Ok i got replies from 3 different people..:
reply 1:
The ionization energy data in electron volts are: 1H(1s) 13.598; 2He(1s2)
24.5874; 3Li(1s2,2s1) 5.3917; 4Be(1s2,2s2) 9.3227; 5B(1s2,2s2,2p1) 8.2980;
6C(1s2,2s2,2p2) 11.2603; 7N(1s2,2s2,2p3) 14.5341. The order of presentation
is the atomic number, symbol, ground state configuration, ionization energy
in E.V.
The "rules" of the correlation of the trends goes as follows:
1. There is a competition between the INCREASED attraction of the nucleus and
electrons as the atomic number increases and the DECREASED attraction due to
shielding by other electrons in a particular atom.
2. Electrons with a lower principle quantum number are more effective in shielding the outermost
valence electron than an electron with the same principle quantum number as
the outermost valence electron. For example, a '1s' electron shields a '2p'
electron than a '2s' electron shields a '2p' electron.
3. A 'ns' electron shields better than a 'np' electron because 's' electrons are "more
penetrating" than 'p' electrons. That is 's' electrons spend more time
closer to the nucleus than do 'p' electrons. Now look at the trends. Without
shielding one would expect the ionization energy of He to be twice that of
H. However it falls about 2 eV shy due to the shielding of the other '1s'
electron. Going from He to Li results in a big decrease in the ionization
energy because the valence electron is in an orbital with a principle
quantum number of '2' so on average it is more distant from the nucleus and
hence more weakly bound by coulombic attraction.
Going from Li to Be there is an increase in nuclear charge but a
simultaneous presence of the other '2s' electron. However, the "rule" is
that the '2s' electron is not very effective in shielding another '2s'
electron because they are both about the same distance from the nucleus. The
second '2s' electron almost "does not see" the other '2s' electron. In
addition, Be has a filled shell '2s2' (spins paired). That results in a
further stabilization of the atomic configuration, so the ionization energy
of Be is further increased. Moving on to B, there is an increase in nuclear
charge which "should" increase the ionization energy; however, that is more
than counter-balanced by the fact that the average distance of '2s'
electrons from the nucleus is smaller than the average distance of '2p'
electrons, so the outer valence electron "sees" less attraction of the
nucleus and the ionization energy actually decreases. For B, C, N the '2p'
valence orbitals are filling without any pairing and the '2p' electrons are
not very effective at shielding one another because that are all about the
same distance from the nucleus. As a result the ionization energies increase
"normally".
While these "rules" appear somewhat arbitrary, they are actually based
upon fairly rigorous, semi-empirical calculations that support their
validity -- called "Slater's Rules". You can find more details as well as
interactive tools that let you make comparisons between atoms at either of
these (and other) sites:
http://www.wellesley.edu/Chemistry/chem120/slater.html
http://basc.chem-eng.utoronto.ca/~paulozz/che200.htm
Be aware that the "rules" become more approximate the more electrons
that are involved. However, it is "fun" to play around with the rules to
generate "explanations" of various chemical observations.
Reply 2:
This is very perceptive of you. You are partly correct in thinking that the
ionization potential has to do with two *opposing* forces: the attraction
that comes from the protons in the nucleus - which, as the number of protons
increase should increase the first ionization energy; and the repulsion from
the electrons that is already present in the atom -which, as the number of
electrons increase should decrease the first ionization energy.
Look up a graph of ionization energy as a function of atomic number. (I
googled for it and this is the first one I could find -not the best- but you
can look for one yourself: http://www.unco.edu/chemquest/4con22aw.htm )
Now if the first ionization energy were only dependent on the number of
protons, then this chart should be linear -and have a positive slope- as a
function of atomic number (the number of protons). On the other hand, if the
first ionization energy were only dependent on the number of electrons, then
this chart would also be linear -but have a negative slope- as a function of
atomic number (the number of equivalent electrons).
The fact that the trend is not smooth, and the fact that there are peaks and
valleys to the graph, suggests that not only are the two factors (proton
attraction and electron repulsion) involved in setting the ionization
energy, but that (1) these two factors are not linear in effect, and (2)
there may be other factors involved.
Let us just stay within electromagnetic forces (not go into quantum
mechanics), and suppose that there are only these two factors (attraction by
protons and repulsion by electrons) - a look at the trend quickly makes us
realize that as we go from left to right along a row on the periodic table,
that the trend for the first ionization potential is to increase - this
should suggest that --within a row of the periodic table-- that the number
of protons is more important a factor. On the other hand, as we go from top
to bottom on any column of the periodic table (check for example the peaks
of each little trend which is headed by He, Ne, Ar, etc.) we see a decrease
in the ionization potential which suggests that as we go from level to
level, the number of electrons present (already there) is more important a
factor.
The preceding is not a complete answer. As mentioned, there may be other
factors then just electrostatic attraction and repulsion. One clue is the
sudden drop in first ionization potential as we go from He to Li, from Ne to
Na, from Ar to Kr. This transition is accompanied by the additional electron
being found in a much higher orbital (principal quantum number "n"; the
numbers you find to the left of most periodic tables and goes from 1 to 7).
Thus, the actual ionization potential is also affected by the "poorer
stability" of an electron on a higher orbital. I leave this to you for
further study.
Reply 3:
Well, ****, I looked up what has been measured.
I tend to believe in learning from the numbers.
H- 13.6eV
He- 24.6eV
Li- 5.4eV
Be- 9.3eV
Na- 5.1eV
Mg- 7.6eV
In each of these pairs, the 1-outer-electron element
has higher ionization potential than the 2-outer-electron element.
Apparently the presence of a 2nd un-charge-canceled proton in the nucleus
more than makes up for any repulsion between 2 outer electrons in the same 's' orbital.
Think of lower electron-shells as making up a hard, neutral ball
around which other electrons may dance
under the influence of unbalanced positive charges in the nucleus..
The negative electrons of the inner shells cancel out the positive charge
of an equal number of the protons in the nucleus,
leaving in these cases 1 or 2 positive charges remaining
to hold outer-shell electrons.
The 2 outer electrons do not seem to repel each other as much as
they each independently witness the increased unbalanced positive charge of the nucleus.
I cannot quite say why.
This requires enough knowledge and enough 3-D computing that you may not find your proper answer in the short term.
I think some of my college classmates had a feel for it before they graduated.
Meanwhile, all these explanations of the orbitals are approximations, hand-wavy descriptions that seem to summarize the real behavior, and shorthand rules so people can make predictions with a minimum of advanced calculation.
Likewise, the inner-shell electrons do exclude the outer electrons
from sharing their inner-shell position,
but they are not usually considered to electrostatically repel the outer electrons;
they and an equal number of protons merely have mutually canceled charges.
The exclusion is by quantum numbers, not by coulombic repulsion.
Electrons trapped in a small place develop a finite number of quantum states,
each of which can be occupied by only one electron.
You might consider this the basis of how matter occupies volume;
it certainly sets the size of an atom.
The positive protons repelling each other is not part of this issue.
Something holds them together, and nobody said the electrons had to do it.
What holds them together is called the "strong nuclear force";
it's one of the four basic forces known in physics:
(1) gravity, (2) electromagnetism, (3) strong nuclear force, (4) weak nuclear force.
Electromagnetism is responsible for orbitals and all chemical energies;
Strong Nuclear Force is responsible for nuclei and all nuclear energies,
which, as you know, are much larger than chemical energies.
The truth is, you may want to learn about electron wave-functions
and orbital quantum numbers and scientific computer use as fast as you can.
Soon you may be able to find a free-ware computer program
to help you numerically integrate in 3-D the attractions and/or repulsions
of the space-charge-densities of the electron clouds.
That way you can see for yourself how these clouds always
interpenetrate each other and yet occasionally act fairly independent.
That way you may be able to see what they mean by the term "shielding",
which seems to sometimes happen and sometimes not.
There are some orbital-shape demos on the Internet now; look at those for an earlier taste of it.
Ill try to get my head around these answers, but meanwhile... here you go.
