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How to determine the movement of electrons between atoms

  1. Apr 13, 2016 #1
    Let's say I want to determine what will happen during a chemical reaction. My current reasoning is that the electrons will be more likely to move from a less electronegative atom first. For instance, cesium is much more reactive than potassium, because its electronegativity is much lower, allowing it to give up electrons more easily. Is it this simple for any chemical reaction between complex molecules and simple atoms, or are there other key factors I should consider? Also, when determining what product a reaction will create, I am guessing that the most stable product will be the one that takes the least energy to make. As an example, if I decompose aluminum hydroxide (Al(OH)3), I believe it will be more likely to decompose into AlHO2 and water rather than aluminum metal, hydrogen gas, and oxygen gas, because it requires the movement of less electrons from the octet-satisfied oxygen ions. All I wish to know is if my understanding is correct, and if there is anything very important I need to know, please do correct me.
     
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  3. Apr 14, 2016 #2

    Borek

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    Staff: Mentor

    This is not trivial - reactions are not driven just by the electron movements, and electrons don't just move to the higher electronegativity elements.

    If it were possible to describe chemistry as a set of such simple rules there would be nothing to study for for years.
     
  4. Apr 14, 2016 #3
    Could you please give some examples, because if I am wrong I don't want to remain that way. I'm sure you understand that as a fellow scientist, and I believe you agree with the ability of people to spread knowledge freely. You don't have to tell me everything in the book, but it would be nice if you actually tried to point me in the correct direction. I trust this community, but sometimes the responses I get are very discouraging to the gaining of knowledge. You must be intelligent enough to correct me if you are calling out my failures. Thank you for your time.
     
  5. Apr 14, 2016 #4

    ogg

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    "You must be intelligent enough to correct me if..."?? Intelligence has little correlation with civility. For instance, implying that any post that doesn't satisfy you isn't "intelligent" is plain rude. But you're no doubt intelligent enough to know that. Either you believe that we should spend our time repeating the things that are easily found in textbooks and courses, or you think that a couple of sentences will "correct" you. If the most "stable" (thermodynamically speaking) state was the end state of every chemical reaction, then all carbon existing on Earth would be carbon dioxide, and obviously it is not. Chemistry can't be understood in *detail* without understanding quantum mechanics. Starting with simple molecules (or atoms), we can think of a chemical reaction as progressing along a "reaction coordinate". That is, a graph of "progress" vs energy. (You can think of "progress" as time, to be even more simplistic) So. a graph with reactants on the left and product(s) on the right. Calculate the energy of the reactants and plot that as a point on the left, do the same calculation for the product(s) and put a point on the far right. There's three ways that those 2 pts can be related: they're at the same energy, reactants are higher, or reactants have lower energy. (What would it mean if a reaction occurred leaving the products with more energy?) Exothermic reactions are far more likely (spontaneous)...Anyway, the key concept is "excited state". Between the reactants and products is an excited state configuration. You can put that point anywhere between the other two on the progress axis, and anywhere on the energy axis (within reason) but... The term "excited state" implies that some extra energy has been added to the system, which isn't too far wrong as a starting point, so the ES should be above the reactants. It can be above the products, too, which would mean the reaction can't proceed without getting extra energy from somewhere... Anyway, there are two important cases: one in which the E.S. is below the Products, and one in which it is above. Acid Base reactions in water occur without a E.S. above the Products. Most organic reactions occur with an E.S. above the Product's energy. From the E.S., the final products are determined by both energetic factors AND steric factors. The drop from the E.S. may be less than the energy used from the Reactants, the same, or more. Only if it is more do we consider the reaction to be exothermic. But as already said, many reactions end with products above the reactants in energy. There are plenty of sources to learn more about chemical kinetics (and chemical thermodynamics).
     
  6. Apr 14, 2016 #5
    Thank you for telling me what I should do. I apologize for being rude, it's just that I am very uncomfortable when I don't know something. I don't think I should have to feel this way, but thank you for telling me that I can find this in textbooks. I was simply unaware, and my fault has been corrected. That's all I wanted.
     
  7. Apr 15, 2016 #6
    This is not true. The "E.S." as you call it -- though, transition state is the more appropriate term -- is never lower in energy (more stable) than the products. As a matter of fact, this is completely counterintuitive! One would think that if there were to be progress along a reaction coordinate, and the system encountered some lower energy state on its way to the products, that this conformation would be the true product. In other words, at the geometry of the "products" in the scheme that you mentioned, there would exist a negative gradient along some coordinate [itex]q_i[/itex], i.e. [itex]\frac{\partial^2f}{\partial q_i^2}<0[/itex]. This is never the case! In fact, according to the variational principle, there should always be a positive gradient along all coordinates at the best structure for the given model.

    Borek is correct. What you are saying is, unfortunately, indeed simplifed. We use supercomputers and high precision experiments to elaborate upon exactly this. The problem is that chemistry is nasty due to the fact that reactivity depends on individual molecules/atoms, which are many-body problems in themselves. When you consider the interaction between just two of these many-body systems, the problem becomes even more challenging. In wave function theory (a 3N-dimensional interacting model), this problem is unwieldy. In density functional theory (a 3-dimensional non-interacting model), we encounter problems such as mixing of quantum states, and necessary treatments of quantum subsystems as being open, fluctuating in their number of electrons. These approximations leads to ugly results -- discontinuities in parts of the framework needed to describe essential parts of the picture, and interpretively challenging physical results such as non-integer numbers of electrons.

    The point is, chemistry is often a subjective science (I was a chemistry major at one point, so I empathize with this), and it is most commonly the job of the theoretical or quantum chemist to worry about the technicalities of the aforementioned issues. If you are a motivated chemist who was as bothered with this as I was, I would learn a bit more about quantum chemistry (see Levine for a popular and accessible description, or Szabo and Ostlund for a more rigorous treatment).

    Importantly, if you want to understand some of these concepts a bit better at the more fundamental level, you will need to understand the definition of some of these terms such as electronegativity, which has been interpreted as the negative of the chemical potential,1 i.e. [itex]\chi=-\mu=-\Big(\frac{\partial E[\rho]}{\partial N}\Big)_{\nu_{ext}}[/itex]

    I would urge you to check out these two papers for more info:
    1: http://scitation.aip.org/content/aip/journal/jcp/68/8/10.1063/1.436185
    2: http://www.pnas.org/content/83/22/8440.full.pdf
     
  8. Apr 15, 2016 #7
    Obviously, I need to do some research, because I do not understand any of your explanations. Thank you for actually explaining how these processes work, and enlightening me on the fact that I am of a lower faculty of understanding.
     
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