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Molecules the hydrogen molecule “All theoretical chemistry is really physics; and all theoretical chemists know it.”—Richard.

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Presentation on theme: "Molecules the hydrogen molecule “All theoretical chemistry is really physics; and all theoretical chemists know it.”—Richard."— Presentation transcript:

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2 Molecules the hydrogen molecule “All theoretical chemistry is really physics; and all theoretical chemists know it.”—Richard P. Feynman

3 From last time: the total energy of an isolated electron and two isolated protons is defined to be 0 eV. “Isolated” means they are infinitely far apart so they don’t interact. The total energy of an isolated proton (an H + ) and an isolated H is eV. That makes sense-- just the energy of the H atom; the proton infinitely far away doesn’t interact with the H. The total energy of H 2 + at the “stable” proton-proton separation of R=2a 0 is eV.

4 The energy difference between H + H + and H 2 + is eV, the bond energy of H 2 +, which is the energy decrease upon bonding of H and H + to form H 2 +. If I gave you the enlarged version of this figure on the quiz, you would say the bond energy is eV If I gave you the enlarged version of this figure on the quiz, you would say the bond energy is eV - (-16.3 eV) If I gave you the enlarged version of this figure on the quiz, you would say the bond energy is eV - (-16.3 eV) = 2.6 eV.

5 8.4 The Hydrogen Molecule For H 2, we can repeat the analysis we did for H 2 +, with minor differences. Two electrons contribute to the bonding. The contribution of the single electron to the bonding in H 2 + is eV. When you add a second electron, you expect it to lower the total energy by an additional 2.65 eV, except that the two electrons now repel each other. So the bond energy is E = 2(2.65 eV) - (electron-electron repulsion energy). You can make an accurate calculation of the bond energy (small approximations are required), or you can measure it and find that E(H 2 ) = eV. The bond energy is 4.5 eV.

6 A note on terminology: H 2 is lower in energy by 4.5 eV than two isolated hydrogen atoms. The total energy of H 2 is eV. The – sign on the total energy tells us H 2 is stable. “Bond energy” is the magnitude of the energy reduction, so the bond energy is 4.5 eV. The words “bond energy” tell you to stick a – sign in front of the 4.5 eV when you use the energy in calculations. For H 2,  S leads to bonding,  A does not. “Hold on just a second there, pal. Didn’t you tell us back there in chapter 7 that electrons in a system are described by antisymmetric wave functions?” “Hold on just a second there, pal. Didn’t you tell us back there in chapter 7 that electrons in a system are described by antisymmetric wave functions?” Did you just make a physics mistake?

7 Sorry—no mistake!  =  s (ignoring the “stationary” time dependence), where s is the spin part of the wave function. It is the total wave function  that must be antisymmetric. Since  is symmetric, it follows that s must be antisymmetric. In other words, in this 2-electron system, the spin function is antisymmetric, and the electron spins therefore must be antiparallel (if they were parallel, you wouldn't be able to tell them apart). Makes sense! If both electrons in the H 2 molecule are to go in the lowest-energy 1s state, the exclusion principle tells us they must go in one spin-up and the other spin-down.

8 We could calculate the total energies for symmetric and antisymmetric hydrogen molecule wave functions, just like we did in the previous section. The calculation (using Schrödinger’s equation) requires approximations, but is nevertheless quite accurate. The result is: Bonding. (Notice R 0 is less than 2a 0.) The absence of bonding for E A is a consequence of the Pauli exclusion principle. That’s (almost) it for Chapter 8! Except…

9 Problem 8.2 The protons in the H 2 + molecular ion are nm apart and the binding energy of H 2 + is 2.65 eV. What negative charge must be placed halfway between two protons this far apart to give the same binding energy? Homework! attractive repulsive Let e be the magnitude of the charge on both a proton and an electron. Let q be the charge placed between the two protons, which are a distance R apart. R/2 +e q

10 attractive repulsive R/2 The “binding energy” is the coulomb energy of this charge configuration. Binding energies are always negative. Copying the last equation to the next slide… +e q

11 …and continuing with the math… One simple step at a time, or I’ll make a math error!

12 Plug in numbers (being ultra-safe and doing everything in SI units): Expressed as a fraction of the charge on an electron,

13 Or about 30% of the charge of an electron. A mathcad solution I had posted for a while had an extra - sign that mysteriously disappeared in the paper and pencil algebra that I did not show. Hopefully that error has been permanently eradicated from my notes etc. The solution in this powerpoint presentation is correct, and “better” because I didn’t skip steps. A note on “e.” It is the magnitude of the charge on an electron. An electron has a - sign, a proton a + sign. The answer says the unknown charge has a magnitude 29.9% of that of an electron, and a negative sign.

14 The next lecture is rather long, so if there is time, I will start it now.


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