Hi all,
I have prepared Pd supported heterogenous catalyst by liquid media assisted reduction method. From the XPS it was observed that the surface is composed of both Pd+2 (as the major species, ~65%) as well as Pd(0). However, in XRD, the Pd+2 peak is not prominent and the intensity is very less compared to the Pd(0) peak intensity, although Pd+2 % is very high. Why it is so?
Thanks in advance...
Dear Ansari Palliyarayil ,
you should know, that the formation of an XRD peak is due to x-ray 'reflection' (better name is: elastic scattering) at so-called lattice planes and subsequent x-ray interference. These lattice planes are set up by the lattices in which the atoms (in your case Pd(0) and Pd(2+)) and others are bound. Due to the fact that the radii of Pd(0) and Pd(2+) are quite different, you will have quite different phases of x-ray diffracting material present at your sample surface, and these phases will have different lattice constants and thus different interplanar spacings, giving rise to different peak positions and different peak heights. The diffraction peaks are not solely 'produced' by Pd(0) or Pd(2+) but also by other atoms in the (local) atomic lattice.
So you can only assign a phase, which keeps Pd(0) or Pd(2+), to an XRD peak but not solely Pd(0) or Pd(2+).
So you should not wonder, when you will get different XRD pattern when Pd(2+) is involved instead of Pd(0) or vice versa...
But you should share your XRD pattern, and some details on how you have measured the pattern (GI-XRD?); so we may comment further on.
Thanks in advance and
best regards
G. Martens
Ansari Palliyarayil
XPS measures the energy that an electron is removed from a specific orbital of an atom. XRD measures the distance between electronic planes in a crystal. If the distance electronic planes in the crystal, the type of crystal Na2 [PdCl4] 3H2O and Pd coincide on the surface of the composite with a thickness of 5 nm, then you will see 1 peak in XRD. If the composite contains 20% Pd (2+), then you will receive 20% Pd (2+) on this surface by XPS.
Ansari Palliyarayil
In addition to the excellent answers provided by Gerhard Martens and Yuri Mirgorod I would also examine what is measured by XPS and XRD in terms of penetration depth and the type of catalyst (I worked with such catalysts in my Ph.D when dinosaurs ruled the earth with both XRD and XPS - indeed the 7th ESCA/XPS system in the world...). In XPS you're looking at maybe the top surface 10 - 15 atomic layers maximum only which is invariably fully oxidized and thus Pd2+ will show strongly from the shift in eV. You'll only start to 'see' the bulk if you (for example) argon ion etch the system. With XRD (dependent on the sample and element) you'll have a great deal more penetration (perhaps, in theory, up to a micron or more). In this latter case you'll see more of the bulk Pd0 metal. In XRD you can distinguish the lattice spacing of Pd from PdO (as Gerhard Martens states) and not the oxidation states. One other technique that would be extremely valuable to you is TPR (Temperature Programmed Reduction). If you have the ability to reduce the system (by hydrogen) in your XPS set up then you'll easily be able to remove the oxide and see the Pd metal reflections only.
Dear Alan F Rawle ,
the penetration/ information depth issue is the reason why I asked above for more details about the XRD experiment.
With grazing incident XRD (GI-XRD) close to (or even in) the total reflection regime one has penetration depths down to a very few 10nm,
but still a lot of atomic layers...
Gerhard Martens
Yes, you're correct. I suspect a powder diffraction sample though for this type of material - Debye-Scherrer camera (maybe Scherrer equation for crystallite size too). But let's wait and see if more details are forthcoming especially a diffractogram. I have a Ph.D's worth of experience on such samples! But in the 1970's when things were very different computer wise - all interpretation (both XRD and XPS) by hand...
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