The first question in number 10 is which compounds contain D localized electrons. D. Localized electrons are pie electrons that can be placed in different locations and still have the octet rule. Be satisfied, and all of the valence electrons be placed appropriately. This can occur if multiple resident structures can be written for the molecule. And it turns out that although many molecules in question number 10 have pi bonds, Onley A and I have pi bonds in a benzene ring that are going to be d localized and could be found between differing pairs of carbon atoms in that benzene ring. The benzene ring is the ring that contains six carbon atoms, which is present in compounds. A And, I question, be asks which compounds contain pie. Bond compounds contain a pie bond but are not D localized pie electrons. Well, that would correspond to any of the other molecules that contain pi bonds present in the Group of nine molecules. So that would be be de E. If you draw the Louis structure, for you will see that there's a double bond between the two carbons. G and H contain Onley, non d localized pie electrons question see is a B, E and H, which is our plainer Let's look at B first. It turns out that B has two carbons that are SP two hybridized and our tribunal plainer those air. The central carbons foreseen most of the molecule to be plainer. However, there are two see single bond. Oh, bonds. We'll call them just si o bonds that can rotate, and they could rotate so that the hydrogen atoms are within the plane, defined by the two carbon atoms or out of the plane, so B could be plainer. If the CEO bonds are rotated so that the hydrogen atoms are in the plane defined by the carbons, then let's look at E. He is always plainer, the carbons RSP to hybridized, and there is no bond rotation that can move the hydrogen atoms out of the plane. And then h is not plainer at all. Because it contains a SP three tetra, he'd RL carbon in the molecule. The next question, then, is locate the Cairo Adams in the Group of Nine molecules, and it turns out that there's only one Cairo Adam that is present in Ah, Compound A and that's it. There are no other chi role. A compound So on Lee A. Contains one Cairo Carbon E asks Compare the hybridization of the sulphur atoms in D and G. Now, this is little challenging. You're gonna have to count up all of the Vaillant electrons, and in doing so, you're going to find out that D contains a lone pair on sulphur in addition to the two pi bonds on sulphur. So that results in three electron groups surrounding sulfur, two double bonds and one lone pair. Therefore, with three electron groups. It is SP two hybridized. We'll do something similar with G, some of all the valence electrons for all the atoms present in the molecule, and then place the valence electrons back into the molecule and you'll find out that each sulfur, in addition to the two single bonds that are present, contained two lone pairs. So each sulfur in G is SP three hybridized having four electron groups to single bonds and two lone pairs. Now for F see is a chlorofluorocarbon responsible for the destruction of the ozone layer. How is C her house f related to see? Well, it turns out that FNC are exactly the same molecule. It's just that F has been rotated in comparison to see or C has been rocket rotated in comparison to f Essentially, they're the same molecule.