Chapter 8. Electron Configuration and Chemical Periodicit. 1s 2s 2p 3s 3p

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1 Energy Chapter 8 described by Wave Function (Orbital) e - filling spdf electronic configuration determined by comprising Core Electrons Valence Electrons Electron Configuration and Chemical Periodicit described by Quantum Numbers which are Orbital Energy Aufbau Rules which involve Pauli Exclusion Hund s Rule basis for Periodic Table which summarizes Periodic Properties Quantum Numbers Principal n = 1,2,3,.. define s Orbital size & energy 4-quantum numbers specify the energy and location of electrons around a nucleus (all we can know). This numbers are the framework for the electronic structure of an atom. defines Name Symbol Permitted Values Property Angular momentum, l defines Orbital shape principal angular momentum n positive integers (1,2,3...) orbital energy (size) l integers from 0 to n-1 orbital shape (0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.) Magnetic ml defines Orbital orientation magnetic m l integers from -l to 0 to +l orbital orientation in space Spin, ms defines Electron spin spin m s +1/2 or -1/2 direction of e - spin Quantum Number n Allowed Values Positive integers 1,2,3, Possible Orbitals 2 3 Schrodinger s equation gives an exact solution for H- atom, but does not for many electron-atoms. Electronelectron repulsion in multi-electron split energy levels. Hydrogen Atom Multi-electron atoms l 0 up to max of n Orbitals are degenerate or the same energy in Hydrogen! m l -l, l Orbital Name Shapes or Boundry Surface Plots 1s 2s 2p 3s 3p 3d Electrons will fill lowest energy orbitals first!

2 Inner core electrons shield or screen outer electrons from the positive charge of the nucleus. Screening Impacts 1) alters the energy levels spacing & ordering in manyelectron atoms. 2) outer e- screened by inner electrons. 4-quantum numbers specify all the information we can know the energy and location of electrons around a nucleus. Chemists call this the electronic structure of an atom. Remember this diagram electronic configurations. inner core electron valence electron The Aufbau Process is used to generate the electronic configuration of elements filling the lowest energy orbitals sequentially. 1. Lower energy orbitals fill first (smaller n). 2. Hund s Rule-degenerate (i.e. orbitals with the same energy) orbitals fill one at a time before electrons are paired in an orbital. 3. Pauli Exclusion Principle: No two electrons in an atom can have same 4-quantum numbers. remember filling order using this device! Chemists use spdf notation and orbital box diagrams to symbolize the ground state electronic configuration of elements. Element H He electron shell principal quantum # spdf Notation 1s 1 1s 2 orbital type angular quantum # orbital box diagram Arrow denotes an electron with spin up or spindown. Remember, no two electrons can have the same 4 quantum numbers! # of electrons in orbital Building electronic configuration using Aufbau and Hund Atomic Number/Element H Orbital Box Diagram Full-electronic configuration 1s 1 Condensed-electronic configuration 1s 1 Atomic Number/Element B C Orbital Box Diagram Full-electronic configuration 1s 2 2s 2 2p 1 1s 2 2s 2 2p 2 Condensed-electronic configuration [He]2s 2 2p 1 [He]2s 2 2p 2 He 1s 2 1s 2 1s 2 2s 2 2p 3 [He]2s 2 2p 3 Li Be 1s 2 2s 1 1s 2 2s 2 [He]2s 1 written with noble gas configuration [He]2s 2 1s 2 2s 2 2p 4 1s 2 2s 2 2p 5 1s 2 2s 2 2p 6 [He]2s 2 2p 4 [He]2s 2 2p 5 [He]2s 2 2p 6

3 Unpaired electrons in orbitals gives rise to paramagnetism and is attracted to a magnetic field. Diamagnetic species contain all paired electrons and is repelled by the magnetic field. Diamagnetic atoms or ions: All e- are paired. Weakly repelled in a magnetic field. Diamagnetic all electrons paired 2p Paramagnetic Paramagnetic atoms or ions: unpaired electrons Unpaired e- exist in an orbital Attracted to an external magnetic field. 2p This periodic table shows the two f-block series (lanthanides 4f-block and actinides (5f) where it really is supposed to be, not removed from the table. Representative or Main Group Elements s-block p-block Transition Elements (d-block) Transition Metals Lanthanides (4f-block) Lanthanides (4f-block) Acthanides (5f-block) s-block f-block d-block p-block Anthanides (5f)

4 ns 1 ns 2 Ground State Electron Configurations of the Elements d 1 d 5 ns 2 d 1 odd behavior d 5 odd behavior d 10 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 What is the electron configuration of Mg? What are the possible quantum numbers for the last (outermost) electron in Cl? 4f 5f 4f 1 4f 2 4f 10 4f14 What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s = 12 electrons Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 Using the periodic table on the inside cover of the text and give the full and condensed electrons configurations, partial orbital diagrams showing valence electrons, and number of inner electrons for the following elements: (a) potassium (K: Z = 19) (b) molybdenum (Mo: Z = 42) (c) lead (Pb: Z = 82) What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p = 17 electrons Last electron added to 3p orbital n = 3 l = 1 m l = -1, 0, or +1 m s =! or -! (a) for K (Z = 19) full configuration condensed orbital diagram (b) for Mo (Z = 42) 36 inner electrons and 6 valence electrons full configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5 condensed [Kr] 5s 1 4d 5 partial orbital diagram (c) for Pb (Z = 82) 78 inner electrons and 4 valence electrons. 1s full configuration 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 2 condensed partial orbital diagram There are 18 inner electrons. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 [Ar] 4s 1 4s 1 3d 5s 1 4d 5 5p [Xe] 6s 2 4f 14 5d 10 6p 2 6s 2 6p 2 4p Metals loose electrons (oxidized) and become cations. Non-metals gain electrons (reduced) and become anions. The electronic configuration of each reflects this change in the number of electrons. Na [Ne]3s 1 Ca [Ar]4s 2 Al [Ne]3s 2 3p 1 Non-metals gain electrons so that anion has a noble-gas outer electron configuration. Na + [Ne] Ca 2+ [Ar] Al 3+ [Ne] H 1s 1 F 1s 2 2s 2 2p 5 O 1s 2 2s 2 2p 4 N 1s 2 2s 2 2p 3 Metals lose electrons so that cation has a noble-gas outer electron configuration. H - 1s 2 or [He] F - 1s 2 2s 2 2p 6 or [Ne] O 2-1s 2 2s 2 2p 6 or [Ne] N 3-1s 2 2s 2 2p 6 or [Ne]

5 Metals and non-metal ions tend to form electronic states closest to their nearest noble gas configuration. Isoelectronic species are two different elements with the same electronic configuration--but not the same nuclear configuration. oxidation Na: [1s 2 2s 2 2p 6 3s 1 ] =====> Na + : [1s 2 2s 2 2p 6 ] = [Ne] oxidation Al: [1s 2 2s 2 2p 6 3s 2 3 p1 ] =====> Al 3+ : [1s 2 2s 2 2p 6 ] = [Ne] reduced N: [1s 2 2s 2 2p 3 ] =====> N 3- : [1s 2 2s 2 2p 6 ] = [Ne] reduced O: [1s 2 2s 2 2p 4 ] =====> O 2- : [1s 2 2s 2 2p 6 ] = [Ne] reduced F: [1s 2 2s 2 2p 5 ] =====> F - : [1s 2 2s 2 2p 6 ] = [Ne] Na +, Al 3+, F -, O 2-, and N 3- are all said to be isoelectronic with Ne as they have the same electronic configuration...all subshells are filled. When a transition-metal cation is formed from an atom of a transition metal, electrons are removed first from the ns orbital, then from the (n-1)d orbital. Write the full electronic configuration the following ions: Sc +3, Zn +2,Co 2+ and Co 3+. Distinguish if each is paramagnetic or diamagnetic. Transition Metal Transition Metal Cation (ns) (n-1)d Fe: [Ar]4s 2 3d 6 Fe 2+ : [Ar]4s 0 3d 6 or [Ar]3d 6 Fe: [Ar]4s 2 3d 6 Fe 3+ : [Ar]4s 0 3d 5 or [Ar]3d 5 Mn: [Ar]4s 2 3d 5 Mn 2+ : [Ar]4s 0 3d 5 or [Ar]3d 5 Write the full electronic configuration the following ions: Sc +3, Zn +2,Co 2+ and Co 3+. Distinguish if each is paramagnetic or diamagnetic. A) What is the electron configuration of Mg and Mg 2+? B) What are the possible quantum numbers for the last (outermost) electron in Cl? C) Is ground state F paramagenetic or diamagnetic? Diamagnetic Sc 3+ Paramagnetic Zn 2+, Co 2+, Co 3+

6 What is the spdf and condensed electron configuration of Mg and Mg 2+? Mg 12 electrons Mg 1s 2 2s 2 2p 6 3s 2 [Ne]3s 2 Mg 2+ 1s 2 2s 2 2p 6 3s 0 [Ne]3s 0 = [Ne] What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p = 17 electrons Last electron added to 3p orbital n = 3 l = 1 m l = -1, 0, or +1 m s =! or -! Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic. (a) Mn 2+ (Z = 25) (b) Cr 3+ (Z = 24) (c) Hg 2+ (Z = 80) Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic. C) Is ground state F paramagenetic or diamagnetic? 9 F 1s 2s 2p Unpaired electron = PARAMAGNETIC Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic. Identify n and l quantum numbers for each of the following. (a) Mn 2+ (Z = 25) (b) Cr 3+ (Z = 24) (c) Hg 2+ (Z = 80) Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic. 3p 4dz 2 SOLUTION: third shell fourth shell (a) Mn 2+ (Z = 25) Mn([Ar]4s 2 3d 5 ) Mn 2+ ([Ar] 3d 5 ) + 2e - paramagnetic (b) Cr 3+ (Z = 24) Cr([Ar]4s 1 3d 5 ) Cr 3+ ([Ar] 3d 3 ) + 3e - paramagnetic (c) Hg 2+ (Z = 80) Hg([Xe]6s 2 4f 14 5d 10 ) Hg 2+ ([Xe] 4f 14 5d 10 ) + 2e - not paramagnetic (is diamagnetic) What neutral element has the following orbital-filling diagram? Gallium = Ga Using spdf notation write the complete electron configuration of O, Cl, Ti, Zn? Periodicity in the chemical reactivity of elements occurs because of periodicity in the electronic structure of valence electrons! Using condensed spdf notation what is the electronic configuration of Br and Br - What are the possible quantum numbers for the last (outermost) electron in Cl? 1-electron outer s-orbital 2-electrons outer d-orbital 5-electrons outer p-orbital

7 We must know these properties and be able to apply them! Here s a different view on one periodic table! Know these properties cold (see homework and examples) Amount of energy to remove 1 mole e - from 1 mole of gaseous atoms or element Amount of energy to add 1 mole e - to 1 mole of gaseous atoms or element ns 2 np 1 Electrons in elements are categorized either as inner-core electrons or outer valence-electrons. Inner core electrons shield or screen outer electrons from the positive charge of the nucleus. 1) Inner core electrons : electrons residing in the lower n shells of an element--located closer to the nucleus. 2. Outer core or VALENCE e - : total number of e - in the highest n-value shell. The number of valence e - is given by the Group Number in the periodic table for Group A representative elements. n =2 n =3 n =4 n =5 n =6 inner core electron valence electron Screening Impacts 1) alters the energy levels spacing & ordering in manyelectron atoms. 2) outer e- screened by inner electrons. Effective nuclear charge (Z eff ) is the electrostatic force felt by the outer valence electrons taking into shielding by internal core electrons. To a good approximation: effective nuclear charge, Zeff, is given by: ns 1 ns 2 ns 2 The number of valence electrons can be found by counting the # of e - in the outer most shell (main-group elements only) d 1 d 5 d 10 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 Z eff = Z core e- d 1 d 5 Effective Nuclear charge # protons in atom # of inner core non-valence electrons Larger Zeff means more pull or electrostatic force between nucleus and electrons. 4f 5f

8 Increasing Atomic Radius Z eff = Z core e- **** Configuration Element Z (p + ) Core Valence Radius Z effective Electrons Electrons (pm) [Ne]3s 1 Na [Ne]3s 2 Mg [Ne]3s 2 3p 1 Al [Ne]3s 2 3p 2 Si [Ne]3s 2 3p 3 P [Ne]3s 2 3p 4 S [Ne]3s 2 3p 5 Cl [Ne]3s 2 3p 6 Ar [Ar]4s 1 K [Ar]4s 2 Ca [Ar]4s 2 3d 1 Sc The effective nuclear charge (Z eff ) ( pull on valence electrons) increases across a period and upward in a group! Know this trend and others follow! increasing Z eff Z eff Increases Atomic radii decrease across a Period because the effective nuclear increases. Down a group the atomic radi get larger! Decreasing Atomic Radius Using only the periodic table rank each set of main group elements in order of decreasing atomic size: Increasing n (a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb Using only the periodic table rank each set of main group elements in order of decreasing atomic size: (a) Ca, Mg, Sr SOLUTION: (a) Sr > Ca > Mg (b) K > Ca > Ga (c) Rb > Br > Kr (d) Rb > Sr > Ca (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb These elements are in Group 2A(2). These elements are in Period 4. Rb has a higher n engery level and is far to the left. Br is to the left of Kr. Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr. The atomic radii of cations are smaller than their ground state atoms, while anions are larger than their ground state atom (i.e. remove e - => smaller, add e - => bigger) Greater cation charge => smaller & vis versa Cations get smaller (greater Z eff) Anions get larger (lower Z eff)

9 Ranking Ions by Size Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ (b) K +, S 2!, Cl! (c) Au +, Au 3+ PLAN: Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. Ranking Ions by Size Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ Sr 2+ > Ca 2+ > Mg 2+ These are members of the same Group 2A(2) and therefore decrease in size going up the group. (b) K +, S 2!, Cl! S 2! > Cl! > K + The ions are isoelectronic; S 2! has the smallest Z eff and therefore is the largest while K + is a cation with a large Z eff and is the smallest. (c) Au +, Au 3+ Au + > Au 3+ The higher the + charge, the smaller the ion. Ionization energy is the minimum energy (kj/mol) required to remove 1 mole of e - from one mole of a gaseous atom in its ground state. Electron affinity is the energy required to add (reduce) 1 mole of e - to an atom in the gas state to form an anion. It s a measure of an atom s ability to accept an e -. higher nuclear charge => smaller diameter => harder to remove e - First ionization energies of the main-group elements. How hard it is to remove an electron Electron affinity is largest (most negative) for chlorine and fluorine (i.e like to gain electroncs = reduced). The increasing effective nuclear Charge (Z eff ) and it s impact on atomic radius can help us understand the trend in ionization energies of elements. higher nuclear charge => smaller diameter => harder to remove e - Easily oxidized metal

10 We can remove more than 1 electron from a ground state atom. It requires more energy to remove subsequent electrons. The ionization energy increases dramatically when an core electron is removed from a nonvalence shell (blue area shows big jumps in IE). I 1 + X (g) X + (g) + e - I 2 + X (g) X 2+ (g) + e - I 1 first ionization energy I 2 second ionization energy I 3 + X (g) X 3+ (g) + e - I 3 third ionization energy 1s 2 2s 1 1s 2 2s 2 1s 2 2s 2 2p 1 1s 2 2s 2 2p 2 I 1 < I 2 < I 3 1s 2 2s 2 2p 3 1s 2 2s 2 2p 4 Ranking Elements by First Ionization Energy Using the periodic table to rank the elements in each set in order of decreasing IE 1 : (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs Ranking Elements by First Ionization Energy Using the periodic table to rank the elements in each set in order of decreasing IE 1 : (a) Kr, He, Ar (c) K, Ca, Rb He > Ar > Kr Group 8A(18) - IE decreases down a group. (b) Sb, Te, Sn Te > Sb > Sn Period 5 elements - IE increases across a period. Ca > K > Rb Ca is to the right of K; Rb is below K. (d) I, Xe, Cs Xe > I > Cs I is to the left of Xe; Cs is further to the left and down one period. Identifying an Element from Successive Ionization Energies Name the Period 3 element with the following ionization energies (in kj/mol) and write its electron configuration: Identifying an Element from Successive Ionization Energies Name the Period 3 element with the following ionization energies (in kj/mol) and write its electron configuration: IE 1 IE 2 IE 3 IE 4 IE 5 IE ,230 IE 1 IE 2 IE 3 IE 4 IE IE 6 22,230 PLAN: Look for a large increase in energy which indicates that all of the valence electrons have been removed. The number valence electrons is reflected in the periodic table for Group A elements...find the group with that number of valence electrons. SOLUTION: The largest increase occurs after IE 5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s 2 3p 3 and the element must be phosphorous, P (Z = 15). The electron configuration is: 1s 2 2s 2 2p 6 3s 2 3p 3.

11 Metallic behavior increases as we move down a group and from left to right on the periodic table. --Metals have low IE -- Tend to be oxidized to metal ions. --Charge is group numbers Main Group (or representative) metals form ionic basic oxides when reacted with oxygen while nonmetals form covalent acidic oxides with oxygen. Ionic Oxides Increasing Basicity 2 4 1A Increasing Acidity Li2O BeO B2O3 CO2 OF2 Na2O K2O 2A CaO 3A 4A (14) Ga2O3 GeO2 6A (16) Covalent Oxides 7A (17) SeO3 Br2O7 5 Rb2O SrO In2O3 SnO2 TeO3 I2O7 Most metallic elements 6 Cs2O BaO Tl2O3 PbO2 Properties of Oxides Across a Period basic acidic