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Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory

Chapter 11, Problem 51b

Determine whether each molecule is polar or nonpolar. b. SCl4

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Hey everyone today, we're being asked to identify the polarity of this molecule O N. C. L. Otherwise known as nitrous silk chloride, nature sil chloride. No, I find the easiest way to drawing out or identifying the polarity of a molecule is by drawing out its lewis structure. And the first step to doing this is identifying how many valence electrons we have. Now, oxygen is a group six element meaning it will have six valence electrons. Nitrogen is a group five element. I have five valence electrons. And Chlorine is a group seven element. It is one of the allergens and will have seven valence electrons. This brings our total to 18 valence electrons. So from here we can go ahead and start drawing out our molecule nitrogen is the least electro negative will be in the center. It will be a central atom. And remember the election negativity trend increases from the bottom left to the top right of the periodic table, meaning nitrogen is least election negative, followed by oxygen. And finally by chlorine. So in the molecules so far we already have two four valence electrons. However we still need 18. So we can go ahead Or we still need 14. Sorry, we can go ahead and fill the octet of both chlorine and oxygen. So that will bring us to a total of eight plus eight 16 valence electrons. But we're still missing two. So we can add those as a lone pair around the central nitrogen. One thing you might have noticed by now is that the central nitrogen only has six valence electrons around it. It doesn't have a full octet. This can easily be amended by using one of the lone pairs from the oxygen and making it a double bond here. So now we have eight electrons around the nitrogen and eight electrons around the oxygen. We can further verify that it was proper to take the electrons or rather share the electrons between the oxygen and the nitrogen rather than the chlorine and the nitrogen. By using formal charge. Remember that formal charge can be calculated by the original number number of valence electrons will write that as v minus the number of non bonding electrons around the atom. Or the number of electrons in lone pairs minus The number of electrons and bonds divided by two. So for oxygen as we've drawn it here, This will be six. Original valence electrons -4, not bonded minus the four in bonds divided by two, Which will give us 6 -4 -2 which will equal zero. So now we have the proper form of charge on the molecule. With that in mind we can now go ahead and redraw the molecule how it would be in space due to the lone pair. So you can redraw this as and bonded to C. L. And O fill this out and finally with our lone pair because of the lone pair around the nitrogen we actually witnessed. This sort of repulsion effect that forces the molecule to be bent furthermore, since both oxygen and chlorine are more election negative than nitrogen, we have a polar bond facing this way, and an even larger, sorry, a larger die pool moment going towards the chlorine. So since we have a bent molecule and the dipole moments that are created are unequal to each other, we can say that this is a polar molecule, does a polar molecule. I hope this helps. And I look forward to seeing you in the next one.