NO Lewis Structure - How to Draw the Lewis Structure for NO (Nitric Oxide)
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- čas přidán 17. 06. 2013
- A step-by-step explanation of how to draw the NO Lewis Dot Structure (Nitric acid).
For the NO structure use the periodic table to find the total number of valence electrons for the NO molecule. Once we know how many valence electrons there are in NO we can distribute them around the central atom with the goal of filling the outer shells of each atom.
In the Lewis structure of NO structure there are a total of 11 valence electrons. NO is also called Nitric acid. The Nitrogen in this Lewis structure will only have 7 valence electrons around it. This is an exception to the Octet Rule but one we must make to draw the NO Lewis structure. We can check the formal charges to be sure we have the most favorable Lewis Structure. See how to calculate formal charges: • Formal Charges: Calcul...
---- Steps to Write Lewis Structure for compounds like NO -----
1. Find the total valence electrons for the NO molecule.
2. Put the least electronegative atom in the center. Note: Hydrogen (H) always goes outside.
3. Put two electrons between atoms to form a chemical bond.
4. Complete octets on outside atoms.
5. If central atom does not have an octet, move electrons from outer atoms to form double or triple bonds.
---- Lewis Resources ----
• Lewis Structures Made Simple: • How to Draw Lewis Stru...
• More practice: • Lewis Dot Structure Pr...
• Counting Valence Electrons: • Finding the Number of ...
• Calculating Formal Charge: • Formal Charges: Calcul...
• Exceptions to the Octet Rule: • Exceptions to the Octe...
Lewis Structures are important to learn because they help us understand how atoms and electrons are arranged in a molecule, such as Nitric acid. This can help us determine the molecular geometry, how the molecule might react with other molecules, and some of the physical properties of the molecule (like boiling point and surface tension).
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A little... well... a lot of added info regarding the triple bond vs double bond. This video is a freshmen level Chem 101 question and does a great job getting everyone in the right direction. The deeper you go into "why" the more complex and theoretical your answer will be. As far as why it's a double instead of a triple we have to consider that lewis structures are not an accurate representation of electron placement. You have to study Molecular Orbital Theory to get a better idea of where the bonds are, how many there are, radial/planar nodes, ect. There is no doubt that triple bonds DO exist as a type of resonance alternative BUT how often the molecule is arranged to represent that is the question. The answer is likely not often. Why? Because that leaves the oxygen with an additional electron. Oxygen is more electronegative than Nitrogen BUT Nitrogens ionization energy is greater so it is less likely to give away that extra electron than Oxygen is likely to take it if that makes sense. In case you are wondering "why" yet again, which is a great thing because it shows you hunger for knowledge (science related at that instead of celebrity gossip but I digress), the answer is because of electron placement within corresponding antibonding and bonding orbitals. If you draw the MO diagram for N and O bonding you'll see that the P orbitals along the x, y, and z axis have 1 electron each (Hund's rule). Oxygen on the other hand has a pair in one p-orbital and 2 unpaired in the others. 3 unpaired p-orbitals are more stable in this instance than 1 paired and 2 unpaired. That's also partly why you see Cu and Cr acting up in the 4th period. I wont get any deeper than that but good luck to everyone on their quest for knowledge.
It’s also useful to think of polar covalent bonds made by more electronegative elements as electron leeching, as it makes more sense electronegatively, oxygen kinda hates forming coordinate bonds(e.g. sharing an entire pair of e-s with N so +1 formal charge) with stuff. But i do completely understand the theory of paired electrons in orbitals repelling each other. (A much simpler answer would be that Oxygen only needs 2 electrons to be unreactive and that coordinate bonds are really only formed by strong partial positive dipoles/full positive charges and lone pairs)
You said that the triple bond gives oxygen an extra electron, i think thats a mistake on your behalf as the triple bond actually increases Os formal charge so its kinda losing an electron, not the other way around, as oxygen is more electronegative than nitrogen, it makes more sense for it to prefer a form that doesn’t compromise its formal charge while sticking closer to the octet/(s,p,p,p 2x4=8)rule. It would be frankly absurd if you would believe that oxygen would be taking an additional electron from N, as it simply cannot bend the rules of the octet if said host element is in the n=2 energy level(oxygen is btw), as no 3D orbitals are available for electron elevation to a higher orbital, expanding the octect.
The right way of forming the triple bond resonance structure would be oxygen sharing a lone pair with N, kinda like a coordinate bond; therefore making O and N obey the octet.
Another resonance structure would be if O had 1 lone pair and an unpaired e- (·O:) and N had 2 (:N:) instead of ·N: and :O: and this is the structure you would want to explain the uncommonness via a MO, as the none bonding π*2p is closer to the energy level(higher contribution) of N’s 2p than O’s 2p. Leading to the conclusion that the none-bonding orbital is residing in N most of the time, resulting in N having the unpaired e-(·N: = :O:)
So in conclusion, oxygen does have (very much)less of a tendency to take the e-s of N, but this is actually due to O being unable to expand its octet due to it not being able to elevate its e- to a higher orbital to form more bonds. (Your statements about ionization energy and electronegativity are correct, just not the main reason). You also did not need to use a molecular orbital diagram to explain ionization energy as a molecular orbital diagram is used to explain exceptions of diamagnetism and paramagnetism via paired(dia) or none paired(para) electrons in the MOs(and also to reflect energy levels). The more legal triple bond (that i suggested) doesn’t really work either cus of oxygen pulling more e-s in itself and the partial positive charges on N not being enough for the coordinate bond to form, as the difference between 3.0 and 3.5 only indicates a middling polar covalent bond(usually strong partial positive/full positive charges are needed for coordinate bonds to form, (e.g. H+
I study for medical school in Brasil. In the tests, the universities apply this knowledge, but it is not taught in the preparation books for the exams. Thank you Dr Breslyn for the great explanation. God bless you!
All the best in medical school! --- Dr. B
I like how I'm studying smth that could take me around 2h , when its literally some kind of basic knowledge on yt
Can we use the molecular orbital theory for this structure? I.e. if we need to find no. Of unpaired electrons
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Thanks, glad I could help and all the best with your chemie course! --- Dr. B
Is the lone electron a "pair" when counting electrons groups
@Wayne Breslyn
Isn't the NO molecule a free-radical? Is there a specific notation for that?
Thanks so much for your videos-- they're terrific!
The top electron in N is the free radical
@@zitscx886 so it can react in free radical subst. reactions?
@@hellsingh3694 i guess so. Tho confirm this with someone else. I will also try to do so.
What is the hybridisation of N in No ?
Help please !
Sir, can NO exist for a long time in atmosphere or does N combines with any other to become octet ?
Can we have a triple bond between the N and O?
You're the GOAT
Really helped me in the exam ❤
Happy to help!
Where does the formula come from? Could you show that? I mean, I can use it properly, but I just memorized it. Great vid!
Can't oxygen donate 1 electron to nitrogen making the dative bond and making it achieve octet in its valence shell?
Sir but triplebond of n and o would give same formal charge which is 0 , triple bond can also be exist???
황진웅 Nah, O will have 9 electrons assigned to it, which is highly improbable.
Can you do a vid of zirconium dioxide?
Is there any dative bond in no
Can we consider NO polar?
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By the way which is more stable -ve formal charge or zero formal charge?
0
Isn't there a dative/coordinate bond present
Gey doctor ...why deosnot this nitric oxide radical get into another chemical reaction to get another 1 electron for the 14-7 N and then we put a negative cahrge on top of NO
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Octet not respected for N ? How come, it is in the second line of the periodic table
Thanks in advance
You're right, we would expect, based on theory, that Nitrogen would need eight electrons. But theory doesn't always work in the lab. This is one of the fairly uncommon cases of N breaking the Octet Rule.
What is interesting to me is that NO is an important molecule, especially biologically (en.wikipedia.org/wiki/Nitric_oxide). Part of that may be due to N not having an octet and being more reactive.
--- Dr. B
@@wbreslyn I'll add that regardless of exceptions of theory versus practice, that result is a mathematical necessity. With an odd number of electrons, you can't make *both* nitrogen and oxygen have eight electrons. If we give nitrogen 8, then we could repeat the same question for oxygen. (But since oxygen it more electronegative, the one in the video would make more sense.)
@@JoonasD6 Agreed!
Thank u sir
is there a resonance structure for this NO molecule?
I suppose the answer has to be NO. This is really the best way to draw the NO Lewis Structure for NO. Any other representation would have non-zero formal charges. --- Dr. B
@@wbreslyn what if there was a single or triple bond? I found combinations where formal charges were also 0 in these cases.
Would you mind solving for NAN03
+kushal Bhetuwal Take a look (search CZcams) at my NO3- video and then my NaCl lewis structure video. Just replaced the Cl- with NO3- in the NaCl video. --- Dr. B
Short and effective :D
Thanks! --- Dr. B
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Any time! --- Dr. B
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No problem! --- Dr. B
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but octate of nitrogen is not completed
I dont get it... there are only 7 dots around nitrogen
shouldnt there be 8?
That is almost always the case for N. But the Lewis structure for NO is an exception. There are actually quite a few exceptions to the octet rule:
czcams.com/video/Dkj-SMBLQzM/video.html
Why don't we just put a negative charge on top of The NO as a sign that this radical needs another 1 electron to get the N to 8 electrons
Why are we putting the 2 oxygen electrons to the other side? Why not nitrogen's? Why are we even rearranging them at all?
We're trying to get both atoms as close to an octet (8 valence electrons as possible). The Oxygen atom is more electronegative so we would expect it to have eight valence electrons in the Lewis structure. Since we only have 11 to start out with, this leaves N with 7. This is a sort of strange structure, it's rare to see Nitrogen with only seven valence electrons. --- Dr. B
Wayne Breslyn but we also see it with the NO2 structure right? :/
Yep, you got it!
czcams.com/video/0XTkO8KypUY/video.html
Oxygen has only 6 *valence* electrons. And 8 electrons in total.
I agree completely. Some of those valence electrons are involved in forming chemical bonds. When Oxygen bonds, it shares valence electrons in order to have an Octet (8 valence electrons in it's outer shell). --- Dr. B
@@wbreslyn is this an exception or what coz octat of n is not complete
@@ManjeetSingh-wy5we Yes, it is an exception. It might be mentioned in this video on exceptions...
Exceptions to the Octet Rule: czcams.com/video/Dkj-SMBLQzM/video.html
Why is it a double bond instead of a single bond?
I'm assuming the most likely answer means you have the least amount of lone pairs while still keeping an octet (exceptions not considered)- is that right?
oxygen has 6 valent electrons
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And those odd number of the Lewis structures will do that!
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