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Correct order of dipole moment in the following strucrures is(a) $mathrm{I}=mathrm{II}=mathrm{III}$(b) $mathrm{I}<mathrm{II}<mathrm{III}$(c) $mathrm{JI}<ma...

Question

Correct order of dipole moment in the following strucrures is(a) $mathrm{I}=mathrm{II}=mathrm{III}$(b) $mathrm{I}<mathrm{II}<mathrm{III}$(c) $mathrm{JI}<mathrm{III}<mathrm{I}$(d) III $<$ II $<$ I

Correct order of dipole moment in the following strucrures is (a) $mathrm{I}=mathrm{II}=mathrm{III}$ (b) $mathrm{I}<mathrm{II}<mathrm{III}$ (c) $mathrm{JI}<mathrm{III}<mathrm{I}$ (d) III $<$ II $<$ I



Answers

Correct order of dipole moment in the following strucrures is (a) $\mathrm{I}=\mathrm{II}=\mathrm{III}$ (b) $\mathrm{I}<\mathrm{II}<\mathrm{III}$ (c) $\mathrm{JI}<\mathrm{III}<\mathrm{I}$ (d) III $<$ II $<$ I

The direction of polarity of the Dye poll will always go towards the most electro negative Adam. And to figure out which is the more Electra negative. You can easily look at a periodic table, which should be in your book. Or you can easily look up and alert Your negativity increases going left to right on the product table in increasing, starting at the bottom, going up a column on the product table. And so if we use our periodic table, we'll see that scoring is a lot more Electra negative than hard region. Therefore, therefore, the result in Die Poll will go towards the flooring and hard region and chlorine. Again. We see that the result resulting die poll goes tours to chlorine because it is more Let your negative negative, as you can see on the periodic table again with hydrogen and bro Mean romain is more Electra negative than hard region. So again, the clarity will be going towards Rumi and then again with heart surgeon in iodide. Iodine is a lot more allegedly negative than Hodgdon so again that our pool will be going the same way

So in this video, we're gonna go over a question 144 from Chapter eight, which says which of the following molecules have not die. Pull moments for the molecules that are polar indicate the polarity of each bond in the direction of the net diaper moment of the molecule. Um, so, in a weird given ch to seal to stage sealed three and cecile for so in ch to seal to our carbon atom is going to bring four valence electrons. Each of our hydrogen atoms brings one and are pouring autumns each bring seven s. So that's a total of 20 electrons to draw Louis structure with. So we draw our to pouring atoms bonded to our central carbon atom, and then our two hydrogen sze. Then we can fill in the octet on our Korean Adams, and that uses a 16 18 20 that uses up all 20 of our electrons in the formal charge on each Adam. And this Louis structure is zero for carbon. It's four minus four. And for flooring at seven, minus six minus one. So it's zero on for all Adams. Um, then for ch deal three are carbon autumn brings four valence electrons. Hydrogen brings wine and the three corn and each bring seven, so we have a total of 26. So we have our three Korean atoms bonded to our central carbon atom and our one hydrogen atom. Then we can fill in our rock tots, and that uses up eight times three plus two electrons, so that uses up all 26 of our electrons. So we essentially have the same structure, but with 11 of our hydrogen atoms replaced with chlorine atom, then in Cecile for carbon brings more surveillance electrons, each pouring out and bring seven. We have a total of 32. We fill in the octet on all four chlorine atoms, and that's going to use up eight times four of our electrons. That's 32 electrons of all of them. So we have our Lewis structures eso In each of these, we have one Adam surrounded by four Adams and zero will impair electrons, so that means we have a touch of federal geometry. So what does that tell us about the polarity? Well, in the case of ch, to see El to, we have to bowler bonds from our carbon tour. Chlorine. Chlorine is more Electra negative than carbon? Um, so we end up with a partial negative on our chlorine atoms and a partial positive on our carbon atom and are met Die Pole moment is pointing in this direction towards our chlorine atoms s o. This is polar. Then we have ch seals three and again. Each of our carbon chlorine bombs is polar. So we have any partial negatives honor chlorine atoms and then a partial positive on our carbon atom. And the not die pole moment is in the direction of our three chlorine atoms in that direction s O that molecules also Kohler. Then we have C zeal for again our carbon chlorine bonds are all polar Barmes. But in this case, they're placed symmetrically in our tetra hydro geometry. Since we have four of them, they're placed symmetrically. And so the overall disciple moment there is no world diaper moment. Um and this is a non polar molecule. So in Cecile for even though we have the same CCL bond that we had in ch to see Ellen ch seal three, the symmetric placement of the carbon chlorine bonds makes this a non polar molecule. And B We're given co two and into O, so our carbon atom is gonna bring four valence electrons and then each of our ox Austin Autumn spring six Valence electrons. So we have a total of 16 electrons to drawer Louis structure with. So our central Adam is gonna be the least elector Negative, Adam. So that's our carbon atom. And then we can draw that bondage both of our oxygen atoms. If we were to just fill in the octet honor oxygen atoms, that would give us a formal charge of minus one on each of our oxygen atoms six minus six, minus one and plus two on her central carbon atoms. So let's not Let's not do that. Um, it'll be better if we draw double bonds to the carpet to the carbon atom. So that way our carbon is making four bond, so the formal charge will be four minus for it'll be zero and then on our oxygen atoms. Once we fill in the OC tats, we have a formal charge of six minus four minus two. So that zero So our formal charges are all zero, and then we have into O s. Oh, it's tempting to make oxygen are central, Adam. But, um, we want to follow the rule that the least Electra. Negative. Adam is usually our central Adam s. So it's actually gonna be nitrogen. Um, so whips one of these is an oxygen, um, so we can satisfy a formal charge of zero for oxygen if we are. If we draw a double bond to nitrogen and fill in our octet, that gives us a formal charge of zero on oxygen. Um, and then we can do the same to the other nitrogen. Um, And what are our formal charges? Well, on our oxygen, it zero on our central nitrogen atom, it's five minus four. So it's plus one. And on the terminal nitrate during Adam, it's five minus for minus two. So it's minus one. Um, you could also have nitrogen. If we move these electrons and and move these electrons out, then we have, um, oxygen with full, locked tight in a single bond and a triple bond to nitrogen. Now, our formal charges are, um, are minus one zero and plus one. So those air to possible resident structures for our into O. R. into a molecule. What about the polarity of these bonds? Well, um, are nitrogen? Nitrogen bond is obviously not a polar bond, but our nitrogen oxygen bond is a polar bond. Our oxygen atom is more Electra negative than our nitrogen Adam. So because of that, this is a polar molecule. We have to bonds pointing in opposite directions. And one of those bonds is non polar, and the other bond is polar. So it's a polar molecule with a net die pole moment pointing towards the oxygen atom. Then in carbon dioxide, we have two polar bonds. We have two carbon oxygen bonds. And because of the Electra negativity difference, oxygen is more Electra negative than carbon on these bonds or polar bonds. But in this linear molecule there pointing in opposite directions so they cancel each other out. There is no net die. Pull moment. Eso carbon dioxide CEO too is non polar. So go to was non polar but into oh was polar. Next we have Ph three and in H three s o. For each of these, both phosphorus and nitrogen bring five valence electrons and then our three hydrogen atoms each bring one valence electrons over both of these. We have eight electrons to work with when we're drawing the Louis structure. So I can go ahead and draw my central phosphorus Adam and then three hydrogen atoms bonded to it. That uses up six of my electrons. Just drawing this three bonds and I have two left over to put on the central atom. That's pretty much the only structure we could have drawn here. So we do the same thing or ammonia for in h three, and we get a loan. Pair of electrons and nitrogen. Um, so what is this molecular geometry going toe look like? Well, we have one Adam with three atoms bonded to it and one lone pair of electrons around it that we didn't have that loan. Pair of electrons, we'd have a tribunal plainer structure. But because we have that loan pair of electrons that slightly displaces the other bonds, um, we end up with this tribunal. Pure middle structure. Um, so we have a tribunal pyramidal structure. Um, and interestingly enough, both bosphorous and hydrogen have about the same Electra negativity. So these bonds aren't really polar, but we still end up with a polar molecule because of the presence of this non bonding electron pair. This'll lone electron pair on our prosperous Adam. That gives us in that diaper moment, pointing towards the towards the electron pair and that makes this molecule polar and the same story for ammonia foreign. Age three we have we have this, uh, electron pair that causes us to have a net disciple moment making the making the molecule bowl.

So to determine of thieves have die pulled dimple interactions. We need to determine whether or not they are polar but CEO to we're gonna look at our Louis structure And there are these lone parents. But around the central carbon, it's gonna be a linear, um, the linear ah, linear molecule. And so this is symmetric and non polar. So it's non polar, does not, does not have a type old eyeball. The other ones are a little trickier. Ano too. If you draw it out, it's gonna be, um, a double bond. And this has resonance. But around our central nitrogen, we can see that there are three electron domains. There's, ah, it's bondage to oxygen's and then has this lone pair. And so that's gonna give us a bent geometry, which is gonna mean it's polar and does have a type hold I pull s 02 is gonna be similar. And if you need to review your, um, Louis structure rules definitely do that in order to be will answer these questions. But this is gonna be a polar molecule as well, because we can see how it's ah, how it also has a bent geometry polar and that is going to give it a type, holds I poll. And then the last one is H two s, which now there are two lone parents around sulfur, and that's gonna even give us another bench geometry and therefore polar di pulled I people. So for determining Diebold's, I'm pulling you to figure out whether or not it's polar on and then that will help determine whether or not taken form a diable die pool.

So now on these molecules which will just be drawing them out and adding some die polls. So firstly, we're looking at sis CHF double bond C H f. So I've drawn out my SB two hybridized carbons. My plane A structure where are carbons? Have three atoms bounce one another because we have to account for that double bond now, because we're drawing the CIS confirmation, the same act ums will reside on the same side and as a result, this will be a polar molecule and we have die poll in the direction of our more election negative atoms which are are florins. So moving on to the Trans, I'll just undo some off the working I've done in order to alter the structure that so now we are drawing the same atoms in the opposite confirmation on Due to this arrangement off, our atoms are equally opposing. Die polls between our two atoms council one another out and that makes this molecule not die poll over all. So this is not Pola. So now let's look at our third example We've got C double bonded to another carbon again. We just got a recipe to hybridize senses that. And we have two protons on one side, two florins on the other. And because of the arrangement of our atoms, we have a General Di poll in the direction of armor Electra Negative atoms, which is our florins. The last example. Here we have four florin atoms and again, like all Trans example because we have the same die polls that are equal in volume. However operating and in opposite directions, that means that we have no overall die Polt within this molecule. So this molecule is not polar. So just to recap my fast on third molecule are Pola whereas my second on fourth are not polar.


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