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Announcements--Exam 3 Oct 3..............Includes chapters 7/8/9/10The excluded items include:1. Classical distinction between energy and matter (p. 217)2. Numerical problems involving the Rydberg equation (equations
7.3 and 7.4)3. Spectral analysis in the laboratory (pp. 226-227)4. Numerical problems involving the Heisenberg uncertainty
principle (p. 231)5. Trends among the transition elements (p. 261)6. Trends in electron affinity (pp. 265-266)7. Pseudo-noble gas configuration (p. 269)8. Lattice energy (pp. 283-285)9. IR spectroscopy (p. 292)10.Numerical problems involving electronegativity (p. 296)11.Electronegativity and oxidation number (p. 297)12.Section 11.3: MO theory and electron delocalization13.All sections in chapter 12 except 12.3
is
having
amplitude energy frequency wavelength
related by
units calleda wave and a particle
related by
Quanta
EM radiation
are
emittedabsorbed
electrons
involve energy changes in
atomsatoms
moleculesdescribed by
Wave functions e- configuration comprising
determined by
Aufbau Rules
E = hv c = !v
e- fillinggives
having quantum uunmbers
Wave Function (Orbital)
described by
described by
which are
spdf electronic configuration
Aufbau Rules
determined by
which involve
comprising
Core Electrons
Valence Electrons
Periodic Table
basis for
which summarizes
Periodic Properties
Hund’s Rule
OribitalEnergy
PauliExclusion
e- filling
Quantum Numbers
Quantum Numbers
Principaln = 1,2,3,..
Angular momentum, l
Magneticml
Spin, ms
defines
defines
defines
defines
defines
Orbital size & energy
Electron spin
Orbitalorientation
Orbitalshape
Chapter 8
Electronic Configuration and Periodicity
I. The Periodic Law and the Periodic Table
• 1864 Newland “Law of Octaves
• 1869 Dimitri Mendeleev and Lother Meyer
When the elements are arranged in order of increasing atomic mass, certain sets of chemical and physical properties recur periodically.
• 1913 Henry Mosely relates X-ray frequency to atomic number
When the Elements Were Discovered
1. Principal Quantum Number (n): Defines the size and energy level of the orbital. n = {1,2,3,4,.....}. Also called a shell (K = 1, L = 2, M = 3, N = 4, .....).
2. Angular Momemtum Quantum Number (l): Defines the “shape” of the orbital which is a volume in space where the electron is likely to be found. Also called a subshell. l = {0,1,2,3...up to n-1} where (0=s, 1=p, 2=d, 3=f)
3. Magnetic Quantum Number (ml): Defines the spatial orientation of an orbital of the same energy. ml = {-l, 0, +l}
4. Magnetic Spin Quantum Number (ms): Defines the orientation of “electron spin”. ms = {+1/2 or -1/2}.
Quantum numbers (n,l,ml,ms) specify “allowed states” or “orbitals” which are regions of space where electrons are likely to be found around the nucleus.
Electronic configuration of the elements: four quantum numbers describe an electron in a ground state atom.
Name
principal
Symbol Permitted Values Property
n positive integers (1,2,3,!)
orbital energy (size)
angular momentum
l integers from 0 to n-1 orbital shape
magnetic ml integers from -l to 0 to +l
orbital orientation in space
spin ms +1/2 or -1/2 direction of e- spin
The lowest energy (ground state) electronic configuration of all elements are constructed by filling lowest energy orbitals sequentially in what is called the “Aufbau Process”.
1. Lower energy (n-quantum number) orbitals fill first.
2. Hund’s Rule-orbitals fill one electron at a time before electrons are paired.
3. Pauli Exclusion Principle: No two electrons can have same 4-quantum numbers)
Electrons fill the lowest energy orbitals first, 2 at a time! 1s
2s
2p
3s3p4s
The order of filling of the orbitals can be remembered using a mnemonic device. Memorize this to help you!
For an Hydrogen atomorbital energy only dependson the n quantum number.
For many electrons atomsthe energy of an orbital or electron depends on bothn and l (3s < 3p < 3d)
Chemists use spdf notation and orbital box diagrams to denote or show the “ground state electronic configuration” of elements.
spdf Notation
orbital box diagram
H
He
1s1
1s2
Element
n principal quantum #
l quantum number # of electronsin orbital
Spin quantum number. An arrow denotes an electron with “spin up” (+1/2) or “spin-down” (-1/2).
The Pauli Exclusion principle states: “No two electrons can have the same 4-quantum numbers”. The spin numbers can not be the same (spin up and spin down allowed only).
Li 1s22s1
Example:
(n, l, ml and ms)
Atomic Number/Element
Orbital BoxDiagram
Full-electronicconfiguration
Condensed-electronicconfiguration
[He]2s1
The order of filling of the orbitals can be remembered using a mnemonic device. Memorize how to write it out as it determines electronic structure.
s-blockp-block
d-block
f-block
transition metals
inner transition metals
main groupmain group
H
He
Li
Be
1s1
1s2
1s22s2
1s22s1
Atomic Number/Element
Orbital BoxDiagram
Full-electronicconfiguration
Condensed-electronicconfiguration
1s1
1s2
[He]2s1
[He]2s2
written with noble gas configuration
Electronic configuration using Aufbau Process
1s22s22p3
1s22s22p4
1s22s22p5
1s22s22p6
1s22s22p1
1s22s22p2
B
C
Atomic Number/Element
Orbital BoxDiagram
Full-electronicconfiguration
Condensed-electronicconfiguration
[He]2s22p1
[He]2s22p2
[He]2s22p3
[He]2s22p4
[He]2s22p5
[He]2s22p6
Odd-filling behavior here!4th and 9th position.
Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p
• Diamagnetic atoms or ions:– All e- are paired.– Weakly repelled in a magnetic field.
• Paramagnetic atoms or ions:– Unpaired e- exist in an orbital– Attracted to an external magnetic field.
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.
Paramagnetic Diamagnetic
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.
Magnetic field off Magnetic field on Magnetic field on
Paramagentic Diamagentic
Fe: [Ar]4s23d6 Fe2+: [Ar]4s03d6 or [Ar]3d6
Mn: [Ar]4s23d5 Mn2+: [Ar]4s03d5 or [Ar]3d5
Fe3+: [Ar]4s03d5 or [Ar]3d5Fe: [Ar]4s23d6
When a 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.
Metals lose electrons so that cation has a noble-gas outer electron configuration.
Na [Ne]3s1 Na+ [Ne]
Ca [Ar]4s2 Ca2+ [Ar]
Al [Ne]3s23p1 Al3+ [Ne]
Non-metals gain electrons so that anion has a noble-gas outer electron configuration.
H 1s1 H- 1s2 or [He]
F 1s22s22p5 F- 1s22s22p6 or [Ne]
O 1s22s22p4 O2- 1s22s22p6 or [Ne]
N 1s22s22p3 N3- 1s22s22p6 or [Ne]
Metals loose electrons (oxidized) to become cations. Non-metals gain electrons to become anions. The electronic configuration of each reflects this change in the number of electrons.
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.
Isoelectronic species are two different elements with the same electronic configuration--but not the same nuclear configuration.
Na: [1s22s22p63s1] =====> Na+: [1s22s22p6] = [Ne]oxidation
oxidationAl: [1s22s22p63s23p1] =====> Al3+: [1s22s22p6] = [Ne]
N: [1s22s22p3] =====> N3-: [1s22s22p6] = [Ne]reduced
O: [1s22s22p4] =====> O2-: [1s22s22p6] = [Ne]reduced
F: [1s22s22p5] =====> F-: [1s22s22p6] = [Ne]reduced
Metals and non-metals form ions with electronic configurations closest to their nearest noble gas configuration.
3A 4A 5A 6A 7A 8A2A1A
Metals and non-metal ions tend to form electronic states closest to their nearest noble gas configuration.
What is the spdf and condensed electron configuration of Mg and Mg2+ ? Mg 12 electrons
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 electronsLast electron added to 3p orbitaln = 3 l = 1 ml = -1, 0, or +1 ms = ! or -!
C) Is ground state F paramagenetic or diamagnetic?
Mg 1s22s22p63s2 [Ne]3s2
Mg2+ 1s22s22p63s0 [Ne]3s0 = [Ne]
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)
(b) for Mo (Z = 42) 36 inner electrons and 6 valence electrons1s22s22p63s23p64s23d104p65s14d5
[Kr] 5s14d5
(c) for Pb (Z = 82) 78 inner electrons and 4 valence electrons.
[Xe] 6s24f145d106p2
condensedpartial orbital diagram
full configuration
condensed
partial orbital diagram
full configuration 1s22s22p63s23p64s23d104p65s24d10
5p66s24f145d106p2
6s2 6p2
5s1 4d5 5p
(a) for K (Z = 19)1s22s22p63s23p64s1
[Ar] 4s1
4s1
condensedorbital diagram
full configurationThere are 18 inner electrons.
3d 4p
Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic.
(a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(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.
SOLUTION:
paramagnetic(a) Mn2+(Z = 25) Mn([Ar]4s23d5) Mn2+ ([Ar] 3d5) + 2e-
(b) Cr3+(Z = 24) Cr([Ar]4s13d5) Cr3+ ([Ar] 3d3) + 3e-
paramagnetic
(c) Hg2+(Z = 80) Hg([Xe]6s24f145d10) Hg2+ ([Xe] 4f145d10) + 2e-
not paramagnetic (is diamagnetic)
Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic.
(a) Mn2+(Z = 25) (b) Cr3+(Z = 24)
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) Hg2+(Z = 80)
Identify n and l quantum numbers for each of the following.
What neutral element has the following orbital-filling diagram?
fourth shellthird shell
Identify n and l quantum numbers for each of the following.
What neutral element has the following orbital-filling diagram?
fourth shellthird shell
3p 4dz2
Gallium = Ga
Many atomic properties show periodicity and trends.
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
Many atomic properties show periodicity and trends.
1) Inner core electrons : electrons filling the lower n shells of an element. They are located closer to the nucleus.
2. Outer core or VALENCE e- : those e- in the highest energy level (highest n-value). The number of valence e- is given by the Group Number in the periodic table for Group A.
--Responsible for chemistry and bonding of elements forming compounds or ions (true for representative but not transition metals--more complex).
Electrons in elements are categorized either as inner core electrons or valence electrons.
Periodicity in the chemical reactivity of elements occurs because of periodicity in the electronic structure of valence electrons!
1-electron outer s-orbital
2-electrons outer d-orbital
5-electrons outer p-orbital
Inner core electrons “shield” outer electrons from the positive charge of the nucleus.
Effective nuclear charge (Zeff) is the electrostatic force felt by the outer valence electrons taking into “shielding” by core electrons.
To a good approximation: effective nuclear charge, Zeff is given by:
Zeff = Z – core e-
# of inner non-valence electrons
# protonsEffective Nuclear charge
Bigger Zeff means more “pull” or electrostatic force between nucleus and electrons.
Configuration Element Z (p+)Core
ElectronsValence
Electrons ZeffectiveRadius(pm)
[Ne]3s1 Na 11 10 1 1 186[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
[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 1 227
[Ar]4s2 Ca 20 18 2 2 197
[Ar]4s23d1 Sc 21 18 3 3 135
Zeff = Z – core e-****
Because of increasing effective nuclear charge across a period, atomic radii decrease across a Period. As n increases down a group so does the radius.
Incr
easi
ng A
tom
ic R
adiu
sDecreasing Atomic Radius
n increases
Many atomic properties show periodicity and trends.
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
Periodicity of Atomic Radius
Group I
Group VIII
Cations get smaller (greater Zeff) Anions get larger (lower Zeff)
The radii of cations are smaller than their parent neutral atoms, while anions are larger than its parent.
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
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
SOLUTION:(a) Sr > Ca > Mg These elements are in Group 2A(2).
(b) K > Ca > Ga These elements are in Period 4.
(c) Rb > Br > KrRb has a higher n engery level and 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.
First ionization energies of the main-group elements. First Ionization Energy
Ionization energy is the minimum energy (kJ/mol) required to remove an 1 mole of electrons from one mole of a gaseous atom in its ground state (!H > 0).
I1 + X (g) X+(g) + e-
I2 + X (g) X2+(g) + e-
I3 + X (g) X3+(g) + e-
I1 first ionization energy
I2 second ionization energy
I3 third ionization energy
I1 < I2 < I3
Ranking Elements by First Ionization Energy
PLAN:
PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE1:
(a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs
IE decreases as you proceed down in a group; IE increases as you go across a period.
Ranking Elements by First Ionization Energy
Ranking Elements by First Ionization Energy
PLAN:
SOLUTION:
PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE1:
(a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs
IE decreases as you proceed down in a group; IE increases as you go across a period.
(a) He > Ar > Kr
(b) Te > Sb > Sn
(c) Ca > K > Rb
(d) Xe > I > Cs
Group 8A(18) - IE decreases down a group.
Period 5 elements - IE increases across a period.
Ca is to the right of K; Rb is below K.
I is to the left of Xe; Cs is furtther to the left and down one period.
1s2 2s2 2p3
1s2 2s2 2p1
1s2 2s2
1s2 2s1
1s2 2s2 2p2
1s2 2s2 2p4
The ionization energy increases dramatically when an core electron is removed from a non-valence shell.
Identifying an Element from Successive Ionization Energies
PLAN:
Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration:
IE1 IE2 IE3 IE4 IE5 IE6
1012 1903 2910 4956 6278 22,230
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.
IE6
22,230
SOLUTION:
The largest increase occurs after IE5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s23p3 and the element must be phosphorous, P (Z = 15).The complete electron configuration is 1s22s22p63s23p3.
Identifying an Element from Successive Ionization EnergiesName the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration:
IE1 IE2 IE3 IE4 IE5
1012 1903 2910 4956 6278
Main Group (or representative) metals form ionic basic oxides when reacted with oxygen while non-metals form covalent acidic oxides with oxygen.
Increasing Acidity
Li2O BeO B2O3 CO2 OF2
K2O
Cs2O
Rb2O
CaO
SrO
BaO
Ga2O3
Tl2O3
In2O3
GeO2
PbO2
SnO2
SeO3
TeO3
Br2O7
I2O7
6A(16)
7A(17)
4A(14)
3A2A1A
2
4
5
6
Increasing Basicity Na2O
IonicOxides
CovalentOxides basic acidic
Properties of Oxides Across a Period