Chapter 8 Electron Configuration and Chemical Periodicity
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.
ELECTRON SPIN & MAGNETISM
Electron Spin: you need to understand magnetism & ionization energy to understand electron spin and
quantum number ms.
Magnetism and what you already know: earth has a magnetic field, with a North Pole and South Pole, as do
metal magnets
Electromagnetism = magnetic field created by electrical current
Three definitions for magnetic substances:
paramagnetic = attracted to magnetic field
diamagnetic = not attracted to magnetic field
ferromagnetic = substance that retains its magnetism after being placed in magnetic field (like Fe, Co & Ni)
Figure 8.19 Apparatus for measuring the magnetic behavior of a sample
ELECTRON SPIN & MAGNETISM
Electrons are spinning charged particles which generate a tiny magnetic field
Only two orientations are possible and they are called spin up or spin down, with clockwise being
This is why the 4th Quantum number, ms, is assigned +½ or -½
ELECTRON SPIN
Data showed that the He atom was not affected by magnetic field - why?
There are 2 e-s present, must be that one is up and one is down, cancelling each others magnetic fields - we
say they are "paired"
It turns out that only an atom with one or more unpaired e-s exhibits paramagnetism
Figure 8.1 Observing the Effect of Electron Spin
ELECTRON SPIN
The lowest energy state for an e- in H is principal level n = 1
There is only one type of orbital at n = 1, because l = 0, which is an s or a spherical orbital
For H, the 1 e- is generally in the 1s orbital
In He, there are 2 e-s, and it turns out they are both in the 1s orbital & they are paired up, or coupled with one
spin up and one spin down
Li or Be: 3rd and 4th e-s are in n = 2, but which type of orbital, s or p?
For this we need to look at data for ionization energies
IONIZATION ENERGIES
Ionization energy (IE): energy required to take an e- away from an atom
First IE removes the e- furthest away from the nucleus
Examples: IE1 = 1.31 x 106 J/mol for H, 1.68 x 106 J/mol for Fe, and 0.50 x 106 J/mol for Na
IONIZATION ENERGIES (IE): IE in MJ/mol
e-#
1
2
3
4
5
6
7
Na
.5
4.6
6.9
9.5
13
17
20
Log 5.7
6.7
6.8
7.0
7.1
7.2
7.3
8
25
7.4
9
29
7.9
F
1.7
Log 6.2
92
8.0
106
8.03
3.4
6.5
6.1
6.8
8.4
6.9
11
7.0
15
7.2
18
7.3
10
141
8.2
11
178
8.3
Na: three energy levels present as shown by big changes in IE (from 1 to 2, 9 to 10)
F: two energy levels with slight difference within level - because of s & p orbital differences (7 to 8)
If all the IEs are mathematically adjusted for the increasing force of attraction between protons and remaining
electrons:
-then we find that the IE1 to IE5 in F are almost equal, meaning that these 5 e-s are "alike"
-and that IE6 = IE7 but > IE1 to IE5, so these two e-s are different
-and that IE8 = IE9 but >>> IE7, so these two e- levels are really different
What does all this mean?
IE data define the energy states and orbitals in the atom
At n = 1, there's one orbital with e-s that are very hard to remove
At n = 2, there are 4 orbitals, but they're different because s has different shape than p
Combine the IE data with the known pairing of e-s into 2e- per orbital from magnetic properties and we
determine that:
s orbital has up to 2 e-s
and each p orbital has up to 2 e-s for a total of 6
If F has 5 e- of similar IE – they must be in p orbitals, and they are the
easiest to remove so they must be outermost
Next 2 e-s are close to same energy, n = 2, but only 2 e-s, so they are in an s orbital
The last 2 e-s are very different & have much higher IE, must be close to nucleus, n = 1, 2 e-s in s orbital
This data reveals basic e- configuration of F: 1s has 2 e-s, 2s has 2 e-s, 2p has 5 e-s.
Now look at Na data:
First IE is < 2nd IE, but IE2 to IE9 are about the same; then big jump to IE10 and IE11
Means that: 1 e- is outermost at n = 3; then 8 e-s in n = 2 (notice slight diff for 2s & 2p);
n=1
IONIZATION ENERGIES
Now we look at first IE vs. atomic number
Noble gases have high IE
All of Group IA atoms have low IE, Group IIA has fairly low IE
Transition metals into same IE - these are d e-s
Periodic trend of IE - highest at He, lowest at Fr
Figure 8.10 Periodicity of first ionization energy
Figure 8.11 First ionization energies of the main-group elements
then 2 e-s in
Table 8.1 Summary of Quantum Numbers of Electrons in Atoms
QUANTUM NUMBERS
We know that e-s pair up into two per orbital maximum
Pauli Exclusion Principle - a statement of the facts: no two e-s in an atom can have the exact same set of four
quantum numbers
He 2 e-s:
n
l
ml
ms
1
0
0
+½
1
0
0
-½
ELECTRON CONFIGURATIONS
The best way to figure out quantum numbers is to know electron configuration, so we will do that first.
There are several “rules” or physical laws based on data like Ionization Energies
If 2p is at higher energy level than 2s, then 3p is higher than 3s
Also find that 3d is slightly higher than 3p
In multi-electron atoms:
- 4s slightly lower energy than 3d, so fill 4s before 3d
- always start at 1s, fill in according to increasing energy levels
Figure 8.3 Order for filling energy sublevels with electrons and a box “illustrating orbital occupancies” that you
draw in here:
Hund's Rule: max number of unpaired e-s will occur in ground state
Two methods: orbital box as seen in previous slide or spectroscopic (spdf)
1s2
Figure page 251 A vertical orbital diagram for the Li ground state (You will do horizontal orbital box notation)
__ __ _______
| | | | | | | | etc.
1s 2s
2p
ELECTRON CONFIG USING PERIODIC TABLE
Remembering all the rules and the order for filling orbitals looks difficult!
It turns out the Periodic Table is layed out in blocks: s block is groups 1 & 2; p block is groups 13 to 18; d block
is transition elements groups 3 - 12; and f block is inner transition.
You can figure out the electron config for the last e- in any element by looking at the periodic table. Then fill in
starting from H or nearest noble gas.
Figure 8.5 A periodic table of partial ground-state electron configurations
Figure 8.6 The relation between orbital filling and the periodic table
Practice: pair up and use a plain (not large) periodic table to do the spdf and orbital box notation for B, Ne and
Mg.
Sample Problem 8.2 (Give the full and condensed electrons configurations, partial orbital diagrams showing
valence electrons, and number of inner electrons for the following elements: (a) potassium (b) molybdenum (c)
lead
ELECTRON CONFIGURATIONS
For many e- atoms we can use a shorthand for either method called "noble gas core designation" or condensed
version
Try for examples: Cl and As
ELECTRON CONFIGURATIONS: VALENCE ELECTRONS
Electrons beyond noble gas core are valence electrons: e-s in outermost principal quantum level of atom
Practice with Na, As, Mn, and Pu
Then determine the principal quantum number of last electron in each of the above
ELECTRON CONFIGURATIONS
Hund’s Rule: max number of unpaired e-s will occur in ground state
Why Cr & Cu don't exactly follow the filling rules:
Cr is more stable with 1 e- per orbital including 4s
Cu is more stable with full d shell and 1 e- in 4s
BACK TO QUANTUM NUMBERS
If you have the electron configuration, you have the first two quantum numbers for each electron, n and l.
An s subshell has only one orbital with only one ml quantum number, 0. A p subshell has three orbitals, so you
have to list each one’s ml: -1, 0, +1. A d subshell has ml ranging from -5.,,,0,…+5; etc.
Within each orbital, the ms for the first e-is by definition is +½. The second e- is assigned –½.
Sample Problem 8.1 Determine a set of quantum numbers for the third electron and a set of quantum numbers
for the eighth electron in a fluorine atom.
My problem: Make a table of all four quantum numbers for every electron in vanadium.
n
l
ml
ms
n
l
ml
ms
Table 8.2 Partial Orbital Diagrams and Electron Configurations for the Elements in Period 3
Figure 8.4 Condensed ground-state electron configurations in the first three periods
Table 8.3 Paartial Orbital Diagrams and Electron Configurations for the Elements in Period 3
Table to remember energy levels IF you don’t have a periodic table handy! (You draw in the arrows.)
1s
2s 2p
3s 3p 3d
4s 4d 4p 4f
5s 5p 5d 5f (5g)
6s 6p 6d 6f (6g)
7s 7p 7d 7f
Follow arrows down and to left to fill in electron configuration.
HISTORY OF PERIODIC TABLE
Origin is based on "periodic properties" and relative masses
Johann Dobereiner grouped triads of elements with similar properties and increasing relative mass
In 1864, John Newlands conceived the idea of octaves, since the chem prop's seemed to repeat for every
eighth element
Current table: Julius Meyer and Dmitri Mendeleev
Periodic Trends are the result of atom's e- configuration - # of e-s or really # of protons since its arranged by
atomic number
Look at Argon's e- density vs. distance from nucleus
Not like a billiard ball!
Radius is "soft" and is affected by covalent bonding, since it can overlap
Cl by itself is 132 pm, but in Cl2 radius is 100 pm
Trend: smallest radii are upper right, largest to lower left in general
Figure 8.7 Defining metallic and covalent radii
Figure 8.8 Atomic radii of the main-group and transition elements
Figure 8.9 Periodicity of atomic radius
Sample Problem 8.3 (Using only the periodic table (not Figure 8.15)m 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
PERIODIC TRENDS
IE: discussed previously, but trend is for highest to upper right, lowest to lower left. Always endothermic
process to remove eElectron Affinity: Trend is most negative EA at F, least likely is Fr, worst are noble gases
Table 8.4 Successive Ionization Energies of the Elements Lithium through Sodium
Figure 8.13 Electron affinities of the main-group elements
Figure 8.14 Trends in three atomic properties
Figure 8.15 Trends in metallic behavior
Sample Problem 8.4 (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
Sample Problem 8.5 (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 6273 22,320
ATOMS TO IONS
What happens to e- configuration if an atom becomes an ion?
Ion size: cations get smaller, anions get bigger
More charge is more effect on size
ISOELECTRONIC: iso = same, electrons, so same # of electrons
We can list isoelectronic series of ions and noble gas atoms that have the same number of electrons.
Figure 8.17 Main-group ions and the noble gas configurations
ATOMS TO IONS
Rank the isoelectronic series on the previous slide by size smallest to largest:
Do abbrev. e- config: Mg2+, Fe2+ and Fe3+, Cu1+ and Cu2+. Determine which will be paramagnetic, which is
more strongly paramagnetic, etc. Then rank them by size.
Sample Problem 8.6 (Using condensed electron configurations, write reactions for the formation of the
common ions of the following elements: (a) iodine (b) potassium (c) indium
Sample Problem 8.7 (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+ (b) Cr3+ (c) Hg2+
Figure 8.20 Depicting ionic radius
Figure 8.21 Ionic vs. atomic radius
Sample Problem 8.8 (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+