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Name: Sang
Status: N/A
Age: 16
Location: N/A
Country: N/A
Date: 12/16/2004

Hello, scientists. Which element has higher first ionisation energy between lithium and beryllium?

I asked my Chemistry teacher at school, and he said that lithium will have lower ionisation energy, as the electron (in respect to helium) has occupied a new orbital (2S) and therefore there will shielding from the 1S orbital, while for beryllium, the higher nuclear charge (i.e. higher number or protons) will make the electron harder to lose.

Now this is where (at least for me) it was confusing. In respect to Helium, the lithium atom has gained a proton, therefore IT can be said to have higher nuclear charge as well. Another point, the 2 electrons in beryllium has slight repulsion from each other AND from the electrons in the 1S shell. It just may be me confused with facts, however, could anyone provide a PROPER explanation for this?

(If my facts were true) I could see that the overall strength of 'pull' from the increasing number of protons is greater to COUNTERACT the repulsion between the electrons, thus explaining the more energy required in the first ionisation energy for beryllium. However, if by this logic, it could be said that lithium has less electrons, therefore has a higher ionisation energy!

From the following NIST website you can find the data that I will refer to, as well as a more complete list of ionization energies and electron energy levels: 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:

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.

Vince Calder

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:

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.

Greg (Roberto Gregorius)

Well, Sang, 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.

Give my regards to your chemistry teacher.
Take AP Chemistry if you can.

Jim Swenson

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