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Explain each of the following observations using principles of atomic structure and/or bonding.
a) Potassium has a lower first-ionization energy than lithium.
b) The ionic radius of N3¯ is larger than that of O2¯.
c) A calcium atom is larger than a zinc atom.
d) Boron has a lower first-ionization energy than beryllium.
1994 D
Use principles of atomic structure and/or chemical bonding to answer each of the following.
The radius of the Ca atom is 0.197 nanometer; the radius of the Ca2+ ion is 0.099 nanometer. Account for
this difference.
The lattice energy of CaO(s) is -3,460 kilojoules per mole; the lattice energy for K2O(s) is -2,240 kilojoules
per mole. Account for this difference.
Ionization Energy
Ca 590
Explain the difference between Ca and K in regard to
their first ionization energies,
their second ionization energies.
The first ionization energy of Mg is 738 kilojoules per mole and that of Al is 578 kilojoules per mole.
Account for this difference.
1993 D
Account for each of the following in terms of principles of atom structure, including the number, properties, and
arrangements of subatomic particles.
The second ionization energy of sodium is about three times greater than the second ionization energy of
The difference between the atomic radii of Na and K is relatively large compared to the difference between
the atomic radii of Rb and Cs.
A sample of nickel chloride is attracted into a magnetic field, whereas a sample of solid zinc chloride is not.
(d) Phosphorus forms the fluorides PF3 and PF5, whereas nitrogen forms only NF3. [actually a bonding moment]
1990 D
The diagram shows the first ionization energies for the elements from Li to Ne. Briefly (in one to three sentences)
explain each of the following in terms of atomic structure.
In general, there is an increase in the first ionization energy from Li to Ne.
The first ionization energy of B is lower than that of Be.
The first ionization energy of O is lower than that of N.
Predict how the first ionization energy of Na compares to those of Li and of Ne. Explain.
1973 D
First ionization Energy
Covalent Radii, Å
The covalent radii decrease regularly from Li to F, whereas the first ionization energies do not. For the ionization
energies, show how currently accepted theoretical concepts can be used to explain the general trend and the two
1976 D
M(s) + Cl2(g)  MCl2(s)
The reaction of a metal with chlorine proceeds as indicated above. Indicate, with reasons for your answers, the
effect of the following factors on the heat of reaction for this reaction.
A large radius versus a small radius for M2+
A high ionization energy versus a low ionization energy for M.
1977 D
The electron affinities of five elements are given below.
12 kcal/mole
32 kcal/mole
17 kcal/mole
48 kcal/mole
87 kcal/mole
Define the term “electron affinity” of an atom. For the elements listed above, explain the observed trend with the
increase in atomic number. Account for the discontinuity that occurs at phosphorus.
a) Response must contain a cogent discussion of the forces between the nucleus and the outermost (or "ionized")
electron. For example, a discussion of "the outermost electron on K..." should include one of the following:
i. it is farther from nucleus than the outermost electron on Li
ii. it is more shielded from the nucleus (or "experiences a lower effective nuclear charge") than the outermost
electron on Li
iii. it is in a higher energy orbital (4s) than tne outermost electron on Li (2s)."
2 points for any one
Notes:"K is larger than Li" earns 1 point.
No points earned for "K electron is easier to remove" (or some other restatement).
b) Nitrogen has one less proton than oxygen 1 point
Nitride and oxide ions are isoelectronic 1 point
In nitride ion the electron/proton ratio is greater, causing more repulsion; thus, nitride is the larger ion. 2 points
c) A Zn atom has more protons (10 more) than an atom of Ca 1 point
Electrons in d orbitals of Zn have a lower principal quantum number; thus, they are not in orbitals that are farther
from the nucleus. 1 point
d) Correct identification of the orbitals involved (2s versus 2p) 1 point
Clear statement that the two orbitals have different energies 1 point
Note: Arguments that "the 2p orbital is farther out than the 2s orbital", or that "the Be atom has a filled subshell,
which is a more stable configuration" earn no explanation point.
General note:For all parts (a) through (d), discussions of position in the periodic table earn no points.
a) two points
Ca2+ has fewer electrons, thus it is smaller than Ca.
The outermost electron in Ca is in a 4s orbital, whereas the outermost electron in Ca2+ is in a 3p orbital.
Note: The first point is earned for indicating the loss of electrons, the second point for indicationg the outermost
electrons are in different shells -- must account for the magnitude of the size difference between Ca and Ca2+.
b) two points
U for CaO is more negative than U for K2O, so it is more difficult to break up the CaO lattices (stronger bonds in
CaO). Ca2+ is smaller than K+, so internuclear separations (between cations and O2¯) are less.
Ca2+ is more highly charged than K+, thus cation-O2 bonds are stronger
Note: understanding what "lattice energy" is earns 1 point; size or charge explanation needed for the second point.
Responses that use Lewis structures or otherwise indicate molecules rather than ionic lattice earn no points.
(c) Explain the difference between Ca and K in regard to:
(i) their first ionization energies.
c) two points
i) Ca has ore protons and is smaller. The outermost electrons are more strongly held by the nuclear charge of Ca.
(ii) their second ionization energies.
ii) The outermost electrons in Ca are in the 4s, which is a higher energy orbital (more shielded) than the 3p
electrons in K.
Note: for (i), the idea of attraction between nucleus and electrons must be present; for (ii), a "noble-gas
configuration" argument must be tied to an energy argument in order to earn credit.
d) two points
the highest energy (outermost) electrons in Al is in a 3p orbital, whereas that electron in Mg is in a 3s orbital.
The 3p electron in Al is of higher energy (is more shielded) than is the 3s electron in Mg.
Note: noting that different orbitals are involved earns the first point; a correct energy argument earns the second
Responses that attribute the greater stability of Ca over K (or K+ over Ca+, or Mg over Al) to the stability of a
completely filled (vs. half or partially filled) orbital earn NO credit.
1993 D
a) Electron configuration of Na and Mg (1 pt)
Any one earns a point:
Octet / Noble gas stability comparison of Na and Mg
Energy difference explanation between Na and Mg
Size difference explanation between Na and Mg
Note: If only Na or Mg is used 1 point can be earned by showing the respective electron configuration and using
one of the other explanations
Shielding/effective nuclear charge discussion earns the third point.
b) one point
Correct direction and explanation of any one of the following:
shielding differences
energy differences
# of proton/ # of electron differences
c) two points
Any one set earns one point:
(i) Ni unpaired electrons. paramagnetic
(ii) Zn paired electrons/ diamagnetic
(iii) Ni unpaired electrons/ Zn paired electrons
(iiii) Ni paramagnetic/ Zn diamagnetic
Orbital discussion/ Hund's Rule/ Diagrams earns the second point.
d) two points
Expanded octet or sp3d hybrid of phosphorous (1 pt)
Lack of d orbitals in nitrogen (1 pt)
nitrogen is too small to accomodate (or bond) 5 Fluorines or 5 bonding sites (2 pts)
1990 D
a) Across the period from Li to Ne the number of protons is increasing in the nucleus hence the nuclear charge is
increasing with a consequently stronger attraction for electrons and an increase in I.E.
b) The electron ionized in the case of Be is a 2s electron wheras in the case of B it is a 2p electron. 2p electrons are
higher in energy than 2s electrons because 2p electrons penetrate the core to a lesser degree.
c) The electron ionized in O is paired with another electron in the same orbital, whereas in N the electron comes
from a singly-occupied orbital. The ionization energy of the O electron is less because of the repulsion between two
electrons in the same orbital.
d) The ionization energy of Na will be less than those of both Li and Ne because the electron removed comes from
an orbital which is farther from the nucleus, therefore less tightly held.
1973 D
The trend in moving across a period is that the first ionization energy, I1, increases from group 1 (Li) to group 7 (F) because
of an increase in effective nuclear charge, atoms get smaller (decrease covalent radii) and less metallic through the period. The
I1 is less for B than Be because the electron to be ionized in B is in a higher energy orbital (2p) than is the electron (2s) to be
ionized in Be. The I1 is less for O than N because the electron to be ionized in O is a paired electron in the 2p orbitals. At N,
the outer sublevel of its atom is half-filled, resulting in a symmetrical spherical electron cloud. The extra electron in O reduces
this symmetry and so less energy is required to remove this electron.
1976 D
As radius increases the heat of reaction decreases (less exothermic).
Less energy released by ionic attraction (lattice energy inversely proportional to distance).
As ionization energy increases the heat of reaction decreases (less exothermic),
More energy required to form M2+ while other factors remain unchanged.
1977 D
Electron affinity -the energy released when a gaseous atom gains an electron to form an ion.
As an electron is added to the same valence shell of an atom, when Z increases, the atomic radius decreases. Therefore,
the added electron in going from Al to Si and from P to S to Cl is closer to the nucleus and more energy is released (electron
affinity greater). Also as the atomic radius decreases, the shielding of the nucleus by the surrounding electrons is less
effective, and the attraction for the added electron is greater.
At P, the outer sublevel of its atom is half-filled, resulting in a symmetrical spherical electron cloud. The extra electron
reduces this symmetry and so less energy is released when it enters the atom to form P -.
Be sure to emphasize the “whys”. Coulomb’s Law, electron repulsions, effective nuclear charge, forces
dissipating with distance [back to Coulomb’s Law], etc. should appear in their answers.
A trend is NOT an explanation!!! Also emphasize that electrons are easier to remove from the f than the d
than the p than the s due to “penetration”—Zumdahl has a great graph that explains that—and obviously
it’s easier to remove a 5s electron than a 4s electron. Also, shielding is a great argument as long as the
electrons are from DIFFERENT principal energy levels. The argument is useless when the electrons are
from the same principal energy level [say a 3s vs. a 3p]. Kids over-use the shielding argument and lose points
since they apply it ALL the time rather than when it is appropriate. [I’m convinced dodge ball is to blame!]
Slater’s rules from an advanced inorganic course yield a full mathematical explanation of this for the
obstinate student!
SUBLEVEL OR ANYTHING ELSE!!! It’s all about reducing electron repulsions.
It requires less energy to remove the p4 electron than the p3 due to the fact that the 4th electron is the first to
pair, the first to experience a violent repulsion [the first 3 also experience repulsions—that’s why they spread
out according to Hund’s rule]—and it wants to get the heck out of Dodge! The increased repulsion means it
is less tightly held and therefore requires less energy to remove.
Lastly, I am guilty of personalizing and anthropomorphizing atoms and electrons. Make a point of telling
kids electrons are not human, animal or even pond scum. They need to avoid using phrases like “get the
heck out of Dodge” or an electron “wants to” or anything remotely similar to that!
This is a critical topic to both multiple choice and free response!!
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