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I found this image on wikipedia:

enter image description here

It got me wondering why the axial bond angles were shorter than the equatorial ones. I found this extract elsewhere on the internet -

Extract: (ww.nrcresearchpress.com)

For $AX_{5}E$ molecules which have a square pyramidal shape the extra repulsion exerted by the lone pair would be expected to cause the neighboring four bonds, i.e., those at the base of the square pyramid to be slightly longer than the fifth bond to the apex. Experimental evidence on molecules of this type is scanty (5). The four square-planar bonds are longer than the single bond to the apex in $SbCl_{5}^{2-}$ but they appear to be shorter in $SbF_{5}^{2-}$.

The extract is from a research paper from 1961. It states that the rule isn't followed in $SbF_{5}^{2-}$, and that there's "scanty" evidence anyway.

Now, given that the paper is pretty old, I was wondering if there has been any new evidence collected on this subject. I essentially would like to understand why the bond length of BrF5 follows this order, and why it isn't followed by $SbF_{5}^{2-}$. I'd appreciate some references as well for evidence and perhaps other researched examples.

Note: This isn't a duplicate of this, since that question doesn't address the deviation in $SbF_{5}^{2-}$.

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    $\begingroup$ If the lone pair repels other electrons in the Br-F bonds by causing the bond angle to decrease to approx 85 degrees, would it not follow that the bond length would also be increased and so reduce repulsion. This lengthening would also reduce repulsion to the axial F atom. With luck you may get an expert answer :) $\endgroup$ – porphyrin May 29 '20 at 7:39
  • $\begingroup$ @porphyrin Thanks, that was helpful. Do you know how to explain the anomaly with SbF5 2−? $\endgroup$ – Nikhil Anand May 29 '20 at 15:30

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