2
$\begingroup$

I am trying to work out the an unknown Ni(II) complex. I don't known the geometry and I have read that broad lines on the nmr spectrum indicate large compounds with symmetric environments. Does this mean that the complex would have tetrahedral geometry as square planner would be symmetrical and octahedral too large?

$\endgroup$
4
$\begingroup$

If the proton spectrum is unbroadened and not shifted out of the ~0-10ppm range, then I would have to say it is a diamagnetic compound.

It could be low spin square planar. Tetrahedral and octahedral d8 cannot be diamagnetic.

enter image description here

Image source: http://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png

Could also be two nickel atoms in the complex that are antiferromagnetically coupled for a net spin of zero.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.