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anonymous
 one year ago
resonance structures
anonymous
 one year ago
resonance structures

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why did I get this wrong?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0LMAO the correct answer is F hahhaha tough luck

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it says that for every question, anyways, a double bond in the entity containing N follows the rules of octet rule, but why is NO3 better?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0do u have a picture of the periodic table

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i can post it just a min...

dan815
 one year ago
Best ResponseYou've already chosen the best response.0im looking at one already nvm

Empty
 one year ago
Best ResponseYou've already chosen the best response.2NO2+ doesn't have its best form as a resonance structure: dw:1444197726202:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0there are 3 equivalent resonant structures for NO3+

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see what your talking about for tsecond pic but the first one is a bit.... confusing

dan815
 one year ago
Best ResponseYou've already chosen the best response.0NO2+ doesnt have a resonant structure

dan815
 one year ago
Best ResponseYou've already chosen the best response.0no matter how u move those 4 bond lines around its the same bonds connected to the same atoms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i thought that since NO2 you can add another electron because ionic charge is negative so there would only be one bonding e opn each O which can make one double bond..

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Can you explain that more, I don't quite understand what you mean. Both oxygens are double bonded to the nitrogen atom in the center.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2What do you mean by 'add another electron' I think this is what I don't understand the most.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like for lewis strucutres, when drawing them, if you have a lets say a negative polyatomic ion you add electrons depeing on the number of the ionic charge... unless that's wrong

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Also @dan815 I wouldn't say that NO2+ doesn't have other resonance structures, I just drew two of them out right here: dw:1444198406616:dw It's just that the contributions of the wave function there is much smaller... (Don't worry about this part @dareintheren )

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh i see so the prob of that state is much lower?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0would the prob of that go up with more kinetic energy in your substance

Empty
 one year ago
Best ResponseYou've already chosen the best response.2@dareintheren Can you show me how you would draw the Lewis structure of NO2+ and I'll help you figure it out. @dan815 Yes to the first and yes to the second. >:D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okdw:1444198700072:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.2@dan815 At low temperatures the ground state is occupied most so think of \(O=N^+=O\) as being one continuous "box" where the electrons can travel freely between O,N,and O and higher energies correspond to 'nodes' in the box which are lower probability since they're higher energy (although we can tunnel to them so they're not zero probability). Now when you heat stuff up (increase average kinetic energy) the electrons don't move to higher energy states exactly. The relative distribution actually becomes more even. So at infinite temperatures you will actually have all states equally populated, which is actually a pretty useful approximation for high temperatures with finite states (not all solutions to the Schrodinger equation have infinite solutions, for instance an electron has only two spin states, +1/2 and 1/2).  Ok back to @dareintheren Let me see what you've drawn here give me a sec.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh so really when they are talking about like these electrons jumping down from higher energy levels and releasing photos, thats actually when these electrons are going back to more stable resonant structures?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i wonder how this photon is really created, like what is this strange thing that must come out of no where

Empty
 one year ago
Best ResponseYou've already chosen the best response.2@dareintheren Are you in organic chemistry or general chemistry right now and are you a chemistry major? I want to make sure I give you the right answer to this question cause there are varying levels of difficulty and I want to make sure you can understand it cause I think I need to explain this to you better since I think you have most of the idea correct.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2@dan815 Make a new question with that photon thing in it and I can describe a little but I don't wanna hijack this dude's question anymore lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm in general, and nope not chem major

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Has your class spoken at all about sigma or pi bonds? (sorry it's been a while and I forget if this is considered advanced or not)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ummm no not yet, but i did read up on it, sigma is end to end and pi is side to side.... forgot what they were for .... overlap/ hyrbdizaton right?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, exactly! A sigma bond is just a single bond by itself, but when you draw a double bond (just two lines) that actually represents a single sigma bond and a single pibond. So I guess I'll lay out how I thought to do this, I looked at the periodic table and found the total number of electrons that should be in the valence shells of the atoms, 5 for Nitrogen 6 for Oxygen, and there are two so 12 in all This gives us a total of 5+12=17 electrons to place on our molecule. But electrons are negative and we see that the molecule is NO2+ so we need to remove an electron, so we are only placing 16 electrons instead of 17. The positive charge comes from the protons in the nuclei of the atoms. So now we can start placing our electrons, generally electrons pair up, so instead of placing electrons, we can place electron pairs. 16/2=8 so 8 is just a more manageable number. Now we know that N is in the center, so let's go ahead and draw it out and start placing: dw:1444199679236:dw Since a bond is just a pair of electrons shared and we have to connect them, I used up 2, and we have 6 pairs left to place. Since the molecule is symmetric, we can imagine splitting up the pairs three to each side and there are two ways we might try to do this: dw:1444199775696:dw However the molecule on the left has Nitrogen with an incomplete octet, no good! So we are stuck with the right structure. From here we can see that by looking at formal charges that the plus charge goes on Nitrogen. So this is how I would solve this problem I hope this helps or if you want clarification I know this gets a little wordy there in the middle but I think your way of doing it is not quite right so I figured I'd just walk through all the steps.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0However the molecule on the left has Nitrogen with an incomplete octet, no good! So we are stuck with the right structure , do you mean nitrogen doesn't have a full octet?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah all good, specifically you can tell because it has 4, only 2 from each of the bonds it has. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but doesn't N have a lone pair? do it only needs one more bonding pair doesn't it?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2It would have a lone pair if we placed one on it, but we didn't, we couldn't actually, even if we had did it lopsided like this:dw:1444200975612:dw See, now the oxygen on the left and the nitrogen have a complete octet, however the oxygen on the right has only 6 valence electrons. So this is no good! We have to push that lone electron pair on nitrogen into a bond to give oxygen a complete octet.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444201104385:dw Maybe this makes it clear, this was the only possible way we could have placed these 16 electrons (8 pairs of electrons) on the molecule so that all atoms have a complete octet. 2 pairs on the left oxygen, 4 pairs into the bonds, and 2 more pairs on the oxygen on the right for a total of 8 electron pairs. Now you're probably used to seeing Nitrogen with a lone pair, in this case it wasn't possible to do that and obey the octet rule. Because of that, Nitrogen has a + charge on it since one of the electrons is being shared with the Oxygen. Keep asking questions if this doesn't seem to answer your question or make sense to you, I have gotten so used to this stuff that I might be skipping over something that is obvious to me that wasn't always obvious to me that I'm taking for granted haha.
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