Ch 8 Atomic Electron Configuration and
Chemical Periodicity
Pauli Exclusion principle
No two electrons in an atom have the same set of quantum numbers.
If two electrons occupy the same orbital, they must have opposite spin.
Hund’s rule
When several orbitals of equal energy are available, as in a given sublevel,
electrons enter with parallel spins.
Electron configuration of atoms
Orbital diagrams of atoms
Electron configuration of atoms
Shows the numbers of electrons which are indicated by a superscript in each
sublevel, describing the arrangement of electron in atom)
1s22s22p63s23p64s23d104p65s14d5
Number of electron assigned
to the designated orbital
4d5
Quantum number set:
n = 4, l = 2, mi = varies,
ms = ± ½
Orbital type (l)
Electron Shell (n)
Orbital diagrams of atoms
Diagram shows how the electrons are distributed among orbitals. Each orbital is
represented by parentheses (), or box
or circle
and the electrons are
shown by arrows written as or depends on spin.
9F
1s
2s
2p
1
SAMPLE PROBLEM
PROBLEM:
Determining Electron Configuration
Using the periodic table, 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)
Use the atomic number for the number of electrons and the periodic table for
the order of filling for electron orbitals. Condensed configurations consist of
the preceding noble gas and outer electrons.
PLAN:
SOLUTION:
(a) for K (Z = 19)
full configuration
1s22s22p63s23p64s1
condensed configuration
[Ar] 4s1 (is also called abbreviated configuration- starts
with the preceding noble gas)
partial orbital diagram
Also called spdf notation
There are 18 inner electrons and one valence electron.
4s1
3d
SAMPLE PROBLEM
4p
continued
(b) for Mo (Z = 42)
full configuration
1s22s22p63s23p64s23d104p65s14d5
[Kr] 5s14d5
condensed configuration
There are 36 inner electrons
and 6 valence electrons.
partial orbital diagram
5s1
4d5
(c) for Pb (Z = 82)
full configuration
5p
comprehensive reading
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
condensed configuration
[Xe] 6s24f145d106p2
partial orbital diagram
There are 78 inner electrons
and 4 valence electrons.
6s2
6p2
2
SAMPLE PROBLEM
Determining Quantum Numbers from Orbital Diagrams
PROBLEM: Write a set of quantum numbers for the third electron and a set for the
eighth electron of the F atom.
PLAN: Use the orbital diagram to find the third and eighth electrons.
9F
1s
2s
2p
SOLUTION: The third electron is in the 2s orbital. Its quantum numbers are
n=
2
l=
0
ml =
0
ms= + or -1/2
The eighth electron is in a 2p orbital. Its quantum numbers are
n= 2
l= 1
SAMPLE PROBLEM
ml = -1, 0, or +1
ms= + or -1/2
Determining Electron Configuration of ions
PROBLEM: Using the periodic table, give the full and condensed electrons
configurations:
(a) Iron (K+, Z = 19)
(b) bromide (Br - Z=35)
Use the atomic number for the number of electrons
PLAN:
the periodic table for the order of filling for electron orbitals.
Condensed configurations consist of the preceding noble gas and outer electrons.
SOLUTION:
(a) for K+ (Z = 19),
full configuration
1s22s22p63s23p64s0
condensed configuration
[Ar] 4s0 (is also called abbreviated configuration- starts with the
preceding noble gas)
partial orbital diagram
3s2
3p6
There are 18 inner electrons
and 0 valence electrons.
3
SAMPLE PROBLEM
continued
(b) for Br - (Z = 35, 35+1 electrons, n= 2, s, p, d and orbitals are occupied)
full configuration
1s2 2s2 2p6 3s2 3p6 4s2 3d104p6
This electron configuration is the same as the noble gas Kr, krypton.
Partial Orbital diagram
4p6
3d10
4s2
SAMPLE PROBLEM
Writing Electron Configurations and Predicting
Magnetic Behavior of Transition Metal Ions
PROBLEM: Use condensed electron configurations to write the reaction for the
formation of each transition metal ion, and predict whether the ion is
paramagnetic (the unpaired electron display paramagnetism, it is attracted
by the external field).
(a) Mn2+(Z = 25)
PLAN:
(b) 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:
(a) Mn2+(Z = 25) Mn([Ar]4s23d5)
(c) Hg2+(Z = 80) Hg([Xe]6s24f145d10)
Mn2+ ([Ar] 3d5) + 2e-
paramagnetic
Hg2+ ([Xe] 4f145d10) + 2enot paramagnetic (is diamagnetic)
4
Apparatus for measuring the magnetic behavior of a sample.
Diamagnetic materials are slightly repelled by a
magnetic field and the material does not
retain the magnetic properties when the
external field is removed.
Diamagnetic materials are solids with all paired
electron resulting in no permanent net
magnetic moment per atom. Diamagnetic
properties arise from the realignment of the
electron orbits under the influence of an
external magnetic field. (copper, silver, and
gold).
Paramagnetic materials are slightly attracted by
a magnetic field and the material does not
retain the magnetic properties when the
external field is removed.
Paramagnetic properties are due to the presence of
some unpaired electrons, and from the
realignment of the electron orbits caused
by the external magnetic field.
(magnesium, molybdenum, lithium, and
tantalum).
Take-home message
Factors Affecting Atomic Orbital Energies
The Effect of Nuclear Charge (Zeffective)
Higher nuclear charge lowers orbital energy (stabilizes the
system) by increasing nucleus-electron attractions.
The Effect of Electron Repulsions (Shielding)
Additional electron in the same orbital
An additional electron raises the orbital energy through
electron-electron repulsions.
Additional electrons in inner orbitals
Inner electrons shield outer electrons more effectively than
do electrons in the same sublevel.
5
Take-home message
Condensed ground-state electron configurations in
the first three periods.
Take-home message
6
Take-home message
Take-home message
7
Atomic ionic radii of the main-group and transition elements.
Atomic radius- can be defined and
measured, assuming the atom is
spherical.
1. Atomic radii decrease across the
period from left to right.
2. Atomic radii increase down a group
from top to bottom.
3. The positive ions are always smaller
than atoms of the same element.
4. The negative ions are always larger
than atoms of the same element.
SAMPLE PROBLEM
Ranking Elements by Atomic Size
PROBLEM: Using only the periodic table, rank each set of main group elements in
order of decreasing atomic size:
(a) Ca, Mg, Sr
PLAN:
(b) K, Ga, Ca
(c) Br, Rb, Kr
(d) Sr, Ca, Rb
Elements in the same group increase in size and you go down;
elements decrease in size as you go across a period.
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 > Kr
Rb has a higher energy 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.
8
SAMPLE PROBLEM
Ranking Ions by Size
PROBLEM: Rank each set of ions in order of decreasing size, and explain your
ranking:
(a) Ca2+, Sr2+, Mg2+
PLAN:
(b) K+, S2-, Cl -
(c) Au+, Au3+
Compare positions in the periodic table, formation of positive and
negative ions and changes in size due to gain or loss of electrons.
SOLUTION:
(a) Sr2+ > Ca2+ > Mg2+
(b) S2- > Cl - > K+
(c) Au+ > Au3+
These are members of the same Group (2A/2) and
therefore decrease in size going up the group.
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.
The higher the + charge, the smaller the ion.
First ionization energies of the main-group elements.
•
ionization energy is the measure of how
difficult to remove an electron from the
gaseous atom.
•
Energy must be absorbed to cause
ionization, so the ionization energy is
always positive (endothermic).
•
The first ionization energy is the energy
change to remove the outer-most electron
for gaseous atom to form a +1 ion.
•
Ionization energy increases across the
periodic table from left to right.
•
Ionization energy decreases down from
top to bottom in the periodic table.
•
Ionization energy has unit of kJ.
9
SAMPLE PROBLEM
Ranking Elements by First Ionization Energy
PROBLEM: Using the periodic table only, rank the elements in each of the following
sets in order of decreasing IE1:
(a) Kr, He, Ar
PLAN:
(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.
SOLUTION:
(a) He > Ar > Kr
Group 8A(18) - IE decreases down a group.
(b) Te > Sb > Sn
Period 5 elements - IE increases across a period.
(c) Ca > K > Rb
Ca is to the right of K; Rb is below K.
(d) Xe > I > Cs
I is to the left of Xe; Cs is further to the left and down
one period.
Electron affinities (EA) of the main-group elements.
•
Electron affinity – the energy change
accompanying the addition of 1 mol of electrons
to 1 mol of gaseous atoms or ions.
•
The first electron affinity – accompanying the
formation of 1 mol of monovalent (1-) gaseous
anions.
Atom (g) + e - → ion - ∆E = EA1 (EA1 is
usually negative since the energy is released in
most of cases.)
•
Reactive nonmetals – the elements in 6A (16)
and 7A (17) has more negative EA and forms
negative ions.
•
Reactive metals – the elements in 1A (1) and 2A
(2) has slightly negative EA and forms positive
ions.
•
Noble gas – the elements in 8A (18) has slightly
positive EA and do not tend to lose or gain
electrons. Only larger atoms (Kr, Xe and Rn)
can form compounds.
10
Electronegativity
(EN) of the elements.
The reverse of
electronegativity, the
ability of an atom to lose
electrons, is known as
electropositivity
•
Electronegativity – measures the ability of an atom to attract to itself the electrons pair forming a
covalent bond.
•
The greater the electronegativity of an atom, the greater its attraction for electrons.
•
Electronegativity increases from left to right in the same period.
•
Electronegativity decreases from top to bottom in the same group.
•
Atoms with similar electronegativities will share an electron with each other to form a covalent bond.
•
If the atoms have significant different electronegativities, electrons will be permanently transferred to
one atom to form ionic bond.
Electronegativity (EN) of elements.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
1
H
2.20
He
2
Li
0.98
Be
1.57
B
2.04
C
2.55
N
3.04
O
3.44
F
3.98
Ne
3
Na
0.93
Mg
1.31
Al
1.61
Si
1.90
P
2.19
S
2.58
Cl
3.16
Ar
4
K
0.82
Ca
1.00
Sc
1.36
Ti
1.54
V
1.63
Cr
1.66
Mn
1.55
Fe
1.83
Co
1.88
Ni
1.91
Cu
1.90
Zn
1.65
Ga
1.81
Ge
2.01
As
2.18
Se
2.55
Br
2.96
Kr
3.0
5
Rb
0.82
Sr
0.95
Y
1.22
Zr
1.33
Nb
1.6
Mo
2.16
Tc
1.9
Ru
2.2
Rh
2.28
Pd
2.20
Ag
1.93
Cd
1.69
In
1.78
Sn
1.96
Sb
2.05
Te
2.1
I
2.66
Xe
2.6
6
Cs
0.79
Ba
0.89
*
Hf
1.3
Ta
1.5
W
2.36
Re
1.9
Os
2.2
Ir
2.20
Pt
2.28
Au
2.54
Hg
2.00
Tl
1.62
Pb
2.33
Bi
2.02
Po
2.0
At
2.2
Rn
7
Fr
0.7
Ra
0.9
**
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Uub
Uut
Uuq
Uup
Uuh
Uus
Uu
Lant
hani
des
*
La
1.1
Ce
1.12
Pr
1.13
Nd
1.14
Pm
1.13
Sm
1.17
Eu
1.2
Gd
1.2
Tb
1.1
Dy
1.22
Ho
1.23
Er
1.24
Tm
1.25
Yb
1.1
Lu
1.27
Acti
nide
s
**
Ac
1.1
Th
1.3
Pa
1.5
U
1.38
Np
1.36
Pu
1.28
Am
1.13
Cm
1.28
Bk
1.3
Cf
1.3
Es
1.3
Fm
1.3
Md
1.3
No
1.3
Lr
11