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Describe the intermolecular forces that are present in each of the following compounds. Which kind of force would have the greatest influence on the properties of each compound? (a) ethyl alcohol, $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ (contains one $\mathrm{C}-\mathrm{O}$ single bond); (b) phosphine, $\mathrm{PH}_{3} ;$ (c) sulfur hexafluoride, $\mathrm{SF}_{6}$.

This question asks us to look at the hydrogen bond formation in propofol. So propranolol is the structure shown below. We know from past questions that hydrogen bonding would occur at the alcohol group. In this circumstance, hydrogen bonding can also occur at any O. H. Any N. H. In any FH So in this case we're looking at right here on the molecule. So when looking at this, we know that oxygen is more electro negative than the hydrogen. So we'll have a partial negative charge over here and a partial pop groups and a partial positive charge over here. This would allow for an attraction to form with another protein, all molecule, which we can draw in like so with the rest of the molecule off to their in a similar partial positive in the partial negative. So in solution, when these professional molecules approach one another, the partial positive area can associate with the partial negative area on another molecule. So in this we would draw out our hydrogen bond as a dash line to distinguish that bond from the Covalin bond, which we drew in black, say like right there as a solid line. So our hydrogen bond can be marked as so hydrogen bonds would form like that across all of the propane. All molecules, which would allow for propane, ought to have relatively strong inter molecular forces. Thank you.

In this problem, we are asked to look at the different kinds of inter molecular forces experienced by a number of substances, even though a isn't listed on this website. If you look in the book, it is listed as boron. Try fluoride. So I went ahead and wrote that in as well. Starting with boron try fluoride. We know that the structure of this um the geometrical structure of this compound is actually tribunal planner, meaning that all of the fluoride ion, all the fluoride atoms are positioned equidistant and um equal angle angular from one another. So they all are positioned around the boron so that they cancel one another out. Due to this cancellation, even though this bond would induce a die poll, all three bonds combined together cancel out. And so no doubt dipole dipole is experienced because there's no dipole dipole. We know that this compound would only experience London dispersion forces as it has a molecular weight, meaning it has electrons and so in a disciple can be induced when two of these molecules are brought close to one another. Moving on to our next one. Looking here, we can see that we have our alcohol right here. So because of this, we see that we have a hydrogen bonded to an oxygen. This hydrogen bonded to the oxygen would exhibit hydrogen bonding. We would have our partial negative charge there and are partial positive charge here, which would be able to interact with another molecule, stay right there and we would have our hydrogen bond right there. Because of this, this will exhibit hydrogen bonding furthermore, because we have more electro negative atoms over here, this molecule would also experience a die poll. These covalin bonds, the electrons would not be shared equally between the atoms and we would have to die pull this way. And because there's no symmetry in the molecule, this would cause overall molecular diet poll this way. So we would also see dipole dipole moments and then finally, because we have a structure with electrons, we would also see the London dispersion forces Moving on to the next one. We have hydrogen iodide here. While this is not a compound that would exhibit hydrogen bonding, it does have a pretty strong dot. Cool iodine is much more electro negative than hydrogen, which would allow for it to have partial negative charge. Will hydrogen is partial positive charge and it would induce a die pool. And so we would have those disciple disciple interactions and then for the same reasons as before, we would also have London dispersion forces, oops, sorry, ignore that. And then finally looking at the atom krypton, which is one of our noble gases and a non metal because it's a non metal, it will exhibit molecular, a molecular solid structure as opposed to a metallic structure. And so the molecular structure will um involved similar in their molecular forces, like our molecules. And so these items will also exhibit London dispersion forces, and they will experience induced I polls when they come together. Okay, thank you.

Okay, so let's discuss the inter molecular forces in each of these. So I too is a non polar covalin compound. So the only forces that it will have our London dispersion forces. So you'll need to overcome those London dispersion forces in order to change it into a gas. The second compound here is puller. So it has London dispersion forces. They all Covalin compounds do since its polar, it also has dipole dipole attractions. And because of this oxygen right here, it has a hydrogen bonded to it and these lone pairs that's a hydrogen bond. So this also has hydrogen bonds. And then in this last one this is a polar compound, right? It's bent. So all Covalin compounds have London dispersion forces and this one will also have dipole dipole horses, but it doesn't have hydrogen bonds. Okay, the hydrogen has to be bonded to an N. And O. Or an F. So just dispersion and dipole dipole

Okay, so let's discuss the inter molecular forces in each of these. So I too is a non polar covalin compound. So the only forces that it will have our London dispersion forces. So you'll need to overcome those London dispersion forces in order to change it into a gas. The second compound here is puller. So it has London dispersion forces. They all Covalin compounds do since its polar, it also has dipole dipole attractions. And because of this oxygen right here, it has a hydrogen bonded to it and these lone pairs that's a hydrogen bond. So this also has hydrogen bonds. And then in this last one this is a polar compound, right? It's bent. So all Covalin compounds have London dispersion forces and this one will also have dipole dipole horses, but it doesn't have hydrogen bonds. Okay, the hydrogen has to be bonded to an N. And O. Or an F. So just dispersion and dipole dipole