Lewis Structures: Shapes and Hybridization & Molecular Orbital Diagramby heykyou Tags: diagram, hybridization, lewis, molecular, orbital, shapes, structures 

#1
Jan407, 09:52 AM

P: 8

I don't think the problems I'm going to post are very difficult, but I'm totally clueless regarding chemistry, so they're pretty difficult for me ^^;; I also have a lot of questions to ask, so I hope you don't mind my sticking them all into one topic. I'm looking forward to your help. You don't have to answer ALL my questions, since I seem to explain them confusingly...but I'd really appreciate explanations. Thanks in advance!
First off is about the Lewis Structures. You have to know the electrons needed and those available and the bonds to be formed, right? But I don't understand the significance of the number of bonds, since it doesn't seem to be followed. (At least this was how we were taught to do LS.) Anyway, while drawing the structure, how would you know which shape to follow? (I've been trying to listen to class as much as possible, but I still can't get it.) And we're also required to determine the hybridization, which is another thing I can't understand. And how do you solve something with a charge? One problem is below: 1. The problem statement, all variables and given/known data Draw the structures and determine the shapes and hybridizations of the following: a. ClO_{3}^{} 2. Relevant equations no. of electrons needed: 4x8=32 no. of electrons available: Cl=7, O=6x3, total=25, +1 (because of the charge?), total=26 3226=6 electrons, 3 bonds 3. The attempt at a solution Shape: (I have no idea...) Hybridization: Cl: (Ne) 3s^{2} 3p^{5} ground state: ↓↑3s ↓↑3p ↓↑3p ↓3p hybrid state: ↓↑3sp^{3} ↓↑3sp^{3} ↓↑3sp^{3} ↓3sp^{3} Is that correct? If the hybridization is correct (hopefully), I have another question about it. For example, the hybridization for BrF_{5}. Br: (Ar) 3d^{10} 4s^{2} 4p^{5} ground state: ↓↑4s ↓↑4p ↓↑4p ↓4p 4d 4d 4d 4d 4d Is the hybrid state: ↓↑4sp^{3}d^{2} ↓4sp^{3}d^{2} ↓4sp^{3}d^{2} ↓4sp^{3}d^{2} ↓4sp^{3}d^{2} ↓4sp^{3}d^{2} ...or: ↓4sp^{3}d^{3} ↓4sp^{3}d^{3} ↓4sp^{3}d^{3} ↓4sp^{3}d^{3} ↓4sp^{3}d^{3} ↓4sp^{3}d^{3} ↓4sp^{3}d^{3} I see both being used, and I don't know how to determine which one to use. Sometimes the ns part is just left as it is, working only with the np and nd, but then sometimes all of them are hybridized or something... I hope you got what I'm blabbering about ^^;; I'd really appreciate help about SHAPES (the seesaw, etc.) because I can't understand them... Well, another question: how do you draw the molecular orbital diagram of something that has a charge? For example, O_{2}^{2}. I only know how to do those without charges, but would it be the same with or without the charge? Another about formal charges: For example for SO_{4}^{2}. Does the structure need to be drawn first before determining the FC of S? And does S have only one formal charge? Thanks very much in advance 



#2
Jan407, 10:42 AM

Emeritus
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P: 11,154

I haven't read past the first problem, so let's start there.
Good attempt on the structure  it's close but you need to keep the following in mind (eveything up to calculating the number of bonds is correct, except for a typo at "O=6X8"): 1. Look at the structure you've drawn and, count (i) the number of electrons provided by each atom (eg: you have 7 electrons from each O atom), and (ii) the total number of electrons around each atom, including shared electrons electrons (eg: around each O atom, there's 8 electrons). The point of this counting excercise is to make sure that the first number matches the valency of the atom (6, for O), and the second number comes out to be 8 (making each atom have a complete octet). But it's not always possible to have a structure where both these criteria are met^{*}. However, whenever there isn't the correct number of valence electrons from a given atom, then the excess or deficit makes up the net charge on the atom. In this case, since O has a valency of 6 and you've got 7 electrons provided by each O atom, that's an excess of 1 electron for each O. That gives each O atom a net charge of 1 (and you need to show this in the diagram by writing "" or "1" next to each O atom). Thus the three O atoms have a total charge of 3. What you need to do, is the same counting procedure for the Cl atom (to see if it has an octet and/or a net charge). And when you've counted the number of electrons provided by the Cl atom, make sure that the total number of electrons (those from the Cl atom + those from the 3 O atoms) provided comes out to be 26. If it doesn't, you've got too many or too few electrons and you need to add or remove some, to make the number right. Finally, make sure that the sum of all the net charges (just as a doublecheck) equals the net chrge on the ion (for ClO_{3}^{}, the total charge is 1). ^{*} NOTE : It may not always be possible to ensure neutral (chargeless) octets for each atom. In that case, the following order tells you the which is the lesser of the alternative evils (topmost being the least evil, or most prefered, and bottommost being the least prefered): I: charged octets (like the O atoms above) II: neutral nonoctets III: charged nonoctets 



#3
Jan407, 06:13 PM

P: 8

Thanks for replying!
Whoops...I corrected the typo ^^; Another try at the structure: (I know this isn't correct yet...) There should be 26 electrons, but in my structure there are only 25...and I don't know where to add the other one...to Cl? I read somewhere that those below C, O, N, F and Ne can have more than an octet or something... So if I add an electron to Cl, do I put the net charge there? Ack...something just hit me. So is this more correct... Thanks again 



#4
Jan407, 11:30 PM


#5
Jan507, 05:04 AM

P: 8

If the charge is moved, do I have to move the one electron from Cl too? Sorry for asking this question, but I'm not sure about it ^^; Can an atom have an odd number of electronsone isn't paired?
And sorry again, but can you explain about the electronegativity? The higher the electronegativity, the more likely an atom is to accept molecules, right? But exactly how I know when it's more likely to accept molecules? (I've been asking my teacher, but my other classmates have been bombarding her with lots of questions already...) Also, I asked her about my tries in solving this problem. She said we didn't even need to draw a Lewis Structure, but just a structure. (The instructions were: "Draw the structures and determine the shapes and hybridizations of the following.") So what's the difference between a LS and just a structure? When drawing the nonLewis structure, do I still have to compute for the electrons needed, available, etc.? (But I still appreciate the help in LS because I can understand it better now, and am still looking forward to any help regarding that :) ) 



#6
Jan507, 10:02 AM


#7
Jan507, 11:10 AM

P: 8

My answer was really funny _;; Kinda stupid to not figure out what I was supposed to do...
But... Wow, everything falls into place! I can't believe that's so hard...and that's just one problem. Well, I'll try solving the other LS on my own, and ask you again when something comes up. Thanks for explaining this topic so well! So, there can never be a single electron (doesn't belong in a pair) sticking to the atom? About the "just structures", I tried to look in the Internet too, but I don't think there are other ways to draw structures except for LS... I guess I'll ask my teacher what she meant when she said that... Umm, can I ask about the hybridization? (Don't mind my first post about it; I already understood that part ^^; Actually I asked my teacher earlier about most of the problems I posted here, but I still can't understand hybridization so...) When doing the hybridization on something with a charge, does the charge affect the electrons on the hybridized atom? What I think: In the example ClO_{3}^{}, the hybridization of Cl won't be affected since the charge is on an O. But my question is, what if the charge is on the central atom? (Or does the charge NEVER land on the central atom, since it's most likely the least electronegative?) 



#8
Jan507, 11:38 AM

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P: 11,154

Charge does not affect the hybridization. As for the "just structures", don't bother asking your teacher  they don't exist. The only correct way to do this is by drawing complete Lewis Structures, and anyone who suggests otherwise, sadly, is only trying to avoid doing the work (of teaching it properly) themselves.
I'll look at the rest of the question sometime over the weekend, if no one else has gotten to it before. 



#9
Jan707, 09:48 PM

P: 8

Thanks so much for the help!
But I have another question... What about BH_{4}^{}? My answer is: I'm not sure which of the two is more electronegative, but putting the negative sign on B seems more correct, since B has 3 valence electrons (and in the diagram has 4). And so, the hybridization would be: ground state: ↑↓2s ↑↓2p 2p 2p hybrid state: ↓ ↓ ↓ ↓ sp^{3} Meaning the charge affected the hybridization? Because if I don't add an extra electron to the hybridization, I'd only end up with sp^{2}, which isn't correct, right? Thanks again and sorry for the trouble ^^; 



#10
Jan707, 10:22 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154




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