Kevin D. McMahon
Reseda Science Magnet
The Quantum-Mechanical Atom
and the
Properties of Elements
I. THE QUANTUM—MECHANICAL ATOM
A. Electromagnetic Radiation
Radio waves, microwaves, visible light, ultraviolet, X-rays, and gamma rays are all
examples of electromagnetic radiation.
Electromagnetic radiation has three primary
characteristics: wavelength, frequency, and speed.
Wavelength (symbolized by the Greek letter lambda,
"") is the distance between two consecutive peaks or
troughs in a wave.
Question #1:
Which form of electromagnetic radiation has the shortest wavelength?
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The frequency (symbolized by the Greek letter nu, "") is defined as the number of waves
(cycles) per second that pass a given point in space. The speed of all electromagnetic
radiation is the speed of light (symbolized by "c") which is equal to 2.9979 X 108 meters
per second.
There is an inverse relationship between wavelength and frequency:
• waves with the smallest wavelength have the highest frequency,
• waves with the largest wavelengths have the lowest frequency.
The mathematical expression:  = c
where  is in meters
 is in cycles/seconds
or Hertz (Hz)
describes the relationship between the primary characteristics of electromagnetic
radiation.
Question #2:
Calculate the frequency of green light if it has a wavelength of 5 X 10-7m.
Question #3:
An FM radio station broadcasts at a frequency of 105.1 MHz. What is the wavelength of the radio
waves?
Hint!
To use the formula  = c, the frequency must be in Hz. 1MHz = 1,000,000 Hz.
Quantum Theory
At the end of the 19th century classical physics held that matter and energy were distinct:
• matter was thought to consist of particles while energy was in the form of a wave (ie:
electromagnetic energy),
• particles were things that had mass and whose position in space could be specified,
• waves were described as massless and delocalized; that is, their position could not be
specified.
Classical physics also held that matter should be able to absorb or emit any quantity of
energy.
Absorbing energy was like a ball being pushed up a ramp.
Losing energy was like a ball
rolling down a ramp. Any amount of energy gain or loss is
possible.
In 1901, the German physicist, Max Planck, explained that the
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radiation by solid bodies heated to incandescence could not be explained by classical
physics.
To account for his observations Planck proposed that energy could be gained or lost only
in whole-number multiples of "h." "h" is called Planck's constant. It was determined by
experimentation to be 6.626 X 10-34 Js (a joule, J, is a unit of energy; 1J = 0.24 calories)
The energy gained or lost is given by the expression:
E = nh
where "n" is an integer (1,2,3...), h is Planck's contant, and  is the frequency of
electromagnetic radiation absorbed or emitted.
This relationship implied that energy was "quantized," that is, not
continuous. This was very different than classical physics. Instead
of the ball rolling up and down the ramp as in classical physics, the
ball would jump up or fall down certain allowed or quantized steps.
This relationship implied that energy was "quantized," that is, not
continuous:
• Energy could only occur in units of h.
• Each "packet" of energy is called a "quantum."
Albert Einstein proposed that electromagnetic radiation is quantized and that it was
packaged in "particles" he called "photons." The energy of each photon is given by:
E photon = h = hc/
Question #4:
The laser in a compact disc player uses light with a wavelength of 7.8 X 102 nm. Calculate the
energy of a single photon of this light.
Einstein also demonstrated that a photon of electromagnetic radiation had an associated
mass which could be determined by: m = h/c. Einstein had previously shown in his
special theory of relativity that: E = mc2 . This expression can be re-written as: m =
E/c2 . In addition, we know that the energy of a photon is given by: E = hc/ ,
therefore, we can substitute E in our first equation with hc/ and obtain:
m = hc/ = h/c
c2
Question #5:
Calculate the mass of the photon of light emitted by the laser from the previous question.
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Remember, the light had a wavelength of 7.80 X 102 nm. Your answer should be expressed in kg.
Light, which had been previously thought of as being
purely wave-like, was now being described by Quantum
Theory as having the particulate properties of matter.
This would ultimately revolutionize our understanding of
the atom.
According to classical physics the light emitted by the sun
possess only wave characteristics. But as we know, the
white light emitted by the sun is made up of numerous
colors. The light emitted by the sun consist of a continuous
spectrum of colored light of different wavelengths.
According to Quantum Theory the light emitted by the
sun consists of a stream of tiny energy packets called
photons. Each photon has an energy equal to h and
depending on its frequency () it will have its own
unique color.
Louis de Broglie (1892- 1987) suggested if energy exhibited particulate behavior then
matter should exhibit wave characteristics. Using Einstein's equation: m = h/lc and
considering a particle with a velocity (v) de Broglie obtained: m = h/v de Broglie
rearranged the expression to obtain:  = h/mv
de Broglie equation demonstrated that a particle with a mass (m) will have a wavelength
(l) when traveling at velocity (v).
• Notice that as mass increases the wavelength decreases.
• de Broglie's equation has been shown to be true through experimentation.
Question #6:
Calculate the de Broglie wavelength for an electron with a velocity of 15% of the speed of light.
The mass of an electron is equal to 9.1 X 10-31kg Hint! To calculate 15% of the speed of light...
(.15) (3 X 108 m/s)
Question #7:
Calculate the de Broglie wavelength for the fastest measured baseball (5.2 oz) with a velocity of
100.8 mph).
Hint!
Okay, but you should know this...
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5.2 oz x 1 lb x 1 kg =
16 oz 2.2 lb
100.8 miles x
1 km x
hr
.62 miles
1000m x
1 km
1 hr x
60 min
1 min =
60 s
The work of Planck, Einstein, and de Broglie led to the postulation of the "Wave-Particle
Duality of Matter" which stated that matter exhibits both particulate and wave
characteristics. Large pieces of matter, such as baseballs, exhibit predominately
particulate properties whereas tiny pieces of matter, such as photons, exhibit
predominately wave properties. Pieces of matter with intermediate mass, such as
electrons, show both particulate and wave properties. Appreciating the dual
characteristics of the electron is fundamental to understanding the structure of the atom
and the properties of the elements.
Quantum Model
When hydrogen gas receives a high-energy spark it absorbs the energy. Hydrogen gas
then releases the absorbed energy as light. When this light is passed through a prism or
diffraction grating the hydrogen emission spectrum has distinct lines rather than being a
continuous spectrum.
The emission spectra of hydrogen shows that hydrogen atoms will absorb and emit only
certain wavelengths of energy.
• This is in accordance with Planck's quantum theory.
• The energy absorbed and emitted can be determined using the equation:
E = hc/
( is equal to the wavelength of light emitted).
In 1913, a Danish physicist named Niels Bohr used the emission spectra of hydrogen to
develop his "Quantum Model of the Atom." Bohr suggested that the electron in a
hydrogen atom moves around the nucleus only in certain allowed circular orbits. An
electron in the orbit closest to the nucleus is in its lowest energy state or its "ground
state."
The electron can absorb a particular photon of energy and
jump to another orbit.
When the electron falls down to
a lower orbit it will emit a
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photon of light.
Bohr's mathematical computation for the energy of various orbits corresponded perfectly
with the energies associated with the emission spectra of hydrogen. Each line in
hydrogens emission spectra responds to a photon of light emitted when an electron
drops from one orbit to a lower orbit.
The most important equation to come from the Bohr model is the expression for the
energy levels available to the electron in the hydrogen atom:
n is an integer (the larger the value of n, the larger the orbit), z is the nuclear charge
which for the hydgrogen atom is 1. The negative sign in the equation means that the
energy of the electron bound to the nucleus is lower than it would be if the electron were
not bound to the nucleus.
The equation can be used to calculate the change in energy of an electron and the wavelength of light emitted or absorbed when the electron changes orbit. E = Efinal - Einitial
Question #8:
Calculate the energy of an electron in a hydrogen atom when n = 6.
Question #9:
Calculate the energy emitted when an n= 6 electron falls down to the n =1 (ground state) level.
Hint. Calculate E then use this value in the formula below:
E = Efinal - (- 6.05 X 10-20 J)
Questions #10:
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Calculate the energy required to excite the hydrogen electron from level n = 1 to level n = 2.
Hint! Calculate the value for E and then use this and the value you calculated previously for E
in the formula: E = Efinal - Einitial
Question #11:
Calculate the wavelength of light that must be absorbed to produce this change.
Hint: Use the formula  = hc/E and use the value of E that you calculated in the previous
problem. Remember, c = 3 X 108 m/s.
When Bohr's Quantum Model was applied to atoms other than hydrogen it did not work.
• The Quantum Model is fundamentally incorrect in that the electrons do not move
around the nucleus in circular orbits.
• The Quantum Model was important because it helped pave the way for other theories.
Bohr was awarded the Nobel Prize in Physics in 1922.
The Wave Nature of Electrons
Erwin Schrodinger, an Austrian physicist, applied de Broglie's concept that the electron
exhibited wave behavior to develop the Wave-Mechanical Model of the Atom. The
electron has significant wave characteristics because of its size. Because of this wave
characteristic the electron's position could not be determined with certainty. Werner
Heisenberg demonstrated mathematically that there was a fundamental limitation to just
how precisely we can know both the position and momentum of a particle at a given
time.
• This principle is known as Heisenberg's Uncertainty Principle.
• This uncertainty is insignificant when dealing with substantial mass, but becomes very
significant when dealing with particles such as electrons.
There was a philosophical implication with Heisenberg's Uncertainty Principle. It meant
that there was a fundamental limitation to how precisely we could know something,
especially like the location of an electron. Given this, Schrodinger recognized that it
wouldn't be possible to know the exact location of the electron. But perhaps he could
determine the "probability" of locating an electron.
Schrodinger's analysis allowed him to predict the probability
of finding the electron at some distance from the hydrogen
nucleus. His calculations showed that the probability of
finding an electron increased moving away from the nucleus
to a maximum of 0.529Å . The probability of finding an
electron diminished as the distance from the nucleus
increased.
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When the probabilities of finding electrons are plotted with respect to their distance from
the nucleus one obtains a nearly sphrical distribution of electrons around the nucleus
with the greatest concentration approximately 0.529 Å away from the nucleus.
There is no way of knowing where the electron is within the sphere because of
uncertainty. Because of its wave nature the electron effectively occupies the entire
volume of the sphere. The sphere has no distinct limit since the probability of finding an
electron at some further distance from the nucleus never goes to zero. Therefore, the size
of the sphere is arbitrarily defined as the radius of the sphere that encloses 90% of the
total electron probability.
The sphere of electron probability surrounding the nucleus is called the "1s" orbital. The
1s orbital corresponds to the innermost orbit of the Bohr Model.
• An "orbital" is not an orbit which has a defined circular path with a specified radius.
• The path or motion of the electron in an orbital is not known.
• The distance from the nucleus of an electron in an orbital is specified by the radial
probability.
This model of the atom is called the Wave-Mechanical Model.
• de Broglie, Schrodinger, and Heisenberg are credited with the development of this
model.
• The Wave-Mechanical Model not only explains the behavior of hydrogen atoms, but
all other known atoms.
Quantum Numbers
When the Schrodinger equation is solved for hydrogen, it is determined that there are
many orbitals that satisfy it. Each orbital is characterized by "quantum numbers" which
describe various properties of these orbitals.
The Principle Quantum Number (n):
The principle quantum number (n) is related to the size and energy of the orbital.
n can have integral values of 1, 2, 3 .... n is sometimes referred to as an energy "level" or
"shell" and corresponds to the Bohr energy level.
As n increases...
• the orbital becomes larger,
• the electron spends more time further from the
nucleus,
• the electron has higher energy.
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The Azimuthal Quantum Number (l):
l relates to the shape of the atomic orbital. l can have integral values of 0 to n-1. The
value of l for a particular orbital is commonly assigned a letter:
• when l = 0, the orbital is called "s."
• when l = 1, the orbital is called "p."
• when l = 2, the orbital is called "d."
• when l = 3, the orbital is called "f."
This system of naming orbitals arises from early spectral studies. l also indicates the
number of orbitals for each value of n:
• when n=1, there is only 1 orbital, ie: the s orbital,
• when n=2, l can be 0 or 1; thus there are two orbitals—s and p,
• when n= 3, l can be 0, 1, or 2; thus there are three orbitals- s, p, and d.
• when n=4, l can be 0, 1, 2, or; thus there are four orbitals- s, p, d, and f.
Each orbital is designated by the number for the principle quantum number and the
letter designation for l.
For example: 2p orbital
Here the 2 indicates that n = 2, and the p indicates that l = 1.
Sometimes l is referred to as a "sublevel" or "subshell."
The Magnetic Quantum Number (ml):
ml relates to the orientation of the orbital in space relative to the other orbitals in the
atom. ml can have integral values between l and -1, including 0.
When l = 0 = s, there is only 1 value of ml, that is, 0.
The sublevel s has only one orbital which has a spherical shape.
The 1s and 2s orbitals are separated by a "node," a region of zero
probability of finding an electron. Here we see the 1s orbital nestled
within the 2s orbital.
When l = 1 = p, there are three values
of ml: -1, 0, 1. The sublevel p consist
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of three orbitals corresponding to ml = -1, 0, 1. There are no p orbitals when n = 1. The
shape of these orbitals are often referred to as looking like dumb bells.
When l = 2 = d, there are five values of ml: -2, -1, 0, 1, 2. The sublevel d has five orbitals.
There are no d orbitals when n = 1 and n = 2. When l = 3= f, there are seven values of ml:
-3, -2, -1, 0, 1, 2, 3. The sublevel f has seven orbitals. There are no f orbitals when n = 1, 2,
or 3.
Electrons can move from one orbital to another. In the lowest energy state, the ground
state, the electron resides in the 1s orbital. If energy is absorbed by the atom, the electron
can be transferred to a higher energy orbital producing an "excited state."
• If the electron drops to a lower energy level the atom emits energy.
• Like the Bohr atom this can account for the spectral lines of hydrogen. But unlike the
Bohr model the Wave-Mechanical model can account for the emission spectra of all other
atoms.
Electron Spin and the Pauli Exclusion Principle:
Samuel Goudsmit and George Uhlenbeck (University of Leyden
in the Netherlands) found that a fourth quantum number was
necessary to account for additional characteristics of emission
spectra. The electron spin quantum number, ms, can have only
two values: +1/2 and- 1/2.
ms = +1/2 or -1/2 is interpreted that the electron can spin in one
of two opposite directions. Since a spinning electric charge
produces a magnetic field, the two opposing spins produce oppositely directed magnetic
fields.
The Austrian physicist, Wolfgang Pauli, proposed his Exclusion Principle which stated
that in a given atom no two electrons can have the same set of four quantum numbers.
Since electrons in the same orbital have the same values of n, l, and ml they must have
different values of ms. Since only two values of ms are allowed, an orbital can only have
two electrons, and they must have opposite spins.
Question #12:
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The dots represent the probable electron distribution for the 1s orbital. Which percent that
corresponds to the outer limit of the orbital?
Question #13:
The quantum number that corresponds to the Bohr orbital is...
a. n
b. l c. ml d. ms
Question #14:
When l = 2, the orbital is...
a. d orbitals
b. f orbitals
c. s orbitals
d. p orbitals
Question #15:
When n = 3, l can be ...
a. 1
b. 0, 1
c. 0,1,2
d. 0,1,2,3
and the corresponding orbitals are...
a. s
b. s, p
c. s, p, d
d. s, p, d, f
Question #16:
Which orbital has the lowest energy?
a. 4d b. 3d c. 4s
d. 4p
Question #17:
When l = 2= d, values of d included:
a. -2, -1, 0, 1, 2
b. 0, 1, 2
c. -1, 0, 1
d. 0
Question #18:
Which of the following are possible sets of quantum numbers for an electron?
a. n = 1, l = 0, ml = 1, ms = +1/2
b. n = 9, l = 7, ml = -6, ms = -1/2
c. n = 2, l = 1, ml = 0, ms = 0
d. n = 1, l = 1, ml = 1, ms = +1/2
Quantum Numbers and the Periodic Table
The Periodic Table can be understood using our know- ledge of the Wave-Mechanical
Model of the Atom. The numbers running down the left hand side of the table
correspond to the principle quantum number "n." The Periodic Table is also divided into
clearly identifiable sections. These sections correspond to the Azimuthal Quantum
number (l) or the orbitals s, p, d, and f.
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As we go across the Periodic Table electrons fill orbitals in accordance with the
rules set forth by the Wave Mechanical Model and the Pauli Exclusion Principle.
Hydrogen has 1 electron which occupies
the 1s orbital, which can be depicted as: The
superscript "1" represents the number of electrons
occupying the 1s orbital. The arrow represents an electron spinning in a particular
direction.
Helium completes the 1s orbital:
Note the arrows are in opposite directions
indicating opposite spin.
The n = 1 energy level is now filled. Electrons must now begin to fill the n = 2 energy
level and orbitals:
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Carbon has one more electron than Boron.
Should the next electron be added to the p
orbital already containing an electron or a
completely empty p orbital?
The German physicist, F.H. Hund proposed a rule stating that atoms seek to be at their
lowest energy state.
• This is achieved when an atom has a maximum number of unpaired electrons.
• All unpaired electrons have parallel spins. Therefore...
Carbon's electron configuration is...
Continuing across the row or period 2
The elements of period 3 (n = 3) add electrons to the orbitals in the same manner as just
described for period 2. The valence electrons are the electrons in the outermost principle
quantum level of the atom. For example, Silicon has the electron configuration:
Si: 1s2 2s2 2p6 3s2 3p2
Silicon has 4 valence electrons: 3s2 3p2
Valence electrons are the most important in chemistry since they are involved in
chemical reactions. The inner electrons (non valence) are called "core electrons." The
core electrons of silicon are: 1s2 2s2 2p6
Because the core electrons have the same configuration as the Group 8A element of the
previous period, electron configuration are often abbreviated using the symbol of the
previous Group 8A element followed by the valence electrons. For example, silicon's
electron configuration could be depicted as Si: [Ne] 3s2 3p2
As we move to the fourth period of the Periodic Table we fill the 4s orbital:
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K: [Ar] 4s1
Ca: [Ar] 4s2
The next element, Scandium, begins a series of elements known as the "transition metals."
The transition metals add electrons by filling the d orbitals. In filling the transition
elements of period 4 (n=4) the 3d orbitals are filled. The 3d orbitals are filled before the
4p orbitals because they are lower in energy.
Use the Periodic Table to observe the electron configuration of the period 4 transition
metals. Note the anomalous configurations of Cr and Cu.
The elements of period 5 add electrons to the orbitals in very much the same manner as
described for period 4. Again, you may use the Periodic Table to review the electron
configuration of these elements.
The element Barium has the electron configuration: Ba: [Xe] 6s2 is followed by
Lanthanum, which has the expected electron configuration: La: [Xe] 6s2 5d1
The element following lanthanum is Cerium. This element belongs to a group of
elements called "lanthanides." The lanthanides consist of 14 elements in which the 4f
orbitals are being filled. The 4f and the 5d energy sublevels are so close that sometimes
electrons will fill the 5d orbital before the 4f. Electrons fill elements of period 7 in a
manner similar to that of period 6. The actinides, like the lanthanides are elements that
are having their f orbitals filled. The Lanthanides and Actinides are sometimes referred
to as the f-transition elements.
The Wave-Mechanical Model of the Atom and the electron configurations we obtain by
its application provides us with the necessary framework to understand why elements
possess certain properties and why these properties are periodically repeated through
the Periodic Table.
Question #19:
The electron configuration for silicon is : [Ne] 3s2 3p2
The electrons fill the orbitals as follows...
a.
b.
c.
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Question #20:
Which of the following is a possible electron configuration?
a. [Kr] 5s2 4d1 b. [Kr] 5s2 5d1
c. [Kr] 5s2 5p1
The element described above is a...
a. transition element b. lanthanide c. actinide
Question #21:
Which element has a ground state electron configuration in which it has three unpaired p
electrons?
a. Boron
b. Argon
c. Nitrogen d. Scandium
Question 22:
Identify the element that has an excited electron configuration of 1s2 2s2 2p5 3s1 .
a. Neon
b. Sodium c. Carbon d. Fluorine
THE PROPERTIES OF ELEMENTS
Periodicity
Mendeleev discovered that when elements were arranged by increasing atomic mass the
properties (such as valence and reactivity) of the elements would periodically repeat. He
called his arrangement of elements the Periodic Table of the Elements.
Modern Periodic Table arranges elements by increasing atomic number. Even with this
fundamental change the elements are still observed to be grouped in families that share
the same or similar physical and chemical properties.
The Wave-Mechanical Model of the Atom allows us to understand the periodicity of the
elements and why families of elements exist. The questions that follow will help you to
explore the relationship between the Wave-Mechanical Model and periodicity.
Question #23:
Which statement best explains the concept of periodicity?
a. Members of the same period or row complete their s orbitals before filling their p orbitals.
b. Members of the same group or column have similar electron configurations.
c. Members of the transition elements have partially full or completely full d orbitals.
d. Elements further down a group have higher principle quantum numbers.
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Question #24:
Which of the following groups of elements would have similar chemical properties.
a. boron, carbon, nitrogen
b. manganese, iron, silver
c. nitrogen, phosphorus, antimony
d. gadolinium, europium, holium
Atomic Radius
The size of an atom cannot be precisely defined because the size of its
orbitals cannot be determined exactly. Atomic radii must be
determined by measuring the distance between atoms in a
compound. The distance between the two nuclei in Br has been
determined to be 2.28Å . The atomic radius of Br is assumed to be
1/2 this distance or 1.14Å .
Atomic radii have been determined for most of the elements of the Periodic Table
Question #25:
Which statement best describes the trend of atomic radii in the Periodic Table?
a. The atomic radius increases going across a period and down a group.
b. The atomic radius decreases going across a period and increases going down a group.
c. The atomic radius increases going across a period and decreases going down a group.
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom:
X(g) + ionization energy --> X +(g) + ewhere the atom is assumed to be in the ground state.
More than one electron can be removed from an atom. Consider the ionization of
aluminum:
kJ/mole
Al
(g) ---> Al+
(g) + e- I= 580
Al+ (g) ---> Al+2 (g) + e- I= 1815
Al+2 (g) ---> Al+3 (g) + e- I= 2740
Al+3 (g) ---> Al+4 (g) + e- I=11,600
The first ionization energy (I1 ) is significantly lower than the second ionization energy
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(I2) because of the increased charge of the nucleus and resultant attraction to the electron.
The difference in ionization energies between I1 and I2 can also be explained because the
first electron to be ionized was a 3p whereas the I2 electron is in the 3s orbital. Much
energy is required to remove the 4th e- because it's a core electron.
Question #26:
What trend do you observe in the Periodic Table with respect to ionization energy?
a. Ionization energy increases going across a period and decreases going down a group.
b. Ionization energy decreases going across a period and increases going down a group.
c. Ionization energy increases going across a period and going down a group.
Question #27:
There are a couple noticeable exceptions to the trend you observed in the previous question. One
of these exceptions suggests that...
a. atoms seek to fill partially full orbitals first.
b. electrons fill orbitals to maximize electron-electron repulsion.
c. it is more stable to have all of your p orbitals partially full.
Electron Affinity
Electron affinity is the energy change associated with the addition of an electron to a
gaseous atom: X(g) + e- ---> X- (g)
The negative values (example, F has an electron affinity of -328 kJ/mole) indicates that
energy is released when the electron is added. The more negative the electron affinity
the more stable the electron configuration achieved when an electron is added. The
Group 7 elements have the most negative electron affinity. This is due to the fact that
when Group 7 elements (ns2 p5 ) add an electron they complete their "p" orbitals (ns2 p6)
which is a stable electron configuration.
Question #28:
Ionization energy tells us how willing an atom is to part with an electron. Electron affinity
suggests how much an atom would like to add an electron. Given this which of the following
would you expect to be true?
a. Atoms tend to want electrons less going across and up the periodic table.
b. Atoms tend to want electrons more going across and down the periodic table.
c. Atoms tend to want electrons more going across and up the periodic table.
Stable Electron Configuration
You will observe by examining the Periodic Table that all of the Group 8 elements have
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their outer shell full of electrons. This is known as a "stable electron configuration."
Elements try to achieve a stable electron configuration if they do not already have one.
They can do this by reacting with other elements who are also trying to achieve stability.
Achieving stability can be accomplished by losing, gaining, or sharing electrons. In
doing this unstable elements combine or bond with other unstable elements through
chemical reactions to form compounds. In compounds each element has achieved a
stable electrons configuration. The Group 8 elements already have a stable electron
configuration and hence have little incentive to react with other elements. Consequently,
these elements are known as "inert." They are also referred to as "Noble" because of their
lack of association with "common" reacting elements.
Question #29:
The alkali metals (Group IA) are very reactive. Given their electron configuration, ionization
energy, and electron affinity which of the following statements best describes how the alkali metals
would achieve a stable electron configuration.
a. Alkali metals attract additional electrons to fill up their s and p orbitals.
b. Alkali metals are in fact already stable and do not need to alter their electron configurations.
c. Alkali metals give up their one valence electron.
Question #30:
The halogens (Group VIIA) elements are also very reactive. Given their electron configuration,
ionization energy, and electron affinity which of the statements best describes how the halogens
might try to achieve a stable electron configuration.
a. A halogen will attract an electron to fill its p orbital thereby achieving a stable electron
configuration.
b. Halogens lose electrons to become stable.
c. Halogens form "metal bonds" a share many electrons between them.
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Question # 31:
Identify the regions of s, p, d, and f orbital fillings.
Question #32:
Write electron configurations for each element identified by the letters a through h.
Question #33:
Use arrows to indicate increasing atomic radii, increasing electron affinity, and decreasing
electronegativity.
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