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Saturday Study Session 2
2nd Class
Atomic Structure and Bonding
Opening Activity
Trend
What It Is
Across a Period Down a Group
Atomic radius
First ionization
energy
Justification
Justification
Justification
Justification
Content Crash Course
• Atomic Structure
• Instrumentation
– Mass Spec
– PES
• Electron Configurations
• Periodic Trends
• Intramolecular Bonds
Atomic Structure
• Small, dense nucleus
– Protons (+) and neutrons (no charge)
– Protons + neutrons = mass number
• Remaining atomic volume
– Electrons (-)
– Number of electrons in a neutral atom = atomic number
• Isotopes have the same atomic number, but different
mass numbers.
• Useful equations:
E = hv
c = λν
1.) Naturally occurring copper is composed
of two isotopes, copper-63 and copper-65.
If copper is 69.1% copper-63 and 30.9%
copper-65, the average atomic mass could
be calculated as
A. [(0.309 x 65) + (0.691 x 63)]/2
B. (63 + 65)/2
C. (65 x 30.9) + (63 x 69.1)
D. (0.309 x 65) + (0.691 x 63)
Question 1 Answer
D
Clue: Average atomic mass is calculated as
the sum of the products of the relative
abundance and atomic mass for each
isotope of a given element.
2.) The photoelectron spectra
show the energy required to
remove a 1s electron from a
nitrogen atom and from an
oxygen atom. Which of the
following statements best
accounts for the peak in the
upper spectrum being to the right
of the peak in the lower spectrum?
A. Nitrogen atoms have a half-filled p subshell.
B. There are more electron-electron repulsions in oxygen
atoms than in nitrogen atoms.
C. Electrons in the p subshell of oxygen atoms provide
more shielding than electrons in the p subshell of
nitrogen atoms.
D. Nitrogen atoms have a smaller nuclear charge than
oxygen atoms.
Question 2 Answer
D
Clue: Less effective nuclear charge means
less energy required to remove an
electron.
Electron Configurations
• Atoms
– # electrons = atomic #
– Fill according to Aufbau Principle, Hund’s Rule, and
Pauli Exclusion Principle
– Use periodic table to “cheat”
• Orbital notation
• Noble gas notation
• Ions
– Cations lose electrons
– Anions gain electrons
– Isoelectronic = same # of electrons
3.) An element Z has the electron
configuration [Xe]6s2. The oxide of this
element will have which formula below?
A. Z2O
B. Z2O3
C. ZO
D. ZO2
Question 3 Answer
C
Clue: The electron configuration indicates 2
valence electrons, which would be given
up to form a +2 ion.
Periodic Trends
Trend
What It Is
Atomic radius
Essentially the
distance from
the nucleus to
valence
electrons
First ionization
energy
Energy required
to remove an
electron from an
atom in the gas
phase
Across a Period Down a Group
Decreases
Increases
Justification
Increased effective
nuclear charge
Stronger force of
attraction between
nucleus and electron
cloud
Justification
Increased principal
energy level
Less strong force of
attraction between
nucleus and
electron cloud
Increases
Decreases
Justification
Increased effective
nuclear charge
Stronger force of
attraction between
nucleus and
electrons
More energy required to
remove an electron
Justification
Increased principal
energy level
Less strong force of
attraction between
nucleus and
electrons
Less energy required to
remove an electron
4.) Of the following, the species that has the
correctly predicted largest radius is
A. Ar
B. Br C. K+
D. Sr2+
Question 4 Answer
B
Clue: Anions exhibit greater radii because of
additional electrons and decreased
effective nuclear charge.
5.) The observation
involving the first ionization
energies of O and N is best
explained by which of the
following?
A. Differences in electronegativity between O and N cause the
first ionization energy of O to be less than N.
B. Repulsion between the two paired electrons in the O orbital
cause the first ionization energy of O to be less than N.
C. Greater effective nuclear charge in N than in O causes the
first ionization energy of O to be less than N.
D. Greater atomic radius in N than in O causes the first
ionization energy of O to be less than N.
Question 5 Answer
B
Clue: Comparing electron configurations of
O and N should allow you to see that O
has paired electrons while N does not.
6.) Which of the following elements would
have the largest second ionization
energy?
A. Hydrogen
B. Fluorine
C. Sulfur
D. Potassium
Question 6 Answer
D
Clue: While both H and K have one valence
electron, only K has more than one
electron period.
Ionization Energies for element X (kJ mol-1)
First
Second
Third
Fourth
Fifth
580
1815
2740
11600
14800
7.) The ionization energies for element X are listed
above. On the basis of the data, element X is
most likely to be
A. Na
B. Mg
C. Al
D. Si
Question 7 Answer
C
Clue: The huge jump between third and
fourth ionization energies means we’re
dealing with three valence electrons.
Intramolecular Bonds
•
•
•
•
Bonds BETWEEN atoms
Not to be confused with intermolecular forces (IMFs)
Result in chemical change
Ionic
– Large differences in electronegativity
– Usually metal and non-metal
• Covalent
– Small differences in electronegativity
– Usually non-metals
• Polar (H-F); dipole moments DO NOT cancel
• Non-polar (F-F); dipole moments DO cancel out
8.) Which of the following particulate
diagrams best shows the formation of
water vapor from hydrogen gas and
oxygen gas in a rigid container at 125º C.
Question 8 Answer
C
Clue: 6H2 (g) + 3O2 (g)  6H2O (g)
Or
2H2 (g) + O2 (g)  2H2O (g)
9.) In which of the following processes are
covalent bonds broken?
A. I2 (s)  I2 (g)
B. C (diamond)  C (g)
C. Fe (s)  Fe (l)
D. CO2 (s)  CO2 (g)
Question 9 Answer
B
Clue: Phase changes involve the
overcoming of IMFs, not
INTRAMOLECULAR bonds. Diamonds
involve network covalent bonds.
Short
Free Response
1.) Using principles of atomic structure and the information
in the table above, answer the following questions
about atomic fluorine, oxygen, and xenon.
(a) Write the equation for the ionization of atomic fluorine
that requires 1,681.0 kJ mol-1.
(b) Account for the fact that the first ionization energy of
atomic fluorine is greater than that of atomic oxygen.
(You must discuss both atoms in your response.)
(c) Predict whether the first ionization energy of atomic
xenon is greater than, less than, or equal to the first
ionization energy of atomic fluorine. Justify your
prediction.
Short Free Response Rubric
(3 points possible)
Free Response
#1
1.) The table above shows the first three ionization energies for atoms
of four elements from the third period of the periodic table. The
elements are numbered randomly. Use the information in the table
to answer the following questions.
(a) Which element is most metallic in character? Explain your
reasoning.
(b) Identify element 3. Explain your reasoning.
(c) Write the complete electron configuration for an atom of element 3.
(d) What is the expected oxidation state for the most common ion of
element 2?
(e) What is the chemical symbol for element 2?
(f) A neutral atom of which of the four elements has the smallest
radius?
Free Response #1 Rubric
(8 points possible)
Free Response #1 Rubric
Free Response #1 Rubric
Free Response #2
2.) Answer the following problems about gases.
(a)
The average atomic mass of naturally occurring neon is 20.18 amu. There
are two common isotopes of naturally occurring neon as indicated in the table
above.
(i) Using the information above, calculate the percent abundance of each
isotope.
(ii) Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally
occurring neon.
(b)
A major line in the emission spectrum of neon corresponds to a frequency of
4.34 × 1014 s−1.
Calculate the wavelength, in nanometers, of light that corresponds to this
line.
(c)
In the upper atmosphere, ozone molecules decompose as they absorb
ultraviolet (UV) radiation, as shown by the equation below. Ozone serves to
block harmful ultraviolet radiation that comes from the Sun.
O3 (g)  O2 (g) + O (g)
(d) A molecule of O3(g) absorbs a photon with a frequency of 1.00 × 1015 s−1.
(i) How much energy, in joules, does the O3 (g) molecule absorb per photon?
(ii) The minimum energy needed to break an oxygen-oxygen bond in ozone is
387 kJ mol-1. Does a photon with a frequency of 1.00 × 1015 s−1 have
enough energy to break this bond? Support your answer with a calculation.
Free Response #2 Rubric
(9 points possible)
Free Response #2 Rubric
Free Response #2 Rubric