What is the favoured geometry of chlorine trifluoride?

[![enter image description here](https://i.stack.imgur.com/pJMQs.png)](https://i.stack.imgur.com/pJMQs.png)


Why is the first structure on the left more stable than the second one on the right?


In the first one the angle between the two pair of nonbonding electrons is about 120° which is a lot less than that in the second structure. And it is also very well known that **lone** **pair**-**lone** **pair repulsions** are worse than **bonding pair-bonding pair repulsions**. So why is the first structure more stable?




**Short Answer**


The structure on the left is "preferred" because the structure on the right cannot exist. You cannot put 4 electrons in a p-orbital.


**Detailed Explanation**


In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized. This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons or bond to a ligand. Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.


The drawing on the right has two lone pairs placed in one p-orbital. **A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.**


**The question really becomes**, does $\ce{ClF3}$ prefer to exist with


* 2 axial fluorines and 1 equatorial fluorine


or


* 1 axial fluorine and 2 equatorial fluorines?


We can use [Bent's rule](https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule/15622#15622) to answer this question. Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital). An electronegative substituent will draw electron density away from the central atom. Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.


Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals. **Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.**


**A final note on the bonding in $\ce{ClF3}$**, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a [hypercoordinated](https://en.wikipedia.org/wiki/Hypervalent_molecule) or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons. See these earlier answers for more detail on hypercoordinate bonding.


* [Why does F replace an axial bond in $\ce{PCl5}$?](https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544)
* [How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?](https://chemistry.stackexchange.com/questions/19433/how-can-iodine-bond-with-5-fluorines-in-iodine-pentafluoride/19437#19437)
* [Hypervalency and the octet rule](https://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule)


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Using the usual ["AXE method"](https://en.wikipedia.org/wiki/VSEPR_theory#AXE_method) all molecules of $\ce{AX3E2}$ type, where


* $\ce{A}$ represents the central atom;
* $\ce{X}$ represents the number of ligands (atoms) bonded to A;
* $\ce{E}$ represents the number of lone electron pairs surrounding the central atom;


are assumed to be [T-shaped](https://en.wikipedia.org/wiki/T-shaped_molecular_geometry). I think it does not matter what the central atom $\ce{A}$ and ligands $\ce{X}$ are.


[![enter image description here](https://i.stack.imgur.com/jKqBi.png)](https://i.stack.imgur.com/jKqBi.png)


OP is right in that lone pairs repulsion is smaller in the second structure, however, we have to consider repulsions between lone pairs and bonding pairs as well. Yes, generally, lone pair - lone pair (lp-lp) repulsions are greater than lone pair - bonding pair (lp-bp) and bonding pair - bonding pair (bp-bp) ones, but the later could not be just neglected.


For the first structure we have:


* One lp-lp repulsion at 120°;
* Two lp-bp repulsions at 120° (one bonding pair in the same plane interacts with two lone pairs);
* Four lp-bp repulsions at 90° (two bonding pairs out of plane interacts with two lone pairs);
* Two bp-bp repulsions at 90° and one at 180°.


For the second structure we have:


* One lp-lp repulsion at 180°;
* Six lp-bp repulsions at 90° (three bonding pairs out of plane interacts with two lone pairs);
* Three bp-bp repulsions at 120°.


So, in the second structure the lp-lp repulsion is smaller, but the overall lp-bp repulsion is bigger while the overall bp-bp repulsion is basically the same. Well, it is difficult to say *a priori*, will the bigger lp-bp repulsion outweigh the smaller lp-lp repulsion, but it assumed in VSEPR theory that it will, so the first structure is more favorable.


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