Presentation on theme: "Atomic Electron Configurations and Chemical Periodicity"— Presentation transcript:
1 Atomic Electron Configurations and Chemical Periodicity Chapter 8Atomic Electron Configurations and Chemical Periodicity
2 Chapter goalsUnderstanding the role magnetism plays in determining and revealing atomic structure.Understand effective nuclear charge and its role in determining atomic properties.Write the electron configuration of neutral atoms and monatomic ions.Understand the fundamental physical properties of the elements and their periodic trends.
3 Electron Spin and the Fourth Quantum Number The fourth quantum number is the spin quantum number which has the symbol ms.The spin quantum number only has two possible values.ms = +1/2 or −1/2ms = ± 1/2This quantum number tells us the spin and orientation of the magnetic field of the electrons.Wolfgang Pauli discovered the Exclusion Principle in 1925.No two electrons in an atom can have the sameset of 4 quantum numbers, n, l, ml, and ms
4 Electron Spin Spin quantum number effects: Every orbital can hold up to two electrons.Consequence of the Pauli Exclusion Principle.The two electrons are designated as havingone spin up ms = +1/2and one spin down ms = −1/2Spin describes the direction of the electron’s magnetic field.
5 Paramagnetism and Diamagnetism Unpaired electrons have their spins aligned or (in diff. orbitals)This increases the magnetic field of the atom.Total spin 0, because they add up.Atoms with unpaired electrons are called paramagnetic .Paramagnetic atoms are attracted to a magnet.
6 Paramagnetism and Diamagnetism Paired electrons have their spins unaligned . (in the same orbital)Paired electrons have no net magnetic field.Total spin = 0, because of cancellation,½ − ½ = 0Atoms with no unpaired electrons are called diamagnetic.Diamagnetic atoms are not attracted to a magnet.
7 Atomic Orbitals, Spin, and # of Electrons Because two electrons in the same orbital must be paired (due to Pauli’s Exclusion Principle), it is possible to calculate the number of orbitals and the number of electrons in each n shell.The number of orbitals per n level is given by n2 (see table at end of chapter 7.)The maximum number of electrons per n level is 2n2 (two electrons per orbital.)The value is 2n2 because of the two paired electrons per orbital.
8 #orbitals Max n shell l subshell ml #e– 1 K s 1 1 2 2 L s 1 2 8 4 1 p s1122Ls12841p–1,0,1363Ms121p18–1,0,13962d-2,-1,0,1,25104Ns121p–1,0,13632162d10-2,-1,0,1,253f-3,-2,-1,0,1,2,3714
9 Atomic Subshell Energies and Electron Assignments The principle that describes how the periodic chart is a function of electronic configurations is the Aufbau Principle.The electron that distinguishes an element from the previous element enters the lowest energy atomic orbital available.
10 Penetrating and Shielding the radial distribution function shows that the 2s orbital penetrates more deeply into the 1s orbital than does the 2pthe 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 nucleusthe 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 effectivelythe result is that the electrons in the 2s sublevel are lower in (more negative) energy than the electrons in the 2p
11 Atomic Subshell Energies and Electron Assignments The Aufbau Principledescribes the electronfilling order in atoms. Thisis product of the effectivenuclear charge, Z*, ZeffFor the same n, Z* is higherfor s orbital: s > p > d > fThen, e− in s is the mostattracted by nucleus and hasthe lowest energy
12 Atomic Subshell Energies and Electron Assignments One mnemonicto remember thecorrect fillingorder forelectronsin atoms is theincreasing(n + ) value
13 Atomic Subshell Energies and Electron Assignments or we can use this periodic chart
14 Atomic Electron Configurations Now we will use the Aufbau Principle to determine the electronic configurations of the elements on the periodic chart.1st row elements
15 Atomic Electron Configurations Hund’s rule tells us that the electrons will fill the p and d orbitals by placing electrons in each orbital singly and with same spin until half-filled. That is the rule of maximum spin. Then the electrons will pair to finish the p orbitals.Electrons inorbitals of or samekind, such as p or dorbitals, in the sameshell (n), have thesame energy; theare said to bedegenerate.
16 Atomic Electron Configurations 3rd row elements…
17 Atomic Electron Configurations 4th row elements…
18 Atomic Electron Configurations 4th row elements…
19 Atomic Electron Configurations 4th row elements…The five d orbitals are degenerate
20 Atomic Electron Configurations 4th row elements…
21 Atomic Electron Configurations 4th row elements…The five d orbitals are degenerate
22 Atomic Electron Configurations 4th row elements…
23 Atomic Electron Configurations 4th row elements… The [Ar] 4s1 3d5 configuration of Cr is more stable than [Ar] 4s2 3d4 (expected)
24 Atomic Electron Configurations 4th row elements… The [Ar] 4s1 3d10 full d configuration of Cu is more stable than [Ar] 4s2 3d9 (expected)
25 Atomic Electron Configurations 4th row elements…
26 Atomic Electron Configurations 4th row elements… (remember Hund’s rule): __ is better (lower energy) than __ __4p p
30 P 1 2 (P–1)d 3 4 5 6 7 (P)p (P)s (P–2)f Ba 56 Be 4 Mg 12 Ca 20 Sr 38 Ra882A1AH1Li3Na11K19Rb37Cs55Fr87Ga31In49Tl81B5Al133AGe32Sn50Pb82C6Si144AAs33Sb51Bi83N7P155ASe34Te52Po84O8S166ABr35I53At85F9Cl177AKr36Xe54Rn86Ne10Ar18He28A12(P–1)d33B4B5B6B7B8B8B8B1B2B212223242526272829304ScTiVCrMnFeCoNiCuZn394041424344454647485YZrNbMoTcRuRhPdAgCd577273747576777879806LaHfTaWReOsIrPtAuHg891041051061071081097(P)pAcRfDbSgBhHsMt(P)s5859606162636465666768697071CePrNdPmSmEuGdTbDyHoErTmYbLu90919293949596979899100101102103(P–2)fThPaUNpPuAmCmBkCfEsFmMdNoLr
31 Valence Electrons They determine the chemical properties of an electrons in shell with highest n, i.e., the outermost electrons, those beyond the core electrons1s2 2s2 2p6 3s11s2 2s2 2p6 3s2 3p21s2 2s2 2p6 3s2 3p6 3d10 4s2 4p61s2 2s21s2 2s2 2p6 3s2 3p6 4s2 3d7They determine the chemical properties of anelement. For the representative elements, theyare the ns and np electrons; for transitionelements they are the ns and (n−1)d electrons.
32 # of valence electrons = 1 4s1 P1A11s11H322s1Li1133s1Na# of valence electrons = 11944s1K3755s1Rb5566s1Cs8777s1Fr
33 # of valence electrons = 2 4s2 Be123s2Mg# of valence electrons = 2204s2Ca385s2Sr566s2Ba887s2Ra
34 # of valence electrons = 3 4s2 4p1 52s2 2p1B133s2 3p1Al# of valence electrons = 3314s2 4p1Ga495s2 5p1In816s2 6p1Tl
35 # of valence electrons = 7 92s2 2p5F173s2 3p5Cl354s2 4p5Br535s2 5p5I856s2 6p5AtFor the representative elements, the # of valence electrons = # of group
36 The element X has the valence shell electron configuration, ns2 np4. X belongs to what group?chalcogensBa56Be4Mg12Ca20Sr38Ra882A1AH1Li3Na11K19Rb37Cs55Fr87Ga31In49Tl81B5Al133AGe32Sn50Pb82C6Si144AAs33Sb51Bi83N7P155ASe34Te52Po84O8S166ABr35I53At85F9Cl177AKr36Xe54Rn86Ne10Ar18He28AHgLa57Sc21Y39893BTi22Zr40Hf721044BV23Nb41Ta731055BCr24Mo42W741066BMnTcRe2543751077BFeOs26Ru44761088BIr4577Co27Rh109Ni28Pd46Pt78Cu29Ag47Au791B30Cd48802BZnAcUnqUnpUnhUnsUnoUne
37 Energy (Orbital) Diagram 4p4s3d3p3sE2p2sBe 1s2 2s21s
42 Transition Metal Cations In the process of ionization transition metalsthe ns electrons are lost before the (n-1)dFe: [Ar] 3d6 4s2 Fe2+: [Ar] 3d6Fe2+: [Ar] 3d6 Fe3+: [Ar] 3d5Cu: [Ar] 3d10 4s1 Cu+: [Ar] 3d10Cu+: [Ar] 3d Cu2+: [Ar] 3d9Fe, Fe2+, Fe3+, Cu, and Cu2+ are paramagnetic
43 Two problems of ions, charge, and electron configuration An anion has a 3− charge and electron configuration1s2 2s2 2p6 3s2 3p6. What is the symbol of the ion?The neutral atom has gained 3e- to form the ion, thenthe neutral atom had 15 e-. In the neutral atom the # e-= # p+ = Atomic number, that is 15. The element is,then, phosphorus (phosphorus). Symbol of ion is P3−.A cation has a 2+ charge and its electronconfiguration is [Ar] 3d7. What is the symbol of the ion?Here, the neutral atom has lost 2e-. It is a transitionmetal, due to the 3d electrons. Remember they firstlylose e-s in 4s orbital. Symbol of ion is Co2+.Neutral atom has = 27 e- = 27 p+ = atomic #[Ar] 3d7 lost
45 Atomic Properties and Periodic Trends Establish a classification scheme of the elements based on their electron configurations.Noble GasesAll of them have completely filled electron shells. They are not very reactive.Since they have similar electronic structures, their chemical reactions are similar.He 1s2Ne [He] 2s2 2p6Ar [Ne] 3s2 3p6Kr [Ar] 4s2 4p6Xe [Kr] 5s2 5p6Rn [Xe] 6s2 6p6
46 Atomic Properties and Periodic Trends Representative Elements arethe elements in A groupson periodic chart.These elements will havetheir “last” electron in anouter s or p orbital.These elements have fairlyregular variations in theirproperties.Metallic character, for expl,increases from right to leftand top to bottom.
47 Atomic Properties and Periodic Trends d-Transition ElementsElements on periodicchart in B groups.Sometimes calledtransition metals.Each metal has d electrons.nsx (n-1)dy configurationsThese elements make the transition from metals to nonmetals.Exhibit smaller variations from row-to-row than the representative elements.
48 Atomic Properties and Periodic Trends f - transition metalsSometimes called innertransition metals.Electrons are being added to f orbitals.Electrons are being added two shells below the valence shell!Consequently, very slight variations of properties from one element to another.
49 Atomic Properties and Periodic Trends Outermost electrons (valence electrons)have the greatest Influence on the chemicalproperties of elements.
50 Atomic Properties and Periodic Trends Atomic radii describe therelative sizes of atoms.Atomic radii increase within acolumn going from the top tothe bottom of the periodic table.The outermost electrons areassigned to orbitals withincreasingly higher values of n.The underlying electronsrequire some space, so theelectrons of the outer shellsmust be further from thenucleus.
51 Atomic Properties and Periodic Trends Atomic radii decreasewithin a row going fromLeft to right on theperiodic table.This last fact seemscontrary to intuition.How does nature makethe elements smallereven though the electronnumber is increasing?
53 Atomic RadiiThe reason the atomic radii decrease across a period is due to shielding or screening effect.Effective nuclear charge, Zeff, experienced by an electron is less than the actual nuclear charge, Z.The inner electrons block the nuclear charge’s effect on the outer electrons.Moving across a period, each element has an increased nuclear charge and the electrons are going into the same shell (2s and 2p or 3s and 3p, etc.).Consequently, the outer electrons feel a stronger effective nuclear charge.For Li, Zeff ~ +1For Be, Zeff ~ — For B, Zeff ~ +3
54 Atomic RadiiExample: Arrange these elements based on their increasing atomic radii.Se, S, O, TeO < S < Se < TeIn the same group atomic size increasesas n (and Z) increases─ Br, Ca, Ge, FF < Br < Ge < Casame group same period
55 Ionization Energy First ionization energy (IE1) The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom to form a 1+ ion.Symbolically:Atom(g) + energy ion+(g) + e EndothermicMg(g) + 738kJ/mol Mg+ + e IE1= 738kJ/mol
56 Ionization Energy Second ionization energy (IE2) The amount of energy required to remove the second electron from a gaseous 1+ ion.Symbolically:ion+ + energy ion2+ + e-Mg kJ/mol Mg2+ + e IE2= 1451 kJ/molAtoms can have 3rd (IE3), 4th (IE4), etc. ionization energies. The values are consecutively getting larger.
57 Ionization Energy Periodic trends for Ionization Energy: 1) IE2 > IE1It always takes more energyto remove a second electronfrom an ion than from aneutral atom.2) IE1 generally increasesmoving from IA elements toVIIIA elements.Important exceptions at Be &B, N & O, etc. due to s and pand half-filled subshells.3) IE1 generally decreasesmoving down a family.IE1 for Li > IE1 for Na, etc
59 Sr < Ca < Mg < Be Ionization EnergyExample: Arrange these elements based on their (increasing) first ionization energies.Sr, Be, Ca, MgSr < Ca < Mg < BeAl, Cl, Na, PNa < Al < P < ClO, Ga, Sr, SeSr < Ga < Se < O
60 Ionization EnergyThe reason Na forms Na+ and not Na2+ is that the energy difference between IE1 and IE2 is so large.Requires more than 9 times more energy to remove the second electron than the first one.The same trend is persistent throughout the series.Thus Mg forms Mg2+ and not Mg3+.Al forms Al3+ and not Al4+.
61 H 1312 Ionization Energies (kJ/mole) HeLiBeBCNOFNeNaMgAlSiPSClArK
62 Electron Affinity (EA) Electron affinity is the amount of energy absorbed or emitted when an electron is added to an isolated gaseous atom to form an ion with a 1- charge.Sign conventions for electron affinity.If EA > 0 energy is absorbed (difficult)If EA < 0 energy is released (easy)Electron affinity is a measure of an atom’s ability to form negative ions.Symbolically:atom(g) + e- ion-(g) EA (kJ/mol)
63 Electron Affinity General periodic trend for electron affinity is the values become more negative from left to right across a period on the periodic chart (affinity for electron increases).the values become more negative from bottom to top at a group on the periodic chart.−Noble gases have EA > 0 (full electron confg)An element with a high ionization energy generally has a high affinity for an electron, i.e., EA is largely negative. That is the case for halogens (F, Cl, Br, I), O, and S.
64 Electron Affinity F (Z= 9) and Cl (Z = 17) have the most negative EA Noble gases, He (2), Ne (10), and Ar (18), EA > 0; also Be, Mg, NThey are all first Electron Affinity. A(g)- + e- A2-(g) EA2(kJ/mol) is the 2nd
65 Electron Affinity Two examples of electron affinity values: Mg(g) + e kJ/mol Mg-(g) EA = 231kJ/molBr(g) + e- Br-(g) kJ/mol EA = -323 kJ/molBr has a larger affinity for e− than Mg. The greater the affinity an atom has for an e− , the more negative EA is, the smaller it is.
66 Ionic RadiiCations (positive ions) are always smaller than their respective neutral atoms. When one or more electrons are removed, the attractive force of the protons is now exerted on less electrons.ElementNa11 p+, 11e-Mg12p+, 12 e-Al13 p+, 13e-Atomic Radius (Å)1.861.601.43IonNa+11 p+, 10e-Mg2+12 p+, 10 e-Al3+13 p+, 10e-IonicRadius (Å)1.160.850.68
67 Ionic Radii nine electrons ten electrons Anions (negative ions) are always larger than their neutral atoms.F 1s2 2s2 2p5 + e− F− 1s2 2s2 2p6 same Znine electrons ten electronsElementN7 p+, 7e-OFAtomicRadius(Å)0.750.730.72IonN3-7 p+, 10e-O2-8 p+, 10e-F−9 p+, 10e-Ionic1.71The three1.26ions are1.19isoelectronic
68 Ionic RadiiCation (positive ions) radii decrease from left to right across a period.Increasing nuclear charge attracts the electrons and decreases the radius.Rb+ and Sr2+ are isoelectronic, same # of e-sIonRb+Z = 37 p+Sr2+Z = 38 p+In3+Z = 49 p+IonicRadii(Å)1.661.320.94
69 Ionic RadiiAnion (negative ions) radii decrease from left to right across a period.Increasing electron numbers in highly charged ions cause the electrons to repel and increase the ionic radius.For these isoelectronic anions…10 e− and 7 p p p+IonN3-O2-F−IonicRadii(Å)1.711.261.19
70 Ionic Radii Example: Arrange these ions in order of decreasing radius. Ga3+, K+, Ca2+K+ > Ca2+ > Ga3+Cl−, Se2−, Br−, S2−Se2− > Br− > S2− > Cl−isoelectronic isoelectronic, same # of electronsSe2−(34 p+) > Br−(35 p+); they have 36 e− each.S2−(16 p+) > Cl−(17 p+); they have 18 e− each.Br− > S2− because Br− is in the 4th period, S2− is in the 3rd.
71 Ionic Radii of isoelectronic species Isoelectronic species have the same number of electrons. Here are some examples with the number of (protons) and + or − chargesN3−(Z=7) > O2−(Z=8) > F−(Z=9) > Ne(Z=10) neutral >Na+(Z=11) > Mg2+(Z=12) > Al3+ (Z=13) all have 10e−The nuclear charge (+) increases from left to right, so does attraction force to electrons: r decreases.S2−(Z=16) > Cl− (Z=17) > Ar0 (Z=18) > K+ (Z=19) >Ca2+ (Z=20) > Sc3+ (Z=21) all of them have 18e−