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Bonding-Antibonding Pair question.

  1. Mar 25, 2007 #1
    Ok, so I know the difference between bonding and antibonding orbitals (one the electrons add destructively and the other they add constructively) but what I don’t understand is how we know which electrons will add destructively and which electrons will add destructively. For this wouldn’t we need to know the phase of the individual electrons?

    For example my chemistry book says that when two H atoms combine the first two electrons will go into the bonding pair…… but why! Doesn’t this mean that the electrons are in phase? And if so how do we know the first two are in phase?

    It then gives a theoretical example of He2 and why it doesn’t exists saying that the first two electrons go into the bonding pair and the second two go into the antibonding pair. How do we know that the third and fourth valence electrons go into the antibonding pair?

    My book gives some nifty looking energy level diagrams for these situations and shows how the antibonding pair is higher in energy than the bonding pair but I still don’t understanding why the electrons fall into the bonding pairs in that order.

    For example why couldn’t the first two go into the bonding pair and then the second two also go into the bonding pair as well?
  2. jcsd
  3. Mar 26, 2007 #2


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    When one atomic energy level from each H are combined into a moleculer energy level, two energy levels are produced. One is lower in energy (bonding) and one is higher (antibonding) in energy. Only two electrons may occupy each energy level and the first two are placed in the lower of the two. If there were more electrons around, they would have to go in the higher (antibonding) energy level. This is where the extra electron find itself in the H2- radical.

    The electrons must be in orbitals... they just don't exist as free electrons roaming around. The electrons are obliged to occupy an orbital that has space available and to occupy the orbital of lowest energy with space available. Thus, in He2 two electrons are obliged to occupy the lowest energy orbital and two the higher energy orbital. The net bond order (MO theory) is zero... exactly the same as two individual He atoms.

    Bond order = # electron pairs in bonding orbitals - # electron pairs in antibonding orbitals.
  4. Mar 26, 2007 #3
    I understand what bond order is and why the He2 molecule doesn't exist I also understand that the electrons will go into the lowest energy level possible and that is why the go into the antisymetric orbital, but i gues what i am asking is "why is the second highest energy level anti-bonding"?

    why is there an bonding energy level and a anti bonding energy level. I undestand the diffrence between the two but not why they exist in the order they do.
  5. Mar 26, 2007 #4


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    You ask about how we know if the electrons from two separate hydrogen atoms are "in phase". Perhaps you should try to understand what the term "in phase" actually represents in molecular orbital theory. In MO theory, electrons don't have an inherent "phase" at all. The term "phase" actually describes the space that the electrons occupy IN A MOLECULAR ORBITAL. Again, "phase" is not a term that has any meaning for the electrons themselves. It cannot be used to describe electrons. In MO theory, electrons do not "add constructively" or "add destructively" either since they don't have "phases".

    "Phase" refers to the functions themselves that describe the space that the electrons occupy. This space is described by a periodic wave function. The Schroedinger wave equation gives an exact solution for this function for hydrogen. These atomic wave functions can be combined in two ways to produce a molecular wave function. The two ways that they can be combined are either "in phase" or "out of phase", "phase" being the sign of the wave function that describes the combined atomic orbitals (molecular orbital). One results in a bonding orbital and the other an antibonding orbital. The electrons must occupy one of these states... they have nowhere else to go but into much higher energy "harmonics" of the solution for this molecular periodic wave function.

    The combination that is "in phase" results in an orbital space that increases electron density between the nuclei of the individual atoms. The combination that is "out of phase" results in an orbital space that decreases the electron density between the nuclei of the individual atoms.
  6. Apr 1, 2007 #5
    Take two normalized 1s atomic orbitals at infinite separation: 1sa and 1sb. Each has identical energy levels Ea and Eb. That is, H|1sa> = H|1sb> = Ea = Eb. In order for the electrons to be bound, these atomic energies must be less than 0.
    Now, bring the two orbitals together and form the two possible normalized wavefunctions:
    Phi1 = c'(1sa + 1sb)
    Phi2 = c"(1sa - 1sb)
    The positive coefficients c' and c" are chosen so that the resulting wavefunctions are normalized.
    H|Phi1> = H|c'(1sa + 1sb)> = c' (Ea + Eb)|Phi1>
    H|Phi2> = H|c"(1sa - 1sb)> = c"(Ea - Eb)|Phi2>
    Since the atomic energies are less than zero, (Ea + Eb) < (Ea - Eb).
    The positive combination of 1s orbitals is "in-phase" because the overlap of the 1s orbitals is positive and gives the bonding comination. Since it has the lower energy, it fills first and gives a total energy below the separated atom limit. Consequently, hydrogen molecule is stable wrt two atoms. In contrast, (Ea - Eb) lies above the separated atom limit and diatomic helium is unstable wrt to separated atoms.
    A final point is that helium can be liquified, but low temperatures are required and there is a much smaller disruption of the atomic structure than is found in a typical covalently bonded diatomic.
  7. Apr 1, 2007 #6
    I would also like to add that a single moleclar orbital (MO) can only hold one or two electrons if they are of different spin. This is because of the Pauli principle which says formally that no two electrons can have the same space and spin coordinates.
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