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Chapter 8 2013 described by Core Chapter 8 Electrons Wave Function e- filling spdf electronic configuration comprising (Orbital) Valence determined by Electrons described by Aufbau basis for Rules Periodic Quantum which involve Table Numbers which summarizes Orbital Pauli Hund’s which are Energy Exclusion Rule Periodic Properties Electron Configuration and Chemical Periodicity Quantum 4-quantum numbers specify the energy and location of Numbers electrons around a nucleus (all we can know). This numbers are the framework for the “electronic structure Orbital size of an atom”. Principal define & energy n = 1,2,3,.. s defines Name Symbol Permitted Values Property principal n positive integers (1,2,3...) orbital energy (size) Angular Orbital angular l orbital shape (0, 1, 2, and 3 momentum, l defines momentum integers from 0 to n-1 correspond to s, p, d, and f shape orbitals, respectively.) Magnetic Orbital magnetic integers from -l to 0 to +l orbital orientation in space defines ml ml orientation direction of e- spin spin m +1/2 or -1/2 Electron s Spin, ms defines spin Quantum Allowed Schrodinger’s equation gives an exact solution for H- Possible Orbitals Number Values atom, but does not for many electron-atoms. Electron- n Positive integers 1 2 3 electron repulsion in multi-electron split energy levels. 1,2,3,4.... Hydrogen Atom Multi-electron atoms 0 0 1 0 1 2 l 0 up to max Orbitals are of n-1 “degenerate” or Energy the same energy in Hydrogen! ml -l,...0...+l 0 0 -1 0 1 0 -1 0 1 -2-1 0 12 Orbital Name 1s 2s 2p 3s 3p 3d Electrons will Shapes or Boundry fill lowest Surface Plots energy orbitals first! Inner core electrons “shield” or “screen” outer 4-quantum numbers specify all the information we can know the energy electrons from the positive charge of the nucleus. and location of electrons around a nucleus. Chemists call this the “electronic structure of an atom”. Screening Impacts Remember this diagram 1) alters the energy electronic configurations. levels spacing & ordering in many- electron atoms. 2) outer e- screened by inner electrons. valence inner core electron electron The “Aufbau Process” is used to Chemists use “spdf notation” and “orbital box generate the electronic configuration diagrams” to symbolize the “ground state of elements filling the lowest energy electronic configuration” of elements. orbitals sequentially. 1. Lower energy orbitals fill first (smaller n). spdf orbital box Element 2. Hund’s Rule-degenerate (i.e. orbitals with Notation diagram the same energy) orbitals fill one at a time Arrow denotes an before electrons are paired in an orbital. electron with “spin 1 up” or “spin- 3. Pauli Exclusion Principle: No two electrons H 1s down”. in an atom can have same 4-quantum numbers. Remember, no two electrons can have 2 remember filling order He 1s the same 4 using this device! quantum numbers! electron shell principal quantum # orbital type # of electrons angular quantum # in orbital Building electronic configuration using Aufbau and Hund Atomic Orbital Box Full-electronic Condensed-electronic Atomic Orbital Box Full-electronic Condensed-electronic Number/Element Diagram configuration configuration Number/Element Diagram configuration configuration B 1s22s22p1 [He]2s22p1 1s1 1s1 H 2 2 2 2 2 C 1s 2s 2p [He]2s 2p He 1s2 1s2 1s22s22p3 [He]2s22p3 1s22s22p4 [He]2s22p4 Li 1s22s1 [He]2s1 1s22s22p5 [He]2s22p5 written with noble gas configuration 1s22s22p6 [He]2s22p6 Be 1s22s2 [He]2s2 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: Diamagnetic – All e- are paired. all electrons paired – Weakly repelled in a magnetic field. 2p • Paramagnetic atoms or ions: Paramagnetic – Unpaired e- exist in an orbital unpaired electrons – Attracted to an external magnetic field. 2p Representative or This periodic table shows the two “f-block series s-block (lanthanides 4f-block and actinides (5f) where it really Main Group Elements p-block is supposed to be, not removed from the table. 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) Ground State Electron Configurations of the Elements 6 What is the electron configuration of Mg? np 1 2 2 3 4 5 1 ns ns np np np np np 2 2 2 2 2 2 2 ns ns ns ns ns ns ns 1 5 10 odd behavior d d d odd behavior 1 5 d d What are the possible quantum numbers for the last (outermost) electron in Cl? 4f1 4f2 4f10 4f14 4f 5f 8.2 What is the electron configuration of Mg? Using the periodic table on the inside cover of the text and give the full and condensed electrons Mg 12 electrons configurations, partial orbital diagrams showing 1s < 2s < 2p < 3s < 3p < 4s valence electrons, and number of inner electrons for the following elements: 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 (a) potassium (K: (b) molybdenum (c) lead (Pb: Z Z = 19) (Mo: Z = 42) = 82) What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ! or -! (a) for K (Z = 19) There are 18 inner electrons. Metals loose electrons (oxidized) and become full configuration 1s22s22p63s23p64s1 cations. Non-metals gain electrons (reduced) and 1 condensed [Ar] 4s become anions. The electronic configuration of each orbital diagram reflects this change in the number of electrons. 4s1 3d 4p Na [Ne]3s1 Na+ [Ne] (b) for Mo (Z = 42) 36 inner electrons and 6 valence electrons Metals lose electrons so 2 2 6 2 6 2 10 6 1 5 full configuration 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d Ca [Ar]4s2 Ca2+ [Ar] that cation has a noble-gas condensed [Kr] 5s14d5 outer electron configuration. Al [Ne]3s23p1 Al3+ [Ne] partial orbital diagram 5s1 4d5 5p H 1s1 H- 1s2 or [He] (c) for Pb (Z = 82) 78 inner electrons and 4 valence electrons. Non-metals gain 1s22s22p63s23p64s23d104p65s24d10 F 1s22s22p5 F- 1s22s22p6 or [Ne] full configuration electrons so that anion 5p66s24f145d106p2 has a noble-gas outer O 1s22s22p4 2- 2 2 6 2 14 10 2 O 1s 2s 2p or [Ne] condensed [Xe] 6s 4f 5d 6p electron configuration. N 1s22s22p3 N3- 1s22s22p6 or [Ne] partial orbital diagram 6s2 6p2 Metals and non-metal ions tend to form electronic Isoelectronic species are two different elements states closest to their nearest noble gas with the same electronic configuration--but not the configuration. same nuclear configuration. oxidation Na: [1s22s22p63s1] =====> Na+: [1s22s22p6] = [Ne] oxidation Al: [1s22s22p63s23p1] =====> Al3+: [1s22s22p6] = [Ne] reduced N: [1s22s22p3] =====> N3-: [1s22s22p6] = [Ne] reduced O: [1s22s22p4] =====> O2-: [1s22s22p6] = [Ne] reduced F: [1s22s22p5] =====> F-: [1s22s22p6] = [Ne] Na+, Al3+, F-, O2-, and N3- 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 Write the full electronic configuration the following atom of a transition metal, electrons are removed ions: Sc+3, Zn+2,Co2+ and Co3+ . Distinguish if first from the ns orbital, then from the (n-1)d each is paramagnetic or diamagnetic. orbital. Transition Metal Transition Metal Cation (ns) (n-1)d Fe: [Ar]4s23d6 Fe2+: [Ar]4s03d6 or [Ar]3d6 Fe: [Ar]4s23d6 Fe3+: [Ar]4s03d5 or [Ar]3d5 Mn: [Ar]4s23d5 Mn2+: [Ar]4s03d5 or [Ar]3d5 Write the full electronic configuration the following A) What is the electron configuration of Mg ions: Sc+3, Zn+2,Co2+ and Co3+ . Distinguish if and Mg2+? each is paramagnetic or diamagnetic. B) What are the possible quantum numbers for the last (outermost) electron in Cl? C) Is ground state F paramagenetic or diamagnetic? Diamagnetic Sc3+ Paramagnetic Zn2+ , Co2+ , Co3+ What is the spdf and condensed electron Use condensed electron configurations to write the reaction configuration of Mg and Mg2+ ? Mg 12 electrons for the formation of each transition metal ion, and predict whether the ion is paramagnetic. 2 2 6 2 2 Mg 1s 2s 2p 3s [Ne]3s 2+ 2 2 6 0 0 Mg 1s 2s 2p 3s [Ne]3s = [Ne] (a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(Z = 80) What are the possible quantum numbers for the last (outermost) electron in Cl? Write the electron configuration and remove electrons starting Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s with ns to match the charge on the ion. If the remaining 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons configuration has unpaired electrons, it is paramagnetic. Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ! or -! C) Is ground state F paramagenetic or diamagnetic? Unpaired electron = 9F 1s 2s 2p PARAMAGNETIC Use condensed electron configurations to write the reaction Identify n and l quantum numbers for each of the for the formation of each transition metal ion, and predict following. whether the ion is paramagnetic. (a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(Z = 80) 2 Write the electron configuration and remove electrons starting 3p 4dz with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic. third shell fourth shell SOLUTION: 2+ 2 5 2+ 5 - (a) Mn (Z = 25) Mn([Ar]4s 3d ) Mn ([Ar] 3d ) + 2e paramagnetic What neutral element has the following orbital-filling (b) Cr3+(Z = 24) Cr([Ar]4s13d5) Cr3+ ([Ar] 3d3) + 3e- diagram? paramagnetic Gallium = Ga (c) Hg2+(Z = 80) Hg([Xe]6s24f145d10) Hg2+ ([Xe] 4f145d10) + 2e- not paramagnetic (is diamagnetic) Using
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