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D orbitals 5 or 6?

  1. Mar 25, 2005 #1
    d orbitals.. 5 or 6???


    i always thought ther wer 5 d orbitals corresponding 2 quantum number m = -2, -1,..., 2

    now i read somwhere that ther wud actually be 6 possible sollutions to schrodigner's equation.. i.e. six d orbitals, and th d z-square orbital is a linear combination of two of these 6, giving effectively 5.
    it also says that this combination is done because the two orbitals that are combined 'have no independant existance'

    firstly,, is this true???? (th place i read this is not exactly reliable)

    secondly, if it is, then how can 2 of these 6 orbitals 'have no independant existance'??? wt does this mean?
    if it does not hav independant existance, then wt meaning is ther 2 calling it a seperate orbital
    also, if ther wer 6 orbitals, shudnt ther be a possibilty of 12 d electrons???

    i thot that every valid solution to schrodingers equation(provided it satisfies those 3-4 constraints, continuos, finite etc) is in fact physicaly possible??

    thnx.. i know nothin of quantum mechanics so sorry if i asked some ****
  2. jcsd
  3. Mar 25, 2005 #2


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    5 should be indeed.


  4. Mar 26, 2005 #3
    you can always make linear combinations of wave functions, because the Schrodinger equation is linear. If you write down two different wave functions psi1 and psi2, with the same energy value (ie, both have the same n value for hydrogenic orbitals), then a*psi1 + b*psi2 is a solution with the same energy value, for any constants a and b.

    The usual px, py, pz orbitals that you see in chemistry textbooks are linear combinations of the l=1 states that you normally get when solving the hydrogen atom.

    Making a linear combination of solutions does not give you "more" solutions, because the solutions form a complete set. You can form any state with l=2 by taking a linear combination of d-orbitals (ie. you can form the states m = +2,+1,0,-1,-2 projected on any axis you want by adding up the usual d-states). So to answer your question, there really are an infinite number of solutions with l = 2, because you can combine solutions. However, you only need five solutions because then you can form any other solution.
  5. Apr 29, 2005 #4
    kanato! you seem like someone getting the picture with the linear algebra involved in orbital theories!

    have you seen any different linear combinations of the d orbitals than the usual xy,yz,zx,x2-y2 and z2 set? Hope this is not too OT... but I would like to see what it looked like if one symmetry-adapted the 5 d-orbitals to e.g. tetrahedral symmetry, i.e. one set (don't remember if it was the t or the e set) having tetrahedral symmetry and the other set actually having no tetrahedral symmetry (which would be nonbonding in the tetrahedral field). In other words I would like to have some sort of picture of what the d orbitals would look like in order to account for the t+e splitting in a tetrahedral ligand field, some sort of purely d-orbital precursors to the t and e set that are obtained after bonding with the ligands.

    Hope my question is understandable. And if it's OT please send me a pm if you can tell me more!
    My thought was just: if on is free to decide which linearly independent set of orbitals one uses to describe, why not explain e.g. tetrahedral bonding with a representation of the d-orbitals which clearly shows one set with tetrahedral symmetry and one set without?
  6. Apr 29, 2005 #5
    What you really should ask yourself is : why are there 5 d-orbitals that each correspond to this quantumnumber ???


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