Chem 31 SI 11/08/09
Periodic Properties
Quantum Numbers Warm Up:
1. Label the s, p, d, and f blocks on the periodic table below.
2. For an electrically neutral oxygen atom, there are 8 electrons, and each has its own unique set of atomic numbers.
Assign a viable set of atomic numbers (n, l, ml, ms) to each electron.
1) n = 1, l = 0, ml = 0, ms = ½
5) n = 2, l = 1, ml = -1, ms = ½
2) n
= 1, l = 0, ml = 0, ms = - ½
6) n
= 2, l = 1, ml = 0, ms = ½
3) n
= 2, l = 0, ml = 0, ms = ½
7) n
= 2, l = 1, ml = 1, ms = ½
4) n
= 2, l = 0, ml = 0, ms = - ½
8) n
= 2, l = 1, ml = -1, ms = - ½
3. For n=3, there are18 possible electrons (2n2). Each electron has a unique set of quantum numbers. Map out the
possible configuration. Instead of l use s,p, or d. Instead of ml use an orientation subscript (example: Px, Py, Pz).
Instead of ms use ↑ or ↓. On the periodic table below, highlight or circle the n=3 ground state electron series.
↑↓
↑↓
↑↓
↑↓
↑↓
↑↓
↑↓
↑↓
↑↓
3s
3px
3py
3pz
3dyz
3dxy
3dz
3dx2-y2
3dz2
Note the Correlated sets of quantum numbers:
3s Block:
n = 3, l = 0, ml = 0, ms = ½
n = 3, l = 0, ml = 0, ms = - ½
3p Block:
n = 3, l = 1, ml = -1, ms = ½
n = 3, l = 1, ml = -1, ms = - ½
n = 3, l = 1, ml = 0, ms = ½
n = 3, l = 1, ml = 0, ms = - ½
n = 3, l = 1, ml = 1, ms = ½
n = 3, l = 1, ml = 1, ms = -½
3d Block:
n = 3, l = 2, ml = -2, ms = ½
n = 3, l = 2, ml = -2, ms = - ½
n = 3, l = 2, ml = -1, ms = ½
n = 3, l = 2, ml = -1, ms = - ½
n = 3, l = 2, ml = 0, ms = ½
n = 3, l = 2, ml = 0, ms = - ½
n = 3, l = 2, ml = 1, ms = ½
n = 3, l = 2, ml = 1, ms = - ½
n = 3, l = 2, ml = 2, ms = ½
n = 3, l = 2, ml = 2, ms = - ½
(# of possible electrons = 2n2 or 2(32) = 18 electrons)
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Chem 31 SI 11/08/09
Periodic Properties
Trends in Atomic Radius:
1. On the periodic table below, use arrows to show the general directions of increase in atomic radius moving
vertically and horizontally along the table.
Note: the distance of the outermost electrons from the nucleus dictates the atomic radius…that distance is
determined by
1) their n-value
2) the magnitude of the attractive positive charge that they experience from the nucleus. We call this the
effective charge or “Zeff”:
Zeff = Z – S
Zeff = effective charge felt by an electron outside the shield
Z = actual charge of the nucleus (number of protons)
S = number of electrons in the shield.
2. Explain why the atomic radius of Helium is smaller than the atomic radius of Hydrogen.
Hydrogen has a nucleus with charge of +1 and one 1s electron, thus the 1s electron experiences a charge of +1
that keeps it attracted to the nucleus.
Helium has a nucleus with charge of +2, and two 1s electrons. These electrons do not shield each other from
the nucleus since they are in the same energy shell, so they both experience a full 2+ charge. They are
therefore more strongly attracted to the nucleus and thus drawn in closer…so the radius is smaller than that
of Hydrogen.
3. Explain why the atomic radius of Oxygen is smaller than the atomic radius of Boron.
2n electrons in Oxygen experience a stronger effective charge through the 1s shield than do 2n electrons in
Boron, so they are drawn more closely to the nucleus thus making the radius smaller. Check it out:
2n electrons in Oxygen: Zeff = 8-2 = +6
2n electrons in Boron: Zeff = 5-2 = +3
+6 > +3 !!
4. Explain why the atomic radius of Helium is smaller than the atomic radius of Neon.
Electrons in orbitals with higher n-values are farther away from the nucleus. He only has electrons in the 1s
orbital, while Neon has electrons in 1s, 2s, and 2p, thus it has a larger atomic radius. This generally accounts
for the fact that atomic radius increases as you move down a column.
5. In what region does a slight deviation exist for the rules discussed above? Why? (see pages 336-337)
Transition Metals. Why? Remember, the presence of electrons in the outermost energy level (the highest nvalue) dictates the atomic radius.
Consider the first series of transition metals from Sc to Zn. This is the 3d series. We have already added
electrons to the 4s orbital before filling the 3d, and these 4s orbitals are farther away from the nucleus than
the 3d. As we move across, each electron added to 3d orbital accompanies an addition of 1 proton to the
nucleus, so the effective charge (Zeff) felt by the 4s electrons essentially stay the same, so the atomic radii
basically stays the same. Everytime you move one element to the right, the Zeff equation looks like this:
Zeff = (Z+1) – (S+1)
so Zeff is relatively unaffected, thus atomic radius stays approx the same.
6. (# 59 from the text) Choose the larger atom from the following pairs: (draw vertical and horizontal arrows
on the table from one element to the other… this should make it obvious)
A) Al or In
B) Si or N
C) P or Pb
D) C or F
7. (#61 from the text) Arrange the following elements in order of increasing atomic radius: Ca / Rb / S / Si / Ge / F
F < S < Si < Ge < Ca < Rb
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Chem 31 SI 11/08/09
Periodic Properties
Trends in Ions and Ionization Energy
1. How does the atomic radius of a cation compare to its corresponding neutral atom and why? Circle the larger
species:
Li vs Li+.
Ca vs Ca2+.
FYI: The atomic radius of a cation is smaller than its corresponding neutral atom.
Note: During 11/9 lecture Dr. Leenstra indicated that this concept would not be tested on the exam. I only
included it in the worksheet because I did not have that information at the time. HW problems are assigned
that tested this concept, that is why I included it in the worksheet.
2. How does the atomic radius of an anion compare to its corresponding neutral atom and why?
2-
3-
Circle the larger species: O vs O .
N vs. N
FYI: The atomic radius of an anion is larger than its corresponding neutral atom.
Note: During 11/9 lecture Dr. Leenstra indicated that this concept would not be tested on the exam.
3. An isoelectronic series is a group of species with the same number of electrons. List the species in the following
isoelectronic series in order of decreasing atomic radius: F- / Ne / O2- / Mg2+ / Na+
Solve by comparing the number of electrons with the number of protons. They all have 10 electrons, O = 8
protons, F = 9 protons, Ne = 10 protons, Na = 11 protons, Mg = 12 protons. Atomic radius from largest to
smallest looks like:
O2- > F- > Ne > Na+ > Mg2+
4. On the periodic table above, use arrows to indicate the increase in first ionization energy vertically and
horizontally.
(NOTE THAT IONIZATION ENERGY TREND IS BASICALLY THE OPPOSITE OF ATOMIC RADIUS)
5. Why does Beryllium have a larger first ionization energy than Magnesium?
In Beryllium, the valence electrons occupy the 2s orbital, so they areheld closer to the nucleus than are the 3s
valence electrons in Magnesium. Thus it takes less energy to remove a valence electron (process of ionization)
from Magnesium than it does Beryllium, so the IE for Be is higher!
6. Why does Beryllium have a larger first ionization energy than Lithium?
Beryllium has a smaller atomic radius for reasons discussed on page 2 #2 and #3 on this worksheet. It
requires more energy to remove electrons situated closer to the nucleus.
7. Consider the image on the top right of the page. The x-axis is atomic number and the y-axis is first ionization
energy. The chart generally follows the trends already described; however, two dips occur between Li and Ne, and
again between Na and Ar. Account for both of these deviations / exceptions.
Deviation #1: Between Beryllium (1s22s2 ) and Boron (1s22s22p1) (or more generally between 2A and 3A)
Because the 2p orbital is slightly shielded by the 2s orbital, 2p electron on Boron is easier to remove (it is less
tightly bound to the nucleus).
Deviation #2: Between Nitrogen (1s22s22p3 and Oxygen (1s22s22p4) (or more generally between 5A and 6A)
This is because adding the fourth electron to the p orbital requires pairing up electrons with opposite
spins…this is slightly less favorable so the fourth electron is more easily “kicked out”.
Nitrogen 2p orbital:
↑
↑
↑
Versus Oxygen 2p orbital: ↑
↓
↑
↑
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Chem 31 SI 11/08/09
Periodic Properties
8. Which would you expect to require the highest amount of energy: the first ionization of sodium or the second
ionization of magnesium? Briefly explain why.
The second ionization would require WAY more energy since we are removing core electrons.
Note: During 11/9 lecture Dr. Leenstra indicated that this concept would not be tested on the exam.
9. (#75 from text) For each of the following elements, predict where the “jump” occurs for successive ionization
energies: Note: During 11/9 lecture Dr. Leenstra indicated that this concept would not be tested on the exam. The
jumps occur when we have ionized the atom to the point that it has a noble gas electron configuration…from
that point on we are removing core electrons as opposed to valence electrons. This requires MUCH more
energy, so the IE series “jumps”
A) Be
C) O
Between IE2 and IE3
Between IE6 and IE7
B) N
D) Li
Between IE5 and IE6
Between IE1 and IE2
10. What element has a higher first ionization energy than any other element on the periodic table?
A. Radon
B. Francium
C. Hydrogen
D. Helium
11. Which of the following statements is true regarding the second ionization energy?
A. The second ionization energy can be greater than or less than the first ionization energy depending on
the electron configuration.
B. The second ionization energy is always positive. (so is the first ionization)
C. The second ionization energy can be positive or negative, depending on the electron configuration.
D. The second ionization energy for an ion with a noble-gas electron configuration is always less than the
second ionization energy for an ion with multiple valence electrons.
Trends in Electron Affinity vs. Metallic Character
1. On the above periodic table, use arrows to show the general direction of metallic character.
2. Although the trends are less consistent, outline the general trends for electron affinity. What exception exists for
the 5A elements and why?
It is almost always favorable to add electrons, so the ∆E for electron affinity is almost always negative. A
more negative ∆E for the electron affinity means that an atom more readily accepts an electron. The TREND
is that ∆E becomes more negative as you move to the right across the periodic table. Think about: who want
an electron more, Na (which tends to form Na +) or Cl (which tends to form Cl-)? Answer: Cl !!!
We do not need to know the 5A exception. It is because we are adding an electron to np3, which requires pairing
electrons in one of the orbitals, which is slighlt unfavorable thus the electron is less readily embraced.
3. (#77 from text): Choose the element with the more negative (more exothermic) electron affinity from each of the
following pairs:
A) Na or Rb
C) C or N
B) B or S
D) Li or F
4. (#81 from text): Arrange the following elements in order of increasing metallic character: Fr / Sb / In / S / Ba / Se
S < Se < Sb < In < Ba < Fr
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