Some Basic Doubts In Chemistry

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

  1. ash64449 Registered Senior Member

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    795
    hello friends,

    i have some doubts...
    While conducting electrical discharge through gas in cathode ray tube,it is required that pressure of gas should be low.

    What is the connection between the flow of charge and pressure? I mean at normal pressures,there is no electrical discharge through gas but reducing the pressure helps to observe electrical discharge. So what is the connection between pressure and conduction of electricity?



    Another doubt: according to a principle in chemistry,electron filling in shells takes place in increasing order of energy levels.(aufbau's principle).

    I have learned that electrons have higher potential energy as distance from the nuclues increases.so energy level increases when we go away from nuclues. So shells nearer to the nuclues should have lower energy level than the one that is far away from it.

    But the 4s shell is the lower energy level than 3d shell.but 4s shell is actually far and 3d shell is nearer. As a result,4s is the one that should have higher energy level than 3d. But why opposite over here?
     
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  3. arauca Banned Banned

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    One thing is the voltage you apply , but the the other point is the the volume of molecules in the discharge , as you reduce the pressure you are reducing the number of molecules and so the voltage ratio per molecule will increase to were the atom will be ionized and the flow ions move to the respective electrodes . I believe different gases have different ionization potential
     
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  5. leopold Valued Senior Member

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    the idea of "low pressure" is to prevent the electron beam from hitting extraneous atoms.
    you want no atoms present inside a CRT.
    it all has to do with ionization.
     
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  7. arauca Banned Banned

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    I forgot to add . You go at about 0.0001mm Mercury you get glow discharge were is used for sputtering coating as you go to a lower pressure to 0.000001 the glow decrease and only electron flow,
     
  8. ash64449 Registered Senior Member

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    795
    ions move? I thought it was the electrons inside the molecules are pushed other electrons and thus the current is produced.

    What is ionization potential?
     
  9. ash64449 Registered Senior Member

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    what are extraneous atomes?

    What is ionization?
     
  10. ash64449 Registered Senior Member

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    sorry, i haven't understood anything from this post
     
  11. ash64449 Registered Senior Member

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    Can someone answer to 2nd doubt? It is the one that is more important!!
     
  12. arauca Banned Banned

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    I believe they do Example : let say you have Argon gas or any of the inert gases, as you reach the potential of the atom say Helium for simplicity at 0.0001 mm Hg. you get a glow in the chamber . The glow is an indication ionization have taken place . Now if you place a gold or any other metal on the cathode and near by a piece of glass . you will deposit the metal on the cathode on the glass. This means the ion collides with the surface on the cathode and dislocate the surface metal and deposits on the glass .

    Now that was an example with Helium gas and I believe it takes 22, volt to remove the electron from S orbital
    If yo take an Argon gas you will ionize the most outer orbital electron. the same will take place if you have Nitrogen gas or ther gases

    Sorry if My explanation is irrelevant or confusing.
     
  13. exchemist Valued Senior Member

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    6,637
    Ash, I'll have a go, though I'm a little rusty. In principle the 3s, 3p and 3d subshells should indeed have lower energy than 4s. In fact, in the hydrogen atom (the simplest case with one electron only) they do. But with multi-electron atoms the situation is complicated because the effective nuclear charge experienced by an electron in one of the outer orbitals is reduced by the presence of the electrons in the inner shells. This is called "shielding".

    The degree to which an outer orbital is shielded depends on the degree to which it "penetrates" the cloud of shielding electrons. Recall that an orbital is a 3D probability distribution of where an electron in that orbital is likely to be found. Also recall that orbitals are like standing waves and have "nodes" i.e. points at which the electron probability is zero. With s orbitals, there is no node at the nucleus, i.e. the electron can go right up to it, and spend a portion of its time exposed to the full nuclear charge. But with p, d, f orbitals etc, there is a node at the nucleus. This means that electrons in s orbitals penetrate the most and are more exposed to full nuclear charge.

    Remember also that all these orbitals are present in all atoms (i.e. the solutions to the wave equation all exist), it is just they are not populated with electrons, until the lower energy orbitals have been filled. As one goes up the atomic numbers, filling shells according to the Aufbau pricinple, the nuclear charge is also going up, which pulls down the energy of the orbitals (stronger electrostatic attraction means lower potential energy). But the s orbitals are pulled down in energy more quickly with increasing atomic number, than the p and d etc, because of their greater penetration.

    So what happens with 4s and 3d is that by the time we get to Potassium, the 4s is actually lower in energy than 3d, due to the greater penetration of 4s, compared to 3d. Only after Calcium has the nuclear charge gone up enough to pull down the energy of 3d to the point where it is filled in preference to 4p. Which is why the d block "transition" elements start where they do in the periodic table. You get an analogous phenomenon, but even more extreme, at the start of the filling of the f orbitals. 4f does not start to be filled until 6s, yes that's 6s, has been filled. This accounts for why the Lanthanide and Actinide series start where they do, rather than as one might have expected from simplistic application of the Aufbau principle.

    Sorry if this is a bit complicated and I hope it is free from errors. If you have further questions or think I've made any mistakes in the explanation do query and I'll do my best to unscramble. It's been 40 years since I learnt this stuff!

    P.S. It's not "Aufbau's Principle", it is "the Aufbau Principle". This is because "Aufbau" is the German word for "building up". It is not named after a Dr. Aufbau!
     
  14. ash64449 Registered Senior Member

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    795
    This Explains it all!! And By the way i am just 15 year old. And now i am going to study these things.

    But there comes lots of question. Why nodes exist? And why the probability of finding electron vanish to zero and further going upwards,again probability of finding electrons increases?

    This is hard for me to understand because if there are probability of finding electrons are there beyond that node,then the affect of Nuclear Charge must exist at the area of node.

    And why no electrons are not found in the area of nodes sounds interesting!!!

    And also only if you answer these questions will i be able to understand the previous question to full potential.
     
  15. ash64449 Registered Senior Member

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    795
    Sorry!! It is because when i was in 10th standard,i didn't learned these things. In fact,i studied in one of the cbse schools.
    But some of my friends studied Aufbau principle in 10th standard itself because they studied in state schools.

    When i am going to +1 this year,i encountered problems so i asked their textbook for me to study and the doubt i asked to you is from theirs textbook.
    In that textbook, they didn't mention the meaning of Aufbau. so i thought it must be a person!!
     
  16. exchemist Valued Senior Member

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    Ash if you are only 15 and understand all this I think you are doing extremely well and I encourage you to consider chemistry or physics at university. It is mind-expanding and I never regretted it for an instant!

    As to why there are nodes at the nucleus in the p, d, f etc wavefunctions but not in the s ones, this is to do with the shapes of the solutions to the wave equation, according to which azimuthal quantum number (l) they have. As you may know the principal quantum number is n and tells you the main shell: 1, 2, 3 etc. However another quantum number also falls out of the maths, which is l. For s orbitals l = 0, for p, l=1, for d l=2 etc. Values of l >0 signify no-zero angular momentum of the electron. I think we may have discussed the shapes of these s, p, d, f etc orbitals elsewhere, but if we haven't there's a summary here:http://en.wikipedia.org/wiki/Atomic_orbital

    From the pictures, you can see the s orbitals are shaped like concentric spheres, the p type have 2 phases, one either side of the nucleus with a node in a plane passing through it, d type has 4 phases, with 2 nodes, in a pair of planes at right angles which intersect at the nucleus, and so on. Don't ask me for the maths, it's too hard, but these are the shapes of the "standing waves" that are possible, i.e. the solutions of the wave equation for an atom.

    The nodes are analogous to the nodes on a vibrating piano or violin string. (Think of the fundamental and harmonics). At a node, the amplitude of the wave is zero. In QM this corresponds to the probability of finding the electron there being zero.
     
  17. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    exchemist´s reply was good but he forgot to tell you that the, s, p & d are the traditional names* for the 0,1,2 possible angular momentum states / magnitudes in an atomic scale, that are quantized. I.e. the “s orbitals” must in some sense pass straight thru the nucleus as they have zero angular momentum. The others can not go thru the nucleus as if they did their angular momentum would be zero while doing so, but it is a constant, just like any planet´s is.

    An interesting little fact for you to file away is that for Earth to hit the sun (have orbit changed to zero angular momentum) takes more energy given to Earth than for sending earth into deep space!

    Good you are here asking questions. I hope you continue to do that. I came here to see if I should accept your request to be my "friend." I will, in a few minutes, not that that will give you much benefit but you will be among a select few.

    *The spectral of single atom lines observed were grouped and named long before their quantum mechanics was understood. The “s” comes from the German word for “singlet” and “d” from German for “doublet” as some lines, for example, the strong yellow lines from a sodium lamp (called the “Sodium D lines”), are actually two slightly different wave length – i. e. a “doublet.”
     
  18. exchemist Valued Senior Member

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    That's fine Ash I wasn't being critical, just thought I should straighten out any misunderstanding while I was at it. What country are you in, by the way (if you don't mind saying, that is)?
     
  19. ash64449 Registered Senior Member

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    795
    These does helps me.. Thank you!
     
  20. ash64449 Registered Senior Member

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    795
    India
     
  21. exchemist Valued Senior Member

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    Ha! Do you know, after writing my second response (which I think you had not seen when you wrote yours) I had to dash off to pick up my son from school. As I was cycling I thought, maybe I should have added your point, that the non-zero angular momentum in orbitals for which l>0 has a classical significance in that it implies the electron is in a sense "in orbit" and hence can't go in to the nucleus. But on reflection I thought this might add confusion, interesting thought the analogy certainly is. Because then one can get dragged back to questions such as why an s orbital electron doesn't "fall into" the nucleus, even though it has no angular momentum.

    It's a weird and wonderful world inside the atom, is it not?

    By the way I googled "cbse school" and they seem to be in India. So it looks as if we have a very interested, Indian scholar of 15 on our hands here. I don't know many 15yr olds who are into this sort of thing in the way he or she is.
     
  22. ash64449 Registered Senior Member

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    795
    I am "he"
     
  23. ash64449 Registered Senior Member

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    Yeah.. I thought of asking this but Billy T was offline.. Can you answer this one?
     

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