SOLVED: Geometry SO2, PF3Cl2, BrF5, ClF3. Draw the Lewis structure, calculate the formal charges, and find the geometries.

Okay, so draw the louis structure and calculate the former charge. So first we need to write down the formula to calculate the former charge. The former charge actually equals to the number of the valence electrons surrounding the standard atom minus the number of the non bond in valence, electrons right and then minus the number, the bonds over 2 point. So now, let’s look at each of them right. The first 1, which is as 2 point so in terms louis structure, the sulphur, is in the sender and forms a bond like this so 12341234 point and the sulphur it has 6 electrons, so actually has another lump hair electrons so to calculate the former charge in This case we find for sulphur. The f c equals to the venus electron right. Sarved has 6 venus electron minus the number of non bonding electron, the true right, true non bonding electrons and the miners that bonding there are 4 bonds were 2 double bond. So there are 4 bonds 4 over 2 point, so the former charge is 2 point for option 2. I’M the same same. The option has 6 valence electrons and has actually 4 pair electrons and the mines has 2 right. True bonds. Oh sorry, i’m wrong here. So the bee is bonding electrons over 2, so they are actually 1 to 48 bonding electrons. So in this case it’s a bonding electrons, so the formal charge actually is 0 and here for oxygen. It has 4 bonding electrons or 1234 and over 2 point. So, in this case, the former charge is also 0 and we can look at the second case, which is a p f, 3 cl 2 pot. So for the s o 2, it actually is a band angle ray the geometry. Is a band angle, m, p, f, c: cl 2. If we work on that, you would find the phosphors actually form 5 bonds and…