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Draw Lewis structures and give the other information requested for the following molecules: (a) $\mathrm{BF}_{3}$. Shape: planar or nonplanar? (b) ClO $_{3}^{-}$. Shape: planar or nonplanar? (c) HCN. Polar or nonpolar? (d) $\mathrm{OF}_{2}$. Polar or nonpolar? (e) $\mathrm{NO}_{2}$. Estimate the ONO bond angle.

Answer for the given cotton, we have given assertion is position is only iron carbon Bondi stand. Yeah. One distances and mhm barrow same are equal. Uh huh. And the reason for the given position is Dubai electrons in the cycle. Oh, bend titanium group of mm kerosene, r. D, localized. So here we have option A is the correct option. R S T, correct explanation. Okay. Oh, baby.

So we've got a skeleton of a molecule here. Okay, if we take all of this and added up, Right, so we've got a total of five hydrogen. So we're gonna try to find its valence electrons to start and then it looks like I've got three carbons which each have four valence electrons. And it looks like I've got five oxygen's and then on nitrogen. So I'm going to need a total of 52 valence electrons. I'm gonna go ahead and give everybody a full octet here and then I'll count and see what I have. So we'll see if we have all single bonds or if we're going to need double bonds here. So if you add all this up, You're going to get 56 electrons. So when we have four extra, it says that we need to double bonds. So we're gonna need to redraw this now with some double bonds. Mhm. We'll start with this side here now. The two most likely places to to put a double bond are going to be right in there and to one of 21 of the N. O bonds. Okay, so go ahead and draw this C double bond. Oh here we've got our owes a double bond here. There are a number of there's a number of resonant structures you could draw. This is simply one of them. You don't have to draw this one. There are others that would work as long as you're adding up to the correct number. So now we have a number of central atoms. Okay, so this carbon in this carbon both have four pairs of terminal atoms and no lone pairs. So they will touch your head roll. So the angles involved there are going to be approximately 19 0.5 degrees. Okay, this carbon here in blue You can see has three things around it. So it's going to be tribunal planer. So we can expect that bond angle to be about 120°.. Okay, these two oxygen's here doing green. You can see that each have two terminal atoms and two lone pairs. That means they are bent. Okay. But there's a total of four pairs of electrons. So again, it's about 1095°.. So and then finally, we've got this nitrogen here at the end. You can see it has three sets of electrons around it. So it's going to be tribunal planer. So the angles around that Are going to be 120°..

You're asked to draw the Lewis structures of the following molecules. So starting with part A, we have clf three, um, in to draw the little structure. We're going to count the number Beyonce electrons that we have. So chlorine has seven valence electrons, and each flooring also has seven. So that gives us a total of 28 bounds electrons to use to draw this molecule chlorine After Central Adam wanted to 123 Florian's. Then we will complete the Ark tat of the florins tees up the rest of our electrons. And after we've done this, we still have four electrons left since we've used 24 of our electrons. So the remaining four will become two lone pairs on the central chlorine atom. For part B, we have each to as eat. Um, selenium has £6. Electrons in each hydrogen has one, so that gives us a total of £8. Electrons. Selenium is our central Adam wanted to to hide regions. And then we will use the remaining four electrons to become two lone pairs on our central Adam report. See, we have an H 20 H. We have 14 bones electrons available to us since each nitrogen has five valence electrons, hydrogen, you tired and as one. And the oxygen has £6 electrons. So we will have nitrogen bonded to to hide regions which is bonded to oxygen bonded to another hydrogen. And we will use the rest of our electrons by completing the octet of our nitrogen and our oxygen. Next, we have p of seal three. We count on the move. In with electrons we have is the total is 32 ounce electrons. We have our phosphorus bonded to an oxygen bonded to three Koreans. And to use the rest of our electrons will complete the octet. Uh, the atoms. Now this gives us the structure that is technically correct. However, it has not turned to minimize because we will have a plus one charge on her phosphorus and a minus one charge on our oxygen. And to minimize the charge, we will take one of the lone pairs off the oxygen and create a double bond. And this is okay because phosphorus can expand a doctor on. And this is the form in which all of the atoms air charge minimized. Next, we have ch three siege to PR. You have 20 films electrons available to us. So we will have a carbon bonded to three hydrogen wanted to another carbon which is bonded to two hydrogen and that carbon is bonded to a bro mean And we will complete the octet of the roaming by adding three lone pairs of electrons. And this uses of all of our electrons. Four part F. We have N. C. L three. Hydrogen has £5 electrons, and each quarry has seven, giving us £26. Electrons, nitrogenous or central atom bonded to 12 three Koreans. You will complete the architect of the chlorine, and we still have a two electrons left. So I'll becoming lone pair on the nitrogen. And lastly, we have C H three and H two. We have 14 valence electrons in the molecule, so we'll have a carbon that has bonded to three hydrogen and that carbon has bonded to a nitrogen which is bonded to two hydrogen in our last two electrons will become a lone pair on the nitrogen

Let's look at the bonding in the molecule. So for diet fluoride, So so far I fluoride is composed of three elements. One sulfur to flow rides the electron configuration of sulfur in quotations. Neon three s 23 p four Straw this in terms of arrow diagrams just for the valence electrons three yes, and the three p. So three s to three p four flooring. That's an electron configuration of helium to us too. Two p five. Drawing this in terms of air diagrams to us, too. Two p five circle the electrons involved in bonding be all of our lone pair our loan electrons here so we would have 32 unpaid electrons and sulfur and the one unpaid electron in each flooring involved in bonding a straw, a three dimensional sketch of SF two showing the overlap of orbital's. So is there a place for sulfur? Yes, he overlapped for Rudolph from one floor een in the overlap of orbital's from another flooring. So we've got three dimensional sketch of SF two and let's look at the bond angles. So according to millions upon theory, the F S F bond is 109.5 degrees. But in SF two, there is lone pair bond repulsion, so the angle is reduced to 98.2 degrees. Here violence. One theory is not compatible to the no experimentally measured bond angle.