Download Chapter 8 2013

described by
Chapter 8
Wave Function
(Orbital)
e- filling
spdf electronic
configuration
Core
Electrons
comprising
Valence
Electrons
determined by
described by
basis for
Aufbau
Rules
Quantum
Numbers
Periodic
Table
which involve
Orbital
Energy
which are
Pauli
Exclusion
Hund’s
Rule
which summarizes
Electron Configuration and Chemical Periodicity
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”.
Quantum
Numbers
Principal
n = 1,2,3,..
Orbital size
& energy
define
s
Name
defines
Angular
momentum, l
Magnetic
ml
Spin, ms
Quantum
Number
n
defines
Orbital
orientation
defines
Electron
spin
Allowed
Values
Positive integers
1,2,3,4....
Orbital
shape
defines
3
2
Symbol
principal
n
angular
momentum
l
Permitted Values
positive integers (1,2,3...)
integers from 0 to n-1
magnetic
ml
spin
ms
integers from -l to 0 to +l
+1/2 or -1/2
ml
0 up to max
of n-1
-l,...0...+l
Orbital Name
Shapes or Boundry
Surface Plots
0
0
0
0 -1 0 1
0 -1 0 1 -2-1 0 12
2s
3s
1s
1
2p
0
1
3p
orbital energy (size)
orbital shape (0, 1, 2, and 3
correspond to s, p, d, and f
orbitals, respectively.)
orbital orientation in space
direction of e- spin
2
3d
Orbitals are
“degenerate” or
the same energy in
Hydrogen!
Electrons
will
fill lowest
energy
orbitals first!
Multi-electron atoms
Energy
Hydrogen Atom
l
Property
Schrodinger’s equation gives an exact solution for Hatom, but does not for many electron-atoms. Electronelectron repulsion in multi-electron split energy levels.
Possible Orbitals
1
Periodic
Properties
Inner core electrons “shield” or “screen” outer
electrons from the positive charge of the nucleus.
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”.
Screening Impacts
1) alters the energy
levels spacing &
ordering in manyelectron atoms.
2) outer e- screened by
inner electrons.
Remember this diagram
electronic configurations.
valence
inner core electron
electron
The “Aufbau Process” is used to
generate the electronic configuration
of elements filling the lowest energy
orbitals sequentially.
Chemists use “spdf notation” and “orbital box
diagrams” to symbolize the “ground state
electronic configuration” of elements.
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.
Element
remember filling order
using this device!
spdf
Notation
H
1s1
He
1s2
orbital box
diagram
Remember, no two
electrons can have
the same 4
quantum numbers!
electron shell
principal quantum #
orbital type
angular quantum #
Building electronic configuration using Aufbau and Hund
Atomic
Number/Element
H
Orbital Box
Diagram
Full-electronic
configuration
1s1
He
1s2
Li
1s22s1
Condensed-electronic
configuration
Be
Atomic
Number/Element
1s2
[He]2s1
[He]2s2
Orbital Box
Diagram
# of electrons
in orbital
Full-electronic
configuration
Condensed-electronic
configuration
B
1s22s22p1
[He]2s22p1
C
1s22s22p2
[He]2s22p2
1s22s22p3
[He]2s22p3
1s22s22p4
[He]2s22p4
1s22s22p5
[He]2s22p5
1s22s22p6
[He]2s22p6
1s1
written with noble
gas configuration
1s22s2
Arrow denotes an
electron with “spin
up” or “spindown”.
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)
ns2np6
ns2np5
ns2np4
ns2np3
ns2np2
ns2np1
d10
odd behavior
d5
What is the electron configuration of Mg?
d5
d1
d1
ns22
ns
ns1
odd behavior
Ground State Electron Configurations of the Elements
What are the possible quantum numbers for
the last (outermost) electron in Cl?
4f1
4f2
4f14
4f10
4f
5f
8.2
What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2
[Ne] 1s22s22p6
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
2 + 2 + 6 + 2 + 5 = 17
electrons
Last electron added to 3p orbital
1s22s22p63s23p5
n = 3 l = 1 ml = -1, 0, or +1 ms = ! or -!
(a) for K (Z = 19) There are 18 inner electrons.
full configuration 1s22s22p63s23p64s1
condensed
[Ar] 4s1
orbital diagram
4s1
3d
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.
4p
(b) for Mo (Z = 42) 36 inner electrons and 6 valence electrons
full configuration 1s22s22p63s23p64s23d104p65s14d5
condensed
[Kr] 5s14d5
partial orbital diagram
5s1
4d5
6s2
6p2
Na+ [Ne]
Ca [Ar]4s2
Ca2+ [Ar]
Al [Ne]3s23p1
Al3+ [Ne]
5p
(c) for Pb (Z = 82) 78 inner electrons and 4 valence electrons.
1s22s22p63s23p64s23d104p65s24d10
full configuration
5p66s24f145d106p2
condensed
[Xe] 6s24f145d106p2
partial orbital diagram
Na [Ne]3s1
Non-metals gain
electrons so that anion
has a noble-gas outer
electron configuration.
Metals lose electrons so
that cation 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 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: [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
atom of a transition metal, electrons are removed
first from the ns orbital, then from the (n-1)d
orbital.
Transition Metal
Write the full electronic configuration the following
ions: Sc+3, Zn+2,Co2+ and Co3+ . Distinguish if
each is paramagnetic or diamagnetic.
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
ions: Sc+3, Zn+2,Co2+ and Co3+ . Distinguish if
each is paramagnetic or diamagnetic.
A) What is the electron configuration of Mg
and Mg2+?
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
configuration of Mg and Mg2+ ? Mg 12 electrons
Mg
1s22s22p63s2
Mg2+
1s22s22p63s0
[Ne]3s2
[Ne]3s0 = [Ne]
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 -!
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.
C) Is ground state F paramagenetic or diamagnetic?
Unpaired electron =
PARAMAGNETIC
9F
1s
2s
2p
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)
Identify n and l quantum numbers for each of the
following.
(c) Hg2+(Z = 80)
(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.
third shell
SOLUTION:
(a) Mn2+(Z = 25) Mn([Ar]4s23d5)
(b) Cr3+(Z = 24) Cr([Ar]4s13d5)
(c)
4dz2
3p
Hg2+(Z
= 80)
Hg([Xe]6s24f145d10)
Mn2+ ([Ar] 3d5) + 2eCr3+ ([Ar] 3d3) + 3eHg2+ ([Xe]
paramagnetic
paramagnetic
4f145d10)
fourth shell
What neutral element has the following orbital-filling
diagram?
Gallium = Ga
+ 2e-
not paramagnetic (is diamagnetic)
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
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
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.
Screening Impacts
1) alters the energy
levels spacing &
ordering in manyelectron atoms.
2) outer e- screened
by inner electrons.
1) Inner core electrons :
electrons residing in the lower
n shells of an element--located
closer to the nucleus.
n =2
n =3
d5
d5
To a good approximation: effective nuclear charge,
Zeff, is given by:
ns2np6
5
ns2np
ns2np4
3
ns2np
ns2np2
ns2np
The number of valence
electrons can be found by
counting the # of e- in the
outer most shell (main-group
elements only)
d1
ns22
ns
Effective nuclear charge (Zeff) is the electrostatic
force felt by the outer valence electrons taking into
“shielding” by internal core electrons.
ns1
n =6
d1
The number of valence
given by the Group Number in
the periodic table for Group A
representative elements.
n =5
1
valence
inner core electron
electron
n =4
e- is
d10
2. Outer core or VALENCE e- :
total number of e- in the highest
n-value shell.
Zeff = Z – core eEffective
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
8.2
Configuration
Element
****
Z (p+)
Core
Electrons
Valence
Electrons
Zeffective
Radius
(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
Atomic radii decrease across a Period because the
effective nuclear increases. Down a group the
atomic radi get larger!
Zeff Increases
Using only the periodic table rank each set
of main group elements in order of
decreasing atomic size:
Decreasing Atomic Radius
Increasing Atomic Radius
Increasing
n
The effective nuclear charge (Zeff) (“pull” on valence
electrons) increases across a period and upward in
a group! Know this trend and others follow!
increasing Zeff
Zeff = Z – core e-
(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
(b) K > Ca > Ga
(c) Rb > Br > Kr
(d) Rb > Sr > Ca
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 Zeff)
Anions get larger
(lower Zeff)
Ranking Ions by Size
Rank each set of ions in order of decreasing size,
and explain your ranking:
Ranking Ions by Size
Rank each set of ions in order of decreasing size,
and explain your ranking:
Sr2+ > Ca2+ > Mg2+
(a) Ca2+, Sr2+, Mg2+
(a) Ca2+, Sr2+, Mg2+
(b) K+, S2!, Cl!
These are members of the same Group 2A(2) and therefore
decrease in size going up the group.
(c)
Au+,
S2! > Cl! > K+
The ions are isoelectronic; S2! has the smallest Zeff and
therefore is the largest while K+ is a cation with a large Zeff
and is the smallest.
(b) K+, S2!, Cl!
Au3+
PLAN:
Compare positions in the periodic table,
formation of positive and negative ions and
changes in size due to gain or loss of
electrons.
(c) Au+, Au3+
Au+ > Au3+
The higher the + charge, the smaller the ion.
First ionization energies of the main-group elements.
“How hard it is to remove an electron”
Ionization energy is
the minimum energy
(kJ/mol) required to
remove 1 mole of efrom 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
and fluorine (i.e like to gain electroncs = reduced).
The increasing effective nuclear Charge (Zeff) 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
We can remove more than 1 electron from a
ground state atom. It requires more energy to
remove subsequent electrons.
I1 + X (g)
X+(g) + e-
I2 + X (g)
X2+(g) + e- I2 second ionization energy
I3 + X (g)
X3+(g) + e-
The ionization energy increases dramatically
when an core electron is removed from a nonvalence shell (blue area shows big jumps in IE).
I1 first ionization energy
1s2 2s1
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
Using the periodic table to rank the elements in
each set in order of decreasing IE1:
Ranking Elements by First Ionization Energy
Using the periodic table to rank the elements in
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
(d) I, Xe, Cs
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
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:
PLAN:
IE1
IE2
IE3
IE4
IE5
1012
1903
2910
4956
6278
IE6
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.
Name 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
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 electron configuration is: 1s22s22p63s23p3.
Metallic behavior increases as we move down a
group and from left to right on the periodic table.
Main Group (or representative) metals form ionic
basic oxides when reacted with oxygen while nonmetals form covalent acidic oxides with oxygen.
Increasing Acidity
Covalent
Oxides
--Metals have low IE
-- Tend to be
oxidized to
1A
Ionic
Oxides
metal ions.
--Charge is group
numbers
Increasing
Basicity
2 Li2O BeO
Properties of Oxides Across a Period
basic
acidic
3A
4A
(14)
B2O3 CO2
6A
(16)
7A
(17)
OF2
Na2O
CaO
Ga2O3 GeO2
SeO3 Br2O7
5 Rb2O SrO
In2O3 SnO2
TeO3 I2O7
6 Cs2O BaO
Tl2O3 PbO2
4 K 2O
Most metallic elements
2A