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I am trying to figure out the NMR spectra of F3P-BH3

I am solving this out using 2nI+1 rule.
IP = 1/2 and IB = 3/2

Hence for this compound,

(2nIP+1)(2nIB+1) = (2x1X1/2+1)(2X1X3/2+1) = (2)(4) = 8 lines (quadraplet of doublet)

This is what I came up with, it would be great if you could correct me if I'm wrong. Trying to solve few problems before the exam =)

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I think you have got the calculation right. However, I would also consider the coupling between F and H because coupling through 3 bonds is also usually visible. So, each of your 8 lines will be split into four by the H-atoms.

That will be the major part of the spectrum. You would also get weak signals with a different splitting pattern due to the other isotope of boron ($\ce{^10B, I=3}$, abundance=20%).

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  • $\begingroup$ Oh yeah right. Thank you! $\endgroup$ – Rehearse Jan 22 at 14:45

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