Which substance in each of the following pairs would you
expect t...

Question

Which substance in each of the following pairs would you
expect to have larger dispersion forces?
(b) HCl or HI

Accepted Answer

Hello. Everyone in this video, we're given two different molecules which is H. 20. And H. Two S. We're seeing which compound has weaker London dispersion forces. So I first like to go ahead and draw the lower structure and seeing maybe if there's a difference than that and if there's anything obvious about the structure that gives us a way that we have a weaker London dispersion force. So first let's go ahead and actually remember what London dispersion forces are. These interactions is the weakest inter molecular force, meaning that all molecules do have this but we still have a stronger and weaker version of this London dispersion force. So let's start off with interpreting our H 20 molecule here. So hydrogen has one valence electron and we have two items of that. So that equals to two total feels electrons contributing from our hydrogen. Now for Austrian doesn't group 68. So we have six valence electrons. So taking the sum of these two will get a total of eight valence electrons. And now for H two S. Again we have two of our hydrogen. So what times two is two. And again, sulfur also does have six fans, electrons. Two plus six is equal to eight. So these two do have the same lower structure. Then we have hydrogen connected to let's just do auction first, oxygen and another hydrogen and we have two lone pairs on the auction. Same thing for our sulfur here. Rh to us. We have sulfur in the center connected to two high regions and has two lone pairs. So we see here that the lower structure is the same and there's no difference here. Besides the central atom which is sulfur and oxygen. So for our auction here that electro negativity is 3.44 and each of the hydrants has the electro negativity of 2.2. Now, for sulfur, of course we just said the hydrogen is 2.2 and sulfur Xyz 2.58. So these values, I've got them from my own sources. But you could have gone in from your textbook, given to you by Professor or even if you have found it online, just use whatever values that you have. So for here now that bringing these values, we can see that for H20. We do have a bigger electro negativity. The difference between our auction and our hydrogen. So blue, let me say that we have a bigger, I want your negativity and because we have electro a bigger electro negativity, we have stronger London dispersion forces. And this of course is just weaker alone. Dispersion forces were just some flood to L. D. F. So the final answer then, which is asking us, which would have weaker than dispersion forces, is that H two S as weaker London dispersion forces. So this right here is going to be my final answer for this problem

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