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$\mathrm{FcCl}_{\text {j }}$ solurion addcd ro $\mathrm{K}_{i}\left[\mathrm{Fc}(\mathrm{CN})_{6}\right]$ givcs (a) while with KSCN gives (b). (a) and (b) respecrively arc (a) $\mathrm{Fe}_{1}\left[\mathrm{Fe}(\mathrm{CN})_{\mathrm{G}}\right]_{3}, \mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{SCN})_{6}\right]$ (b) $\mathrm{\Gammae}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}, \mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{CNS})_{6}\right]$ (c) $\mathrm{Fc}_{3}\left[\mathrm{Fc}(\mathrm{CN})_{i}\right]_{2}, \mathrm{Fc}(\mathrm{CNS})_{3}$ (d) $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}, \mathrm{KFe}(\mathrm{CNS})_{3}$

Hello and welcome to this video solution of numerous. Here we are given a link comprehension. So here in statement one were given that and iron or a on roasting with sodium carbonate and lying in the presence of air gives to compound PNC. And you're from option the second part the solution be in concentrated. It's alan reaction with production furrows and it gives a blue color on precipitated of compound their cost solution of C. On treatment with considerate statistical gives a yellow colored compound E. And the compound even treat it with a seal gives an orange red compound F. Which is used as an ox raise anything. So you have to make certain predictions to come into the solution. So let me show you how it's made. So first of all we take for if he oh she had to poetry thus eight and a two freedom global neck And with air that is given seven or 2 good line gives you to a free two or three that's eight the name so Seattle four. That's it. It's a frustration. Next we have if we two or three reacting with concentrate sales that is given gives you who official three Last three H 2. Right now this official tree reacts with who before if he 10 6 that is protection fellow sign a great this gives you the blue solution pushing blue solution of if the four if he CN six whole trade plus 12 cases. Mm Next From three we have two any he wanted to see our well for That's 8/10 of food gives you we need to the odd 47 is the yellow color solution. Greatness in A. Two so four. That's it too. No you have in A to see our two or seven. Okay, last case here is the orange color substance of people pr two or seven that's 2010. So this is the so let us state what is A what is B. The first we have me is If you see a 23, this is a Next we have every two or 3 SCB or Yes. So this suits also be ready And in in 204 this is C. Right BNC. That is mentioned. Yes. Now this pushing blue books. Mhm. Yeah. Yeah. Whoa it is. Mhm. Okay. To see again this election with the this one your local er this is E and we have orange. What are some common? It does if Yeah. Right. I hope this is clear to you and have a very good rest of the day. Thank you.

The geometry of the coalition and it can be predicted Officer managing behavior isn't? Which of the following? Is that correct? Yes, it is correct because magnetic behavior can be seen by the general first is the correct And I see him for two minutes involved the S two hi creation. Yes and I plus to have electronic conference and that is really 840 and seeing is that a strong field again that will pairing. So it will be the yesterday for all of us and it is within a group. Do you cannot value follows. It is incorrect because they don't know what they do for us five days then for me then P. D. C. LLC and for whom I am very scared. Ford is also correct. So correct. Option will be 1, 2 and four Palestine, correct.

Hi everyone. So in this question they ask The electron configuration of the four elements uh given here. And it's this element in the correct order of magnitude of their electron affinity. So, uh first is two years to do before you. Second is three years 23 P 3 30 is two years to before and 43 years. 23 before. Select the correct answer. Using the goals. One and that So told Smarter than first. Want to dance again. Smarter than one. The next third smaller than for smaller than one. Smaller than second. Then third. Once more than seconds rather than force more than third than these. Second. Smarter than once. More than four. Smarter than third. So option B is correct. Mhm of shut. Mhm B is all right. Okay. Cold smaller than four. Smalling than one. Yeah. Yeah. Cook And smaller than four. Smaller than one. Smaller than second. Mhm. Yeah. Second. Yeah. Yeah. Mm seeking electronic sided mm three. Pure beetle. But mhm experiences. Let's republish. Mhm. And more energy is given out mom. And now she It's given on. Thanks a lot.

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.


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