(b) Using these partial charges and the atomic radii given in Figure 7.8, estimate the dipole moment of the molecule.

Video transcript

Hi everyone for this problem. We're being told using the partial charges and the atomic radio. I calculate the dipole moment of sodium chloride. So we're looking for di pool moment here and we're going to be using partial charges and atomic radio I. So let's start off by writing out what is the electro negativity for sodium and chlorine. So our electoral negativity for sodium is 0.9 and our electoral negativity for chlorine is three. So we need to calculate the relative electoral negativity for both. Okay, so our relative electro negativity for sodium is going to be the electro negativity of sodium over the the some of the electro negativity for both. So 0.9 plus three. So this gives us a relative electro negativity of 0.23. And for chlorine we do the same thing. It's going to be the electro negativity of quarrying over the some of the electro negativity ease 0.9 plus three. So we get a relative electro negativity of 0.77. So out of both of these are chlorine is going to be the more electro negative. Okay, so with that being said, if we multiply the relative electoral negativity of chlorine by two electrons We get 1.54. And so that implies if we're comparing that to a neutral chlorine it's . more than a neutral chlorine. So this implies a negative .54 charge on chlorine and a positive .54 charge on sodium. So I'll write that up here. So we have a negative 0. Charge on on chlorine. Let me move this down. This is for chlorine negative .54 and it's a positive .544 sodium. So we used that we calculated that by first calculating their relative electro negativity and since chlorine is more electro negative, we multiplied it by two electrons. A neutral chlorine is one. And we see it here that it was 1.54. So that means it's .54 more than a neutral chlorine. So now that we know the charges, we need to next figure out our ionic radius. Okay, so our ionic radius For both for sodium it is 1.16 Angstrom and for chlorine It is 1. Angstrom. And so the sum of these is going to give us the separation. So our separation is going to equal the sum of these. So we have 1.16 plus 1.67 gives us 2. angstrom. And that's going to be important. So now we can go ahead and calculate the dye pull moment and we can do that using this equation, our di pole moment is going to equal our charge Times our radius. Okay, so we said we calculated our charge to be 0.54 or 0.54. That's our charge. Okay, so let's write that here, zero 5 electrons. Okay, so we need to go from we need to calculate the charge of this. And the way we're able to do that is by Knowing the charge of one electron. So the charge of one electron is 1.60 times 10 to the negative 19 Kalume. Okay. So now our electron cancels and we're in charges of Kalume And we need to go from, we need to cancel out this Kalume and we can do that using our separation. That's an angstrom. So we calculated our separation to be 2.83 angstrom. So we're going to use that here. So we have we're going to multiply this by 2.83 angstrom. Okay. And now we can go from angstrom two m using our conversion one angstrom is equal to one times 10 to the negative 10 m. Okay, so let's make sure our units cancel. We have our angstrom cancels and we're left with Kalume here and meter. So we want to cancel out and go from Kalume meters to die pole moment. Okay, so we can do that using our di pole conversion. So that is one die pole is equal to that looks like a P. Let me fix that one die pole is equal to 3.34 times 10 to the negative 30 Kalume meters. Okay, so now our Kaloum cancels and our meter cancels and we're left with the unit of di pole, which is what we want. Okay. So now we have everything that we need to solve. And when we do, we get a final answer of 7.32 typo Okay. And that is the answer to this problem, and that's the end of this problem. I hope this was helpful.