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I had to draw the molecular shape for $\ce{NHF2}$ as a Trigonal Planar like this
NHF2 Structure Diagram

But when I check to make sure it was right, I came across a source saying

NHF2-difluoroammonia--tetrahedral, almost trigonal (Source 1)

After looking some more I found a source that now said it was

Trigonal pyramidal, Bond angle of 107 degrees Bond angle of F-N-F > F-N-H (Source 2)

Can someone clarify which one it is so I can understand.

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  • $\begingroup$ Consider the presence of an electron pair on nitrogen; that should guide you to predict the shape of such as molecule. $\endgroup$
    – mannaia
    Feb 27, 2015 at 7:26

2 Answers 2

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You can look up the molecule on chemspider, where you have a little applet for the 3D structure. Or you can download a coordinate file from NIST and view it in a molecular viewer, like Avogadro. Or keep on reading for some deeper insight.

As Klaus already pointed out, if VSEPR is a valid concept, one would arrive at the conclusion, that the molecule is trigonal pyramidal, like ammonia. I performed a DF-BP86/def2-SVP calculation on this to analyse the electronic structure a bit more.
molecular structure of HNF2    enter image description here
One reason why VSEPR does fail (a lot) is that lone pairs are more stable in orbitals with high s character. Another point that is often observed is that orbitals with higher p character are directed to more electronegative elements (Bent's Rule).[1,2] As a result of that, the bond angle between electronegative elements gets smaller. This can be observed here, too. From the ideal sp³ arrangement we would expect an 109.5° angle, but here it is about 5° smaller. As a consequence also the hydrogen-nitrogen-fluorine angle has to decrease. In this case the bond angle becomes much smaller because of an intramolecular hydrogen bond.[3]
The NBO analysis consequently builds the same picture: $$\begin{array}{llll}\hline \text{type}&\text{main contribution}&\text{other contributions}\\\hline \sigma\text{-Bd}~\ce{N-F}& 64\%~\ce{F}~(14\% s + 86\% p) & 36\%~\ce{N}~(15\% s + 85\% p) \\ \sigma\text{-Bd}~\ce{N-H}& 68\%~\ce{N}~(25\% s + 75\% p) & 32\%~\ce{H}~(100\% s) \\ \sigma\text{-LP}~\ce{N} & 100\%~\ce{N}~(52\% s + 48\% p) & \\ \sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(75\% s + 25\% p) & \\ \sigma\text{-LP}~\ce{F} & 100\%~\ce{F}~(10\% s + 90\% p) & \\ \sigma\text{-LP}~\ce{F} & 97\%~\ce{F}~(100\% p) & 2\%~\ce{N}~(7\% s + 93\% p) & 1\%~\ce{F2}~(15\% s + 85\% p)\\\hline \end{array}$$

The planar (trigonal) structure on this level of theory is about 33 kcal/mol higher in energy. It is the transition state for the nitrogen inversion, which is therefore very slow.

Notes:

  1. Bent's rule on wikipedia
  2. Bent's rule on our network: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?
  3. The bond length is much smaller than the sum of the corresponding van der Waals radii, suggesting a hydrogen bond. Compare A. Bondi J. Phys. Chem., 1964, 68 (3), pp 441–451.
    $r_{\mathrm{vdW}}(\ce{HF})=r_{\mathrm{vdW}}(\ce{H})+r_{\mathrm{vdW}}(\ce{F})= (106 + 140)~\mathrm{pm}= 246~\mathrm{pm}$
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  • $\begingroup$ what programs did you use in the calculation/analysis for this post? $\endgroup$
    – permeakra
    Feb 27, 2015 at 10:38
  • $\begingroup$ @permeakra Gaussian 09, rev. D.01 and the built-in NBO 3.1; pictures are chemcraft. $\endgroup$
    – Martin - マーチン ♦
    Feb 27, 2015 at 10:46
  • $\begingroup$ I favorited the question because of this answer! $\endgroup$
    – M.A.R.
    Feb 27, 2015 at 12:24
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    $\begingroup$ +1 Martin's work would have been a nice publication a few years back: pubs.acs.org/doi/abs/10.1021/ja00427a013 $\endgroup$
    – DavePhD
    Feb 27, 2015 at 13:31
  • $\begingroup$ @DavePhD Haha, that's true, but a 'few' years back you would not have done it in an hour ;) Thanks for the link, I was looking for a publication, but not quite able to find one in the allotted time. $\endgroup$
    – Martin - マーチン ♦
    Feb 28, 2015 at 13:05
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Assuming for a moment that VSEPR is a valid concept, difluoramine ($\ce{NF2H}$) is represented by $\bf{AX_3E_1}$, and so is ammonia ($\ce{NH3}$). Try to walk your way down from there.

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