In kJ/mol) Figure 8.19 © McGraw Hill Behavior Patterns for IE and EA Reactive nonmetals have high IEs and highly negative EAs. These elements attract electrons strongly and tend to form negative ions in ionic compounds. Reactive metals have low IEs and slightly negative EAs. These elements lose electrons easily and tend to form positive ions in ionic compounds. Noble gases have very high IEs and slightly positive EAs. These elements tend to neither lose nor gain electrons. © McGraw Hill 3.4 Electron Configurations of Ions: 1. Ions with Noble Gas Configuration When forming ions, elements at either end of a period gain or lose 1-3 electrons to attain a filled (closed shell) outer level: 1𝑠 2 𝑜𝑟 𝑛𝑠 2 𝑛𝑝6 𝑛 ≥ 2 The resulting ion will have a noble gas electron configuration and is said to be isoelectronic with that noble gas. Na(1s22s22p63s1) → e– + Na+([2He]2s22p6) [isoelectronic with 10Ne] Br([18Ar]4s23d104p5) + e– → Br- ([18Ar]4s23d104p6) [isoelectronic with 36Kr] Metals tend to lose electrons and form cations. Examples: K+, Mg2+, Al3+ Non-metals tend to gain electrons and form anions. Examples: F-, O2-, N3- ©McGraw-Hill Education. Electron Configurations and Orbital Diagrams of Main Group Ions or [10Ne] Z=13, 10 e- or [18Ar] Z=16, 18 e- © 2017 Pearson Education, Inc. Main-group Elements Whose Ions Have Noble Gas Electron Configurations Figure 8.25 ©McGraw-Hill Education. The nearest noble gas electron configuration. Isoelectronic ions All ions within a row are isoelectronic because they have the same number of electrons. ©McGraw-Hill Education. Electron Configurations of Ions: 2. Ions 3+without a Noble Gas Configuration 3- Except for Al and N , all the other metals and non-metals of groups 3A – 5A do not form ions at all (covalent bonding is preferred), or do not form ions with noble gas configuration. A pseudo-noble gas configuration is attained when a metal atom empties its highest energy level but leaves untouched the (n – 1)d10 sublevel. Sn (Z = 50): [36Kr]5s24d105p2 → 4e– + Sn4+([36Kr]4d10) - Other ions in this category: Ga3+, In3+ A metal may lose only the np electrons to attain an inert pair configuration: – The ion has a filled ns2 and (n – 1)d10 sublevels. – Sn([36Kr]5s24d105p2) → 2e– + Sn2+([36Kr]5s24d10) – Other ions in this category: Pb2+, Bi3+, Ga+, In+, Tl+ ©McGraw-Hill Education. Sample Problem 8.6 - Problem and Plan Writing Electron Configurations of Main-Group Ions Using condensed electron configurations, write equations representing the formation of the ion(s) of the following elements: a) Iodine (Z = 53) b) Potassium (Z = 19) c) Indium (Z = 49) PLAN: We identify the element’s position in the periodic table and recall that: – Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17)are isoelectronic with the nearest noble gas. – Metals in Groups 3A(13) to 5A(15) lose the ns and np electrons or just the np. ©McGraw-Hill Education. Sample Problem 8.6: Solution a) Iodine is in Group 7A(17), so it gains one electron, and I−is isoelectronic with xenon: I ([Kr]5s24d105p5) + e– → I– ([Kr] 5s24d105p6) a) Potassium is in Group 1A(1), so it loses one electron; K+ is isoelectronic with argon: K ([Ar] 4s1) → K+ ([Ar]) + e– b) Indium is in Group 3A(13), so it loses either three electrons to form In3+ (with a pseudo–noble gas configuration) or one to form In+ (with an inert pair): In ([Kr] 5s24d105p1) → In+ ([Kr] 5s24d10) + e– In ([Kr] 5s24d105p1) → In3+([Kr] 4d10) + 3e– ©McGraw-Hill Education. Electron Configurations of Ions: 3. Transition Metal Ions Transition metal ions rarely attain a noble gas configuration. When they do, they lose all of the ns2 and (n – 1)dx electrons. Examples: Sc3+, Y3+, Ti4+. A transition metal typically forms more than one cation by losing all ns2 and some of the (n – 1)d electrons. The lowest oxidation states of the transition metal ions are formed by losing the ns2 electrons first, before losing the (n – 1)dx electrons! ©McGraw-Hill Education. Electron Configurations of Transition Metal Cations in Their Ground State The iron atom has two outer electrons: Fe (Z=26): 1s22s22p63s23p64s23d6 When iron forms a cation, it first loses its outer electrons: Fe2+: 1s22s22p63s23p63d6 [4s0] It can then lose another 3d electron: Fe3+ : 1s22s22p63s23p63d5 Exercise: give the Zn2+ electron configuration © 2017 Pearson Education, Inc. Summary of Ion Electron Configurations ❑ Main group s-block metals and Aluminum lose ALL electrons with the highest n value (outer electrons) to attain a closed shell electron configuration of the previous noble gas. ❑ Main group p-block metals other than Al and Ga may lose np electrons before the ns electrons. ❑ Main group p-block metals other than Al attain some form of closed subshell electron configuration: either a pseudo-noble gas configuration (when they lose ALL outer ns and np electrons), or an inert pair configuration (when they only lose the np electrons). ❑ Some non-metals (always in the p-block) gain electrons to attain the following noble gas configuration. Non-metals known to gain electrons and form anions: all halogens, O, S, Se (few compounds), N (1 compound). ❑ Transition (d-block) metals lose the ns electrons before the (n – 1)d electrons. ©McGraw-Hill Education. Magnetism of Atoms and Ions A species with one or more unpaired electrons exhibits paramagnetism – it is attracted by a magnetic field. – Ag (Z=47) [36Kr] 5s1 4d10 A species with all its electrons paired exhibits diamagnetism – it is not attracted (and is slightly repelled) by a magnetic field. – Zn (Z = 30): [18Ar] 4s2 3d10 ©McGraw-Hill Education. Measuring the Magnetic Behavior of a Sample The apparent mass of a diamagnetic substance is unaffected by the magnetic field. The apparent mass of a paramagnetic substance increases as it is attracted by the magnetic field. Figure 8.27 ©McGraw-Hill Education. Sample Problem 8.6 - Problem and Plan Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions PROBLEM: Use condensed electron configurations and partial orbital diagrams to predict whether the following ions are paramagnetic or diamagnetic: a) Co3+ (Z = 27) b) Cr3+ (Z = 24) c) Hg2+ (Z = 80) PLAN: We first write the condensed electron configuration of the atom, recalling the irregularity for Cr. Then we remove electrons, beginning with ns electrons, to attain the ion charge. If unpaired electrons are present, the ion is paramagnetic. ©McGraw-Hill Education. Sample Problem 8.6 - Solution ©McGraw-Hill Education. Ionic Radii: cations of groups 1A-3A Cations are much smaller than the parent atoms. Cation sizes increase down the column (group). Cation sizes decrease from left to right within a row because the cation charge increases in the same direction (isoelectronic cations). In cations, the effective nuclear charge on the noble-gas core electrons is higher than in parent atoms. © 2017 Pearson Education, Inc. Ionic Radii: anions of groups 6A-7A Anions are much larger than the parent atoms. Anion radii increase down the column. Anion radii decrease across the row from left to right. © 2017 Pearson Education, Inc. Isoelectronic ions All ions within a row are isoelectronic because they have the same number of electrons, corresponding to the nearest noble gas closed shell. ©McGraw-Hill Education. Isoelectronic Ions The effect of effective nuclear charge is very visible across isoelectronic series of ions. Isoelectronic ions are cations and ions that have the same noble gas electron configuration in the same orbitals: S2- (184 pm) > Cl-(181 pm) > 18 e18 e16 p+ 17 p+ © 2017 Pearson Education, Inc. K+ (133 pm) 18 e19 p+ > Ca2+(99 pm) 18 e20 p+ Sample Problem 8.7 - Problem and Plan Ranking Ions by Size PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking: a) Ca2+, Sr2+, Mg2+ b) K+, S2−, Cl− c) Au+, Au3+ PLAN: We find the position of each element in the periodic table and apply the ideas presented in the text. ©McGraw-Hill Education. Sample Problem 8.7 - Solution SOLUTION: a) Mg2+, Ca2+, and Sr2+ are all from Group 2A(2), so their sizes decrease up the group: Sr2+ > Ca2+ > Mg2+. b) The ions K+, S2−, and Cl− are isoelectronic. S2− has a lower Zeff than Cl−, so it is larger. K+ is a cation and has the highest Zeff, so it is smallest: S2− > Cl− > K+. c) Au+ has a lower charge than Au3+, so it is larger: Au+ > Au3+. ©McGraw-Hill Education.