Periodic Trends
(May 4, 1999)
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It is impossible to analytically solve the Schrödinger equation (or use the Bohr model for that matter) for a system involving more than 2 particles: e.g. all multi-electron atoms/ions.

Slater came up with an idea that would "approximate" the solution for the Schrödinger equation.  He suggested that the most important electrons, those in the valence shell, could be treated as though they experience an "effective nuclear charge" of the nucleus due to the number of the protons in that species as well as any effects of "shielding" by the core electrons present.

Consider the following cartoon:
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The valence 2s electron of lithium is "shielded" to some extent, from feeling the entire 3+ charge of the nucleus by the "core" 1s electrons.  It is important to note that the electrons DO NOT move in orbits as illustrated, but they do exhibit an "average" distance from the nucleus which is what the circles represent.  The 1s electrons, on average, are much closer to the nucleus than the 2s electron of Li and hence "shield" it.

The result is the following equation:
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En = -2.18 x 10-18 J (Zeff2/n2)
En = -1310 kJ/mol (Zeff2/n2)
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Recalling that "ionization" is tantamount to promoting an electron from it's ground state to the n = "infinity" energy level (infinite distance from the nucleus) we can calculate Zeff (the effective nuclear charge experienced) for the 2s electron in Li.

Example:
    The ionization energy for Li is listed as 520 kJ/mol.
What is Zeff for the 2s electron?

Applying the second form of the equation and noting that
DE = Einf. - Egs = 0 - Egs = -E2s.

520 kJ/mol = 1310 kJ/mol(Zeff2/22)
0.397 = Zeff2/4
1.59 = Zeff2
1.26 = Zeff


This gives rise to an interesting observation. Since flourine has a much larger 1st ionization energy than lithium it must be "smaller" in a sense, since the valence electrons (2p) are closer, on average, to the nucleus.  IP1(F) = 1681 kJ/mol leading to a calculated Zeff of 2.26 for the 2p valence electrons of fluorine.  This means that, while fluorine is formed by adding six electrons to the valence shell and six protons to the nucleus of lithium (Zeff(2s) = 1.26), the valence electrons are not shielded as well from the much larger nuclear charge present in fluorine.

What am I saying?

The nucleus of F has a much higher charge density (9 protons) compared to Li (3 protons), and the electrons in the valence shell do not shield each other effectively from experiencing it.  Consider the following cartoon:
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Again, the circles here do not actually represent "orbits" but instead depict the average distance a subshell of electrons is from the nucleus.
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Clearly, the 2p electrons, being spatially well separated ("occupying" different px, py, pz orbitals), do not shield each other from the nucleus very well.  The 2s electrons, although close in energy to the 2p electrons, are in an orbital with greater "penetration" (probability of being near the nucleus) and hence are at somewhat lower energy.
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Energy is proportional to distance.
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High energy electrons:
= easily ionized
= further from the nucleus.
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Larger size = smaller ionization energy

We arrive at several important observations about the periodic table.

Periodic Trends: Size

1st, as one travels from left to right across a period, the "size" of an atom (or similarly charged ion) DECREASES.  That's right.  Despite the fact that we are adding electrons to the valence shell and protons to the nucleus, the size of the atoms to the right in the same period actually decreases.  Reason: the electrons (and their motion) are responsible for the vast majority of the atom's "size" and the valence electrons do not "shield" each other well, Zeff increase from left to right.
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Size: Li > Be > B > C > N > O > F > Ne
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2nd, as one travels DOWN the same family of elements, the size increases for atoms or similarly charged ions.  One can show that Zeff actually increases slightly as we go down a column, but this is a subtle increase.  In contrast, n increases by one unit with each successive period.  Since En = -2.18 x 10-18 J (Zeff2/n2) the valence electrons are increasing in energy.
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example (atomic radius):
Rn > Xe > Kr > Ar > Ne > He
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Consider the following table:
 
element
valence e-
n
Zeff (calculated)
IP1
(kJ/mol)
Li
2
1.26
520.2
Na
3
1.84
495.8
K
4
2.26
418.8
Rb
5
2.77
403.0
Cs
6
3.21
375.7
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Zeff does increase, but n increases faster.  Another informative way of remembering this trend is to consider the idea of filling shells, where each shell has an average distance from the nucleus (as illustrated by the circles in the cartoons for both fluorine and lithium).  A large increase in size occurs as we jump from one shell to the next: the noble gas exhibits the smallest "atomic" radius in each period, and the next alkali metal created by the Aufbau Principle fills the next shell and hence is much larger.

3rd the size of an ion depends very much on its charge.  Increasing positive charge translates to greater attractive forces experienced by the remaining electrons, and hence smaller size.
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Si4+ < Si3+ < Si2+ < Si+ < Si

For isoelectronic species (species containing the same number of electrons) size depends on the charge of the nucleus in a similar manner:
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species: Al3+ < Mg2+ < Na+ < Ne < F-
# protons 13 12 11 10 9
# electrons 10 10 10 10 10
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Note here that the same number of electrons are present for each species, but that a greater number of protons are present for Al3+, hence it is much smaller than F-.

Periodic Trends: Ionization Energies

Recall that the ionization energy is the amount of energy required to cause a gaseous atom/ion to eject its outermost (valence) electron.  If the electron is closer to the nucleus it is lower in energy, and hence harder to promote to "infinite" separation as required by the ionization process.
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Shown above is a plot of 1st ionization energy versus atomic number which reflects the periodic trends we are after.

Trend 1:  As we move from left to right across a period the ionization energy tends to INCREASE.  This is for the same reasons as was stated earlier: Zeff is increasing, while n is unchanged.  The valence electrons experience greater effective nuclear charge, are closer to the nucleus, and hence are more diffcult to remove.

Trend 2: As we move down a family (same column of the periodic table) ionization energy DECREASES.  Again, this is directly related to size as before.  One can visually see this on the graph by looking only at the trend for the noble gases (which are the peaks) or the the alkali metals (which are the valleys).  As one moves from He -> Ne -> Rn the ionization energy follows a smooth curve to lower values.

Other trends:

As one increases the charge on a species it becomes more difficult to remove an electron.  Again, size can be used as an indicator: small size = hard to remove the valence electron.

Consider the following table:
 
Ionization Energies of the 3rd Period Elements in kJ/mol
Na Mg Al Si P S Cl Ar
I1
 495.8 737.7 577.6 786.5 1012 999.6 1251.1 1520.5
I2
 4562 1451 1817 1577 1903 2251 2297 2666
I3
 7733 2745 3232 1912 3361 3822 3931
I4
 11580 4356 4957 4564 5158 5771
I5
 16090 6274 7013 6542 7238
I6
 21270 8496 9362 8781
I7
 27110 11020 12000
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As one goes down column 4, we see that each ionization event becomes more energetically costly than the previous one.  We can easily rank the ionization energies for the following species:

Si4+ > Si3+ > Si2+ > Si+ > Si
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Note that there is always a big jump in ionization energy as one attempts to remove a core electron.  These values are designated in pink and are considerably larger in magnitude than the ionization energy directly above.

If one compares the following processes:
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Process
DHion (exptl)
Na+  --->  Na2+ + e-
4562 kJ/mol
Mg+ --->  Mg2+ + e-
1451 kJ/mol
Al+ ---> Al2+ + e-
1817 kJ/mol
Si+ ---> Si2+ + e-
1577 kJ/mol
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One can readily see that the first process is most energetically costly, since it corresponds to removal of a core electron.
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One can calculate the ionization energies for single electron species such as Be3+, Li2+, He+, and H by using the Bohr equation:
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En = -1310 kJmol (Z2/n2).
DE = Einfinity - E1 = -E1
= 1310 kJ/mol (Z2/n2)
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In all 4 cases the last electron remaining is a 1s electron, so n = 1. Quite clearly, greater positive charge on the ion/species translates to increased ionization energy.
.
species
Z
n
DHion (calc)
 DHion (exptl)
H
1
1
1312 kJ/mol
1312 kJ/mol
He+
2
1
5248 kJ/mol
5250 kJ/mol
Li2+
3
1
11808 kJ/mol
11815 kJ/mol
Be3+
4
1
20992 kJ/mol
21006 kJ/mol

Significance of the ionization energy trends

1st of all, since it is very hard, comparatively speaking, to remove a core electron, it makes sense that the oxidation state exhibited by the alkali metals in ionic compounds is always +1.  Similar arguments can be used to explain the +2 oxidation state exhibited by the alkaline earth metals in ionic compounds, and also the +3 oxidation state exhibited by the Gp3A metals in ionic compounds.

Next, it should be readily apparent that the elements to the right (chalcogens: Gp6A, halogens: Gp7A) are difficult to ionize at all, and as it turns out these elements tend to form anions instead.  This explains their tendency for non-metallic behavior and the fact that, as elements, they are not reducing agents (electron donors) but instead are oxidizing agents (electron acceptors).


Electron Affinity

Electron Affinity is defined as the heat associated with the following process (wherein a species binds an electron):

E + e- --> E-

It is often referred to as electron attachment enthalpy: DHEA.

The trends associated with electron affinities are those expected based on the sizes of the atoms and the relative energies of the orbitals which will be populated by the newly added electron.

As one moves from left to right across a period the electron affinity INCREASES (more energy is released to the surroundings upon binding an electron).

1st electron affinities are always negative, or at worst, 0.  They are zero for atoms which have filled subshells: the alkaline earth metals and the noble gases.  All other elements have negative first electron affinities.

In general, as one moves down a column, the 1st electron affinity DECREASES.  Notable exceptions include the 2nd period elements whose small size make it energetically costly to take on entire -1 charge.  Thus: EA follows the trend
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Cl > Br > I > At, but F actually fits between Br and Cl.
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atom
electron affinity
(kJ/mol)
F
-328
Cl
-349
Br
-324.6
I
-295.2
At
-270
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2nd electron affinities are always endothermic (costs energy to place another electron on a species which is already negatively charged).  This is complimentary to the trend observed for ionization energies. It is not "too costly" to keep adding electrons as long as you are filling the same valence shell.

So the electron affinity of  N > N- > N-2 > N-3 where only the first species actually releases energy upon binding an electron. It is not theoretically possible to measure the electron affinities of the noble gases or the anions whose electronic configuration mimics them, but it should be clear that placing an electron in the next highest shell would not be nearly as advantageous as placing it in the valence shell.  O can form O-2, but not O-3.  N cannot form N-4.  F does not form F-2.

This explains why halogens form -1 anions in ionic compounds, and never -2 anions.  Likewise it explains why oxygen and its related family of elements only form -2 anions in ionic compounds.

Click here to review the last section of Chapter 9.