Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur

tetrafluoride monoxide 1OSF42 according to the following

unbalanced reaction:

SF41g2 + O21g2¡OSF41g2

The O atom and the four F atoms in OSF4 are bonded to a

central S atom.

(e) For each of the molecules you drew in part (d), state how many

fluorines are equatorial and how many are axial.

Video transcript

All right. Hi, everyone. So this question says to consider the compound Seoc four which contains an oxygen atom and four chlorine atoms bonded to a central selenium atom determine how many chlorine atoms are equatorial and how many are Axio in the molecule? And here we have four different answer. Choices labeled A through D proposing how many chlorine atoms are axial and equatorial. OK. So first, we have to go ahead and draw out the Lewis structure to get a sense on how many groups can be axial and equatorial. So our first step is to count the number of valence electrons from each atom in the actual molecule. Starting off. Here, we have selenium and oxygen, both of which are in group six A. So they're going to contribute six valence electrons each. In addition to this, we have chlorine which is in group seven A. So thats seven valence electrons. And here there are going to be four chlorine atoms in this molecule, the seven multiplied by four equals 28 total electrons from all of our chlorine atoms in total. And so the sum of 66 and 28 is ultimately going to be equal to a total of 40 valence electrons to distribute in our Lewis structure. So now let's go ahead and start with our general structure or our more basic structure. Our central atom is seino. So lets go ahead and start with one bond, two each atom. So that's a single bond between selenium, each of our chlorine atoms and oxygen just to start off. So here it's always best to make sure that you have the appropriate amount of valence electrons on the central element. So here considering the valence electrons of selenium because it has only five bonds, it has five valence electrons at this point. So here we can go ahead and add one more bond just to go ahead and satisfy its requirements. So here if you recall, oxygen usually has two bonds to make it neutral. So we can go ahead and add a double wand in between selenium and oxygen. And now we can go ahead and consider how many valence electrons are remaining. So at this point, weve used a total of 12 valence electrons to create the six bonds around the central atom. And so 40 subtracted by 12 means that there are 28 electrons that have yet to be added. And so now that selenium, which is the central atom has all of its possible bonds, we're going to go ahead and add lone pairs to the outside atoms to complete their octets. So here chlorine being a halogen is going to prefer three lone pairs. So that's three lone pairs per chlorine atom and oxygen prefers two lone pairs. And so counting the number of lone pairs that we've just added, that's three multiplied by four chlorine atoms. So it's 12 and two more from oxygen equals 14 lone pairs. So 14 multiplied by two is equal to 28 which means that this low structure is complete. So now let's talk about geometry. In this particular case, we have five atoms attached to the central atom and there are no lone pair on the central atom. So our geometry can be represented by the symbol A X five which corresponds to trigonal bi pyramidal geometry. And so here, the way that our Lewis structure is currently drawn can represent this geometry. So no modifications are needed in that sense. So now to go ahead and figure out how many atoms are Axio or equatorial recall that axial atoms are those that are along the vertical axis. Whereas the equatorial atoms are those that are on the side here. So given our current Lewis structure, we have two chlorine atoms in the axial position and two more in the equatorial position. Oxygen is also in the equatorial position. But it's also worth mentioning that another configuration is possible in which the oxygen is axial instead. And so if we swap oxygen with one of the axial chlorine, another configuration is possible which results in one axial chlorine and three equatorial chlorine because oxygen in our other configuration would be occupying the other axial position and there you have it. So based on this, our answer is going to be option C in the multiple choice because we can have either two Axio chlorine and two equatorial chlorine or one Axio chlorine and three equatorial chlorine. And so with that being said, thank you so very much for watching. And I hope you found this helpful.