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Hybridization and Bonding

  1. Aug 8, 2011 #1
    Question 1:

    When hybridization takes place, how do all the hybridized orbitals have the same energy?

    Here is an extract from my book:

    sp2 Hybridisation in C2H4: In the formation
    of ethene molecule, one of the sp2 hybrid
    orbitals of carbon atom overlaps axially with
    sp2 hybridised orbital of another carbon atom
    to form C–C sigma bond. While the other two
    sp2 hybrid orbitals of each carbon atom are
    used for making sp2–s sigma bond with two
    hydrogen atoms. The unhybridised orbital
    (2px or 2py) of one carbon atom overlaps
    sidewise with the similar orbital of the other
    carbon atom to form weak π bond, which
    consists of two equal electron clouds
    distributed above and below the plane of
    carbon and hydrogen atoms.

    The last time I checked, carbon was sp3 hybridized, and what is the "unhybridized orbital"? Why does the hybridization of carbon change with the formation of multiple bonds?

    Question 2:

    PCl5 has a trigonal bipyramidal structure, I do not understand how this is possible. The electronic configuration of the valence shell of P is 3s2 3p3. In its exited state, (I think) one electron from the 3s orbital will jump to the 4s orbital. So, totally, there are 2 s orbitals and 3 p orbitals taking part in bond formation with the 5 Cl atoms. The p orbitals will be bonded mutually perpendicularly to each other, where as the 2 s orbitals will be bonded in a position in which there is minimum repulsion from the other bonded pairs of electrons. But, the trigonal bipyramidal structure does not satisfy this condition. Why?

    Question 3:

    What is the difference between axial and equatorial bonds?

    Question 4:

    Can someone give me a formula to calculate the hybridization which will work for all the compounds? (I know 3 formulae, but each one has limitations)

    Thanks in advance.
  2. jcsd
  3. Aug 8, 2011 #2
    1. Carbon isn't always sp3 hybridized. It's dependent on what it's bonded to. What about in CO2, where it shares a double bond with each oxygen atom? In that case, it's sp1 hybridized, as there are only 2 bonding regions.

    A carbon with 4 bonding regions is sp3, 3 is sp2, 2 is sp1.

    2. Have you looked at the molecule? Trigonal bipyramidal is the optimal structure. If you move any chlorines, you'd be decreasing bond angles.


    3. Take glucose, for example-


    In the alpha confirmation, the bond between the rightmost carbon and the OH group is perpendicular to the bonds in the middle of the molecule. That's an axial bond. On the right, you'll see that the carbon-OH bond is parallel to the central bonds, which is equatorial.

    I cannot answer 4.
  4. Aug 9, 2011 #3


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    Science Advisor

    Carbon "isn't" hybridized. It is a model to describe the bonding. Usually, you can describe bonding with many different models. E.g. it is equally possible to describe bonding in ethylene (C2H4) using sp3 hybrid orbitals on C.
  5. Aug 9, 2011 #4
    The structure of CH4 is not square planar but tetrahedral, because px, py and pz are mutually perpendicular to each other, and there is no specific direction for s orbital, so it is formed where there is minimum bond pair repulsion. I was trying to use a similar logic for PCl5.
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