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RormuaLoudsStructureFormulaLmis Studere _NO:NHPO NH-OHPBraHow many bonds does nitrogen typically form? Does this make sense based on the number valence electrons ni...

Question

RormuaLoudsStructureFormulaLmis Studere _NO:NHPO NH-OHPBraHow many bonds does nitrogen typically form? Does this make sense based on the number valence electrons nitrogen alom? Explain Your answerDoes phosphorus typically form the same number of bonds as nitrogen? Use the periodic table to explain why Youanswer makes senssCircle the structures below that represent typical bonding pattemn for nitrogen:eNINH;

Rormua LoudsStructure Formula Lmis Studere _ NO: NH PO NH-OH PBra How many bonds does nitrogen typically form? Does this make sense based on the number valence electrons nitrogen alom? Explain Your answer Does phosphorus typically form the same number of bonds as nitrogen? Use the periodic table to explain why Youanswer makes senss Circle the structures below that represent typical bonding pattemn for nitrogen: eNI NH;



Answers

Explain why nitrogen bonds with hydrogen to form $\mathrm{NH}_{3},$ but not $\mathrm{NH}_{2}$ or $\mathrm{NH}_{4}$ . Use Lewis dot structures to support your argument.

So Hey, we're just using some o se ri with R NH three Orbital's. So we have ammonia here, and we know that ammonia is a para middle structure, but we have bonding angles of approximately 107.5 degrees, and then we have that lone pair on the top. So now, if we fill or electrons in our molecular orbital's, we have 5678 electrons to fill 12345678 And now to identify our bonding and anti bonding. These are bonding orbital's and then above. Those are anti bonding orbital's. And we know this because on an energy level, diagram are anti bonding are always higher energy than our bonding Orbital's.

If we drop the atomic orbital diagram for each nitrogen atom, we see that there are three um paired electrons and the P orbital's. This then allows for three of P orbital's overlap each other to form the end to molecule forming three bombs. This could also be described using SP. Hybridization would have to be SP instead of SP three hybridization because we would need at least two on hybridized p orbital's in order to form to live on.

9000 forms and but phosphorus is converted took me for from people. That is enough for this. The reason is this that yeah mm The pipe bowl is very big. Mhm. Yeah come on. Yes because 9%. This is that B block any make right so it cannot expand its balancing. That's why people people bowling is possible only in the end but in the past for us it can form easily by by bonding so the pipe in the morning is very big. That's why Nigel can form only enter. But phosphorus can form also people because of it can expand its valence because of presents of that the ornaments so backed up so I'm gonna be

So that's what we're gonna talk about. Question 108 in Chapter 20 which says many structures of phosphorous containing compounds are drawn with P, equals zero bonds. These bones are not the typical pie bonds were considered, which involved the overlap of to pee or medals. Instead, they result from the overlap of a D orbital on Prospero's with a P or bill on oxygen. This type of prime bonding is sometimes uses an explanation for what H three people before has the first structure below rather than the second. Some of the first structure we have a double bond to an ox, a general. In the second, we put a lone pair of electrons on the central phosphorus Adam also. Presumably, it's the overlap between a peanut doable that forms this pi bond this double long between phosphorus and oxygen infrastructure. Um, and then we're asked to draw a picture showing how a D orbital and a P or really overlap to form a pi bon. So before we do that, I have drawn P and D Orbital's. We'll look at first so Rp orbital's are sort of bumbles aligned along the X y and Z axes are dear bottles, on the other hand, are sets of two dumbbells that air staggered with the access is so they're diagonal rather than a lined up and down and across. Um, and there, since we have shoes that's been there in a plane in the XY plane, Lex y Z plane or the X Y plane rather than just the line along one access s. So how can we draw these overlapping? Well, we have our P orbital, which is straight up and down, and R d orbital, which is staggered or tilted on a diagonal. Um, and they overlap like this.


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