H2 Chemistry Atomic Structure Questions
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Hi, I would be grateful for some help with regards to the following questions:
1. Statement: “4s orbitals have lower energy than 3d orbitals for elements lighter than copper.”
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- Inference: That empty 4s orbitals have lesser energy than 3d orbitals. This was found in other sources, however, that filled 4s orbitals have more energy than that of filled 3d orbitals.
- Why does empty orbitals have energy? Are orbitals not just regions that have a high probability of locating electrons of specific energy levels, and therefore, shouldn’t no boundary exist to ‘separate’ different orbitals? If so, what would possess this energy?
- Why the emphasis ‘lighter than copper’ ?
- Why does 4d orbitals have lower energy than 3d orbitals? Is the average distance of an unfilled 4d orbital from the nucleus greater than that of 3d orbital, given the higher principal quantum number? Also, what would cause the change in relative energy levels, that the 4d orbital would posses more energy after being filled? In particular, is it the action of filling the 3d, or 4d orbital, that sets the change?
2. What is meant by orbitals becoming more diffused (e.g. when principal quantum number increases)?
3. Why does the energy of the electrons increase as the principal quantum number increases?
4. Does outer shell electrons exert shielding on inner shell electrons, or does the term ‘shielding’ only specifically refer to the repulsive forces exerted by inner shell electrons on outer/ valence shell electrons? Additionally, if outer shell electrons were to exert a repulsive force on inner shell electrons, would that not increase the effective nuclear charge, since the force should repel the electron closer towards the nucleus?
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First and foremost, you're asking conceptual questions that require a more involved discussion, either with your JC teacher, or your personal/private tuition teacher. It is not feasible to discuss these at length on an online forum.
As far as the A level H2 syllabus is concerned, you do not need to (and it can be counterproductive to) concern yourself with some of the more involved concepts and details.
If you're interested anyway, you could try posting your questions on the Chemical Forums, for potentially more involved discussions.
Brief comments :
1.1. Ignore 'filled' vs 'empty'. Not only is it not required, it will be counterproductive and confusing to you (ie. an A level student, assuming that's what you are) to dwell on this.
1.2. It is not the orbitals that possess energy, but electrons occupying these orbitals that possess said energy.
1.3. A moot and irrelevant point. Since elements heavier than copper are (as far as the H2 syllabus is concerned) either period 4 'p' block elements (whose chemistry do not depend on 4s and 3d orbitals) or period 5 'd' block elements (whose chemistry depends on 5s and 4d orbitals, and are beyond the syllabus anyway).
1.4. Is this a typo on your part? 4d electrons certainly possess more energy than 3d electrons. Do you mean 4s electrons? If so, here's the key to this classic riddle :
On average, 4th quantum shell orbital electrons do indeed have more energy than 3rd quantum shell orbital electrons. But at the same time, it's also true that d orbital electrons have more energy than s orbital electrons (for the same quantum shell). Hence, these two 'opposing' trends counteract each other, and it just so happens (in this universe at least), the latter trend slightly outweighs the former trend. That's really all there is to it.
2. It means less concentrated or focused. Think of it this way : each orbital can only hold 2 electrons. Since 3rd quantum shell orbitals are much larger than 2nd quantum shell orbitals, yet both orbitals hold the same number of electrons, hence if electrons are visualized as lights, then a filled 2nd quantum shell orbital would be much brighter (ie. lights are more concentrated or focused) compared to a 3rd quantum shell orbital, even though both orbitals have exactly the same number of lights (ie. 2 electrons).
3. Think of it this way : wouldn't you need to possess more energy in the first place, to resist the electrostatic pull of the nucleus and remain further away from the nucleus, compared to an inner shell electron? So whether the chicken or egg came first, all these are beyond the concerns of the A level H2 syllabus.
4. The term ‘shielding’ only specifically refer to the repulsive forces exerted by inner shell electrons on outer/ valence shell electrons. As to your hypothesized, "if outer shell electrons were to exert a repulsive force on inner shell electrons, would that not increase the effective nuclear charge, since the force should repel the electron closer towards the nucleus?", it is a moot point for four reasons :
Firstly, the effect of the electrostatic attraction of the positively charged nucleus far outweighs the relatively negligible effect of your hypothesized phenomenon.
Secondly, chemical reactions concern only the valence electrons, and thus as a practical functionality, the concept of effective nuclear charge applies specifically to these valence electrons which take part in chemical reactions. You are probably arguing that your hypothesized phenomenon would indirectly reduce the atomic/ionic radius and hence, increase the electrostatic attraction of the nucleus for valence electrons. If so, that brings us to the 3rd point.
Which is, thirdly, even if that is the case, it would apply to all atoms/ions everywhere in any case, and consequently, we can equally discount it's contribution towards meaningful comparisons across atoms/ions.
Fourthly and finally, the practical functional definition of effective nuclear charge is nuclear charge + shielding effect. And since shielding effect has already been defined (ala circular logic) as that of inner shell electrons repelling outer shell electrons, to say shielding effect increases effective nuclear charge would run counter to the point of a useful concept of shielding effect. More pertinently, as mentioned in point #1, your hypothesized effect is really rather negligible.
It's like, are you running out of a burning building more because you don't wish to be burnt alive, or more because you also happen to have a craving for a McSpicy meal at the same time as you're running out of the building... deliciously hot spicy firey stuff.
Originally posted by Cre8ion:Hi, I would be grateful for some help with regards to the following questions:
1. Statement: “4s orbitals have lower energy than 3d orbitals for elements lighter than copper.”
-
- Inference: That empty 4s orbitals have lesser energy than 3d orbitals. This was found in other sources, however, that filled 4s orbitals have more energy than that of filled 3d orbitals.
- Why does empty orbitals have energy? Are orbitals not just regions that have a high probability of locating electrons of specific energy levels, and therefore, shouldn’t no boundary exist to ‘separate’ different orbitals? If so, what would possess this energy?
- Why the emphasis ‘lighter than copper’ ?
- Why does 4d orbitals have lower energy than 3d orbitals? Is the average distance of an unfilled 4d orbital from the nucleus greater than that of 3d orbital, given the higher principal quantum number? Also, what would cause the change in relative energy levels, that the 4d orbital would posses more energy after being filled? In particular, is it the action of filling the 3d, or 4d orbital, that sets the change?
2. What is meant by orbitals becoming more diffused (e.g. when principal quantum number increases)?
3. Why does the energy of the electrons increase as the principal quantum number increases?
4. Does outer shell electrons exert shielding on inner shell electrons, or does the term ‘shielding’ only specifically refer to the repulsive forces exerted by inner shell electrons on outer/ valence shell electrons? Additionally, if outer shell electrons were to exert a repulsive force on inner shell electrons, would that not increase the effective nuclear charge, since the force should repel the electron closer towards the nucleus?
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Hi UltimaOnline, thanks for your help in answering these questions! I am aware that the answers to these questions would require higher level knowledge (e.g. Quantum Mechanics and the like) and deeper discussions with tutors and teachers, but as it is, both resources are not available to me (ie. I do not have any chemistry tution classes, and I am hesitant about asking my own JC teacher), and I (to my own detriment, perhaps) do get a little paranoid when I memorize things I don't understand, which thus lead me to surfing the net for answers. I have, however, purchased a copy of the book you recommended, and it is helping me supplement my own school notes very well, so I am grateful for your help on that as well.
However, I have a question to your answer 1.4, that is, why would the fact that the 4th quantum orbital electrons possessing more energy than 3rd quantum shell orbital electrons run contrary to the fact that p orbital electrons have more energy than s orbital electrons in the same quantum shell, since I would assume, that in the order of increasing energy level, it would spell out as something like 3s,3p,...4s, 4p. Also, would it be correct to deduce that the reason that p orbital electrons have more energy than s orbital electrons, because p orbitals are further (or less penetrating?) than s orbitals from the nucleus, in the same principal quantum shell?
Once again, I appreciate your help immensely :)
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Right, I understand. Sure no prob. However, as I said previously, should you find the discussions here lacking (since not many others respond to Chem queries here, and I am unwilling to go too far beyond the syllabus*), you can always try discussing in more depth at the Chemical Forums, for future topics (throughout your JC life).
(* With the exception of areas I personally find worth going into for H2 students, H3 stuff like formal charges, resonance, and organic mechanisms.)
would it be correct to deduce that the reason that p orbital electrons have more energy than s orbital electrons, because p orbitals are further (or less penetrating?) than s orbitals from the nucleus, in the same principal quantum shell?
Yes, that's correct. And that's all (actually, that's already an overkill) you need to concern yourself with, for the H2 syllabus.
why would the fact that the 4th quantum orbital electrons possessing more energy than 3rd quantum shell orbital electrons run contrary to the fact that p orbital electrons have more energy than s orbital electrons in the same quantum shell
It doesn't really run contrary. I reiterate these two non-contrarian facts :
On average (ie. averaging out the energies of the different orbitals within a quantum shell), 4th quantum shell orbital electrons do indeed have more energy than 3rd quantum shell orbital electrons.
But at the same time, it's also true that d orbital electrons have more energy than s orbital electrons (only for the same quantum shell).
Therefore, it shouldn't be surprising (nor does it run counter to either of the 2 facts above), that 3d orbitals have slightly higher energy compared to 4s.
Originally posted by Cre8ion:Hi UltimaOnline, thanks for your help in answering these questions! I am aware that the answers to these questions would require higher level knowledge (e.g. Quantum Mechanics and the like) and deeper discussions with tutors and teachers, but as it is, both resources are not available to me (ie. I do not have any chemistry tution classes, and I am hesitant about asking my own JC teacher), and I (to my own detriment, perhaps) do get a little paranoid when I memorize things I don't understand, which thus lead me to surfing the net for answers. I have, however, purchased a copy of the book you recommended, and it is helping me supplement my own school notes very well, so I am grateful for your help on that as well.
However, I have a question to your answer 1.4, that is, why would the fact that the 4th quantum orbital electrons possessing more energy than 3rd quantum shell orbital electrons run contrary to the fact that p orbital electrons have more energy than s orbital electrons in the same quantum shell, since I would assume, that in the order of increasing energy level, it would spell out as something like 3s,3p,...4s, 4p. Also, would it be correct to deduce that the reason that p orbital electrons have more energy than s orbital electrons, because p orbitals are further (or less penetrating?) than s orbitals from the nucleus, in the same principal quantum shell?
Once again, I appreciate your help immensely :)
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I am sorry if I am a being a little dull here, but why would the fact the p orbital having more energy than s orbital for the same principal quantum shell explain how it is possible for 3d orbital to have more energy than 4s orbital, given that it is of a lower principal quantum number?
Thanks again!
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Another question, though a little off-topic here since it's on redox:
NH4NO3 + NaOH -> NH3 + NaNO3
Is there any redox reaction taking place here, since the oxidation state of N changes ie. OS of N in NH4 is -3, while that in NO3 is +5? Is this considered a change in OS, since the OS of the respective elements in their respective polyatomic ions remain unchanged (ie. OS of N in NO3 remain unchanged before and after, and OS of N in NH4 and NH3 remains the same)? (Please ignore the fact that the equation is unbalanced)
Thank you! :)
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Mmm... I'm not sure I can explain it any clearer than I already have. Could someone (eg. ChemGuide) please step in to explain it in another way?
Originally posted by Cre8ion:I am sorry if I am a being a little dull here, but why would the fact the p orbital having more energy than s orbital for the same principal quantum shell explain how it is possible for 3d orbital to have more energy than 4s orbital, given that it is of a lower principal quantum number? Thanks again!
No, this is not a redox reaction. It is a Bronsted-Lowry acid-base proton transfer reaction, with no change in the OS of the N atoms involved. To understand this properly, you have to visualize or draw out the mechanism, to appreciate that the individual N atoms in this reaction do not experience any change in OS (where OS = formal charge + electronegativity consideration).Originally posted by Cre8ion:Another question, though a little off-topic here since it's on redox:
NH4NO3 + NaOH -> NH3 + NaNO3
Is there any redox reaction taking place here, since the oxidation state of N changes ie. OS of N in NH4 is -3, while that in NO3 is +5? Is this considered a change in OS, since the OS of the respective elements in their respective polyatomic ions remain unchanged (ie. OS of N in NO3 remain unchanged before and after, and OS of N in NH4 and NH3 remains the same)? (Please ignore the fact that the equation is unbalanced)
Thank you! :)
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Hmmm, ok thanks for your help anyway!
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First of all, my bad for the typo. I meant 'd' orbital (not 'p' orbital, even though that would still be true) having more energy than 's' orbital, for the same shell. If your confusion was just over the typo, then it's all cleared up now?
Apart from the typo, try this analogy to understand why 3d orbitals have more energy compared to 4s orbitals :
Class A (4th quantum shell) has, on average, better grades than Class B (3rd quantum shell).
But the highest (energy) student (d orbital) in class B has better grades (more energy) the lowest (energy) student (s orbital) in class A.
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Hmmmm, would it be possible to deduce from that then, that a (unfilled, at least) 4s orbital, is in some ways, closer to the nucleus than the 3d orbital (either because a d-orbital is less penetrating <<< I'm not sure about the accuracy/ and relevance of this fact though), but on average, a higher principal quantum shell number orbital will be further away from the nucleus than one with lower principal quantum shell number?
Thanks :)
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Sorry, I have another question!
When they state that electrons are added to an orbital in a particular manner (for e.g. a 4s orbital before 3d orbital), do they mean the adding of electrons across the period to (ie. from B to C the next electron is added to another 2p orbital), or does it also refer to the how electrons will be added to a particular element (during the formation of ionic bonds or covalent bonds)?
Really thanks a lot for your time and help, UltimaOnline! :')
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Originally posted by Cre8ion:
Hmmmm, would it be possible to deduce from that then, that a (unfilled, at least) 4s orbital, is in some ways, closer to the nucleus than the 3d orbital (either because a d-orbital is less penetrating <<< I'm not sure about the accuracy/ and relevance of this fact though), but on average, a higher principal quantum shell number orbital will be further away from the nucleus than one with lower principal quantum shell number?
Yes, you could reasonably make that deduction, though (it wouldn't surprise you that) the truth is a little more complex than this, and totally would not be asked (at all) at 'A' levels in any case.As for your other question about adding electrons, as far as the A level H2 syllabus is concerned, you can take it to mean both in isolation, ie. meaning as added across a period, as well as in the context of the formation of covalent/ionic bonds.
Again (by now this certainly wouldn't surprise you) that the truth is a little more complex than this, and totally would not be asked at 'A' levels (at all) in any case.
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'A' levels is a preparatory bridging course between O levels (kids play) and Uni level (adult work), and it should be evident by now, that your conceptual thoughts and musings are not examinable at 'A' levels, which is focused more on the practical application of chemical principles (eg. how to simply calculate the pH of a solution, without pondering over the philosophical or quantum physical implications). So while deeper thinking should certainly be encouraged, but for 'A' level purposes, it might be counter-productive and miss the target (ie. requirements to excel in the exams) to focus too much on these non-examinable components, and so a balance will be required.
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Thanks UltimaOnline, I will try to keep that in mind.
One other question though, (I'm revising for my CTs, so I have lots of questions, sorry!), is "What does it mean by an effective orbital overlap?" (in the chapter of Chemical Bonding - Covalent Bonds)
Thank you, as always! :)
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Originally posted by Cre8ion:
Thanks UltimaOnline, I will try to keep that in mind.
One other question though, (I'm revising for my CTs, so I have lots of questions, sorry!), is "What does it mean by an effective orbital overlap?" (in the chapter of Chemical Bonding - Covalent Bonds)
Thank you, as always! :)
Perhaps the following BedokFunland JC questions and answers of mine, will help you understand "effectiveness of orbital overlap"
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For more reading on "effectiveness of orbital overlap", refer to the following book (which all H2 Chem students should buy... have you bought yours yet?).
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For the benefit of other visitors to this thread, here's my
Full List of Recommended Books for H2 Chemistry students
http://www.infinity.usanethosting.com/Tuition/#Books_for_H2_Chemistry -
Thanks for your help as always, and yes, I have bought the book :)