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    I am having a minor difficulty in grasping certain parts of the electronic configuration required for Unit 1.1.
    Potassium (K) has the electronic configuration 1s2 2s2 2p6 3s2 3p6 4s1.
    Why does it skip the 3d sub-shell? I was wondering whether it could be something to do with the fact that the energies of 4s and 3d overlap.
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    This took me a while to accept. Basically, you have to accept that the 4s is of a lower energy than the 3d energy level.

    Therefore the 4s fills before the 3d energy level.

    However, it gets more complicated when you're removing electrons, but this only involves transition metal in Module 5, so you don't need to know about that yet.
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    1s
    2s 2p
    3s 3p 3d
    4s 4p 4d 4f
    ...

    Pretend there are arrows going down from 1s, then another one through 2s, then another one through 2p and 3s etc etc. I know you are asking why but thats because its at a lower energy level as stated above. But if you get confused then you can use this and it will be quick at telling you what orbital fills first
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    Statements such as 'the 4s is lower in energy than the 3d' are actually of very little use. The precise energy of the 4s and 3d orbitals depends principally upon the nuclear charge and is thus dependant on the atom in question.

    As the nuclear charge increases, the s-orbitals are lowered in energy more relative to the d-orbitals as a result of their precise form, specifically because the wavefunction for an s-orbital is non-zero at the nucleus.
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    (Original post by Fly.)
    Statements such as 'the 4s is lower in energy than the 3d' are actually of very little use. The precise energy of the 4s and 3d orbitals depends principally upon the nuclear charge and is thus dependant on the atom in question.

    As the nuclear charge increases, the s-orbitals are lowered in energy more relative to the d-orbitals as a result of their precise form, specifically because the wavefunction for an s-orbital is non-zero at the nucleus.
    Dude..i admire your knowledge..but i think it is way and above the op's query!
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    (Original post by Fly.)
    Statements such as 'the 4s is lower in energy than the 3d' are actually of very little use. The precise energy of the 4s and 3d orbitals depends principally upon the nuclear charge and is thus dependant on the atom in question.

    As the nuclear charge increases, the s-orbitals are lowered in energy more relative to the d-orbitals as a result of their precise form, specifically because the wavefunction for an s-orbital is non-zero at the nucleus.
    it mustn't be forgotten that the theoretical energy levels of the orbitals are mathematically calculated from atoms with 1 or 2 electrons. They are theoretical concepts and have no "real" meaning until occupied (when the mathematical models that defined them become invalid - paradox but true).

    So there is a guideline of the order of energies, theoretically (or mathematically) calculated and then there are the observed variations of which orbitals are entered from atom to atom ascending in atomic number.

    It has been deduced from experimental (spectroscopic etc) observation that the 4s orbitals are entered before the 3d.

    Fine - and chemists are then left with the job of reasoning why.

    Logically, the model for energy levels produced from solutions of the Schroedinger equation for the hydrogen atom and the helium ion cannot hold for larger atoms where the electrons themselves play a big part in the actual energies of the "orbitals" (in inverted commas because orbitals being regions of space they cannot actually possess energy - only the electrons that occupy them can)

    So, without wishing to bore the pants off anyone, it becomes a case of observation and acceptance - and at the same time trying to logically explain why - these explanations are mere hypothesis ... and as the very things they are trying to explain are used as the evidence for the actual hypotheses... they are consequently worthless, scientifically speaking.

    en fin...

    Just learn the order of filling of the orbitals (Aufbau - 4s before 3d).. and
    learn that the 4s orbital is evacuated by ionisation before the 3d
    and you will be able to sleep well at night

    I hope...
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    That is quite a lot of information but it has really helped thanks . As for sleeping well at night...it will be a while before I can do that!
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    (Original post by Fly.)
    Statements such as 'the 4s is lower in energy than the 3d' are actually of very little use. The precise energy of the 4s and 3d orbitals depends principally upon the nuclear charge and is thus dependant on the atom in question.

    As the nuclear charge increases, the s-orbitals are lowered in energy more relative to the d-orbitals as a result of their precise form, specifically because the wavefunction for an s-orbital is non-zero at the nucleus.
    What I said is taken from the AQA A-level book, as this is the way I learnt it.

    Clearly, you don't accept this as a "useful" answer, but for some people (like me) it helps them through the syllabus.

    I believe over-indulging yourself in certain parts of the syllabus isn't "useful", as questions about why the orbitals fill up in the way they do rarely come up in our exams. I like to think about concepts as simply as they come, and not to overcomplicate them. It got me through my A-levels, so it works for me. Each to their own ...
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    (Original post by Death)
    I am having a minor difficulty in grasping certain parts of the electronic configuration required for Unit 1.1.
    Potassium (K) has the electronic configuration 1s2 2s2 2p6 3s2 3p6 4s1.
    Why does it skip the 3d sub-shell? I was wondering whether it could be something to do with the fact that the energies of 4s and 3d overlap.
    initially, 4s has a lower energy level than the 3d sub-shell this means that the 4s sub-shell becomes filled before the 3d sub-shell...

    however, when electrons are filled in the 3d sub-shell, the energy level of the 3d sub-shell drops below that of the 4s sub-shell.

    Remember that the electrons that are held further from the nucleus are at higher energy level and so are lost more easily

    and also that the outmost electrons are lost first.

    This is why the electrons are lost from the 4s sub-shell before the 3d sub-shell.

    This means that electrons in transition metal are lost first from the 3d sub-shell (and then from the 4s sub-shell) in the formation of transition metal ions

    charco, talk about overcomplicating things, i understand that what we learn things in science that can just be unsubstantiate theories of explaining certain phenomenon, however, being suspicious of the reliability of such theories doesn't get you marks for A-level! So it's probably just best to learn A-levels as 100% truth and treat the syllabuses as law! I don't agree with learning things parrot fashion but this is what you must do to a certain extent (particularly true for bio).

    Anywayz... i'm done with A-levels (woot!), just thought i'd be helpful and sort out some of this confusion.
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    (Original post by Revenged)
    initially, 4s has a lower energy level than the 3d sub-shell this means that the 4s sub-shell becomes filled before the 3d sub-shell...

    however, when electrons are filled in the 3d sub-shell, the energy level of the 3d sub-shell drops below that of the 4s sub-shell.

    Remember that the electrons that are held further from the nucleus are at higher energy level and so are lost more easily

    and also that the outmost electrons are lost first.

    This is why the electrons are lost from the 4s sub-shell before the 3d sub-shell.

    This means that electrons in transition metal are lost first from the 3d sub-shell (and then from the 4s sub-shell) in the formation of transition metal ions

    charco, talk about overcomplicating things, i understand that what we learn things in science that can just be unsubstantiate theories of explaining certain phenomenon, however, being suspicious of the reliability of such theories doesn't get you marks for A-level! So it's probably just best to learn A-levels as 100% truth and treat the syllabuses as law! I don't agree with learning things parrot fashion but this is what you must do to a certain extent (particularly true for bio).

    Anywayz... i'm done with A-levels (woot!), just thought i'd be helpful and sort out some of this confusion.
    Rev: My post was a response to my perceived overcomplication in the post by Fly.... it was not intended to be a further overcomplication!

    if you read my last paragraph you will see...

    "en fin...

    Just learn the order of filling of the orbitals (Aufbau - 4s before 3d).. and
    learn that the 4s orbital is evacuated by ionisation before the 3d
    and you will be able to sleep well at night"


    Which is in total agreement with your post!! :ditto:
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    (Original post by charco)
    Rev: My post was a response to my perceived overcomplication in the post by Fly.... it was not intended to be a further overcomplication!

    if you read my last paragraph you will see...

    "en fin...

    Just learn the order of filling of the orbitals (Aufbau - 4s before 3d).. and
    learn that the 4s orbital is evacuated by ionisation before the 3d
    and you will be able to sleep well at night"


    Which is in total agreement with your post!! :ditto:
    The funny thing is that it's not just 4s before 3d. Its more like 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f. I have to memorise that right?
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    Dunno which board you are studying Death but most only require a knowledge of configs up to Krypton...
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    (Original post by Death)
    The funny thing is that it's not just 4s before 3d. Its more like 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f. I have to memorise that right?
    Yes, you do. You also must remember the maximum number in each orbital
    Remember : s2 f14 d10 p6.
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    (Original post by charco)
    Dunno which board you are studying Death...


    lol...comedy, wasn't in my chem A-level
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    (Original post by Death)
    The funny thing is that it's not just 4s before 3d. Its more like 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f. I have to memorise that right?
    Got to this thread fairly late, and it's mostly been answered, but here is something else to confuse you, but it's important to remember.

    With regards to chromium and copper, they are not [Ar] 4s2 3d4 and [Ar] 4s2 3d9 as you'd expect, but [Ar] 4s1 3d5 and [Ar] 4s1 3d10. Same applies for molybdenum and tungsten at [Kr] 4s1 3d5 and [Xe] 4s1 3d5, and silver and gold at [Kr] 4s1 3d10 and [Xe] 4s1 3d10.

    These anomalies occur due to the extra stability of half full and full sets of "d" orbitals respectively.

    As already pointed out, you will probably not be expected to know about f electron shells at the level you are discussing, but they hold 14 electrons in a full shell, compared with 10 of the d. For the lanthanides, they are generally [Xe] 4fn 5d0 6s2. I have listed the exceptions and the reasons below.

    For example cerium is [Xe] 4f1 5d1 6s2 not [Xe] 4f2 6s2, because of the sudden contraction and reduction in energy of the 4f orbitals immediately after lanthanum is not yet sufficient to avoid occupancy of the 5d orbital.

    Gadolinium is [Xe] 4f7 5d1 6s2 not [Xe] 4f8 6s2, which reflects the stability of the half filled shell as discussed earlier.

    Lutetium [Xe] 4f14 5d1 6s2. The shell has been filled

    Marcus
 
 
 
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