Some Basic Doubts In Chemistry

Discussion in 'Chemistry' started by ash64449, May 15, 2013.

  1. exchemist Valued Senior Member

    Messages:
    6,623
    Oh dear! All I can say is that we are dealing with wave-particles and wave-particles are not like particles, because of the additional rules imposed on their behaviour by their wave-like nature! The wave equation only has certain solutions, corresponding to standing wave patterns, and nothing else is possible. So the electron can only have certain energies, none of which is as low as it would be if it fell into the nucleus.

    Paradoxically, you can have an excited state of the hydrogen atom, with the electron in, say, the 2s orbital, with no angular momentum because l=0, yet it can emit a photon and drop down to a lower energy state, say 1p, in the process gaining a unit of angular momentum (l=1)! This serves to illustrate how the "orbiting electron" Rutherford-Bohr model of the atom does not work, as we discussed in an earlier thread.

    I'd be happy if any other reader has a more satisfying way to visualise what is going on, but the way I've always seen it, this is just why we need "wave" or "quantum" mechanics.
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. ash64449 Registered Senior Member

    Messages:
    795
    yeah... I see. You modified your post to change 2p to 1p!! I thought so!!

    Edit: i think you were meaning 1s. There is no 1p. And as a result,l=0.
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. exchemist Valued Senior Member

    Messages:
    6,623
    Sorry quite right. I should have said 3s -> 2p. This is what I originally intended but I made a typo, which I then corrected wrongly, making things even worse! Getting careless because trying to write atomic physics and prepare supper at the same time - too hard.

    I wanted to say s->p firstly because only transitions with a change in l of +/- 1 are electric dipole allowed and secondly because I wanted an example in which energy is emitted, i.e. lost, but there is nevertheless a gain in angular momentum. 3s -> 2p, as I originally intended, seems to do that.

    Apologies again for any confusion.

    The best short book I recommend for all this stuff, by the way, is Herzberg's classic text, "Atomic Spectra and Atomic Structure". It was what we were given by our physical chemistry tutor to read in our first term at university. You can find it here for example:http://www.amazon.co.uk/Atomic-Spectra-Structure-Dover-Physics/dp/0486601153

    It is old but nothing it says is now out of date, so far as I know.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    Great & thanks for the infro - Perhaps with sicfourm´s help he will be another Raman (not sure of the spelling) but a very world class physicist to come out of India.
     
  8. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    I´m back now. On question:
    Why an s orbital electron doesn't "fall into" the nucleus, even though it has no angular momentum?

    The most honest answer is: Because that is the way nature is. (Nature does not and on the atomic scale often does not, need to conform to human expectations, which are (until transformed by some high level education) based only on human experiences at the macro level.) It is not much more of an explanation, but makes people feel like they have one to say: “Energy levels of isolated atoms are quantized.” I.e. the energy level of an electron in the 1s state is fixed or can only be greater in an “excited state, not less by falling into the nucleus.

    These energy level can be changed - changed by bringing other atoms near one whose energy level you are considering. This is why solid and liquids, with rare exceptions (like X-ray line spectra), there is no line spectra emitted by dense mater. Because of these mutual interactions, amazingly, the radiation emerging from a tiny hole in the wall of a hot uniform temperature chamber is exactly the same regardless of what the chamber is made of or the internal cavity shape!

    That radiation is called "black body" radiation. There was quite a mystery about why it had that particular intensity profile - Plank solve the problem (Mystery was called back then "the ultra violet catastrophe" as classical physic firmly said: “Every possible mode / wave lengths / even harsh UV, should be with the same energy - the "equal partition of energy" law) The solution was to simply postulate energy levels were quantized. I.e. Planck launched quantum theory, but may not have been fully aware of it at the time.
     
    Last edited by a moderator: May 16, 2013
  9. Trippy ALEA IACTA EST Staff Member

    Messages:
    10,890
    The answer is to do with probability.

    What you're looking at when you look at a probability density plot is the probability of finding an electron at a given point in space.

    What we're interested in for the purposes of this discussion is a different value called the radial probability. The radial probability is (essentially) the probability of finding an electron at a given radius.

    Think of it this way, we can divide the space occupied by a 1s orbital into a bunch of really small cubes.
    The probability density - the spherical shape we associate with the s orbitals; is the probability of finding an electron in any one of these really small cubes.
    The closer to the nucleus we look, the higher the probability of finding an electron in one of these cubes.
    At any given radius we can define a volume made up of all of the cubes at that distance.
    The closer the nucleus we look, the fewer cubes we see.

    These two trends oppose one another, the closer we look to the nucleus, the higher the probability of finding an electron in a given cube, but the fewer cubes there are.

    The net result of that is that if you measure the radial probability, rather than the probability density, we find that the radial probability - the probability of finding an electron at a specific distance from the nucleus from the nucleus. As it happens, the radial probability curve peaks at (approximately - they differ by 0.1%) the Bohr radius. This is a good thing, because the Bohr radius is where we (more or less) expect a 1s electron to be (in its ground state, in a neutral hydrogen atom).

    Here's a page which may be of some use to you: Why doesn't the electron fall into the nucleus? (How the battle of the infinities saves the electron from its death spiral)

    It gives a less watered down explanation than I have offered here, but, it should still be readily accessable. Hope this all helps.
     
  10. Trippy ALEA IACTA EST Staff Member

    Messages:
    10,890
    Here's an illustration of what I mean (I whipped it up in excel, forgive the crudeness).

    Please Register or Log in to view the hidden image!


    Radial v Density
    The probability density is the highest in the center, where the nucleus is, each number represents the probability of finding an electron in that cell.
    You can tell this by looking at the table, and it is confirmed by looking at the bar graph on the right, beneath the table.
    The numbers at the top right are the radial probability. They represent the the probability of finding the electron at that distance from the nucleus, the shading is the same for both tables.
    The second graph is the radial probability function.

    As you can see, while the problability of finding an electron in any given cell is the highest in the middle, and decreases as you move outward, you are most likely to find an electron at a radius of '2' or '3' - the electron spends a little over half its time at that distance.
     
  11. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    Trippy´s post 26 link about probability density is a useful, but not the only, answer to the question. I prefer the one Bohr gave instead because it tends to explain why all the energy levels (not just the n= 1 level with highest probability density) exists. All particles are also waves. Even atoms, like silver atoms, not just subatomic particles are waves too. The famous Stern-Gerlack experiment proves that by causing a beam of silver atoms to interfere – not be just classical particles.

    Please Register or Log in to view the hidden image!



    Bohr´s knew the wave length associated with an electron of any kinetic energy. He reasoned that this wave length, wrapped all the way around the nucleus, had to have its "head" constructively interfere with its on "tail." When this logic is applied you get not only the n=1 "Bohr Radius" but the radius of the n= 2, the n= 3 etc. orbits too. I.e. this goes a long way towards explaining why quantization of energy levels exists. Why it does is the real question.

    If they did not, the electron would spiral into the nucleus very quickly. Many years ago, I used rate of radiation that an electron charge would have if accelerated in the Bohr orbit (that circumferential acceleration is balancing the force of nuclear attraction). I forget my answer, but it was a very tiny fraction of a second, before all matter in the universe would cease to exist, at least as we know it does.
     
  12. IncogNegro Banned Banned

    Messages:
    210
    I have a question over the 4s 3s energy part. If the shells overlapped a bit, say the 4s reached closer to the center than the 3s but it also reached further, the peak point of power would be observed close to the nucleus and the electron would have the lowest energy as it passed through the 3s grouping?
     
  13. exchemist Valued Senior Member

    Messages:
    6,623
    Ash, to tidy up some loose ends from your earlier questions, Ionization Potential is what in my era we called what I gather is today more frequently called Ionization Energy. This is the energy required to remove an electron completely from its atom. It can be thought of as the limit of convergence of the energy levels of successive shells of atomic orbitals, as the principal quantum number, n, -> ∞. (In classical terms one would say it is the energy needed to just give the electron "escape velocity" from its orbit - but we've now discussed in several contexts the serious defects of the orbiting electron model of the atom, so please don't attempt to take the analogy any further).

    Once an electron has been removed, we have an atom which has a net +1 positive charge. This is an ion, a cation, to be precise. There are atoms (e.g. halogen atoms) which are very happy to accept an extra electron, into a partly filled low energy orbital, thus becoming ions with a charge of -1, called anions. With enough energy you can strip more electrons from an atom and there are atoms that will happily accept more than one extra electron, giving rise to ions with multiple net charges on them.

    Looking at the Periodic Table, the ionization energies of Group I, or "alkali" metals is notably low, so they form cations very easily. This is because they have their outermost electron in the s orbital of a brand new principal shell. By the time you get across to the other side of the period, at the halogens, the nuclear charge has gone up a lot, pulling the orbital energies down, so that ionisation energies are far higher, making it rare to find a halogen cation. On the contrary in fact, the energy level of the one remaining unfilled p orbital is so low that if you pop an extra electron into it, you get a big energy release - meaning that halogens have a high "electron affinity", in other words form anions very easily.

    This illustrates one of the (many) ways in which considering the electronic configuration, the nuclear charge, penetration and shielding effects, etc, give deep insights into why the various elements have the chemical properties that they do - in other words why the Periodic Table looks the way it does.

    You can look all these terms up on Wiki or in other sources for more information.
     
  14. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    Ash: again exchemist has given you an excellent post. I can only added that usually the ionization potential, IP, is stated in eVs - the energy a electron gets by "falling" thru two points in potential (voltage) field that differ by one volt. For example, the IP of the hydrogen atom is 13.6eV and relatively low as he said, but the neutral atom (lithium I think from memory) with 3 protons and 3 electrons is even easier to ionize at there are two electrons in the n= 1 "shell" and the third is further from the nucleus in the n=2 shell. That not only makes it take less energy to strip that third electron off starting from location further way from the nucleus, but the inner two electrons are "screening" out much of the nuclear attraction force holding it in the atom. I.e. it is sort of like there were less than three protons attracting it. *

    Hydrogen is by far the least complex atom. In fact the energy levels and the energy required to ionized from any given level is given by:
    13.6[1/(m^2) - 1/(p^2)] where m & p are the principle quantum levels. (Usually designated by n but I can´t use n as I am speaking of pairs of levels.)

    For example, when p >>> 1, a level far from the nucleus, the atom is essentially ionized and if from the ground state level (m=1) by 13.6eV and if from the first excited state (m= 2) by 13.6/4 = 3.4eV etc. This simple formula also tells the energy difference between any two levels.

    For example an electron falling from p= 3 down to m = 2 level will have energy 13.6[1/4 -1/9] = 13.6 x 0.1388888 = 1.88888 eV This is the first, and lowest energy photo of the "Balmer series." It is a pretty deep red.

    If the electron falls down to the ground state (m= 1) from only the first excited state (p= 2) its energy will be 13.6[1 - 1/4] = 10.2 eV.

    This is the longest wavelength of the Lyman series. It is invisible - harsh UV and all the other lines of the Lyman series are even harsher UV.
    The only visible radiation hydrogen atoms can emit is the Balmer series (upper state transitions that end on the first excited state. The higher states falling down to the m= 2 level are Balmer series too, but after the first few Balmer lines, their photons have too much energy to be seen by humans (bees can, but that is another story) as they are in the "near UV."

    * As your first "homework" assignment from me, tell if my memory that the third element is lithium is correct or not, and find its IP and post that. I am more than 5 times older than you, but usually my memory is quite good. I rarely ever look things up - just pull out of my memory. (All the math above for example - and it is seldom given, I think, as clearly as I did above.) Ask if part is not clear.
     
    Last edited by a moderator: May 17, 2013
  15. Trippy ALEA IACTA EST Staff Member

    Messages:
    10,890
    Billy, it's true of all energy levels. If you were paying attention, you would have realized this.

    For example, if you had followed to the page by Richard Bader, a Quantum Chemist, you would have seen this image:

    Please Register or Log in to view the hidden image!


    Which is the radial probability distribution for the 2s and 3s orbitals.

    Note that, as with the 1s orbital, all of these show a 0% probability of finding the electron at the nucleus.

    Having studied both quantum mechanics as a physics subject; and quantum chemistry, the application of quantum mechanics to chemistry; I'm well versed in this, thankyou.

    The Stern-Gerlach experiment wasn't about interference, it demonstrated that the angular momentum of a silver atom is quantized. In the case of a silver atom it is quantized as ±½, but this isn't neccessarily true of all atoms. It's always quantized, but, some have more angular momentum than that, some have less.

    The probabilistic explanation also explains this things, as illustrated by Richard Baders ilustrations.

    This leaves me wondering if you actually read the page I linked to, in full.

    For starters, there is the point that radial distributions predict a 0% probability of finding an electron at radius zero, then there's the interplay of potential and kinetic energy as the nucleus is approached - also discussed on the page. In short, the probabilistic explanation does not predict the death of the universe, it predicts the opposite, it predicts that we should find electrons at the Bohr radius.

    Addendum: Please note that the resource I linked to in post 26 was developed by someone with a PhD in Physical Chemistry, who taught for 34 years (now retired), and was developed as part of a reference text for general chemistry
     
  16. Trippy ALEA IACTA EST Staff Member

    Messages:
    10,890
    Regarding the Hydrogen emission series, these images may be of some help:

    An illustration of the Balmer series as the lie within the visible portion of the spectrum:

    Please Register or Log in to view the hidden image!



    Two representations of the Balmer series and where they fit in relation to other transitions:

    Please Register or Log in to view the hidden image!



    Please Register or Log in to view the hidden image!

     
  17. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    I don´t see it that way as its starting point is the quantum orbital distribution and then it shows where the probablility density max is. I.e. it is an "after the fact" justification more than an explaination, I think. I will admit to being wrong if without any prior knowledge of the orbitals, probability density analysis can derive the n=2 "Bohr radius."

    Thanks for posting the nice illustration of Balmer, Lyman etc. Hydrogen lines. I am some what surprised that the analysis in my post 31 got 3.4eV, not the 3.03eV for the Balmer red line. Do you know why this 10% "error" happened? Your third post33 figure show the energy difference for that red lines levels as 328 - 146= 182 KJ yet when I go here: http://www.convertunits.com/from/kJ/to/eV to convert 182KJ to eV it gives numerically very different from 3 (1.136) and made me realize there is some huge factor omitted in your third figure´s numbers. There is not even a micro Joule difference between the n=3 to n=2 levels

    Yes, you are correct, the Stern-Gerlack experiment was a poor choice. Its results only looks like an interference pattern. The inhomeogenious magnet field is just selecting with different force between the spin up and spin down atoms.

    Later, by edit: I just notice the first figure´s 3.03 is not in eV but third figure´s numbers still seem hugely wrong.
     
    Last edited by a moderator: May 18, 2013
  18. ash64449 Registered Senior Member

    Messages:
    795
    I think this says that angular momentum has nothing to do with the energy of electron or it is just because Electron is not just particle but can also act like a wave!

    Or if energy is related to angular momentum,that means it must be because of wave-particle duality..

    But i cannot visualize how that kind of nature helps to explain decrease in energy leads to gain in angular momentum... How can it deal with that?

    Yeah.. the world of atoms is really confusing!! 2s orbital has less energy than 3s but they still have zero angular momentum.
    2p has lower energy than 3s but electron in 2s has angular momentum!!

    How can you explain this result with the help of wave-particle duality??
     
  19. exchemist Valued Senior Member

    Messages:
    6,623
    It's a good point. I think there must be energy associated with the angular momentum. The total energy of the electron in any orbital is comprised of potential plus kinetic. So those with non-zero angular momentum presumably have their energy partitioned between the two a bit differently from an s orbital electron. But I am vey rusty on this and would welcome a comment from another knowledgeable reader. Failing that, I'll have to dig out my old copy of Atkins' Molecular Quantum Mechanics and see if I can find the answer there for you

    Please Register or Log in to view the hidden image!

    .
     
  20. ash64449 Registered Senior Member

    Messages:
    795
    What is this spectral lines of atoms and how it relates to the different orbitals? Are you saying singlet represents only one spectral line of atoms and and 's' orbital has only single spectral line of atoms?

    Likewise doublet? What is this spectral line?
     
  21. ash64449 Registered Senior Member

    Messages:
    795
    Thank you!! Hope you won't have any problem...
     
  22. ash64449 Registered Senior Member

    Messages:
    795
    Well,potential energy of electron decreases when electron jumps from outer orbital to inner orbital.

    Does kinetic energy of that orbital too decreases?
     
  23. ash64449 Registered Senior Member

    Messages:
    795
    So this arrangement of electrons in different orbitals is done with the help of radial probability and not probability density?
     

Share This Page