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

l 0 up to max 0 0 1 0 1 2 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 spdf notation write the complete Periodicity in the chemical reactivity of elements electron configuration of O, Cl, Ti, Zn? 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 2-electrons 5-electrons outer s-orbital outer d-orbital outer p-orbital We must know these properties and be able to apply Here’s a different view on one periodic table! Know them! 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

Electrons in elements are categorized either as Inner core electrons “shield” or “screen” outer inner-core electrons or outer valence-electrons. electrons from the positive charge of the nucleus.

1) Inner core electrons : Screening Impacts electrons residing in the lower 1) alters the energy n shells of an element--located levels spacing & closer to the nucleus. ordering in many- electron atoms. 2. Outer core or VALENCE e- : 2) outer e- screened total number of e- in the highest n =2 by inner electrons. n-value shell. n =3 n =4 valence electron The number of valence e- is n =5 inner core electron given by the Group Number in n =6 the periodic table for Group A representative elements.

Effective nuclear charge (Zeff) is the electrostatic The number of valence force felt by the outer valence electrons taking into electrons can be found by 6 np 1

- 2 counting the # of e in the 2 4 ns

“shielding” by internal core electrons. ns np np np np np 2 2 2 2 2 2 outer most shell (main-group 2 ns ns ns 3 ns ns 5 ns elements only) ns 1 To a good approximation: effective nuclear charge, 1 5 10 d d Zeff, is given by: d 1 5 d d

Zeff = Z – core e-

Effective Nuclear # protons # of inner core charge in atom non-valence electrons

Larger Zeff means more “pull” or electrostatic 4f force between nucleus and electrons. 5f

8.2 Z = Z – core e- The effective nuclear charge (Zeff) (“pull” on valence eff **** electrons) increases across a period and upward in Core Valence Radius Configuration Element Z (p+) Z Electrons Electrons effective (pm) a group! Know this trend and others follow! [Ne]3s1 Na 11 10 1 186 1 Zeff Increases [Ne]3s2 Mg 12 10 2 2 160 [Ne]3s23p1 Al 13 10 3 3 143 [Ne]3s23p2 Si 14 10 4 4 132

[Ne]3s23p3 P 15 10 5 5 128 eff [Ne]3s23p4 S 16 10 6 6 127 [Ne]3s23p5 Cl 17 10 7 7 99 [Ne]3s23p6 Ar 18 10 8 8 98 [Ar]4s1 K 19 18 1 227

1 increasing Z [Ar]4s2 Ca 20 18 2 2 197 [Ar]4s23d1 Sc 21 18 3 3 135

Atomic radii decrease across a Period because the Using only the periodic table rank each set effective nuclear increases. Down a group the of main group elements in order of atomic radi get larger! decreasing atomic size: Decreasing Atomic Radius

(a) Ca, Mg, Sr Increasing (b) K, Ga, Ca n (c) Br, Rb, Kr

(d) Sr, Ca, Rb Increasing Increasing Radius Atomic

Using only the periodic table rank each set of The atomic radii of cations are smaller than their main group elements in order of decreasing ground state atoms, while anions are larger than their atomic size: ground state atom (i.e. remove e- => smaller, add e- => (a) Ca, Mg, Sr bigger) Greater cation charge => smaller & vis versa

(b) K, Ga, Ca Cations get smaller Anions get larger (greater Z ) (c) Br, Rb, Kr eff (lower Zeff) (d) Sr, Ca, Rb SOLUTION: These elements are in Group 2A(2). (a) Sr > Ca > Mg These elements are in Period 4. (b) K > Ca > Ga Rb has a higher n engery level and (c) Rb > Br > Kr is far to the left. Br is to the left of Kr. (d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr. Ranking Ions by Size Ranking Ions by Size Rank each set of ions in order of decreasing size, Rank each set of ions in order of decreasing size, and explain your ranking: and explain your ranking:

(a) Ca2+, Sr2+, Mg2+ (a) Ca2+, Sr2+, Mg2+ Sr2+ > Ca2+ > Mg2+ These are members of the same Group 2A(2) and therefore (b) K+, S2!, Cl! decrease in size going up the group. (b) K+, S2!, Cl! S2! > Cl! > K+ + 3+ 2! (c) Au , Au The ions are isoelectronic; S has the smallest Zeff and + therefore is the largest while K is a cation with a large Zeff and is the smallest. PLAN: Compare positions in the periodic table, formation of positive and negative ions and (c) Au+, Au3+ Au+ > Au3+ changes in size due to gain or loss of The higher the + charge, the smaller the ion. electrons.

Ionization energy is First ionization energies of the main-group elements. the minimum energy “How hard it is to remove an electron” (kJ/mol) required to remove 1 mole of e- from one mole of a gaseous atom in its ground state. higher nuclear charge => smaller diameter => harder to remove e-

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-.

Electron affinity is largest (most negative) for chlorine The increasing effective nuclear Charge (Zeff) and and fluorine (i.e like to gain electroncs = reduced). 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 We can remove more than 1 electron from a The ionization energy increases dramatically ground state atom. It requires more energy to when an core electron is removed from a non- remove subsequent electrons. valence shell (blue area shows big jumps in IE).

+ - I1 + X (g) X (g) + e I1 first ionization energy

2+ - I second ionization energy I2 + X (g) X (g) + e 2

3+ - 1s2 2s1 I3 + X (g) X (g) + e I3 third ionization energy 1s2 2s2 1s2 2s2 2p1 1s2 2s2 2p2 1s2 2s2 2p3 I1 < I2 < I3 1s2 2s2 2p4

Ranking Elements by First Ionization Energy Ranking Elements by First Ionization Energy Using the periodic table to rank the elements in Using the periodic table to rank the elements in each set in order of decreasing IE1: each set in order of decreasing IE1: (a) Kr, He, Ar He > Ar > Kr (a) Kr, He, Ar Group 8A(18) - IE decreases down a group.

(b) Sb, Te, Sn (b) Sb, Te, Sn Te > Sb > Sn Period 5 elements - IE increases across a period. (c) K, Ca, Rb (c) K, Ca, Rb Ca > K > Rb (d) I, Xe, Cs 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 Identifying an Element from Successive Ionization Energies Ionization Energies Name the Period 3 element with the following ionization Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration: energies (in kJ/mol) and write its electron configuration:

IE1 IE2 IE3 IE4 IE5 IE6 IE1 IE2 IE3 IE4 IE5 IE6 1012 1903 2910 4956 6278 22,230 1012 1903 2910 4956 6278 22,230

PLAN: Look for a large increase in energy which indicates SOLUTION: that all of the valence electrons have been The largest increase occurs after IE , that is, after removed. 5 the 5th valence electron has been removed. Five The number valence electrons is reflected in the periodic table for Group A elements....find the electrons would mean that the valence 2 3 group with that number of valence electrons. configuration is 3s 3p and the element must be phosphorous, P (Z = 15). The electron configuration is: 1s22s22p63s23p3. Metallic behavior increases as we move down a Main Group (or representative) metals form ionic group and from left to right on the periodic table. basic oxides when reacted with oxygen while non- metals form covalent acidic oxides with oxygen. Covalent Increasing Acidity Oxides --Metals have low IE -- Tend to be 1A 2A 3A 4A 6A 7A oxidized to Ionic (14) (16) (17) metal ions. Oxides 2 Li2O BeO B2O3 CO2 OF2 --Charge is group Increasing numbers Na2O Basicity

4 K2O CaO Ga2O3 GeO2 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