For each of the following molecules that contain polar covalent bonds, indicate the positive and negative ends of the dipole, using the symbol ⇸ . Refer to Table 1.3 as needed.
So, for this problem we need to go ahead and figure out the direction of the dipole of each of these molecules, so what I'm going to do is actually draw out our chem 7 to give us a visual aid. So, see your CNOF, Cl Br and iodine, let's fix that a little bit, okay? So, we know that our electronegativity increases as we move up and to the right of the periodic table and we don't actually need to know more values for this as until the very last one, okay? we also know that our hydrogen is way over here to the left, So, which one do you think is more electronegative hydrogen or chlorine according to the flow to the trend that we have here we know that it's definitely chlorine. So, our dipole is going to be pulled to the right, okay? Now, what about HI same thing it's a halogen, right? You can assume that halogens are very electronegative, we've actually got it pulling to the right water? Well, we've got oxygen, we've got hydrogen, oxygen is one of the most electronegative atoms and our hydrogen is going to be pulling, our dipole is going to be pulled this way and this way. Okay, we're going to be pulling toward that oxygen. Now, for this four HOCl, right? You might recognize this as part of bleach, what's going on here is our oxygen actually has an electronegativity value of 3.5. Notice that our fluorine is about, is our most electronegative atom and each of these is a space away, right? So it's kind of hard to tell but chlorine has an EN value of 3.0. So, actually oxygen is more electronegative than chlorine, so we'd expect to see our dipole pulling in this direction towards that oxygen as well, just not as strong, the dipole won't be as strong as the difference between H and O versus Cl and O, cool? So this is the answer to our problem, let me know if you have any questions.