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There is an increase in bond length of about $40\rm~pm$ from $\ce{NH3}$ to $\ce{PH3}$, but only by about $20\rm~pm$ from $\ce{NF3}$ to $\ce{PF3}$. I understand that the bond length will increase with the larger $\ce{P}$ atom, but I cannot see why the changes are so different when the atom is changing in the same way.

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    $\begingroup$ I'm sorry I don't have time for a fuller answer, but the simplest explanation is that both nuclei in a bond affect the bond length, and it is roughly related to the nuclear charge. Since there are two factor interactions here, the change from H->F cannot be isolated from the change of N->P when trying to figure on trends. $\endgroup$ – Lighthart Apr 25 '16 at 16:46
  • $\begingroup$ Related question with very related answer. Another related answer. $\endgroup$ – Jan Apr 26 '16 at 13:41
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This answer of mine basically leads the way to answering your question but it is focussed on bond angles, not lengths. So let me rewrite it to focus on lengths.

All four compounds would ideally want a set of $90^\circ$ angles which nitrogen cannot achieve due to steric stress; instead, the angles are expanded to $102^\circ$ ($\ce{NF3}$) and $107^\circ$ ($\ce{NH3}$). This seems couner-intuitive, because the smaller atoms require a larger bond angle. But looking at it closely, we realise that the $\ce{N-H}$ bond is simply shorter so the hydrogens are closer together hence more strained and hence further apart. Hydrogen being a very small atom the bond lengths cannot be extended well.

The fluorine atom on the other hand is much larger, allowing not only for a greater standard bond length but also for greater bond length variations. It can extend the bonds when compared to $\ce{NH3}$ to allow for a smaller bonding angle and in turn allow for a larger s-contribution to nitrogen’s lone pair. Lengthening (that is, weakening) the bonds slightly allows for a greater stabilisation on the central atom and thus a more stable overall molecule.

Another important issue is the fact that a smaller bond length in $\ce{NF3}$ would bring the fluorine atoms closer together causing electrostatic repulsion of their lone pairs (although that is what is commonly meant with steric stress). Hydrogen, having no lone pairs, does not have this issue.

In phosphorus’ case, both bond angles are almost perfectly $90^\circ$, so we can expect an almost unstrained molecule. The smaller hydrogen still calls for an equally short bond but the larger fluorine can now come closer making the bonds stronger, because the angles are still near-perfect and phosphorus’ lone pair has a very high s-contribution. In fact, the $\ce{P-F}$ bonds are rather strong; much stronger than their $\ce{N-F}$ counterparts.

Rationalising the entire thing we come up with:

  • The difference between $\ce{NH3}$ and $\ce{PH3}$ ($40~\mathrm{pm}$) is approximately equal to the larger phosphorus atom’s size. (The difference between the empirical atomic radii is $35~\mathrm{pm}$.)

  • In $\ce{NF3}$’s case, the bonds are lengthened by some $15$ or $20~\mathrm{pm}$ to accomodate for steric stress both. This weakens the bonds but stabilises the atoms, especially nitrogen. Also, steric stress between the fluorine atoms is reduced.

  • $\ce{PF3}$ has the bond lengths one would expect.

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  • $\begingroup$ I saw your other answer but could not relate it to bond lengths... but this makes sense now! Thank you. $\endgroup$ – chemicool Apr 27 '16 at 5:36

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