fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
Solid-State NMR Interaction
Hey,
I'm just in the preparations for an exam and I have a question that pops up in every exam in the last years. I can't remember that it was mentioned in
the lectures and I can't find any aswer in the power point either:
So you always get a solid state NMR of for example Sodium, Boron, Silicon, whatever ... then you get some coordinations like Atoms in an octahedron or
tetrahedron and you need to match them to the signals. This is not really a problem but then there is always the same question in there.
With Sodium for example:
"Name the dominant solid state interaction of the Sodium-nuclei with the examined Alumosilicate"
The same question exists for Aluminium as well...
Now I am not sure what they want to hear here. There are only few interactions, Dipolar and Quadrupolar ones and I bet they are heading for the
quadrupolar ones since they are new. But on the other hand they are not talking about the interaction between the Spectrometer or the radiation and
the nucleus but the interaction of the nuclei... this is what confuses me a little.
I mean this could also be any interaction between two atoms in a solid state compound even if no NMR is carried out.
And the next question is to name a parameter for the ( I cannot find a translation for this term anywhere. For Crystals there can be a close structure
and a far structure. Amorphous compounds only have close structures and very perfect crystals have far structures as well). The question is to name a
close structure parameter that defines a strong interaction (interaction is the one from the question above).
Well this confuses me even more. What parameter could this eventually be ? Quadrupolar moments have interaction streghts as well like a coupling
constant. A stronger interaction would cause a bigger coupling constant and the interaction is defined by the coordination since the quadrupolar
moment can act as a gradient. The gradient part becomes bigger for energy gradients in the coordination and coordination is an indication for the
structure.
So maybe they want to here about the structure and coordination around said atom. Which causes a strong or weak quadrupolar interaction.
Any suggestions ? I'd be very thankful for any idea.
[Edited on 26-6-2016 by fluorescence]
|
|
|
Metacelsus
International Hazard
Posts: 2531
Registered: 26-12-2012
Location: Boston, MA
Member Is Offline
Mood: Double, double, toil and trouble
|
|
23Na has a spin of 3/2, so there will be quadrupolar effects (and they will be almost certainly larger than the dipole effects). I'm not
too experienced with solid-state NMR, but you might be able to tell based on line width - doesn't quadrupolar coupling cause a lot of broadening?
[Edited on 6-26-2016 by Metacelsus]
|
|
|
fluorescence
Hazard to Others
Posts: 285
Registered: 11-11-2013
Member Is Offline
Mood: So cold outside
|
|
I think so. Very strong quad. interactions require a lot of special techniques which all have the aim to reduce the width so probably yes. I think
they really want to hear it. We cannot solve them by hand so they will just give us solutions and tell us what it could be and we h ave to descide on
it. And since this is all coordination based and coordination goes with the quadrupolar inteaction it will probably be it.
The question is just a bit strange. In the question itself NMR isnt included. They ask for an interaction in solid states of a nucleus with the rest
of the crystal and parameters this could be much more than NMR.
|
|
|
Quibbler
Hazard to Self
Posts: 65
Registered: 15-7-2005
Location: Trinidad and Tobago
Member Is Offline
Mood: Deflagrated
|
|
It's a strange question. But as you said the obvious answer is the quadrupole interaction.
Problem is as the sodium is an ion in the structure the quadrupole inteaction will be very small. The size of the inteaction is determined by the
nuclear qudrupole moment (Q) and the electric field gradient (efg).
For an ion the electrons around are very symmetrical (filled shell and no bonds) so the efg is very small and determined by the symmetry of
surrounding groups.
The dipole-dipole inteactions will also be small as most of the other nuclei have zero spin (unless there is water of crystallization adding protons
in the mix).
|
|
|
Texium
|
Thread Moved
23-8-2016 at 08:37 |
![[*]](images/xpblue/default_icon.gif){kind=icon}
