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Chapter 7 Summary – Most of this chapter was covered in honors chemistry. You will need to recall
what you have learned in the past as we build on this foundation by adding depth and a few new things.
This summary is intended assist you in this process.
History of the Atom (review from chapter 2)
People: Lavoisier, Proust, Dalton, J.J. Thomson, Rutherford, Bohr, and Schrödinger.
What they did:
Lavoisier: Verified the Law of Conservation of Mass (mass is neither created nor destroyed) and
showed that combustion involved oxygen.
Proust: Determined the Law of Definite Proportions (a given sample of a compound always contains
the exact same proportion of elements by mass). This law established constant composition of a sample
of compound. In other words, all the salt in a saltshaker has the formula NaCl. It is not simply a random
pile of sodium and chlorine.
Dalton: Determined the Law of Multiple Proportions (when two elements form a series of compounds,
the ratios of the masses of the second element that combine with 1 gram of the first element can always
be reduced to small whole numbers).
Example:
Compound A
Compound B
Compound C
Mass of N2 That Combines w/ 1 g
of O2
1.750 g
0.8750 g
0.4375 g
By this law the proportions of the masses should combine to form simple, whole number ratios.
Dalton’s Original Atomic Theory:
1.) Each element is made up of tiny particles called atoms.
2.) The atoms of a given element are all identical; atoms of different elements are different in some way
or ways.
3.) Compounds form when atoms of different elements combine. A given compound has the same
relative type and number of atoms.
4.) Chemical reactions involve the reorganization of atoms, the way they are bonded changes. The atoms
themselves are not changed.
5.) Everything is made of atoms.
J.J. Thomson: Used cathode ray tubes to identify the electron. He reasoned that all atoms must contain
electrons and he further reasoned that there must be something positively charged in atoms because all
atoms had a neutral charge. The result was his plum pudding model.
Rutherford: Rutherford and his famous “gold foil” experiment demonstrated that the nucleus is a very
small part of the atom and that the nucleus is positively charged. In short,  particles were shot at a gold
foil. Some of the particles were deflected while others passed through. The new atomic structure was
one with a central positively charged nucleus surrounded by electrons.
Bohr: Determined that electrons can only travel around the nucleus of an atom in certain allowed
circular orbits. Classical physics could not explain this phenomenon. Classical physics dictated that the
electron should eventually spiral into the nucleus as it loses energy. Bohr rationalized that certain values
for angular momentum (speed, mass and orbital radius) would permit the electron to not fall into the
nucleus. It was not clear why this happened, but it did explain the emission spectrum of hydrogen. The
Bohr model, however, did not work for any other atom.
Schrödinger: Determined that electrons do not travel in defined paths, but that they are found a
“spherical cloud” within a given distance from the nucleus. He also concluded that the exact position of
an electron couldn’t be determined. Instead, only probable location can be determined. This is the
current model of the atom. This is referred to as the quantum mechanics model of the atom.
The darker areas indicate the highest
probability of electron location in a
helium atom. This is not an actual
atom, but a computer model using the
Schrödinger equation in a 2D
perspective.
Electron energy levels: atomic spectra, quantum numbers, and atomic orbitals.
The purpose of this section is to introduce you to quantum numbers and provide a brief review of
electron configurations and orbital shapes.
Principle Quantum Number (values 1-8): Denoted by the variable n, the principle quantum number
indicates the energy and size of an orbital. The larger n is the larger the orbital is and the more time the
electron spends farther away from the nucleus. A larger n also means a larger energy because the
electron is less tightly bound to the nucleus.
Angular Momentum Quantum Number (values 0 to n-1 for each n): This quantum number is related to the
shape of the orbital and is denoted by the variable l. Each numerical value of l is assigned a letter that
corresponds to an orbital shape.
Value of l
0
1
2
3
4
Letter Used
s
p
d
f
g
This system of letters is remnant of old spectral studies.
Magnetic Quantum Numbers (values from l to –l): Denoted Ml, this quantum number indicates the
orientation of the orbital in space relative to the other orbitals in the atom.
This chart shows the allowed
quantum numbers for all elements.
Note the presence of g orbitals.
These orbitals exists, but electrons
only move their when excited.
Orbital Shapes and Energies
The shapes of the orbitals are pictorially represented on the next page. What is important to note is
which orbital shapes are found in each principle quantum energy level:
Principle Quantum Number
Orbital Shapes Present
1
s
2
s,p
3
s,p,d
4
s,p,d,f
Beyond the 4th energy level all levels will use all four (s,p,d, and f) orbital shapes. The aforementioned g
shaped orbital can be observed in special laboratory conditions that are outside the scope of the class.
The fourth quantum number is the electron spin quantum number (ms). It arises from the known
phenomenon that as a charged object spins it produces a magnetic field. Electron behavior demonstrates
two possible magnetic moment (or magnetic field) orientations. This led to the Pauli Exclusion Principle
that states that in a given atom no two electrons can have the same set of four quantum numbers. This
also dictates that any orbital shape can only hold two electrons and they must have opposite spins.
Electron Energies
An electron’s energy is dependent on three values: (1) the kinetic energy of the electron as it moves
around the nucleus, (2) the attraction between the electron and the nucleus, and (3) the repulsion of
other electrons. This results in energy differences between the different orbital shapes (s, p, d and f)
within the same energy level (n). The result is as follows:
Ens < Enp < End < Enf
E = energy
n = principle quantum number
This occurs because the lower energy orbitals penetrate closer to the nucleus. A consequence of this is
electron shielding where inner electrons of lower energy shield the outer electrons from the positive
charge of the nucleus.
The final principle for electron orbitals is the Aufbau Principle. This principle states that as protons are
added one by one to the nucleus to build up the elements, electrons are similarly added to these
hydrogenlike orbitals.
Hydrogen’s one electron occupies the 1s orbital. The short hand of this is written 1s1. Another way of
writing this is to use arrows to show the electrons in each orbital. Below are example of how the first 8
elements fill their orbitals. It is important to notice that each orbital can only handle two electrons and
that each orbital shape has a different number of orbitals within it.
Shape
# of
Orbitals
# of
electrons
s
1
2
p
3
6
d
5
10
f
7
14
The order in which the electrons fill the orbitals is governed by
Hund’s rule which states that the lowest energy configuration
for an atom is the one having the maximum number of unpaired
electrons allowed by the Pauli principle in a particular set of
degenerate (equal) orbitals. Again, the below picture will
demonstrate both this and the Aufbau principle.
Electron configurations for elements beyond argon.
The order of the orbitals does not follow the counting numbers. For example the electron configuration
for iron is [Ar] 4s23d6. This may seem a bit backwards that the 3d orbital would come after the 4s orbital.
Experimental data show that the 3d actually has more energy then the 4s and so comes after it when
writing electron configurations. The below picture of the periodic table illustrates how the orbitals fill.
As mentioned, there are some discrepancies when dealing with the d and f orbitals. For example, the
predicted electron configuration of manganese is [Ar] 4s23d4. Experimental data, however, show the
electron configuration to be [Ar] 4s13d5.
Periodic Trends
Using the information about electron configurations we can make some general conclusions regarding
the ordering of the elements on the periodic table. It appears that the elements on the periodic table are
ordered by mass, when they are in fact ordered by increasing number of protons. This also plays an
important role in the conclusions we make about the periodic table.
Ionization Energy
Ionization energy is the energy required to remove an electron from a gaseous atom or ion:
X(g)  X+ + ewhere the atom or ion is assumed to be in its ground state. Below are the energies needed to remove
successive electrons from aluminum.
Al(g)  Al+(g) + e-
IE1 = 580 kJ / mol
Al+(g)  Al2+(g) + e-
IE2 = 1815 kJ / mol
Al2+(g)  Al3+(g) + e-
IE3 = 2740 kJ / mol
Al3+(g)  Al4+(g) + e-
IE4 = 11,600 kJ / mol
A few things are clear from the above data. As each electron is removed, the amount of energy needed to
remove it increases. This is not surprising given what you know about electron configurations. Electrons
furthest from the nucleus (in this case the 3p electron) require the least amount of energy to be removed
because they spend less time near the positively charge nucleus than the inner electrons. Notice also the
large increases in energy between the 1st and 2nd electron and the 3rd and 4th electron. These large
differences are caused by a few relationships. First, the outermost electron is in the 3p orbital and
spends very little time near the nucleus. Therefore, the nucleus does not exert as strong a pull on that
electron. Second, the inner electrons do spend more time near the nucleus, which “shields” the
outermost electron from the pull of the nucleus (this phenomena is called electron shielding). Finally, the
inner electrons due to all electrons having a negative charge repel the inner outermost electron.
The second conclusion you can make is based on the energy different between the 1st and 2nd electrons
and the 2nd and 3rd electrons. The difference between the energies required to remove the 2nd and 3rd
electron is smaller than that between the 1st and 2nd. This is due to electron repulsion. Electrons in the
same orbital (the second and third electrons are both in the 3s orbital) repel one another due to their like
negative charges. This decreases the amount of energy needed to remove an electron.
The ionization energy trend is easy to understand given the above idea on electron configuration.
Ionization energy increases across a period. This is because the nucleus increases its effective charge as
protons are added to the elements and the electrons are being added to the same principle quantum
level. So, the charge on the nucleus is increases, but the added electrons are not being added at a further
distance. Therefore, the electrons will experience a greater pull from the nucleus. (This phenomena is
referred to as effective nuclear charge.)
Ionization energy decreases down a group. Again, given what is known about electron configuration this
is not surprising. Traveling down a group one sees a great increase in the number of protons, but that is
offset by the huge shielding effects and larger principle quantum energy levels. Cesium, for example, has
one of the lowest first ionization energies. This makes sense given that Cesium’s outermost electron is in
the 7th principle quantum energy level, a long distance from the nucleus, and shielded by 54 other
electrons.
Below is a chart showing the first ionization energies of the first 6 periods. There are some exceptions in
the trends (between Be and B for example), and these will be discussed in the first week of school.
Summing up:
-Outermost electrons are the easiest to remove because: they are feel less pull from the nucleus because
of their distance from it, the feel less pull from the nucleus because they are “shielded” from it by inner
electrons, and they are repelled by the inner electrons.
-The periodic trend is ionization energy increases across a period and decreases down a group.
Atomic Radii
Like the Ionization Energy trend, the trend for atomic radii is caused by the relationship between an
atom’s protons and electron configuration. Atomic radii decrease across periods and increase down
groups. Atomic radii decrease across periods for two reasons: (1) increasing effective nuclear charge as
the number of protons increases with each element and (2) electrons are being added to the same
principle quantum energy level. Point one is easy to understand, more protons means a greater pull on
the electrons, but point two is implying that the electrons are not being added at successively greater
distances from the nucleus. Since electrons are added to the same energy level, there is not any (or at
least not any significant) increase in electron shielding.
Atomic radii increase down groups because of increased electron shielding and decreased nuclear pull
due electrons being added to higher and higher principle quantum energy levels. Again, looking at
Cesium, its outer electrons feel little nuclear pull due to their distance from the nucleus (larger principle
quantum energy level) and greater electron shielding from the large number of core electrons. These 2
factors more than offset the increased effective nuclear charge moving down a group.
It is important to note that the measurements for atomic radii are an estimate. Since electrons are
constantly moving it is impossible to exactly measure an atom’s radius. Instead an atomic radius is
defined as one-half the distance between the two nuclei in two adjacent metal atoms or a diatomic
molecule as a gas.
The distance of ionic radii (the radius of a cation or anion) is also important. The size of an ion’s radius is
helpful in characterizing compounds since ionic radius affects the physical and chemical properties of
ionic compounds.
When a neutral atom is converted to an ion it is expected that the radius would change. The loss or
addition of electrons changes the effective nuclear charge and the electron shielding. As expected, the
loss of an electron results in a decrease of atomic radius and gaining an electron causes the atomic radius
to expand. Ions with a charge greater than one (gaining more electrons) have a greater radius than those
only a charge of negative one. Conversely, ions with a positive charge (loss of electrons) greater than one
will have a smaller radius than ions with a charge of positive one.
Electron Affinity
Electron affinity is the exothermic energy change that occurs when an electron is accepted by an atom in
the gaseous state to form an anion.
X(g) + e-  X-(g)
-??? kJ / mol
The more negative the energy is, the greater the affinity of the element to gain an electron. This also
means that the ion that will be created will be very stable. Generally, electron affinity increases across a
period (except for noble gases) and is decreases down a given group.
Self Assessment (includes some review of naming and solubility rules)
Questions 1-3 refer to atoms of the following elements.
(a) Lithium
(b) Carbon
(c) Nitrogen
(d) Oxygen
(e) Fluorine
1. In the ground state, have only one electron in each of the three p orbitals
2. Have the smallest atomic radius
3. Have the smallest value for first ionization energy
4. Which formula would be expected for a binary compound between barium and nitrogen?
(A) Ba3N2
(B) Ba2N3
(C) Ba2N
(D) BaN2
(E) BaN
Questions 5-8 refer to atoms of the following elements.
I. Sodium
II. Strontium
III. Uranium
IV. Bromine
V. Bismuth
5. The atom with the largest radius is
(A) I
(B) II
(C) III
(D) IV
(E) V
6. There are only two liquid elements at room temperature and atmospheric pressure. One of these is
(A) I
(B) II
(C) III
(D) IV
(E) V
7. The element that, when an ion, carries a charge of 1- is
(A) I
(B) II
(C) III
(D) IV
(E) V
8. The element that contains 38 protons is
(A) I
(B) II
(C) III
(D) IV
(E) V
9. In which pair of elements is the larger atoms listed first?
(A) K, Ca
(B) Na, K
(C) Cl, S
(D) Mg, Na
(E) O, N
10. Chemical properties of elements are defined by the
(A) electrons
(B) ionization energy
(C) protons
(D) neutrons
(E) electronegativity
Questions 11-13 refer to the following variables
I. n
II. s
III. ml
IV. ms
V. f
11. Which of the symbols represents an orbital?
(A) I and II
(B) II and V
(C) I, II, and III
(D) III and IV
(E) V
12. Which may be occupied by an electron from calcium in an excited state?
(A) I and II
(B) II and V
(C) I, II, and III
(D) III and IV
(E) V
13. Which of these will tell you the orientation of the orbital that the electron occupies?
(A) I
(B) II
(C) III
(D) IV
(E) V
14. Which of the following elements has the greatest number of valance p electrons?
(A) C
(B) Si
(C) Fe
(D) Cl
(E) As
15. Which electron configuration corresponds to that of a noble gas?
A) 1s22s22p63s23p64s1
(B) 1s22s22p63s23p4
(C) 1s22s22p63s23p6
(D) 1s22s22p63s1
(E) 1s 2s 2p 3s 3p 4s
16. Which quantum number describes the shape of an orbital?
(A) n
(B) l
(C) ml
(D) ms
(E) s
Questions 17-20 refer to the following ions.
I. ammonium ion
II. calcium ion
III. bromide ion
IV. iron (III)
V. phosphate ion
17. Which of these carries a single negative charge?
(A) I
(B) II
(C) III
(D) IV
(E) V
18. Which of these will form a compound?
(A) I and II
(B) II and IV
(C) III and V
(D) I and IV
(E) II and V
19. How many different compounds can be made from these ions?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 9
20. Which compound will be insoluble?
(A) I and V
(B) III and I
(C) II and III
(D) III and IV
(E) IV and V
21. Which of the following compounds is soluble?
(A) MgCO3
(B) Al(OH)3
(C) Cr2S3
(D) K2CrO4
(E) NiSO3
22. The correct name for Fe(NO3)3 is
(A) iron nitrite
(B) iron (II) nitrate
(C) ferrous nitroxide
(D) iron (III) sulfate
(E) iron (III) nitrate
23. Which of the following is NOT a correct chemical formula?
(A) SrBr2
(B) Ca2O3
(C) Mg3N2
(D) Na2S
(E) AlI3
24. Which of the following is a correct formula?
(A) NH4SO3
(B) CaC2H3O2
(C) Na2ClO4
(D) Ba(CO3)2
(E) KH2PO4