# Lewis Structure of OF+

Would the Lewis Structure for OF+ be drawn so that the more electronegative atom has the more negative formal charge, or would it be drawn to satisfy the octet rule?

Here are my thoughts thus far:

For one, we could make the formal charge on the oxygen +1 and the formal charge on the fluorine 0, and give them a single bond, which would break the octet rule, but make sure that the more electronegative atom (fluorine) had a less positive formal charge. Oxygen would have 4 free electrons, and fluorine would have 6 free electrons.

However, we could also make the formal charge on the oxygen 0, and the formal charge on fluorine +1, and give them a double bond, which would be wrong in terms of electronegativity, but would mean full octets. Oxygen would have 4 free electrons, and fluorine would also have 4 free electrons

How do I decide?

• Carbon monixide is commonly rendered with a triple bond that gives the electronegative oxygen a positive formal charge, and its dipole is negative on carbon. Nov 25, 2019 at 1:12

## 2 Answers

A strategy different from the one described by Yusuf Hasan would be to look at the neutral isoelectronic molecule. The oxygen atom is isoelectronic to the fluorine cation, so $$\ce{OF+}$$ is isoelectronic to $$\ce{O2}$$.

It is difficult to write a nice Lewis structure for $$\ce{O2}$$, see https://chemistry.stackexchange.com/a/15061. Experiments show that dioxygen is paramagnetic, i.e. it has unpaired electrons. If you try to write this as Lewis structure, you end up with 7 electrons on each oxygen (counting electrons in bonds twice), i.e. no octets either.

For $$\ce{OF+}$$, you should be able to write the same Lewis structure (and the same MO diagram). Because fluorine has a different number of protons than oxygen, however, the energy levels will be different and "the best" Lewis structure might be different as well.

As Yusuf Hasan mentioned, the struggles to write a "nice" Lewis structure for this species hints at high reactivity, so I think this is the main lesson of trying to write a Lewis structure. Which of the Lewis structure is best at approximating the real situation becomes secondary, and it makes sense to move on from a Lewis structure model to more comprehensive models of bonding for species such as this one.

• Thanks for mentioning about the MO diagram,I was a little skeptical that I could draw it exactly parallel to O2. However,you would still admit that the π* antibonding orbitals are going to destabilize the π bonding orbitals,and so a π bond in the Lewis structure is going to be a little less favored? Nov 23, 2019 at 22:50
• @YUSUFHASAN Yes, I think there are at least two strikes against the double bond. Nov 23, 2019 at 23:00

A good way to think about stuff like this, is to imagine a possible synthesis for the given intermediate. One way of making this species would be to cleave the H-OF bond of HOF against it's electronegativity, that is, heterolytically cleaving the bond to make H- and [OF]+

As we are going against electronegativity, you get the idea that the intermediates are quite unstable, and so your structure must represent a reactive species

Then, we can simply find out the structure by the curly arrow diagram, assuming that making the structure of HOF is easier than [OF]+

Another reason why your double-bonded structure won't work is that due to the close electronegativities of O and F, the MO picture would probably resemble that of O2 and F2, and so the pi bonding orbitals would experience at least some cancellation by filling of the antibonding pi orbitals(and even the O-F bond would be destabilized a little by the so-called 'antibonding effect')

As a final note, if you are not willing to evoke MO theory, then you can simply say that due to small size and close electronegativities of O and F, a pi bond will mostly not be favorable between them.

• @JoshuaRonis What I meant was, OF+ has 16 electrons, so it's MO will resemble O2 with just the MOs rearranged by electronegativity(as Karsten pointed out below). Check out MO diagram of O2; since electrons enter into pi* orbitals there, hence the pi bond of O2 weakens, moreover the sigma bond also weakens due to the 'antibonding effect'(check it out as well). Happy to help :) Nov 26, 2019 at 20:14
• Explanation for the downvote would be appreciated Aug 8, 2020 at 15:32
• what do you mean? Aug 8, 2020 at 15:51
• @joshuaronis Oh sorry, I wasn't addressing you, actually, somebody has downvoted my answer some time ago, so I wanted to ask the reason from whomever it was. My bad :) Aug 8, 2020 at 16:15