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Symmetrical Group of a molecule

  1. Sep 3, 2007 #1
    [SOLVED] Symmetrical Group of a molecule

    I'm trying to study PCl5, it have a bi-pyramidal shape, with The P atom at the center.
    I was able to find out the group it's D3h.

    1)To what irreducible representations does The "p" orbital of Phosphor ( P ) and the five "1s" orbitals of Chloride belongs to.

    2)Will the "d" orbital of P will recover ( I don't know the excat english word for it ) with the "1s" of Cl ? I need to find out if the Integral I is equale to 0 or 1 ( if the representation Belongs to A1 )

    The problem is I'm studying this on my own , ANd I'm not able to read the Group table correctly to understand it , for example ( http://www.webqc.org/symmetrypointgroup-d3h.html ) , I'm not able to find out which orbital belongs to which irreducible representation..
    any help ?
     
    Last edited: Sep 3, 2007
  2. jcsd
  3. Sep 3, 2007 #2

    Gokul43201

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    I believe the question must specify which particular p-orbital (p_x, p_y or p_z?) and d-orbital it is asking about.

    The only way I know to find the irreps is to take all 6 orbitals in question, and find the matrices (representatives) that produce the transformations of the various symmetry elements under D_{3h}. You then need to reduce each 6X6 matrix to its irreducible forms. If you are using the p_z orbital for P, then clearly there is a 1-dimensional irrep for that orbital. The remaining 5X5 matrix, I believe, will turn out to be the direct sum of a 2X2 matrix (for the Cl-1s orbitals above and below the [itex]\sigma_h[/itex] plane) and a 3X3 matrix (for the 3 orbitals lying in the [itex]\sigma_h[/itex] plane). You should then try to diagonalize the matrices to see if they can be further reduced.

    If you are intimately familiar with this procedure, you will be able to guess some of the irreps by inspection.

    PS: The word you are looking for is "overlap".
     
  4. Sep 3, 2007 #3
    Ok , first of I Made a mistake in the first question it's not the "p" orbital , it was "1s".
    As for the second one , the question did not mention which "d" orbital.
    now lets forget about the second.
    What I had in mind to solve the first part is:
    I'm trying to see to what the "s" orbital of The P and 5 Cl atoms irreducible representation they belongs too.
    The methode how I was thinking to solve it is the following:
    I try to find each orbital to what representation it belongs to...
    let's say for example the 1s of P belong to A1 and all the rest orbitals "s" of Cl belongs to A2.
    A'1 1 1 1 1 1 1
    A'2 1 1 -1 1 1 -1
    A'2 1 1 -1 1 1 -1
    A'2 1 1 -1 1 1 -1
    A'2 1 1 -1 1 1 -1
    A'2 1 1 -1 1 1 -1
    then Gamma = 1 1 -1 1 1 -1
    Gamma = aA'1 + bA'2 + cE' + ....
    then I proceed and calculate a b c, ...
    by multiplying Gamma by A'1 and A'2 and then divide by the group order.
    If that's the way to go, I 'm having problem identifying to which representation belongs each orbital
     
  5. Sep 4, 2007 #4
    *bump*
     
  6. Sep 8, 2007 #5

    Gokul43201

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    ziad, I dont know very much about this area, so my method may be unnecessarily long. If there is a quicker way to arrive at the irreps, I'm not aware of it.

    This is how I would do it:

    First you draw a picture of the molecule and assign labels to the orbitals that you care about. I'm going to call them: p (Phos), u (Cl-atom on upper side of [itex]\sigma_h[/itex] plane), d (Cl-atom on lower side of [itex]\sigma_h[/itex] plane), 1, 2, 3 (the 3 Cl atoms on the [itex]\sigma_h[/itex] plane).

    We can make a vector space with the state of each atom forming a basis vector. We start with a vector u=(p,u,d,1,2,3) and see how it transforms under each of the symmetry elements of D3h.

    For instance, under C3, we have :

    [tex](p,u,d,1,2,3) \xrightarrow{C_3} (p,u,d,3,1,2) [/tex]

    We can figure out the tensor that achieves this transformation: [itex]u = Du' [/itex]

    In the above case, we see that since all orbitals of interest are s-orbitals, there os no change of sign to worry about and the representative matrix can be figured out easily. By inspection, you arrive at:


    ([tex](p,u,d,1,2,3) = (p,u,d,3,1,2)~\left( \begin{array}{cccccc}
    1 & 0 & 0 & 0 & 0 & 0\\
    0 & 1 & 0 & 0 & 0 & 0\\
    0 & 0 & 1 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 1\\
    0 & 0 & 0 & 1 & 0 & 0\\
    0 & 0 & 0 & 0 & 1 & 0\\
    \end{array} \right)
    [/tex]

    In a similar manner, the representations of all the other symmetry elements of D3h (i.e., [itex]3C_2~,~\sigma_h[/itex]) can be computed.

    This is the first step.

    The next step is to reduce these representations to their irreducible forms.

    At this point, I'll wait for you to calculate the representations of the other symmetry elements before I proceed ... or someone else may provide a quicker solution to this that uses some handy "look-up" scheme.
     
    Last edited: Sep 8, 2007
  7. Sep 14, 2007 #6
    I got it, thx.
    I was just confused between 2 things.
    Thank you for your help
     
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