We can explain why the bond angle of $\ce{NF3}$ (102°29') is lesser than $\ce{NH3}$ (107°48') by the VSEPR theory, since lone pair lone pair repulsion is greater than lone pair bond pair repulsion. Then for $\ce{PH3}$ and $\ce{PF3}$, also, it is expected that the bond angle of $\ce{PF3}$ will be smaller. But I found that the thing is just reverse. Why is this so? Is it an exception?
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3$\begingroup$ I don't see how VSEPR can explain the bond angles in $\ce{NF3}$ and $\ce{NH3}$. Take a look at this question for the correct reason. $\endgroup$– bonCommented Aug 21, 2015 at 17:52
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1$\begingroup$ related: Bent's rule. See chemistry.stackexchange.com/q/15671/4945 and chemistry.stackexchange.com/q/15620/4945 $\endgroup$– Martin - マーチン ♦Commented Aug 21, 2015 at 18:31
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3 Answers
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VSEPR works by accident in the cases of $\ce{NH3}$ and $\ce{NF3}$. In reality, there are much more things to consider as shown in this answer. All four compounds should have bond angles of $90^\circ$ if there were no other effects present.
For both nitrogen compounds, the effects are the short $\ce{N-X}$ bonds which lead to steric clash of the three substituents. Thus, the bond angle is broadened from the original, hypothetical $90^\circ$ to whereever it may seem comfortable to the molecule. Since fluorines are larger atoms than hydrogens are, the $\ce{N-F}$ bond is longer than the $\ce{N-H}$ bond ($137~\mathrm{pm}$ versus $102~\mathrm{pm}$) and therefore the atoms can agree on a smaller bond angle (less crowding).
For phosphorous, we can initially assume that the bond angle of $\ce{PH3}$ is close enough to $90^\circ$ for no hybridisation to be necessary (semi-proven by this answer of Martin) and likewise for $\ce{PF3}$. The bond lengths are $142~\mathrm{pm}$ and $156~\mathrm{pm}$, respectively. The difference in bond lengths is only half of that of the nitrogen compounds ($14$ versus $35~\mathrm{pm}$). Therefore, we can assume the bond length to be mainly dictated by the central phosphorous atom, while the bond angles must be dictated more by the outer atoms. Since again fluorine is larger than hydrogen, this time the better way to stabilise the molecule is not to lengthen the bonds but to expand the angle.
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$\begingroup$ Whats the reason to conclude "dictated by the central phosphorous atom, while the bond angles must be dictated more by the outer atoms. " Isnt bond length greater due to size of F greater than H? P remains same for both $\endgroup$ Commented Mar 29, 2022 at 17:46
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The answer can not be determined using VSEPR theory. The number of lone pairs in both compounds is the same. The answer lies in electronegativity.
In $\ce{NH3}$ nitrogen is more electronegative than hydrogen and therefore it will pull the electrons towards it. Therefore the electrons will be closer to each other hence more repulsion.
But in $\ce{NF3}$ fluorine will pull electrons towards itself therefore the electrons are further comparatively and hence repulsion will be less and so is the bond angle.
But in case of $\ce{PF3}$ and $\ce{PH3}$ back bonding is possible in $\ce{PF3}$ therefore it has a larger bond angle .
answered Aug 21, 2015 at 18:52
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1$\begingroup$ What type of backbonding are you envisaging? $\endgroup$– bonCommented Aug 21, 2015 at 20:21
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$\begingroup$ Fluorine is also larger than hydrogen so sterics have a contribution. $\endgroup$ Commented Aug 21, 2015 at 20:36
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$\begingroup$ Can you please explain why back bonding makes the repulsion increase in $PH_3$? $\endgroup$– AneekCommented Aug 25, 2015 at 16:37
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1$\begingroup$ Actually it forms a partial double bond between P and all the three flourine . $\endgroup$ Commented Aug 25, 2015 at 19:54
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8$\begingroup$ Phosphorus does not access d-orbitals. $-1$ $\endgroup$– JanCommented Nov 11, 2015 at 14:53
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In PH3 and PF3 bond angle of PF3 is greater as in PF3 back bonding takes place. Back bonding is possible in PF3 as P has vacant d orbital(as its atomic no. is 15 therefor its electronic configuration is 1s(2e) 2s(2e) 2p(6e) 3s(2e) 3p(3e) this shows it has vacant 3d orbital which can perform back bonding with fluorine as fluorine has pair of electrons. as backbonding takes place the lonepair of P will be used in doublebond formation with F. and obviously if their is no lonepair then their is no one to repell. there for more bond angle also in PH3 no backbonding can take place as hydrogen has no pair ofelectron(e) for backbonding to take place. May this solution help you.:)
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$\begingroup$ That's not true - significant usage of d-orbitals in p-block element's compounds is outdated theory $\endgroup$– MithoronCommented Jan 8, 2016 at 13:07
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$\begingroup$ lol this is the logic outdated theory all concepts of chemistry are years old and you creating your own theories I would say go for it $\endgroup$– vedantCommented Jan 8, 2016 at 13:19
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$\begingroup$ Chemistry marches on see for ex. chemistry.stackexchange.com/a/18544/9961 for newer description, as this was experimentally disproved. $\endgroup$– MithoronCommented Jan 8, 2016 at 15:17