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The relationship between ionization potentials and effective nuclear charges in atoms through a case study of hydrogen (h), helium (he), and lithium (li). It delves into the concept of effective nuclear charge, which is used to approximate the attraction between an electron and the nucleus in a many-electron atom. The document also discusses the one electron approximation and the shielding rules for estimating effective nuclear charges. It provides insights into the trends in ionization potentials and atomic radii for various elements.
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Now that we understand how the periodic table arises, letāslook at some chemical properties.The
chemistry
of any given element is related to the ease
with which it can give up (or share) its electrons withanother atom.One measure of the willingness to āshareā is the amount ofenergy it takes to remove an electron from the atom. Thisenergy is referred to as
Ionization Potential, I
. i
It is always a positive number; it is a measured number.The H atom is simple; I
= 2.178 x 10 1
J = 13.6 eV
ā¢Instead of asking how much energy it takes to remove anelectron, letās use a ātestā electron to explore what it seesfrom a distance and how attractive the He
+^ is. As the ātestā
electron (1) looks at the He
+^ ion from a distance, what is the
effective charge it sees? How much does electron 2 āshieldāthe +2e nuclear charge? Clearly 1<Z
<2.eff
Electron 1ātestā e
electron 2
He nucleus withZ=2; q = +2e
ā¢As the ātestā electron approaches the nucleus theinteraction with the other electron now becomes veryimportant. Letās make a crude calculation on how largethat interaction is. ā¢In H the average distance from the nucleus for the electron in the 1s orbital is (3/2)a
o
or about 0.
Ēŗ. In He
with Z=2 it is about half that.
Electron 1ātestā e
electron 2
He nucleus withZ=2; q = +2e
ā¢We know of course that the He
+^ ion has an energy of
E(He
+)= -54.4eV. ā¢If we neglected the e-e interaction entirely, then theenergy of the He atom would be 2(-54.4eV)= -108.8 eV.ā¢So,
= E(Hei
+) āE(He)
= -54.4eV ā[2(-54.4)eV + 30eV)] ā^ 24 eV
ā¢This is remarkably close to the value for I
= 24.6 eV. 1
(But of course, we āinfluencedā that by our guess of 30eV for the e-e repulsion. Still it was not an unreasonableguess).
e-e repulsion
eff
ā¢So what about Z
? We saw that the e-e repulsion waseff
nearly half as big as the energy of the single electron inthe He
+^ ion. So as a crude estimate one might guess that
Zeff
for He as seen by our ātestā electron might be something like 1.5. ā¢If we used that number in our H-atom like formula wefind
E1s
(He)
-13.6eV (1.5)
2 = -30 eV
ā¢It turns out the actual number is more like -40eV. Thatmeans the shielding is not quite as strong as Z
=1.5eff
might suggest. Actually, Z
eff^
ā^ 1.7 gets you pretty close
to actual value for E
(He) = -40 eV. (Check it out.)1s
Letās see what a Z
eff^
= 1.7 gives us.
But E
1s^
is just the energy of each electron, E
and E 1
(since 2
they are equivalent, they must each see the same Z
.)eff
So, we can calculate the first ionization energy of He.
I^ (He) = E(He^1
+) ā E(He)
-54 eV ā 2(-40 eV)
26 eV
2
2
1
s^
H
2 electrons
The experimental value for I
(He) is 24.6 eV. We got 1
roughly 26 eV. Pretty good but this is a very simplesystem. Q: Why canāt we simply say that I
40 eV?
A: Because
= E(ion) ā E(neutral).i
In the case of He, this gives a significantly differentresult. However, in many cases I
(particularly for 1
valence electrons). This is known as KoopmanāsTheorem: the ionization energy is basically given by thetopmost orbital energy.
I= 13.6^1
-13.
1s
1s He
H^
Li
2s 1s
I= 5.4 eV^1
each
electron shields the other
electron in same shell by about (1/3) e
shell n by about 0.85 e
Each electron in shell (n-2) shields the electrons inshell n by 1e
We ignore shielding of electrons in n by electronsin n+1.
5.^
One also needs to take into account
penetration
effects.
1
-^
-^
3+
-^
4+
-^
3 valence electrons
10 core electrons
eff
1
ā¢As the ātestā electron approaches the Al
+^ ion from a
distance it basically sees a Z
eff^
close to 1 and it behaves like
an H-like atom. Of course, as the ātestā e
-^ gets closer and
closer to the Al
+^ things get more complicated, but the
general ideas remain valid.
āTestā e
12 electrons Al nucleus withZ=13; q = +13e
eff
2
2
,
j^
H^
j eff
j
2 0
2
j
in^ Ēŗ
;^
is the average
radius of the orbitalcontaining the j
th^ electron. r^ j
The j subscripts have been omitted in this equation.
The idea of Z
eff^
as outlined above just provides a rough rule of
thumb for understanding. The idea can be made much moreprecise using computational methods and high powered theory. Anextension of this idea leads to the Self Consistent Field (SCF)method which is the basis of our information on many-electronatoms. In our shielding rules above we ignored the effect of any electronsin shell n on the Z
eff^
for electrons in shell n-1. This is not a serious
approximation. Using the above two equations you should be able to predict grosstrends in ionization potentials and in atomic and ionic radii. From these two pieces of information you can predict typicaloxidation states and thus a lot about the chemistry of the elements.
