(c) The measured dipole moment of BrCl is 0.57 D. If you assume the bond length in BrCl is the sum of the atomic radii, what are the partial charges on the atoms in BrCl using the experimental dipole moment?

Video transcript

Welcome back everyone in this example, we have the compound iodine bromide which has an experimental dipole moment of 0.70 Dubai's assuming that the bond length of iodine bromide is a sum of the atomic radi, determine the partial charges on the atoms in aydin mono bromide using experimental dipole moment. So we want to recall our electro negativity trend on our periodic tables which is increasing as we go from the bottom left towards the top right of our periodic table. And we also want to make note of the specific electro negativity values attributed to our atoms in our compound iodine mono bromide recognize that for the election negativity of iodine We have a value of 2.5. And for the electro negativity of bromine, we have a value of 2.8 and we want to make note of the fact that this is a covalin A. K. A. A molecular compound because these are both nonmetal atoms based on their position on the periodic table. And so they're going to be sharing these electrons. But unevenly based on these differences in electro negativity between our atoms. And so we can see that we have a higher electro negativity associated with our bromine atom, meaning that the electrons of our bromine atom are closer in the bond towards the roman atom. Now we know we want to recall our formula to determine charge of an atom. And so we would recall that to calculate the charge of an atom, recall that that is represented by Q. And it's set equal to the term mu in the numerator, which is then divided by our radius and recognize that mu here refers to our dipole moment of our bond. Whereas in our denominator we have the term R. For radius. Sorry that's radius within you. And this is referring to the ready of our atoms according to the covalin bond between them. And so we want to make note of specifically our covalin radius is first of our iding atom which we would see Is equal to a value of 1.39 ampules when we look this up in our textbooks or online. And then in terms of our Adam Bro mean it would have a co violent radius of a value of 1.20 when we look it up online or in our textbooks and the units are ampules here or sorry, Angstrom. So what you should imagine is that in our denominator since this is the radi I of our atoms making up our Covalin molecule, we want to actually get the sum here of these two Covalin Grady. And so we would say that the value for R is equal to 1.39 angstrom added to 1. angstrom. And this is going to give us a sum of 2. angstrom. And so now that we know our value for the radius of both of our atoms. We can calculate the dipole moment of our bond by taking the experimental dipole moment of our compound and dividing it by the radi of our atoms. And so we would say for our numerator that to calculate mu and will actually make it the color purple here. To calculate mu. We're going to take our Experimental dipole moment from the prompt of 0.70 Dubai's. So this d unit is units of Dubai for dipole moment. And we're going to divide these 0.70 Dubai's by our atomic red eye that we calculated of two .59 Extremes. And so This is going to give us a value equal 2.270- Dubai's per angstrom as our dipole moment of our covalin bond. And I didn't wanna bromide. And now we want to use this association to get to our charge of our atoms which we should recognize is going to be in units of electrical charge. So that would be units of just lower case E. Units for charge can also be columns. But in this example, because we're determining the charges of our atoms, we want them to be units of electrical charge. So getting into our calculation for Q. We're going to begin with our dipole moment of our bond, which we calculated to be .270- Dubai's per angstrom. And we want to again end up with lower case E. Or electrical units as our final unit. So we're going to multiply by our first conversion factor to get rid of that unit angstrom. So we're going to recall the conversion factor that one angstrom is equivalent to 10 to the negative 10th power meters. So now we can go ahead and cancel our units of angstrom. And now that we're at meters, our next focus is to cancel out meters by recalling the conversion factor that we have, where one Dubai unit is equivalent to our column meter of 3.34 times 10 to the negative 30th power columns, times meters. And so you can see that our units of Dubai's are aligned appropriately so we can cancel out our Dubai units as well as our unit of meter. And now we're left with plumes, which we need to cancel out to get to our final term being electrical units. So we're going to multiply by our last conversion factor to go from columns to electrical units. By recalling the conversion factor that for one electron We have an equivalent of 1.60 times 10 to the negative 19th power columns. And now we can finally get rid of that column term leaving us with electrical units or electrons as our final unit here for charge. And we would calculate in our calculators that our charge is equal to a value Of 0.056 electrical units. And so now that we have the value of our charge of both of our atoms in our molecular compound here. As we stated above, we have a greater electro negativity associated with our Adam Bromine, which has a value of 2.8 as its electro negativity. And it's also further up higher in period four of our periodic table around here, meaning it follows the trend of a greater electro negativity. And because of that, as we stated, it's going to share more negative character between the electrons, making up our covalin bond between bromine and and our adam idea. So we're going to say that we have a -0.056 Charge on bro mean and a positive 0.056 electrical charge on iodine. And sorry, this should say electrical charge. So I hope that that's clear because again, bromine is more electro negative. So it's going to bear more of the negative character in the co valent bond between bromine and iodine because iodine is more is less electro negative, it bears more of the positive character of the electron charge. And so for our final answer, we have calculated the two charges of our atoms making up our iodine bromide molecule. So I hope everything I went through is clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video