Download "ALICE" CHAPTER 12 MODERN VIEW OF ATOMIC STRUCTURE

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ACTIVE LEARNING I N C HEMISTRY E DUCATION
"ALICE"
CHAPTER 12
MODERN VIEW
OF ATOMIC
STRUCTURE
(PART 1)
Atomic Structure & Spectra
Nuclear Notation
Orbital Notation
Configuration Notation
12-1
©1997, A.J. Girondi
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All rights reserved. No part of this document may be reproduced or transmitted in any form by any means,
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© 1997 A.J. Girondi, Ph.D.
505 Latshmere Drive
Harrisburg, PA 17109
[email protected]
Website: www.geocities.com/Athens/Oracle/2041
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©1997, A.J. Girondi
SECTION 12.1
Early Discoveries About the Atom
You have already learned about the concept of atoms and how they can combine to form
compounds. You have also seen that each element has its own characteristic set of properties which help
to distinguish it from all other elements. In this chapter we will study the structure of atoms and the laws
governing the behavior of the particles that make up atoms. This knowledge will lead to an explanation of
the properties of the elements and of their tendencies to form compounds.
John Dalton regarded the atom as a particle with no internal parts. He believed an atom to be the
smallest possible particle. However, certain experiments were being performed which gave definite
indications that Dalton's view was not correct and that there was some sort of internal structure to the atom.
It became apparent that atoms consisted of particles that had electrical charges and that these particles
interacted according to the laws of electromagnetism. Charged particles carry either a positive (+) or
negative (-) charge. We call two negative charges or two positive charges "like" charges. The laws of
electromagnetism state that two like charges {1}_______________ each other, while unlike charges,
{2}_____________ each other.
Experiments performed during the late 1800's and early 1900's by chemists and physicists made
it clear that atoms could, indeed, be broken into smaller parts, contrary to the ideas of {3}_____________.
In 1897, J.J. Thomson discovered that atoms could be "taken apart" when he studied the effects of
electrical discharge on atoms of various gases. In his experiments he concluded that atoms were coming
apart by yielding a stream of negatively charged particles with very small masses (compared to the masses
of the atoms). These small negative particles became known as electrons. Thomson is credited with the
discovery of electrons which were present in the atoms of all of the different gases that he examined.
Another scientist, Ernest Rutherford, and his students performed experiments in England during
the first decade of the 20th century in an attempt to determine the size of atoms. In 1906, Rutherford had
his students direct a beam of positively-charged subatomic "alpha" particles at a very thin sheet of gold
metal. It was known as the alpha scattering experiment." Since they believed that matter was mostly
empty space, they expected the particles to pass through the thin sheet unhindered. To their surprise,
they found that a small fraction of the particles bounced right back! This led Rutherford to believe that in
the center of the atom was a small but very dense "nucleus" with which some of the alpha particles must
have collided. He concluded that most of the mass of the atom was contained in the {4}__________________.
He also concluded that the nucleus was {5}______________ charged since it repelled the positivelycharged alpha particles. He later said that "It was quite the most incredible event that has ever happened
to me in my life. It was almost as if you fired a 15- inch shell into a piece of tissue paper and it came back
and hit you."
Rutherford realized that electrons were located at a considerable distance from the nucleus. If this
were an accurate description of an atom and we could inflate the nucleus of a hydrogen atom to the size of
a basketball, the electron would orbit this "basketball" nucleus at a distance of more than 15 miles away!
Visualizing the atom like this enables you to realize that most of the atom is, indeed, nothing more than
empty space! So much for the athlete who thinks he is "solid muscle!"
With regard to the nucleus itself, it became obvious to scientists that the nucleus was composed
of small particles. One of the particles in the nucleus is the proton. In 1914, Rutherford was given credit
for discovering protons. A proton carries a charge equal to the charge of an electron, but opposite in
character. The electron carries a charge of -1 while the proton carries a charge of {6}_______. In the
nucleus of a neutral atom (that is, an atom with no overall electric charge), there must be an equal number
of protons to balance the charges carried by the electrons. Unlike the electron, the proton is a particle with
a relatively large mass, in atomic terms. The proton has a mass equal to 1,836 times that of the electron!
Using the common unit of the gram to measure masses, the electron has a mass equal to 9.1 X 10-28
grams, and the proton has a mass equal to 1.673 X 10-24 grams.
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©1997, A.J. Girondi
A second component of the nucleus was discovered in
Dalton’s Atomic Theory
1932 by another Englishman, James Chadwick. This particle
(1803)
became known as the neutron. A neutron is an electrically neutral
particle, meaning that it carries no electric charge. The neutron was
very difficult to discover. Because it has no charge of its own, it is
Thomson Discovers the Electron
neither attracted to nor repelled by an electrical charge. A neutron
(1897)
has a mass slightly larger than that of the proton, equal to 1.675 X
10 -24 grams. The presence of this particle accounted for the
observed masses of atoms, which were found to be greater than
that predicted if only protons were present in the nucleus. Figure Rutherford Discovers the Nucleus
(1906)
12.1 and the accompanying table present an overall summary of the
locations and properties of the components of an atom.
Each element differs from all others in that atoms of each
element contain a specific number of electrons, protons, and
{7}_______________.
Indeed, the number of protons in the
nucleus determines the actual identity of an element. Determining
the number of electrons, protons, and neutrons in any given
element is a relatively simple process. The atomic number of an
element is equal to the number of protons found in the nucleus.
The element with atomic number 19, potassium, has 19 protons.
For atoms to be neutral, they must have equal numbers of positive
(protons) and negative (electrons) charges. This means that
potassium must have 19 protons and {8}________ electrons.
Bohr Develops the Planetary
Model of the Atom
(1913)
Rutherford Discovers the Proton
(1914)
Chadwick Discovers the Neutron
(1932)
Table 12.1
Location and Properties of Subatomic Particles
Particle
Charge
Comparative Mass
electron
-1
1/1836
proton
+1
1
neutron
0
1
Location
outside nucleus
inside nucleus
inside nucleus
If an element has 10 protons in its nucleus, how many positive charges does it have?{9}________.
If an atom with 10 protons is neutral, how many electrons must it have?{10}__________
The atomic number of oxygen is 8. How many protons does it have?{11}__________
How many electrons does oxygen have?{12}___________
SECTION 12.2
Mass Number, Atomic Mass, and Isotopes
As you learned in an earlier chapter, the atomic masses of the elements that appear on the
periodic table are really average masses. We will use the element hydrogen as an example. Most
hydrogen atoms (99.85% of them) have no neutrons. A few hydrogen atoms (0.15% of them) have one
neutron. These two slightly different kinds of hydrogen are often referred to as light and heavy hydrogen.
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©1997, A.J. Girondi
This situation is fairly typical of most elements. Elements exist in nature as a mixture of isotopes.
Isotopes are atoms that have the same number of electrons and protons, but different numbers of
{13}______________. As a result, {14}________________ of an element are atoms that have different
masses. The atomic masses that appear on the periodic table are merely "weighted averages" of the
masses of all the naturally-occurring isotopes. Weighted averages are determined by including the
relative amounts of each isotope found in nature when calculating the average mass.
On the periodic table the atomic mass of carbon is 12.011. Actually, no carbon atoms have this
mass. Some have more mass than this, and some have less. The 12.011 is the {15}_____________ mass
of carbon atoms in nature. The number of protons in the nuclei of atoms of a specific element is constant,
but the number of neutrons can vary. Therefore, the mass numbers (the sum of the protons and
neutrons) of atoms of an element can vary. Atoms of a particular isotope of an element are identified by
writing the mass number of the isotope after the name of the atom. For example, the two naturallyoccurring isotopes of hydrogen are hydrogen-1 and hydrogen-2. The most common form of carbon is
carbon-12, while less common isotopes include carbon-13 and carbon-14. The difference in these mass
numbers is a result of differences in the number of {16}_______________ in these atoms. The sum of the
{17}________________ and {18}_________________ in the nucleus of a particular isotope of an atom is
known as its {19}_______________________.
While the atomic mass of an atom of a particular isotope of an element is not found on the periodic
table, it is very close to the sum of the masses of the neutrons and protons in the atom.
{20}______________ have such small masses that they are not even considered when atomic masses are
calculated. This is the case even if an atom contains 100 or more electrons! The mass of a proton is
expressed as 1 amu (atomic mass unit). Since the proton and the neutron have almost the same mass,
the approximate mass of a neutron must also be 1 amu. (Actually, it is a tiny bit more than 1.) So, what is
the mass of an isotope such as oxygen-16?
Oxygen-16 is an atom of a particular kind of oxygen containing a total of 16 protons and neutrons.
Remember, the sum of the protons and neutrons in the nucleus of an atom is called its mass number.
Thus, the mass number of this atom is 16. Since each proton and each neutron has a mass of about 1
amu, you would expect that an atom of oxygen-16 would have a mass of about {21}_________amu.
However, the actual mass of oxygen-16 atoms is 15.99491 amu. Why this tiny difference? This difference
is a result of the fact that when protons and neutrons combine to form a nucleus, some matter is converted
into energy. This is a kind of nuclear reaction. Therefore, whenever a nucleus forms, a bit of matter will be
lost. Thus, chlorine-37 has an atomic mass of 36.96590 amu.
You can determine the mass number of a specific isotope of an element if you round off its atomic
mass to the closest integer (whole number). So, to get the mass number of chlorine-37 just round
36.96590 to {22}_______. This gives you the sum of the protons and neutrons in the nucleus of this
specific form (isotope) of chlorine.
Since the atomic mass of an element given on the periodic table is an average, rounding it off will
usually give you the mass number of the most common isotope of the element. For example, on the
periodic table the average mass given for carbon is {23}____________. Rounding will yield 12, which is
the mass number of the most common isotope of carbon, which is carbon-12. Using this method, what if
the mass number of the most common isotope of element #12, magnesium? {24}___________. What is
the name of the most common isotope of element #15, phosphorus?{25}_______________________.
Since the mass number of an atom is equal to the total number of protons and neutrons it
contains, you can determine the number of neutrons in an atom by subtracting the atomic number
(number of protons) from the mass number (number of protons + neutrons).
atomic number = number of protons
mass number = atomic mass rounded to closest integer
neutrons = (mass number — atomic number)
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©1997, A.J. Girondi
The "nuclear notation" of an atom includes both the atomic number and the mass number of the
atom. The atom's symbol is written with its mass number expressed as a superscript at the upper-left
location and the atomic number written as a subscript at the lower-left location. For example:
23
11 Na
Nuclear notation of sodium -23:
Here are a couple of others:
hydrogen-1:
hydrogen-1:
1
1H
means mass number
= 23 and atomic number = 11
means 1 proton and 0 neutrons in the nucleus
(atomic number = 1, and mass number = 1)
2
1H
means 1 proton and 1 neutron in the nucleus
(atomic number =1, mass number = 2 )
Problem 1. For practice, determine the number of electrons, protons, and neutrons in the atoms of
aluminum and cesium listed below.
a. aluminum–27;
27
13 Al
protons:
electrons:
neutrons :
b. cesium–133;
133
55 Cs
protons:
electrons:
neutrons :
Problem 2. The element uranium has three different isotopes. In the blanks below, write the number of
electrons, protons, and neutrons for each isotope of uranium.
a.
234
92 U
protons:
electrons:
neutrons :
b.
235
92 U
protons:
electrons:
neutrons :
c.
238
92 U
protons:
electrons:
neutrons :
Problem 3. Supply the missing information in the table below.
Isotope
Nuclear
Notation
At.No.
Mass No.
No. e's
No. p's
No. n's
a. aluminum-27
_____
_______
_____
_____
_____
b. bismuth-209
_____
_______
_____
_____
_____
c. calcium-40
_____
_______
_____
_____
_____
d. copper-64
_____
_______
_____
_____
_____
e. ____________
__2__
___4___
_____
_____
_____
f. ____________
_____
__207__
__82_
_____
_____
g. ____________
__8__
_______
_____
_____
__8__
h. ____________
_____
_______
_____
__50_
__69_
i. ____________
_____
_______
__30_
_____
__36_
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©1997, A.J. Girondi
Here is an example which shows how atomic masses on the periodic table are calculated:
Sample
Problem: Oxygen exists in nature in three forms. Oxygen-16 (99.759%), oxygen-17
(0.037%), and oxygen-18 (0.204%). Oxygen-16 atoms have a mass of 15.99491 amu; oxygen-17 atoms
have a mass of 16.99914 amu, and the mass of oxygen-18 atoms is 17.99916 amu. What is the average
mass of oxygen atoms in nature?
To solve this problem, assume that you have a 100 atom sample of oxygen. In that case, since
percent means parts per hundred, 99.759 atoms in the sample would be oxygen-16 (even though atoms
cannot exist in fractional parts); 0.037 atoms would be oxygen-17, and 0.204 atoms would be oxygen-18.
Atomic masses such as these
are determined by experiment.
The total mass of the oxygen-16 atoms in the sample is:
99.759 atoms X
15.99491 amu
1 atom
= 1595.64 amu
The total mass of the oxygen-17 atoms in the sample is:
0.037 atom X
16.99914 amu
1 atom
= 0.63 amu
The total mass of the oxygen-18 atoms in the sample is:
0.204 atom X
17.99916 amu
1 atom
= 3.67 amu
The total mass of all 100 atoms in the sample would be:
1595.64 amu + 0.63 amu + 3.67 amu = 1599.94 amu.
To get the average mass, we divide by the number of atoms in the sample:
1599.94 amu
100
= 15.9994 amu
Does this answer agree with the atomic mass of oxygen on the periodic table?____________
In the sample problem above, the mass of an atom of oxygen-16 is listed as 15.99491 amu. Assuming
that protons and neutrons weigh 1 amu each, why doesn't oxygen-16 weigh 16 amu?
{26}____________
______________________________________________________________________________
Problem 4. In nature, copper is found to exist in two forms: copper-63 and copper-65. Copper-63
atoms have a mass of 62.93 amu, while copper-65 atoms have a mass of 64.93 amu. Naturally-occurring
copper contains 69.40% copper-63. Calculate the atomic mass of naturally-occurring copper atoms.
Show all of your work.
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©1997, A.J. Girondi
Does your answer to Problem 4 compare favorably to the atomic mass of copper appearing on the periodic
table? _________
SECTION 12.3
The Spectrum of Visible Light
Now that you know something about the makeup of atoms in terms of electrons, protons, and
neutrons, we will take time to examine some of the experimental evidence that has contributed to our
present understanding of atomic structure. In particular, we will be studying the light emitted by atoms
when we add energy to them. Such light is called the "emission spectrum" of an atom. To understand the
significance of emission spectra from atoms, we need to learn more about light spectra in general.
In Figure 12.1, is a box that represents a continuous light spectrum. The numbers above the
names of the colors represent the wavelengths of light in units called angstroms. Angstroms are denoted
by the symbol Å which is a capital A with a small circle above it. An angstrom is a unit of length equal to 1 X
10 -10 m (which is very small). The wavelengths in Figure 12.1 range from 3900 Å to 7700 Å and comprise
the continuous spectrum of visible light. These are the wavelengths of light that our eyes can see. When
all of the colors (wavelengths) are present, they appear to us as white light. The numbers below the
spectrum in Table 12.1 represent nanometers (nm). A nanometer is 1 X 10-9 meter ; therefore, a
nanometer is ten times larger than an Angstrom. Both units are commonly used to measure wavelengths
of light.
<----- shorter wavelengths (higher energy)
3900 Å
4400 Å
5100 Å
violet
blues
greens
390 nm
440 nm
510 nm
Figure 12.1
(lower energy) longer wavelengths ----->
5600 Å
6300 Å
yellow
560 nm
7700 Å
reds
630 nm
770 nm
Components of a “Continuous” Spectrum of Visible White Light
A spectroscope is a laboratory instrument that separates the continuous spectrum of white light
into its component wavelengths (colors). The spectrum is said to be continuous because all of the
wavelengths from violet to red are present in white light.
Light is a form of energy. The violet end of the spectrum represents light of higher energy than
light in the red region, which represents visible light of the lowest energy. With this in mind, looking again
at Figure 12.1, you can conclude that as the wavelength of the light gets {27}_______________, the
energy content of the light gets smaller.
What does all this have to do with atomic structure? We shall soon see. When atoms absorb
energy we describe them as being "excited." Excited atoms are capable of emitting light. However, atoms
do not just give off light all by themselves. Something has to be done to them, first. Fluorescent lights do
not emit light until an electrical current is passed through them. The sun emits light through the process of
nuclear fusion which causes hydrogen atoms to combine to form helium. Electricity and nuclear fusion
both involve energy changes in a system. When energy is added to a system (that is, through fusion or
passing an electrical current through it), the atoms absorb that added energy. This "overload" of energy
causes the electrons in the atoms to become very unstable. By unstable we mean that at least some of
the electrons in the atoms have too much energy and they "want" to change their position or their motion
in some way in order to get rid of the excess energy and, thereby, become more stable. Because excited
electrons are so unstable, they will give off this added energy at the first opportunity. This energy that is
emitted by the excited electrons may at least partly be in the form of visible light.
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©1997, A.J. Girondi
In fluorescent bulbs, a stream of electrons flows between the metal ends of a glass tube which
contains a mixture of argon gas and mercury vapor. The electron stream excites the {28}______________
of the mercury atoms which then emit energy as invisible ultraviolet rays. These rays excite the atoms in
the coating of the glass tube which then emit visible light. The light emitted by specific atoms (or more
accurately, by excited electrons in specific atoms) does not compose a continuous spectrum (containing
all wavelengths). The light emitted by excited electrons of a particular atom is composed of certain specific
wavelengths and is called a bright-line spectrum, since only certain wavelengths of light appear as colored
lines. Each element emits a different characteristic set of colored bright lines (wavelengths). No two
elements will emit the same set of wavelengths of light.
Atoms of different elements have
{29}____________________ bright-line spectra. Figures12.2 through 12.6, show the visible portion of
the emission spectra of 5 different elements. Each black line represents a wavelength of light which is
emitted by atoms of that element. If this document were printed in color, the lines would have a different
colors.
ACTIVITY 12.4
The Emission Spectra of Elements
(This activity may be done by all lab groups in your class at the same time . Ask your instructor.)
Procedure - Part A. Now you are going to actually look at the light emitted by excited electrons in atoms
of several metallic elements. A dark or dim room should be available for this activity.
1. Obtain a burner and a platinum or nichrome wire with a small loop at the end. Pour some of each of the
six solutions provided into separate wells of a dropping plate. Each solution contains a different metal ion
including Sr 2+, Li 1+, Ba2+, Ca2+, Cu, 2+ and Na1+. Pour some dilute hydrochloric acid (HCl) into a seventh
well on the plate. Make note of which solution is in each well.
2. Dip the loop of the wire into the lithium solution and then touch it to the edge of the flame of the burner.
This should impart some color to the flame. The burst of color may last only a fraction of a second. For
some solutions it may last longer. What color of light does the lithium produce?{30}________________
Clean the wire well by running tap water over it and then dipping it into the hydrochloric acid. Follow this by
rinsing the wire with tap water again. Touch the clean wire to the flame. If it is clean, the color will have
disappeared. If not, repeat the cleaning procedure.
3. Repeat the procedure above with the other solutions, noting the color emitted by each solution (test
the sodium solution last). The metal solutions which may include copper, strontium, calcium, barium, or
others. Note your observations below. Since sodium is not easy to remove from the wire, clean the loop
well several times before returning it to the materials shelf. The color which you saw emitted by each
metallic element in the flame was actually a blend of various wavelengths of light. These wavelengths of
light compose what is known as the bright-line spectrum or emission spectrum of each element.
Metal in Solution
Color Emitted
_________________
_______________________
_________________
_______________________
_________________
_______________________
_________________
_______________________
_________________
_______________________
sodium
_______________________
4. Discard the solutions and rinse the dropping plate. Return the loop and all materials to the proper
place.
Scientists have made quantitative observations of bright-line spectra by measuring the
wavelengths of the light waves emitted by elements. This is how we got the information revealed in
Figures 12.2 through 12.6. Other references are also available that can provide you with the spectra of
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©1997, A.J. Girondi
other elements. There may be a "spectrum chart" available in your classroom. Use the chart in your
classroom or Figures 12.2 through 12.6 to complete problem 5 below.
Problem 5. Study the emission spectrum of sodium vapor in Figure 12.6, and provide a rough estimate
of the seven wavelengths (in nm) emitted in the spaces provided below. Each line in Figure 12.6
represents a different wavelength of light being emitted.
a. __________
b. ___________
c. __________
e. __________
f. ___________
g. __________
d. __________
If you were to actually observe the emission spectrum of sodium vapor through a spectroscope, you
would probably see only one or two bright lines instead of seven. The reason for this is that the yellow
lines located near 570 nm are so bright, that they "wash out" the other lines. They are so bright, in fact,
that sodium vapor lamps are commonly used as street lights in areas where strong illumination is needed
such as high-crime neighborhoods.
Procedure - Part B.
Every element emits characteristic wavelengths, and no two
elements emit the same set of wavelengths. In this way, spectra are
like fingerprints. By studying atomic spectra, we can identify
elements. This is how scientists know which elements are present
in the sun. Wavelengths of the sun's light have been measured,
separated, and matched with spectra of known elements to identify
the contents of the sun. Your teacher may have equipment which is
designed to "excite" atoms of gases, so that their bright-line spectra
can be observed using small hand-held spectroscopes. Before
going on, ask your teacher if this equipment is available in your lab. If
so, your teacher will assist you in its use. Caution: high voltage
equipment will be used. (See Figure 12.7)
Most light is a blend of many wavelengths. The
spectroscope you will use is a device that contains a plastic "grating"
which separates the wavelengths emitted by a light source so that
you can observe them separately. List below the gases (or vapors)
that you observed using this high-voltage equipment.
Figure 12.7
Spectra Observed:_______________________________________________________________
_____________________________________________________________________________
Ask your teacher if any kinds of "vapor" lights are used to illuminate any of the rooms in your
school. If so, use a spectroscope to examine the light emitted by them, and see if you can identify the
element in the lights by comparing the spectra they emit to those in Figures 12.2 through 12.6 or those
found on a spectrum chart in your classroom. If you do have such lights in your school, go to the proper
location and observe the light using your hand-held spectroscope. Supply the information requested
below.
Location of lights in school:__________________________.
Symbol of element revealed by emission spectrum:________.
Check with your instructor to see if you identified the unknown element(s) correctly.
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©1997, A.J. Girondi
4000 Å
400 nm
4500 Å
450 nm
5000 Å
500 nm
Figure 12.2
4000 Å
400 nm
4500 Å
450 nm
4000 Å
400 nm
4500 Å
450 nm
4000 Å
400 nm
4500 Å
450 nm
4000 Å
400 nm
4500 Å
450 nm
SECTION 12.5
6000 Å
600 nm
6500 Å
650 nm
5500 Å
550 nm
6000 Å
600 nm
6500 Å
650 nm
5500 Å
550 nm
6000 Å
600 nm
6500 Å
650 nm
Emission Spectrum of Neon
5000 Å
500 nm
Figure 12.6
5500 Å
550 nm
Emission Spectrum of Calcium
5000 Å
500 nm
Figure 12.5
6500 Å
650 nm
Emission Spectrum of Mercury
5000 Å
500 nm
Figure 12.4
6000 Å
600 nm
Emission Spectrum of Hydrogen
5000 Å
500 nm
Figure 12.3
5500 Å
550 nm
5500 Å
550 nm
6000 Å
600 nm
6500 Å
650 nm
Emission Spectrum of Sodium
The Bohr Model of the Atom
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Niels Bohr developed a theory to account for the location
of electrons around the nucleus of an atom. He believed his theory
would explain the bright-line spectra emitted by excited atoms. He
proposed that electrons followed specific paths or orbits around
the nucleus. These paths or energy levels, as they are also called,
are numbered starting with the lowest one (closest to the nucleus)
as 1, the next farther from the nucleus as 2, the next as 3, and so
forth. Figure 12.8 illustrates 4 of the energy levels of a n atom.
Bohr's model resembled a planetary system like our solar
system in which he suggested that the electrons revolve around
the nucleus. The first energy level, nearest the nucleus, is
represented as number 1. Each level thereafter is increased by
one. A total of 7 energy levels are needed to explain the structure
of all of the elements.
1
2
3
4
Figure 12.8 Energy Levels
The orbits around the nucleus are called energy levels because there are different and very
specific energies associated with each level. Bohr knew that energy was being added to an atom when it
was being heated or when an electrical current was passed through an element. This extra energy has to
go somewhere. The added energy is absorbed by the electrons that are in the outermost orbit (farthest
from the nucleus). Since there are specific amounts of energy associated with each energy level, the
electron that absorbs all of this extra energy can no longer stay in the orbit (energy level) in which it
normally belongs which is its "ground state." Instead, it will move to another energy level. It is now in an
"excited state" and is what we earlier referred to as an "excited" electron.
This electron is not doomed to spend the rest of its time in this higher energy level. Indeed, the
excited electron is now very unstable. Because this is an unstable state, the electron will soon return to its
ground state. This is where the atomic spectra enters into the picture. The energy that the electron emits
when it returns to its ground state is in the form of light and heat. Figure 12.9 gives a general picture of
this process. In "A" , the electron is excited and jumps to a new energy level that is further from the
nucleus. As the electron falls back to its ground state in "B", energy is given off in the form of light.
LIGHT
EMITTED
A
B
Figure 12.9 Emission of Light from an Excited Electron
As electrons move to lower energy levels, the energy they emit is given off in a "burst" or quantity
of energy with a well-defined wavelength. The quantities of emitted energy are called quanta. Bohr's
theory of the atom gave birth to the quantum theory, a name that reflects the notion that atoms must
absorb and emit energy in specific amounts. Therefore, we say that the energy is "quantized." A
quantum of energy can be defined as the amount of energy needed to move an electron from one energy
level to the next higher one. Similarly, it can be defined as the amount of energy emitted when an
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©1997, A.J. Girondi
electron moves from its present energy level to a lower one.
In his theory, Bohr proposed that electrons were only "allowed" to exist at certain distances from
the nucleus. These distances became his energy levels. He believed that electrons were not "allowed"
to exist between these levels. Although the reasons for this behavior were unclear, the idea did explain
the existence of bright-line spectra. The spectrum of hydrogen contains a specific set of bright lines. Of
these, only 3 or 4 are clearly visible to the naked eye. (See Figure 12.2) So Bohr asked, why only this set
of lines? Why does the excited hydrogen electron emit only this specific set of wavelengths?
According to Bohr, the fact that there was always the same set of bright lines in the hydrogen
spectrum was evidence that only certain energy changes were possible for the hydrogen electron. The
electron could only make certain "jumps" and, therefore, could only emit certain wavelengths of light. The
number of lines was limited because there were only a few "excited" energy levels to which the electron
was "allowed" to move. He explained that the electron was - for whatever reason - not "permitted" to exist
between these levels.
Since every element has a specific number of electrons, different wavelengths of light will be
emitted by excited electrons of atoms of different elements. Even without a spectroscope you can see
that the glow of a neon sign is quite different from the fluorescent lights in your classroom. You probably
did not realize that what you were seeing was the result of excited electrons at work.
To better understand how an electron behaves as it returns to its stable ground state, consider
how you would jump down a staircase. Just as an electron can only stop at certain energy levels, you can
only stop at certain levels (steps). The electron cannot exist between energy levels, and you cannot stop
between the steps.
To get down to the bottom of the staircase, 4
you could jump all the way down at once; or one
step at a time; or one step followed by a two-step 3
jump; or perhaps a two-step jump followed by one
step, and then by a three-step jump. A variety of 2
different jumps is possible, each yielding a different
amount of energy.
Likewise, suppose an electron jumps out to 1
level 4 in Figure 12.10. There are ten different sizes
of jumps possible as it returns to its ground state.
So how many different wavelengths of light could be
Ground State
emitted by this electron?{31}__________ A sample
containing many excited atoms of this element
Figure 12.10 Energy Level Diagram
would emit all possible wavelengths.
SECTION 12.6
The Theory of Quantum Mechanics
Bohr's model worked well in explaining the emission spectra of hydrogen. But, it really did not
work well for atoms with more electrons in them. It even failed to explain some of the complexities of the
spectrum of hydrogen atoms. In addition, it could not explain why electrons are not pulled into the
nucleus of the atom (which would cause the atom to collapse).
We know, today, that electrons do not travel around the nucleus as planets do around the sun.
The failure of Bohr's model led to the development of the theory of quantum mechanics. Quantum
mechanics is a very complex theory. It involves describing the approximate location of any electron in any
atom. The theory provides an explanation for the behavior of electrons in atoms. The details of quantum
mechanics, while beyond the scope of this course, will be discussed briefly later in this chapter. This will
12-13
©1997, A.J. Girondi
provide you with an appreciation and a general knowledge of the principles of the theory.
In these next sections, you will see how each of the energy levels is divided into sublevels. We
will concern ourselves with how the electrons distribute themselves among the energy levels and
sublevels of atoms.
Each energy level can hold a maximum number of electrons. The number of electrons in any
given energy level is limited to 2n2. This expression means 2(n2), not (2n)2! For example, the energy level
3 can be occupied by 2(3)2, or 18, electrons at the most.
Problem 6. Calculate the maximum number of electrons that could occupy the first four energy levels in
Energy Level
Maximum No. Electrons (2n2)
1
__________
2
__________
3
__________
4
__________
Energy levels are sometimes called "shells" in which case each one is designated by a letter, K
through Q (see Figure 12.13). You may see this notation in your future studies in chemistry, so it is
important that you know something about it.
Each main energy level consists of one or more sublevels. The number of sublevels is equal to
the number of the energy level. Therefore, the third energy level has three sublevels, and the fifth
energy level has five sublevels. Which energy level has four sublevels?{32}_________. Which energy
level has two sublevels? {33}__________. The energy level diagram can now be redrawn. The main
energy levels and the sublevels within each energy level are labeled in Figure 12.11. Each sublevel is
designated by a small letter (s,p,d,f).
By overlapping we mean that a higher energy level begins before a lower one ends. Notice on
figure 12.11 how the 4th energy level begins before the third is full. Notice the overlapping of the main
energy levels that occurs as you go beyond the third energy level. This means that a higher energy level
begins before a lower one ends. Clearly, atoms are more complicated than a staircase. Certainly, the third
step on a staircase cannot begin before the second step ends! However, you must keep in mind that
energy levels are not "physical things," whereas staircases are real physical things. Energy levels are
conceptual things – regions of space. The complexity of overlapping energy levels becomes more
noticeable as you progress from the third energy level to the seventh.
The first sublevel of each energy is designated as an "s" sublevel. This is followed by a "p"
sublevel, the a "d", then an "f", depending upon how many electrons there are.
Note Figure 12.13, in which the energy sublevels s, p, d, f are each shown with one or more
circles. These circles represent orbitals. The number of circles represents the number of orbitals in a
particular sublevel. At this point, you can think of an orbital as a region of space where up to two electrons
can coexist. We think of energy levels as a staircase, and electrons can exist only in the steps, not
between them, right? We think of orbitals as "houses" where electrons live. Each of these orbital
"houses" can contain a maximum of two electrons, and – as you can see – there are different numbers of
orbital "houses" on different steps.
12-14
©1997, A.J. Girondi
6d
7th Energy Level
5f
7s
6p
6th Energy Level
5d
4f
6s
5p
5th Energy Level
4d
5s
4p
4s
3d
4th Energy Level
3p
3rd Energy Level
3s
2p
2nd Energy Level
2s
1s
1st Energy Level
Nucleus
Figure 12.11 Overlapping of Energy Levels
We will learn more about orbitals later in this chapter, and we will
change the definition somewhat at that time. Looking at Figure 12.13,
notice that every s sublevel has only one orbital. Look at the p sublevels
in figure 12.13.
How many orbitals does the p sublevel
have?{34}__________.
How many orbitals does the d sublevel
have?{35}__________.
How many orbitals does the f sublevel
have?{36}__________ Since each orbital can hold a maximum of two
electrons, what is the capacity of an "s" sublevel?{37}__________ Of a
"p" sublevel?{38}__________ Of a "d" sublevel?{39}__________ Of an
"f" sublevel?{40}__________. To help you remember the order in which
the energy levels and sublevels fill, you should become familiar with the
diagonal rule. It is shown in Figure 12.12. By simply following the arrows
along the diagonals, you have the correct order of filling. Use this
diagonal rule to see if the levels and sublevels are properly arranged in
Figure 12.13. The order of filling is:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d
In theory there are other orbitals such as 5g, 6f, etc., but they are never
used since there are no atoms that have enough electrons to fill them.
12-15
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
Figure 12.12
The Diagonal Rule
©1997, A.J. Girondi
ORBITALS
Energy Level 7
(Q Shell)
6d
5f
Energy Level 6
(P Shell)
7s
6p
5d
Energy Level 5
(O Shell)
4f
6s
5p
4d
Energy Level 4
(N Shell)
5s
4p
3d
Energy Level 3
(M Shell)
4s
3p
3s
2p
2s
Energy Level 1
(K Shell)
Energy Level 2
(L Shell)
1s
Figure 12.13 Atomic Energy Levels and Sublevels
SECTION 12.7
Orbital Notation of Electrons
Now that we have the sublevels and orbitals identified, we can use this information to describe
where the electrons are located in the various atoms represented on the periodic table. To do this, you
must keep several basic rules in mind:
1. Electrons fill up the energy sublevels.
12-16
©1997, A.J. Girondi
2. The lowest energy sublevel must be completely filled before the next higher sublevel can begin to be
filled. For example, according to the diagonal rule, the 4s sublevel must be filled before the 3d sublevel
can be used. Caution: note that this rule refers to sublevels, not energy levels.
3. Each orbital (represented by a circle in figure 12.13) can hold a maximum number of two electrons. We
will represent electrons as slashed lines. The symbol (/) represents a single electron in an orbital. A full
orbital would contain two electrons and would be represented like this: (X). This method of representing
the electrons in an atom is called orbital notation.
4. Electrons repel each other and will not pair up in a single orbital in any given sublevel until all orbitals in
that sublevel are half full (have one electron in them). The incorrect and correct way to represent a 2p
orbital containing three electrons is shown below:
Incorrect 2p Orbital Notation
2p
(X)(/)( )
Correct 2p Orbital Notation
2p
(/)(/)(/)
Problem 7. You should now fill in the correct orbital notation to the right of the incorrect notation shown
below for a 3d sublevel with 5 orbitals that contain 6 electrons:
Incorrect 3d Orbital Notation
3d
(X)(X)(X)( )( )
Correct 3d Orbital Notation
3d
( )( )( )( )( )
Orbital notation can be easily written for each element on the periodic table. For example, to write
the orbital notation for an atom of iron, we must first look up the atomic number of iron on the periodic
table. The atomic number of iron is {41}_________. How many electrons does iron have?{42}__________.
The orbital diagram shown below illustrates the order in which the sublevels and orbitals are filled for iron
according to the diagonal rule.
Iron:
1s
(X)
2s
(X)
2p
(X)(X)(X)
3s
(X)
3p
(X)(X)(X)
4s
(X)
3d
(X)(/)(/)(/)(/)
Each "X" represents two slashed lines. Is the number of slashed lines shown equal to the number of
electrons in iron? {43}__________
In Table 12.2, the orbital notations for the first ten elements are
illustrated using this set of rules. Notice that even empty orbitals are shown if a sublevel contains at least
one electron (see boron and carbon).
Table 12.2
Orbital Notation of the First Ten Elements
Element
1s
2s
2p
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
(/)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(/)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
12-17
(/)( )( )
(/)(/)( )
(/)(/)(/)
(X)(/)(/)
(X)(X)(/)
(X)(X)(X)
©1997, A.J. Girondi
Table12.3 summarizes much of the important information about the location of electrons in the
first four energy levels.
Table 12.3
Data Summary for Energy Levels 1 Through 4
Energy Level
1st
2nd
3rd
4th
Shell
Shell Capacities
No. of Sublevels
Sublevel Types
Sublevel Capacities
K
2
1
s
2
L
8
2
s,p
2,6
M
18
3
s,p,d
2,6,10
N
32
4
s,p,d,f
2,6,10,14
No. of orbitals per sublevel: s = 1; p = 3; d = 5; f = 7
Problem 8. Now it is time for you to use what you have learned. Fill in the electrons for the elements
below. Follow the rules listed in section 12.7 as well as the diagonal rule. The finished product will be the
orbital notation for an atom of each element.
Element
1s
2s
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
beryllium
magnesium
nitrogen
silicon
calcium
potassium
argon
sodium
chlorine
copper
phosphorus
nickel
)
)
)
)
)
)
)
)
)
)
)
)
2p
)
)
)
)
)
)
)
)
)
)
)
)
(
(
(
(
(
(
(
(
(
(
(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)(
3s
)(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)(
)
)
)
)
)
)
)
)
)
)
)
3p
4s
3d
( )
(
(
(
(
(
(
(
(
(
)
)
)
)
)
)
)
)
)
(
(
(
(
)(
)(
)(
)(
)(
)(
)(
)(
)
)
)
)
(
(
(
(
)(
)(
)(
)(
)(
)(
)(
)(
)
)
)
)
( )
( )
( )
( )( )( )( )( )
( )
( )( )( )( )( )
Note: The 3d sublevel is a "higher energy" level than the 4s because electrons in the 3d have more
energy than those in the 4s. Electrons fill the "lower energy" levels first. However, the lower energy 4s
sublevel is farther from the nucleus than the higher energy 3d. A "higher" energy sublevel is not
necessarily farther from the nucleus than a "lower" energy sublevel. The distance of the energy sublevels
from the nucleus is:
Increasing Distance from Nucleus----->
1s 2s,2p 3s,3p,3d 4s,4p,4d,4f 5s,5p,5d,5f 6s,6p,6d,etc.
(closest)
SECTION 12.8
(farthest)
Electron Configuration Notation
There is another form of notation which reveals which energy level and sublevel the electrons of
an atom are in, but it does not show the individual orbitals. This method is known as electron configuration
notation. A superscript is used to reveal the number of electrons in each sublevel. For example, the
configuration notation for lithium, element number 3, is: 1s22s 1. This means that there are 2 electrons in
12-18
©1997, A.J. Girondi
sublevel "s" of energy level 1, and there is one electron in sublevel "s" of energy level 2. Configuration
notation does not reveal whether the electrons are alone or paired in their orbitals like orbital notation
does. Note that the sum of the superscripts should equal the number of electrons in the atoms. (A lithium
atom contains three electrons.)
If an atom has a configuration notation of 1s22s 22p 3, how many electrons does the atom contain?
{44}_________ How many of these electrons are found in the second energy level?{45}___________
How many are found in the first energy level?{46} __________ More examples of configuration notation
are shown below.
Element
Configuration Notation
1s1
1s2
1 s 22 s 1
1 s 22 s 2
1 s 22 s 22 p 1
1 s 22 s 22 p 63 s 23 p 64 s 23 d 5
1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 4
1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 3
hydrogen
helium
lithium
beryllium
boron
manganese
selenium
antimony
Problem 9. Give the electron configuration notation for the elements listed below.
Element
Atomic Number
Configuration Notation
a. C
6
________________________________________________
b. N
7
________________________________________________
c. O
8
________________________________________________
d. F
9
1 s 22 s 22 p 5
e. Ne
10
________________________________________________
f. P
15
________________________________________________
g. Ar
18
________________________________________________
h. Ca
20
1 s 22 s 22 p 63 s 23 p 64 s 2
i. Mn
25
________________________________________________
j. Zn
30
________________________________________________
k. Kr
36
________________________________________________
12-19
©1997, A.J. Girondi
SECTION 12.9
Review Problems
Problem 10. Naturally-occurring iridium (Ir) consists of two isotopes. Iridium–191 (atomic mass =
190.9992 composes 37.58% of it, while iridium–193 (atomic mass = 192.9913) composes the remaining
62.7%. Calculate the average atomic mass of iridium atoms.
Problem 11. Write the nuclear notation for the following isotopes:
a. iridium–191
b. chlorine–37
c. chromium–50
Problem 12. Indicate the number of protons, electrons and neutrons in each of the following:
protons
electrons
neutrons
iridium-191
______
_______
_______
chlorine-37
______
_______
_______
chromium-50
______
_______
_______
Problem 13. Write the orbital notation for:
a.
sulfur:
1s
( )
2s
( )
2p
( )( )( )
3s
( )
3p
( )( )( )
4s
( )
3d
( )( )( )( )( )
b.
vanadium: 1s
( )
2s
( )
2p
( )( )( )
3s
( )
3p
( )( )( )
4s
( )
3d
( )( )( )( )( )
Problem 14. Write the configuration notation for:
a. sulfur: __________________________________________________________
b. vanadium: _______________________________________________________
ACTIVITY 12.10
An Optional Enrichment Activity The Colors of Transition Metal Chemistry
The elements with atomic numbers 21 through 30 exhibit many properties different from those of
other elements in the same period (row). The elements directly below in the next two periods, however,
have similar properties. Together these elements are known as the transition metals. The purpose of this
activity is to compare the properties of some substances which contain transition metals with those which
12-20
©1997, A.J. Girondi
do not contain non-transition metals. We will then try to relate any unique properties of the transition
metals to the electrons which they contain.
Sc Ti
V Cr Mn Fe Co Ni Cu Zn
Figure 12.14 The Transition Metals of Row Four
Materials: 96-well microplate; plastic micropipets; toothpicks; 0.1M solutions of compounds containing
period 4 metals: KNO 3, Ca(NO3)2, NH4VO 3, Cr(NO3)2, Mn(NO3)2, Co(NO3)2, Fe(NO3)3, Ni(NO3)2, Cu(NO3)2,
Zn(NO3)2; 6.0M ammonia (NH3); 1M potassium thiocyanate, KSCN; 6M hydrochloric acid (HCl); one sheet
of white paper.
Procedure:
1. Place a 96-well microplate on a sheet of white paper. The numbered columns should be at the top with
the lettered rows to your left.
2. Place 5 drops of KNO3 in each of wells A1, B1, C1, and D1.
1 = KNO3
2 = Ca(NO3)2
3 = NH4VO 3
4 = Cr(NO3)3
5 = Mn(NO3)2
6 = Co(NO3)2
1
2
7 = Fe(NO3)3
8 = Ni(NO3)2
9 = Cu(NO3)2
3
4
5
6
10 = Zn(NO3)2
7
8
9
10
NH3(aq) -----> A
KSCN ------> B
HCl ------> C
Control ------> D
Figure 12.15
Microplate
3. Place 5 drops of Ca(NO3)2 in each of wells A2, B2, C2, and D2.
4. Continue to repeat this process of placing 5 drops of solutions of metal compounds in subsequent
12-21
©1997, A.J. Girondi
columns of the microplate. Use chemicals in the order in which they are given in the materials list above.
The last drops will be in column 10.
5. Add 5 drops of 6M NH3 to each well in row A. Mix well with a toothpick. Rinse the toothpick with water
before moving from well to well. CAUTION: ammonia solution is caustic and has strong vapors.
6. Add 5 drops of KSCN solution to each well in row B. Mix well with a toothpick, rinsing the toothpick
between uses as before.
7. Finally, add 5 drops of HCl to each well in row C. CAUTION: Do not mix HCl and KSCN. Clean up all
spills immediately. Mix well. Use row D as a control for comparing with the other rows.
Data and Observations:
1. Observe row D of your microplate. Record your observations - particularly colors, if any - in Table 12.4.
2. Compare the solutions in the wells in rows A, B, and C in each column to the solution in row D in that
column. Where there was a change, record what you observe.
A1_______________ A2_______________ A3_______________ A4_______________
A5_______________ A6_______________ A7_______________ A8_______________
A9_______________ A10_______________
B1_______________ B2_______________ B3_______________ B4_______________
B5_______________ B6_______________ B7_______________ B8_______________
B9_______________ B10_______________
C1_______________ C2_______________ C3_______________ C4_______________
C5_______________ C6_______________ C7_______________ C8_______________
C9_______________ C10_______________
D1_______________ D2_______________ D3_______________ D4_______________
D5_______________ D6_______________ D7_______________ D8_______________
D9_______________ D10_______________
TABLE 12.4 Transition Metal Chemistry
Analysis and Conclusions:
Comment on the order of the metal ions in the wells from left to right in the rows of your microplate when
compared to the positions of the metals in the periodic table:{47}_______________________________
Did you observe any colors in columns 1 and 2 of your microplate? {48}_________________________
The metals used in columns 1 and 2 of your microplate are not transition metals. With respect to sublevel
12-22
©1997, A.J. Girondi
(s,p,d,f), what kind of electrons do the transition metals have that these metals do not have?{49}________
Review what you observed in columns 3 through 9 of your microplate. What do the results in these
columns have in common?{50}________________________________________________________
Did the metal in column 1 and 2 produce any colors?{51}__________ Did the metal in column 10 produce
any colors?{52}___________ Is the "d" sublevel of electrons completely filled in the metal atoms found in
columns 3 through 9 of your microplate?{53}___________
What is true about the distribution of electrons in the transition metals in columns 3 through 9 that is
responsible for the fact that they produce colored products? {54}_______________________________
______________________________________________________________________________
With respect to color, the results in column 10 should have been different from the results in columns 3
through 10. Why? {55}______________________________________________________________
What property would help to identify a compound as one which contains a transition metal ion? { 56}______
______________________________________________________________________________
SECTION 12.11 Learning Outcomes
This is the end of chapter 12. Check off each of the learning outcomes on the next page which
you have completed. When you have mastered them all, arrange to take the chapter 12 exam, and go to
chapter 13 which is part 2 of your study of atomic structure.
_____1. List the subatomic particles and describe their location and properties.
_____2. Determine the number of electrons, protons, and neutrons in an atom of any element given
information such as its atomic number, mass number, or atomic mass.
_____3. Define isotope, give an example of one, and calculate the atomic mass of an element given the
masses of its isotopes and their abundance in nature.
_____4. Describe the Bohr model of the atom.
_____5. Explain how atomic spectra help to explain the behavior of electrons.
_____6. Explain what is meant by energy levels, sublevels, orbitals, and electron configurations.
_____7. Give the capacities of energy levels and sublevels.
_____8. Write the electron configuration notation of atoms.
_____9. Write the orbital notation for atoms.
12-23
©1997, A.J. Girondi
SECTION 12.12 Answers to Questions and Problems
Questions:
{1} repel; {2} attract; {3} Dalton; {4} nucleus; {5} positively; {6} +1; {7} neutrons; {8} 19; {9} 10
{10} 10; {11} 8; {12} 8; {13} neutrons; {14} isotopes; {15} average; {16} neutrons; {17} protons;
{18} neutrons; {19} mass number; {20} electrons; {21} 16; {22} 37; {23} 12.01; {24} 24;
{25} phosphorus-31; {26} Some matter is converted into energy when an atom of oxygen forms;
{27} longer; {28} electrons; {29} different (or unique); {30} reddish; {31} 10; {32} fourth; {33} second;
{34} 3; {35} 5; {36} 7; {37} 2; {38} 6; {39} 10; {40} 14; {41} 26; {42} 26; {43} yes; {44} 7; {45} 5;
{46} 2; {47} the same; {48} no; {49} d; {50} they have color; {51} no; {52} no; {53} no;
{54} They have an incomplete (unfilled) "d" sublevel; {55} Elements in column 10 on the microplate have
a filled "d" sublevel; {56} color
Problems:
1. a. 13,13,14
b. 55, 55, 78
2. a. 92, 92, 142
b. 92, 92, 143
3.
Nuclear
Notation
Isotope
27
Al
13
209
a. aluminum-27
b. bismuth-209
c. calcium-40
83
40
20 Ca
64
d. copper-64
29
207
82
f. _lead-207____
16
8
50
66
30
Pb
O
119
h. _tin-119_____
i. _zinc-66_____
Cu
4
2 He
e. _helium-4____
g. _oxygen-16__
Bi
Zn
Sn
c. 92, 92, 146
At.No.
Mass No.
No. e's
No. p's
No. n's
_13__
__27___
_13__
_13__
_14__
_83__
__209__
_83__
_83__
_126_
_20__
__40___
_20__
_20__
_20__
_29__
__64___
_29____ _29__
_35__
__2__
___4___
__2__
__2__
__2__
_82__
__207__
__82_
_82__
_125_
__8__
__16__
__8__
__8__
__8__
_50__
__119__
__50_
__50_
__69_
_30__
__66___
__30_
__30_
__36_
4. 63.54
5. These are estimates: 490 nm, 495 nm, 568 nm, 572 nm, 592 nm, 595 nm, 615 nm
6. 2, 8, 18, 32
7.
3d
(X)(/)(/)(/)(/)
12-24
©1997, A.J. Girondi
8. Element
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
9.
beryllium
magnesium
nitrogen
silicon
calcium
potassium
argon
sodium
chlorine
copper
phosphorus
nickel
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
1s
2s
2p
3s
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)
(X)(X)(X)
(/)(/)(/)
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)
(X)
(X)
(X)
(X)
(/)
(X)
(X)
(X)
(X)
3p
4s
(/)(/)( )
(X)(X)(X)
(X)(X)(X)
(X)(X)(X)
(X)
(/)
(X)(X)(/)
(X)(X)(X)
(/)(/)(/)
(X)(X)(X)
3d
(X)
(X)(X)(X)(X)(/)
(X)
(X)(X)(X)(/)(/)
1s 22s 22p 2
1s 22s 22p 3
1s 22s 22p 4
(given)
1s 22s 22p 6
1s 22s 22p 63s 23p 3
1s 22s 22p 63s 23p 6
(given)
1s 22s 22p 63s 23p 64s 23d 5
1s 22s 22p 63s 23p 64s 23d 10
1s 22s 22p 63s 23p 64s 23d 104p 6
10. Check periodic table for atomic weight of Iridium (#77). Answer may be a little off due to rounding.
11. a.
12.
13.
14.
191
77 Ir
b.
37
17 Cl
c.
50
24 Cr
iridium-191: 77 protons, 77 electrons, 114 neutrons
chlorine-37: 17 protons, 17 electrons, 20 neutrons
chromium-50: 24 protons, 24 electrons, 26 neutrons
a.
sulfur:
1s
(X)
2s
(X)
2p
(X(X)(X)
b.
vanadium: 1s
(X)
2s
(X)
2p
(X)(X)(X)
3s
(X)
3s
(X)
3p
(X)(/)(/)
3p
(X)(X)(X)
4s
( )
3d
( )( )( )( )( )
4s
(X)
3d
(/)(/)(/)( )( )
sulfur: 1s 22s 22p 63s 23p 4
vanadium: 1s22s 22p 63s 23p 64s 23d 3
12-25
©1997, A.J. Girondi
SECTION 12.13 Student Notes
12-26
©1997, A.J. Girondi