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Hybridization/Effective Nuc. charge questions

  1. Jan 5, 2009 #1

    Ser

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    1) I don't generally have problems with hybridization but when I happened to look at https://www.physicsforums.com/showthread.php?t=282170 I saw something that confused me. Poopcaboose said: "Hybrid orbitals work only when bonds are made, they only require the s and p orbitals since the d orbitals are inside of both. The sp3 hybrid orbital means there s orbital has combined with one of the p orbitals to create 3 sp orbitals and one p orbital. This makes way for three sigma bonds and a pi bond".
    1a) Isn't the first part incorrect- there are hybrid orbitals that use the D orbital? Not in carbon's case (the subject of the linked topic), but there are instances I believe.
    1b) Doesn't sp3 mean that there is 1 S orbital combining with 3 P orbitals to form 4 identical sp3s? This is good for 4 sigma bonds, I believe what he is describing is sp2, right?

    2) My elementary understanding of effective nuclear charge says that it remains more or less the same when going down a group. However, when I look at WebElement's website they are using the Clementi-Raimondi method of calculating effective nuclear charge and it appears that it increases, contrary to most Chemistry 101 sites/my textbook. Why is this? With this being so how is the atomic radii trend explained? I suppose that the increase in effective nuclear charge - decreasing the size- is outweighed by the extra energy level's size addition.

    Thanks for advance, this is greatly confusing me as I look over some stuff in prep. for Chem. 102.
     
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  3. Jan 6, 2009 #2

    chemisttree

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    You are correct, Poop had it wrong.
    You are correct, Poop had it wrong.

    Because as you go down a group, the atomic number increases. As the atomic number increases, the atomic charge increases as does the effective nuclear charge. A 1S electron sees a single proton in hydrogen but sees 19 protons in potassium. Of course as you move down a group you are adding additional shells of electrons so the size of the atom increases for that reason.
    Could you be misapplying the term 'down' to mean 'to the right'? 'Groups' are vertical colums while 'Periods' are horizontal rows. Yes, atomic radii increase as you move down a 'group' and they decrease as you move across a 'period'. Pretty straightforward, yes?
     
  4. Jan 6, 2009 #3

    Ser

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    I did mean down. I guess what I mean is this: people have generally taught, in my experience, that when going down there is no change in nuclear charge. For example, with Hydrogen and Potassium since Hydrogen has 1 proton and 1 electron with no shielding inbetween I could say it has an effective nuclear charge of 1. Since Potassium has 19 protons with 18 shielding it also has 1. I was also taught that when going across a period the protons also increase with the electrons, of course, but since the electrons are added to the same shell they don't have as much shielding ability and that's why the effective nuclear charge decreases, which lets the size increase.
    What I meant about explaining the atomic radii trend is that since the effective nuclear charge increases going down a group - greater pull - it should shrink somewhat. But the trend is that it increases.
    Ex:
    [Nucleus]---|---|
    [Nucleus]--|--|--|
    This example obviously isn't really accurate but I'm trying to use - to show distance and | to show a shell. In the second case, further down a group, the shells are pulled closer together because of the effective nuclear charge increase but since there's an entirely additional shell it still ends up larger.

    Also, to clarify my question 1b just a bit. When I said "I believe what he is describing is sp2, right?" I wasn't agreeing with "The sp3 hybrid orbital means there s orbital has combined with one of the p orbitals", that should be 1 s and 3 p for sp3, 1 s and 2 p for sp2. I was agreeing with (thing in brackets I was trying to imply) "create 3 sp[2] orbitals and one p orbital. This makes way for three sigma bonds and a pi bond" in sp2s case. This understanding is correct, right?

    Thanks a lot for taking the time to answer my questions, school doesn't start for a bit so there's no one to really ask.
     
  5. Jan 6, 2009 #4

    chemisttree

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    Nope. The overall charge of the most stable ion is +1 by that logic... not the effective nuclear charge. In the case of potassium, the 1S electrons have no shielding electrons, therefore those protons see the full nuclear charge. That certainly isn't the case for the 2S electrons and likewise the 2P electrons. They each 'see' a different effective nuclear charge. That said, the size of the atom is determined by the outermost electrons and they do 'see' a somewhat diminished effective nuclear charge than they would if you neglect the shielding effect but the size of the additional shell electrons completely outweigh this smaller effect.

    You were probably taught just the opposite and you remembered it backwards.

    You are ignoring the much larger effect of the additional shell. Atomic radii increases as you move down a group and the atomic size decreases as you move across a row. If you use Slater's rule for the elements in the first period you get the following:
    Z*H = 1.0, Z*Li = 1.3, Z*Na = 2.2, Z*K = 2.2, Z*Rb = 2.2, Z*Cs = 2.2

    You can see that the effective nuclear charge (Z*) increases very slowly down a period.

    It is right in that it allows (makes way for) 3 sigma bonds and one pi bond.
     
  6. Jan 6, 2009 #5

    Ser

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    I forgot to mention that I was talking about the outtermost electrons. The simplified method I'm using is talked about here: http://dl.clackamas.cc.or.us/ch104-06/efffective_nuclear_charge.htm
    I don't go to college there but it has the same method I was using/taught. I'm not very far in Chemistry so I'm not onto the accurate calculation of nuclear charge (ex. Slater's rules, method mentioned in original post, etc). Using the method from the link things in the same period are always the same effective nuclear charge on the outtermost electron which, looking at actual numbers, is apparently wrong. I can accept that the method linked is a simplification and not correct and I do now understand the explanation of the downward part of the trend, I just wanted to point out what I was saying/the method I was using.

    Indeed I was and I have no idea why I typed that.

    As a final question, what did you mean when you said
    Or more to the point, I see that what I did does indeed calculate the most stable ion in that case (+1 for group 1). My question is then, this is not a valid method for ion calculation, right? Using the same technique I get 7 for Fluorine which is obviously not the most stable ion, positive or negative. Or am I missing something.

    I apologize for my denseness, your help and time is much appreciated. It's quite annoying to not understand something sometimes.
     
  7. Jan 8, 2009 #6

    Ser

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    Period should read GROUP, my mistake. I can't find an edit button for some reason - I believe it was here the last time I asked a question. Excuse my mistake if I've simply missed it.
     
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