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Berylium and Lithium

  1. Dec 15, 2004 #1
    Hello all

    I was asked a question on the first ionisation energy of Lithium and berylium, and it confused me.

    It seemed obvious to other people in the class that Berylium has a higher ionisation energy than Lithium (which according to my chemistry teacher is the right answer)

    Now This baffled me. Lithium has 2 electrons in its 1S shell and a single electron in its 2S shell, while Berylium has 2 electrons in its 1S shell and 2 electrons in its 2S shell.

    If looking at the repelling property or electrons (to each other, that is) shouldn't Berylium have a lower ionisation energy than Lithium?
    My chemistry teacher said that because of the increase in nuclear charge (from lithium) Berylium will have a higher ionisation energy, but we must also consider the increase in the number of electrons from lithium to berylium.

    In berylium, not only will the 2S electrons repell each other, but they wil lsti.ll have repulsion from the 1S shell as well!

    Please forgive me if this was a dumb question (i.e. i just forgot a bit of detail)
     
  2. jcsd
  3. Dec 15, 2004 #2

    FZ+

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    Well, electrons are not really 'repelled' by others in orbit in the classical sense. The big effect of the 1S shell is not its charge repulsion, but its effect of shielding the additional electrons from the effect of nuclear attraction. The 2S electron has little effect, especially when compared to the new proton, which boosts nuclear charge by a third.
     
  4. Dec 15, 2004 #3
    Could it be good to explain it like this?

    As Beryllium have 2 electrons in its 2nd shell, there needs to be more energy to remove one electron as both are as close to the nucleus as each other. One electron will be attracted so much but 2 electrons will be attracted more and so more energy is needed to ionise Beryllium than Lithium.

    I hope that is the way to see it but if it isn't, please correct me as well.

    The Bob (2004 ©)
     
  5. Dec 15, 2004 #4
    ..

    Thanks for the replies. FZ+ and Bob.

    I've considered both of your replies, and has been explained to me...to a extent.

    FZ+, when you mean not repelled in the classical sense, do you mean the way Magnets are repelled when the same polls approach each other?
    If this is so, it does make a huge difference to me, as i have always thought of electrons that way.

    The Bob '2 electons attracted more..' bit was a bit confusing to me. If i look at the electrons in a bar magnet sense, the electrons in the same orbital would repell each other. However, as i said before, if this ISN'T the case (i.e. it isn't the bar magnet) then it does make sense to me.

    So is it plausible for me to presume that the increase in the number of protons (i.e. 1+ for berylium in respect to lithium) increases the nuclear charge to a extent that the shielding from the electrons becomes redundant?
     
  6. Dec 15, 2004 #5
    I mean that the 2 electrons are attracted stronger to the nucleus than one electron. However I am unsure that I am right because that doesn't make sense to me either.

    Having said that. Lithium has a Relative Atomic Mass Number of 7 and an Atomic Number of 3. Beryllium has a Relative Atomic Mass Number of 9 and an Atomic Number of 4. Therefore the nucleus is bigger and so attracts the electrons more.

    The Bob (2004 ©)
     
  7. Dec 15, 2004 #6

    FZ+

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  8. Dec 15, 2004 #7
    Be has one more proton and electron than Li has. The added proton attracts all the electrons, making its atomic radius smaller, therefore more energy is required to remove an electron from the Be. It's that simple.
     
  9. Dec 15, 2004 #8
    You see I thought that but I have been told off before for suggesting the simple. I got that idea from http://www.colorado.edu/physics/2000/applets/a2.html because as you increase the number of protons the electrons in the table get lower, even in the same shell.

    The Bob (2004 ©)
     
  10. Dec 15, 2004 #9
    Well I think for this case it is that simple, yes it can get more difficult with different scenarios, but in this case that explanation is just fine.
     
  11. Dec 15, 2004 #10
    ..

    So i can say that the repelling effect between the electrons is relatively negligible compared to the increased number of protons?
     
  12. Dec 15, 2004 #11

    DB

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    Bladibla: "If looking at the repelling property or electrons (to each other, that is) shouldn't Berylium have a lower ionisation energy than Lithium?"

    Bladibla, its quite simple.

    Li has a charge 3p+ and 1 Valence electron. It very much so would like to give away its outer electron and become a stable atom with a full outer shell, in this case 1s. Li is very reactive becase it has such a low ionization energy. Be has a nuclear charge of 4p+ and 2 valence electrons. It is aswell very reative, but a bit less than Li because it has a higher ionization energy. In other words think of it this way: You would need more energy to pull out 2 electrons than 1. A higher ionization energy.

    The Bob: "I mean that the 2 electrons are attracted stronger to the nucleus than one electron. However I am unsure that I am right because that doesn't make sense to me either."

    It all cancels out, thats the beauty of the Periodic table. There is no stable atom with different number of protons and electrons. Li has 3 protons and holds 3 electrons in it's shells, Be has 4 protons and hold 4 electrons in it's shells. Its all proportional. The reason why 2 electrons are attracted to a stronger nucleus is because protons have a larger mass than electrons a have an effect on its attraction. Li and Be are both very reactive, Li is just a bit more reactive because its has a lower nuclear charge and 1 valence electron to remove instead of 2. Be is still reactive, but it requires a higher energy to remove 2 electrons from its slightly higher nuclear charge. Therfore, once again it has a higher ionization energy.
     
  13. Dec 16, 2004 #12
    This is the reason I hate explaination. I can think of what I want to say and I am unable to say it. I was talking about the size of the nucleus but clearly it is the mass that matters in this case.

    Thanks DB. :biggrin:

    The Bob (2004 ©)
     
  14. Dec 16, 2004 #13
    An atom never wants to "give" an electron away, energy is always required to remove that electron.
     
  15. Dec 16, 2004 #14

    I might be wrong here, but i think you are begging the question. (And i do apologize if this wasn't the case, and i have mistaken)

    ''Li has a charge 3p+ and 1 Valence electron. It very much so would like to give away its outer electron and become a stable atom with a full outer shell, in this case 1s. Li is very reactive becase it has such a low ionization energy.''

    Now compare this with Berylium. I think you are talking as if the atom was actually alive to give away electrons to become more stable. But the point is, it cannot ignore the repelling properties of the electrons.

    We can see that the shielding from the 1S electrons, the single 2S electron is easier to lose ( in respect to helium)
    However, it can also be said for berylium, the increase in the number of electrons (+1 from lithium) has more repelling effect and therefore easier to lose a electron (i.e. first ionisation energy) than lithium.

    The only 'proper' explanation for this, in my opinion, is that the increase in the number of protons (+1) has increased the nuclear charge to a extent that the shielding and repelling from the electrons in the 1S and 2S shells is not a singificant amount to counteract. (again i apologize if this isn't the case)
     
  16. Dec 16, 2004 #15
    well, i just got a bit more...complicated answer from the trusty scientists.

    Should i post it?
     
  17. Dec 16, 2004 #16

    DB

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    Cantari:"An atom never wants to "give" an electron away, energy is always required to remove that electron."

    Ya. Absolutely. But we never find Li in nature alone, always in a molecular form. When I said "give" it away, I meant that it's ionization energy is so low compared to atoms to the right of it, that with a slight amount of energy (heat) its will pretty much "give away" it's valence electron (with another atom to bond with of course).

    Bladibla:Should i post it?

    Ya sure. I'm not an expert on quantum physics, but I'm pretty sure an eletron doesnt even "see" it's fellow electrons in an atom. They are never at a position to repel eachother, they're kinetic energy (i guess i could use that term) prevents them from staying in the same place. We don't really know the path of an electron, we measure the probabilities of an electron being at a certain point, at a certain time, in it's orbital (shell). I have never heard about electron-electron repullsion within an atom, but I could be wrong.
     
  18. Dec 16, 2004 #17

    Astronuc

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    Staff: Mentor

    Perhaps these would be useful.

    www.physics.ohio-state.edu/~lvw/handyinfo/ips.html

    http://www.chemistrycoach.com/ionization_potentials_f.htm

    Plotting the IPs against Z might be helpful.

    The electrons arrange in pairs with opposite spin when 2 electrons occupy a subshell. This contributes to stability as manifest in higher ionization potential of the Group 2A elements.

    Boron (B, Z=5) has a lower ionization energy than Be (Z=4), but carbon (C, Z=6) has a higher ionization energy than either Be or B.

    Clearly ionization energies increase with Z, generally speaking, but the first P electron has a lower ionization energy than the second S electron in the first four periods, and only slightly higher ionization potential in the fifth period.

    With in a group, ionization energy energies decrease as the atoms get larger. This trend reverses between periods 6 and 7 for the Group 1A and 2A elements, and between periods 5 and 6 for the transitional metals.
     
  19. Dec 16, 2004 #18
    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.
     
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