Electron Configuration and Chemical Periodicity
Orbital approximation: the e − density of an isolated many-electron atom is
approximated by the sum of the e − densities of each of the individual e − taken
separately.
1. Every e − in an atom has a set of four quantum numbers (n, l, ml, ms) that describe
its spatial distribution and spin state.
2. Every e − dwells in an atomic orbital with a characteristic size, shape and energy
and has a spin (up or down).
3. In many-electron atoms the energies of the orbitals do not depend solely on n as
in the one-electron atoms, but they depend on both n and l, but not on ml)
2p orbital has higher energy than a 2s
3d orbital has higher energy than a 3p
A group of orbitals with exactly equal energies comprise a subshell
Subshells having similar energies make up a shell of orbitals
Electrons with parallel spins ( ↑ ↑ ) tend to stay apart better than those with paired
spins ( ↑↓ ),
When e − - e − interactions become large the orbital approximation tends to break down.
Pauli Exclusion Principle : No two e − in an atom may have the same set of the four
quantum numbers (n, l, ml, ms).
Electron Configuration: list of the occupied orbitals and the number of e − in each.
Ground-State Configuration: electronic configuration of lowest energy.
Factors Affecting Atomic Orbital Energies
1. The Effect of Nuclear Charge (Zeffective)
Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing
nucleus-electron attractions
2. The Effect of Electron Repulsions (Shielding)
Additional electron in the same orbital (raises the orbital energy through electronelectron repulsions.)
1
Additional electrons in inner orbitals (shield outer electrons more effectively than do
electrons in the same sublevel. Shielding by inner electrons greatly lowers the Zeff felt
by outer electrons.)
Penetration and orbital energy
Order for filling energy sublevels
with electrons.
Aufbau Principle (aufbau means build-up in German)
We use it when we are placing electrons into orbitals in the
construction of polyelectronic atoms
This principle states that in addition to adding
protons and neutrons to the nucleus, one simply
adds electrons to the hydrogen-like atomic orbitals
Once more: Pauli exclusion principle: No two electrons
mayhave the same quantum numbers. Therefore, only
two electrons can reside in an orbital (with opposing spin)
p subshells have three orbitals with the same energy
d subshells have five orbitals with the same energy
f subshells have seven orbitals with the same energy
Each of these orbitals may accommodate a maximum of
two electrons.
2
Filling order of the Periodic Table
Orbitals are filled starting from the lowest energy.
•
Example: Hydrogen
1s1
•
1s
2s
Example: Helium (Z = 2)
2p
1s2
1s
2s
2p
3
•
Lithium (Z = 3)
1s22s1
1s
•
2s
2p
Berillium (Z = 4)
1s22s2
1s
•
2s
2p
Boron (Z = 5)
1s22s22p1
1s
•
2s
2p
Carbon (Z = 6)
1s22s22p2
1s
2s
2p
Hund’s Rule: Lowest energy configuration is
the one in which the maximum number of unpaired electrons
are distributed amongst a set of degenerate orbitals.
•
Nitrogen (Z = 7)
1s22s22p3
1s
2s
2p
4
A vertical orbital diagram for the Li ground state
Determining Quantum Numbers from Orbital
Diagrams
PROBLEM:
PLAN:
Write a set of quantum numbers for the third electron and a set
for the eighth electron of the F atom.
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
ml = -1, 0, or +1
ms= + or -
Oxygen (Z = 8)
1s22s22p4
•
1s
2s
2p
Fluorine (Z = 9)
1s22s22p5
1s
•
2s
2p
Neon (Z = 10)
1s22s22p6
1s
2s
2p
full
5
1
2
•
Sodium (Z = 11)
1s22s22p63s1
Ne
[Ne]3s1
3s
•
Argon (Z = 18)
[Ne] 3s23p6
Ne
3s
3p
6
Condensed ground-state electron configurations in the first three periods
7
A periodic table of partial ground-state electron configurations
The relation between orbital filling and the periodic table
8
Keep in mind: Elements in a group have similar chemical properties because they have
similar outer electron configurations.
Categories of electrons
1. Inner (core) electrons are those seen in the previous noble gas and any
completed transition series. They fill all the lower energy levels of an atom.
2. Outer electrons are those in the highest energy level (hi ghest n value). They
spend more of their time farthest from the nucleus.
3. Valence electrons are those involved in forming compounds. Among the maingroup elements, the valence electrons are the outer electrons. For the transition
elements, all the (n -1)d electrons are counted among the valence electrons also,
even though the elements Fe (Z = 26 through Zn (Z = 30) use only a few of them
in bonding as we will see later towards the last weeks of the course.
Key information is embedded in the periodic table
1. Among the main-group elements (A groups), the group number equals the
number of outer electrons (those with the highest n). Chlorine (Group 7A) has 7
outer electrons; Tellurium (Group 6A) has 6 outer electrons.
2. The period number is the n value of the highest energy level.
3. The n value squared (n2) gives the total number of orbitals and 2n2 gives the
maximum number of electrons in the energy level.
Unusual Configurations: Transition and Inner Transition Elements
Periods 4,5, 6 and 7 incorporate the d – block transition elements. The general trend is
to fill the (n -1)d orbitals between the ns and np orbitals. Period 5 follows the same
general pattern as Period 4. In Period 6, the 6s sublevel is filled in Cs and Ba and then
La (Z = 57), the first member of the 5d transition series, occurs. (Inner transistion
elements). For these elements filling of the f orbitals intervenes.
f orbitals: l = 3 ml = -3,-2,-1,0,+1,+2,+3 (7 orbitals).
The Period 6 inner transition series fills the 4f orbitals and consists of the lanthanides
(or rare earths) because they occur after and are similar to La. The other inner transition
series holds the actinides which fill the 5f orbitals that appear in Period 7.
In both series the (n – 2)f orbitals are filled after which the (n -1)d orbitals proceeds.
Period 6 ends proceeds with filling the 6p orbitals but Period 7 is incomplete because
only four elements with 7p electrons have been synthesized so far,
Anomalies Cr: 4s13d5 Mo: 5s14d2 Cu: 4s13d10 Ag: 5s13d10 Au: 6s13d10
9
Determining Electron Configuration
PROBLEM:
Using the periodic table on the inside cover of the text give the full
and condensed electron configurations, partial orbital diagrams
showing valence electrons, and number of inner electrons for the
following elements:
(a) Potassium (K; Z = 19)
PLAN:
(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.
SOLUTION:
(a) for K (Z = 19)
1s22s22p63s23p64s1
full configuration
condensed configuration
[Ar] 4s
1
There are 18 inner
electrons.and 1 valence
electron
partial orbital diagram
(b) for Mo (Z = 42)
1s22s22p63s23p64s23d104p65s14d5
full
condensed configuration [Kr] 5s14d5
partial orbital diagram
There are 36 inner electrons
and 6 valence electrons.
(c) for Pb (Z = 82)
full
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
condensed configuration
partial orbital diagram
[Xe] 6s24f145d106p2
There are 78 inner electrons
and 4 valence electrons.
10
Trends in three key atomic properties
All physical and chemical behavior of the elements is based ultimately on the
electron configurations of their atoms.
Atomic size: how closely one atoms lies next to another.
Defining covalent and metallic radii
The metallic radius is ½ the distance between
nuclei of adjacent atoms in a crystal of the element.
The covalent radius is ½ the distance between
bonded nuclei in a molecule of the element.
C; in a covalent compound, the bond length and
known covalent radii are used to determine other
radii.
Bond length of C – Cl : 177 pm
Covalent radius of Cl: 100 pm
∴ covalent radius of C: 177 – 100 = 77 pm
Trends among main-group elements
1. Changes in n: as n increases the probability that
the outer electrons spend more time farther from the
nucleus increases and the atoms are larger.
2. Changes in Zeff : as the Zeff - the positive charge
“felt” by an e − - increases, outer e − are pulled
closer to the nucleus, the atoms are smaller.
a) Down a group, n dominates. Atomic radius
generally increases in a group from top to bottom.
b) Across a period Zeff dominates. Atomic radius
generally decreases in a period from left to right.
Trends among the transition metals
As we move from left to right, size shrinks through
the first two or three transition elements because of
the increasing Zeff. But from then on, the size
remains relatively constant because shielding by the inner d electrons counteracts the
increasing Zeff.
More on this when we will discuss the transition metal compounds.
11
Atomic radii of the main-group and transition elements
Periodicity of atomic radius
Trends in Ionization Energy
The ionization energy (IE) is the energy in kJ required for the complete removal of 1
mol of electrons from 1 mol of gaseous atoms or ions.
Many-electron atoms can lose more than one electron. The first ionization energy (IE1)
removes an outermost electron (highest sub-level) from the gaseous atom:
Atom (g) → ion+ (g) + e − ∆E = IE1
The second ionization energy (IE2) removes a second electron:
Ion+ (g) → ion2+ (g) + e − ∆E = IE2
(IE2 > IE1)
12
Periodicity of first ionization energy (IE1)
The lowest values occur for the alkali metals and the highest for the noble gases.
First ionization energies of the main-group elements
In general, IE
a) Decreases down a group
b) Increases across a
period (since Zeff increases
and atomic size decreases)
However there are several
small dips as shown in the
adjacent figure
For Be:
13
EXAMPLE
Identifying an Element from Successive
Ionization Energies
PROBLEM: 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.
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.
Trends in Electron Affinity
The electron affinity (EA) is the energy change in kJ accompanying the addition of 1
mol of electrons to 1 mol of gaseous atoms or ions.
The first EA refers to the formation of 1 mol of monovalent (1 –) gaseous anions:
Atom (g) + e − → ion- (g) ∆E = EA1
In most cases energy is released when an e − is added because it is attracted to the
atom’s nuclear charge. The second EA2 must always be positive because energy must be
absorbed to overcome electrostatic repulsions and add an e − to a negative ion.
14
Electron affinities of the main-group elements
Despite the irregularities
a) elements in group 6A and
especially in 7A have high IE and
high EA. They lose e − with
difficulty but attract e − strongly. In
their ionic compounds they form
negative ions.
b) Elements in groups 1A and 2A
have low IE and slightly negative
EA. They lose e − readily but attract
e − very weakly. In their ionic
compounds they form positive ions.
c) Noble gases (Group 8A) have
high IE and slightly positive EA.
They tend not to lose or gain e − .
Trends in three atomic properties
Trends in metallic behavior
15
Example
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
These three elements are all in Group 8A(18), IE
decreases down a group.
(b) Te > Sb > Sn
These are all in Period 5, 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 furtther to the left and
down one period.
Acid-base behavior of the element oxides
• Most main-group metals transfer e − to oxygen, so their oxides are ionic. In water
the oxides act as bases producing OH − ions from O 2− and reacting with acids.
• Non-metals share e − with oxygen, so non-metal oxides are covalent. In water
they act as acids producing H+ ions and reacting with bases.
• Some metals and many metalloids form oxides that can act as acids or as bases in
water. They are called amphoteric (Al2O3)
The trend in acid-base behavior of element oxides
As the elements become more metallic down
a group, the oxides become more basic (blue)
As the elements become less metallic across a
period, their oxides become more acidic (red)
Sb2O5: weakly basic, SiO2, As2O3: weakly
acidic
16
Writing Electron Configurations of Main-Group
Ions
PROBLEM: Using condensed electron configurations, write reactions for the
formation of the common ions of the following elements:
(a) Iodine (Z = 53)
(b) Potassium (Z =19)
(c) Indium (Z = 49)
PLAN: Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) are
usually isoelectronic with the nearest noble gas.
Metals in Groups 3A(13) to 5A(15) lose the np and ns or just the np
electrons.
SOLUTION:
(a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be
I- ([Kr]5s24d105p6)
isoelectronic with Xe: I ([Kr]5s24d105p5) + e(b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic
K+ ([Ar]) + ewith Ar: K ([Ar]4s1)
(c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three
In+ ([Kr]5s24d10) + eelectrons: In ([Kr]5s24d105p1)
2
10
1
In ([Kr]5s 4d 5p )
In3+([Kr] 4d10) + 3e-
The Period 4 crossover in sublevel energies
For main-group, s-block metals remove all e − with the
highest n value
For main-group, p-block metals remove np e − before ns e −
For transition (d-block) metals, remove ns e − before (n-1)d
For non-metals, add e − to the p orbitals of highest n value
Magnetic Properties
A species with unpaired e − exhibits paramagnetism, attracted by an external magnetic
field. A species with all electrons paired exhibits diamagnetism, it is not attracted and,
in fact, is slightly repelled by a magnetic field.
17
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.
(a) Mn2+ (Z = 25)
PLAN:
(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:
(a) Mn2+(Z = 25) Mn ([Ar] 4s23d5)
Mn2+ ([Ar] 3d5) + 2e-
(b) Cr3+(Z = 24) Cr ([Ar] 4s13d5)
Cr3+ ([Ar] 3d3) + 3e-
(c) Hg2+(Z = 80) Hg ([Xe] 6s24f145d10)
paramagnetic
paramagnetic
Hg2+ ([Xe] 4f145d10) + 2e-
not paramagnetic (diamagnetic)
18
Ionic radius
Ionic vs. atomic radius
• Ionic size increases down a group
• Ionic size decreases across a period but increases from cations to anions
• Ionic size decreases with increasing positive charge in an isoelectronic species
and the opposite
• Ionic charge decreases as charge increases for different cations of the same
element
Example
Rank each set of ions in order of decreasing size, and explain your ranking:
(a) Ca2+, Sr2+, Mg2+ (b) K+, S2-,, Cl- (c) Au+, Au3+
(a) Sr2+ > Ca2+ > Mg2+ These are members of the same Group 2A(2), and decrease in
size going up the group.
(b) 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.
(c) Au+ > Au3+ The greater the + charge, the smaller the ion
19
The Periodic Table
Types of Chemical Bonding
1. Metal with nonmetal: electron transfer and ionic bonding
2. Nonmetal with nonmetal: electron sharing and covalent bonding
3. Metal with metal: electron pooling and metallic bonding
20
Lewis Electron-Dot Symbols
For main group elements → The A group number gives the number of valence electrons
→ Place one dot per valence electron on each of the four sides of the element symbol
→ Pair the dots (electrons) until all of the valence electrons are used
For Periods 2 and 3
Example
Use partial orbital diagrams and Lewis symbols to depict the formation of Na+ and O2ions from the atoms, and determine the formula of the compound the ions form.
21
The Ionic Bonding Model
Central idea: transfer of e − from metal atoms to nonmetal atoms to form ions that come
together in a solid ionic compound.
Three ways to represent the formation of Li+ and F − through electron transfer
The total number of e − lost by the metal atoms equal the total number of e − gained by
the nonmetal atoms.
Energy consideration: the importance of lattice energy
Consider,
Li (g) → Li+ (g) + e − IE1 = 520 kJ
F (g) + e − → F − (g) EA = - 328 kJ
The two-step electron-transfer process by itself requires energy:
Li (g) + F (g) → Li+ (g) + F − (g) IE1 + EA = 192 kJ
The total energy needed is even greater than this because metallic Li and diatomic
fluorine must be first converted to separate gaseous atoms, which also requires energy.
o
The ∆H f of LiF (s) is – 617 kJ/mol, which means that 617 kJ of energy is released
when 1 mol of of LiF (s) forms from its elements.
This means that there must be some exothermic components large enough to overcome
the endothermic component discussed earlier. This component arises from the strong
attraction between many oppositely charged ions.
Li+ (g) + F − (g) → LiF (s) ∆H o = −755 kJ
And more energy is released when the gaseous ions coalesce in the crystalline structure
because each ion attracts others of opposite charge, ∆H o = −1050 kJ. The negative of
this value is 1050 kJ and it is called the lattice energy, the enthalpy change that occurs
22
when 1 mol of ionic solid separates into gaseous ions. It indicates the strength of ionic
interactions that influence melting point, solubility and other properties.
Periodic Trends in Lattice Energy
Coulomb’s law
charge A x charge B
Electrostatic force ∝
distance
energy = force x distance
therefore,
cation charge x anion charge
Electrostatic energy ∝
cation radius + anion radius
α ∆H0lattice
1. Effect of ionic size: as we move down a group in the periodic table, the ionic
radius increases and the electrostatic energy between cations and anions
decreases.
2. Effect of ionic charge:
LiF : 1050 kJ/mol, (1 x1 charge), (Li is +1, F is -1)
MgO: 3923 kJ/mol (2 x 2 charge) (Mg is +2, O is -2)
Lattice energy of MgO is about 4 times as of LiF.
23
Covalent Bond (sharing of e − between atoms)
Covalent bond formation in H2
Distribution of electron density of H2
24
Electronegativity – Linus Pauling (the greatest American chemist!)
It is a measure of the ability of an atom to attract e − to itself.
Robert Mulliken (U of Chicago) (1934):
electronegativity ( χ ) ∝
1
(IE1 + EA)
2
The Pauling electronegativity (EN) scale
Polar covalent bond: (dipole moment) and percent ionic character
Electron density distributions in H2, F2, and HF
dipole moment, µ
µ = (eδ) R
To denote polarity,
the arrow should point
toward the negative
end.
25
The ionic character of chemical bonds
Determining Bond Polarity from EN Values
PROBLEM:
PLAN:
(a) Use a polar arrow to indicate the polarity of each bond:
N—H , F—N, I—Cl.
(b) Rank the following bonds in order of increasing polarity:
H—N, H—O, H—C.
(a) Use above figure to find EN values; the arrow should point
toward the negative end.
(b) Polarity increases across a period.
SOLUTION: (a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0
N-H
F-N
I - Cl
(b) The order of increasing EN is C < N < O; all have an EN
larger than that of H.
H─C < H─N < H─O
26
Oxoacids and their strength
They contain the group
–X–O–H
Oxoacids of the same structure show increasing acid strength as the electronegativity of
the central atom increases. Their strength with a given central element increases with
the oxidation state of the central atom, or equivalently, with the number of lone oxygen
atoms attached to the central atom.
Example
Which is stronger? H2SO3 or H2SeO3
H2SO3 because S is more electronegative than Se.
HIO3 or HIO?
HIO3 = IO2(OH) - 2 more O
HIO =I(OH)
Example
Write the electron configuration of the first excited state of F-, O2F: 1s22s22p5
F-: 1s22s22p6
Excited state F-: 1s22s22p53s1
O: 1s22s22p4
O2-: 1s22s22p6
Excited state O2-: 1s22s22p53s1
27